NCERT Solutions Class 11 Chemistry Chapter 7

Equilibrium NCERT Solutions – Class 11 Chemistry

Chemical equilibrium plays a key role in a variety of biological and environmental processes. For example, in the transport and delivery of O2 from our lungs to our muscles, equilibria involving O2 molecules and the protein haemoglobin play a critical role. Class 11 students will study concepts such as equilibrium laws, characteristics, equilibrium constants, and so on in this chapter. 

Equilibrium, Chapter 7 of the NCERT Chemistry textbook for Class 11, has a number of challenging and interesting questions that help students gain a better understanding of the chapter . As a result,  comprehensive NCERT Solutions Class 11 Chemistry Chapter 7 is an important resource for students looking for answers to NCERT textbook questions to excel in school and competitive exams. 

NCERT Solutions for Class 11 Chemistry Chapter 7 

Extramarks Class 11 Science Chemistry Chapter 7 NCERT Solutions are curated by subject experts which makes them stand out. All the answers are explained in detail and in an easy-to-comprehend manner while ensuring they are accurate. Students can access the solutions on Extramarks’ website or app. .

Access NCERT Solution for Chemistry Class 11 Chapter 7 – Equilibrium

Solutions 

NCERT Solutions for Class 11 Chemistry Chapter 7 

NCERT Solutions for Class 11 Chemistry Chapter 7 can be used as reference material by students while preparing for exams. The solutions have answers to all the questions that are asked at the end of the Chapter 7 in the NCERT Chemistry book. 

NCERT Solutions Class 11 Chemistry Chapter 7

The NCERT Solutions for Chapter Equilibrium in Class 11 is a reliable study material available to students. It provides answers to all of the questions in Chapter 7 of Chemistry.  Let’s look at the topics that are covered in Chapter 7 solutions:

Section Number Section Title
7.1 Equilibrium in Physical Processes
7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium
7.3 Law of Chemical Equilibrium and Equilibrium Constant
7.4 Homogeneous Equilibria
7.5 Heterogeneous Equilibria
7.6 Applications of Equilibrium Constants
7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G
7.8 Factors Affecting Equilibria
7.9 Ionic Equilibrium in Solution
7.10 Acids, Bases and Salts
7.11 Ionisation of Acids and Bases
7.12 Buffer Solutions
7.13 Solubility Equilibria of Sparingly Soluble Salts

Chapter 7 Equilibrium

The chapter on equilibrium in chemical and physical processes gives a brief overview of the various concepts of equilibrium in chemical and physical processes, as well as details on how equilibrium is dynamic. This chapter also covers the law of mass action, various factors affecting equilibrium, and the equilibrium constant based on Le Chatelier’s principle. Because it explains how objects behave, equilibrium is the most important part of chemistry.. Equilibrium is explained as chemical theories and models in the NCERT Chapter 7 CBSE Chemistry Class 11 book. 

7.1 Equilibrium in Physical Processes

This topic teaches students about the equilibrium between different physical properties when the chemical composition remains constant. Solid-Liquid equilibrium, Liquid-Vapour Equilibrium, and Solid-Vapour Equilibrium are all examples of equilibrium. 

7.2 Equilibrium in Chemical Processes

Equilibrium in Chemical Process or Dynamic Equilibrium is a process in which the forward and reverse reactions of a chemical equation occur at the same rate with no net change in the product and reactant ratio. 

7.3 Law of Chemical Equilibrium and Equilibrium Constant

The topic discusses concepts such as the connection between the concentration levels of reactants and products in an equilibrium mixture. The topic addresses questions such as:

  • What is the relationship between the concentration of reactants and the products in an equilibrium mixture?
  • How can we calculate equilibrium concentrations from initial concentrations?
  • What factors can be used to change the composition of an equilibrium mixture?
  • How can an equilibrium mixture’s composition be altered?

7.4 Homogeneous Equilibria

Homogeneous Equilibrium refers to a reaction in which the reactants and products are in the same physical state. 

For instance, 2SO2 (g) + O2 (g) = 2SO3 (g) 

Sulphur and oxygen are in the same state as each other, which is gaseous. 

7.5 Heterogeneous Equilibria

Heterogeneous Equilibrium is a reaction in which the reactants and products are in different physical states. 

For instance, 3Fe (s) + 4H2O (g) = Fe3O4 (s) + 4H2O (g) 

Iron and iron oxide are solid in this state. However, the state of water and hydrogen gas is gaseous. 

7.6 Applications of Equilibrium Constants

Students learn how to use the equilibrium constant to calculate equilibrium concentrations, predict the extent of a reaction based on its magnitude, and predict the direction of a reaction. The calculation of equilibrium concentrations is covered in this topic. 

7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G

The Equation Constant K, the Reaction Quotient Q, and the Gibbs Energy G have a relationship. The study of such quantitative relationships between different types of energy is known as thermodynamics. J.W. Gibbs, who is discussed in this section, represents a mathematical expression of this thermodynamic view of equilibrium. 

7.8 Factors Affecting Equilibria

This topic discusses the various factors that influence equilibria as well as the factors that affect equilibria. 

7.9 Ionic Equilibrium in Solution

For weak electrolytes, the ionic equilibrium is the equilibrium established between the unionised molecules and the ions in a solution. Electrolytes include acids, bases, and salts, which can act as both strong and weak electrolytes. 

7.10 Acids, Bases and Salts

This topic reviews the concepts of acids, bases, and salts, as well as three acid and base-based theories. Arrhenius Acids and Bases, Brönsted-Lowry Acids and Bases, and Lewis Acids and Bases are those three theories. 

7.11 Ionisation of Acids and Bases

When a neutral molecule is exposed to a solution, it undergoes ionisation, which is the process by which it splits into charged ions. Students will study Arrhenius’ theory, which states that acids are compounds that dissociate in an aqueous medium to produce hydrogen ions, or H+ (aq). This section also covers the differences between dissociation and ionisation. 

7.12 Buffer Solutions

In this section, students will learn what a buffer solution is, its types, the mechanism of buffering action, preparation of acid and base buffer solutions, the Henderson-Hasselbalch Equation, Buffering capacity and much more. 

7.13 Solubility Equilibria of Sparingly Soluble Salts 

When studying a salt’s properties and the properties of the solution it forms with a particular solvent, its solubility in that solvent is always an important factor to consider. Students will study solubility product, the differences between solubility product and ionic product, their significance and more. 

Class 11 Chemistry Chapter 7: Distribution of Marks

The question-wise weightage  varies for different questions related to Chapter 7. Each question is graded according to the importance of the topic. The chapter contains a total of 73 questions, which are divided into three categories: short type questions, long type questions, and numerical problems. These questions are based on a variety of equilibrium concepts and laws of equilibrium experiments. 

There are one-mark questions with very short answers. Then there are the two-mark short-answer questions, which may include definitions as well. The three-mark questions may include definitions as well as numerical data. Some of the long-answer questions are worth 5-marks. 

Benefits of Chapter Equilibrium Class 11 NCERT Solutions

The NCERT Solutions Class 11 Chemistry Chapter 7 is an important reference material from an exam preparation perspective. Here are some of the benefits of referring to Extramarks solutions:

  • The solutions will assist students in finding accurate answers to the textbook questions which will provide them with a better understanding of the chapter. 
  • Students in class 11 can use the NCERT Solutions for Class 11 Chemistry Chapter 7 to gain detailed knowledge of the topics and questions covered in the chapter. 
  • The solutions are outlined in accordance with the recent syllabus for class 11, chapter 7 of chemistry. 
  • Subject-matter experts have prepared the solutions.
  • NCERT Solutions are written in a straightforward manner so that the students can grasp the fundamentals of chemistry quickly. Additionally, these cover all subjects and chapters, with important questions and answers explained thoroughly. 

Q.1 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium?

PC l 5 (g) ⇌ PC l 3 (g)+C l 2 (g) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaaiaadcfacaWGdbGaamiBamaaBaaaleaacaaI1aaabeaakiaacIcacaWGNbGaaiykamaaoOaaleqabaaakiaawcCicaGL9gcacaWGqbGaam4qaiaadYgadaWgaaWcbaGaaG4maaqabaGccaGGOaGaam4zaiaacMcacqGHRaWkcaWGdbGaamiBamaaBaaaleaacaaIYaaabeaakiaacIcacaWGNbGaaiykaaaa@5231@

Ans.

At equilibrium, let us assume that the concentration of PCl3 and Cl2 as x mol L-1

PC l 5 (g) At equilibrium 0 .5×10 −1 mol L −1 ⇌ PC l 3 (g) x molL −1 + C l 2 (g) x molL −1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8145@

The value of equilibrium constant Kc = 8.3 x 10-3

Now we can write the expression for equilibrium as:

⇒ [ PC l 3 ][ C l 2 ] [ PC l 5 ] = K c ⇒ x⋅+x ( 0.5× 10 −1 ) =8.3× 10 −3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6D32@

⇒ x2= 4.15 x 10-4

⇒ x = 2.04 x 10-2

= 0.0204

= 0.02 (approx)

At equilibrium, we can write as-

[PCl3] = [Cl2] = 0.02 mol L-1

Q.2 One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

FeO( s )+CO( g ) ⇌ Fe( s )+C O 2 ( g ), K p =0.265 at 1050KMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaamOraiaadwgacaWGpbWaaeWaaeaacaWGZbaacaGLOaGaayzkaaGaey4kaSIaam4qaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaamOraiaadwgadaqadaqaaiaadohaaiaawIcacaGLPaaacqGHRaWkcaWGdbGaam4tamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiaacYcacaWLjaGaam4samaaBaaaleaacaWGWbaabeaakiabg2da9iaaicdacaGGUaGaaGOmaiaaiAdacaaI1aGaaCzcaiaadggacaWG0bGaaCzcaiaaigdacaaIWaGaaGynaiaaicdacaWGlbaaaa@65FE@

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: Pco=1.4atm and Pco2 = 0.80 atm?

Ans.

Pco=1.4atm and Pco2 = 0.80 atm ?

FeO( s ) Initially + CO( g ) 1.4 atm ⇌ Fe( s )+ C O 2 ( g ) 0.80 atm Q p = p C O 2 p CO = 0.80 1.4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7EEE@

= 0.571

Given Kp = 0.265 at 1050 K

Qp > Kp

Thus, the reaction will proceed in the backward direction.

To achieve the equilibrium, the pressure of CO will increase and the pressure of CO2 will decrease.

Let us assume that the increase in pressure of CO = decrease in pressure of CO2 = p.

K p = p C O 2 p CO ⇒ 0.265= 0.80−p 1.4+p ⇒ 0.371+0.265p=0.80−p ⇒ 1.265p=0.429 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@784A@

⇒ p=0.339 atm

At equilibrium, partial pressure of CO2,

pC O 2 =0.80−0.339=0.461 atm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaamiCamaaBaaaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH9aqpcaaIWaGaaiOlaiaaiIdacaaIWaGaeyOeI0IaaGimaiaac6cacaaIZaGaaG4maiaaiMdacqGH9aqpcaaIWaGaaiOlaiaaisdacaaI2aGaaGymaiaabccacaqGHbGaaeiDaiaab2gaaaa@5441@

And at equilibrium, partial pressure of CO,

pCO= 1.4 + 0.339 = 1.739 atm

Q.3 Equilibrium constant, Kc for the reaction

N 2 ( g )+3 H 2 ( g ) ⇌ 2N H 3 ( g ) at 500K is 0.061 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaamOtamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaaiodacaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaaikdacaWGobGaamisamaaBaaaleaacaaIZaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiaabccacaqGHbGaaeiDaiaabccacaaI1aGaaGimaiaaicdacaWGlbGaaeiiaiaabMgacaqGZbGaaeiiaiaabcdacaqGUaGaaeimaiaabAdacaqGXaaaaa@5FAC@

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L-1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Ans.

N 2 ( g ) At a particular time: 3 .0 molL −1 + 3 H 2 ( g ) 2 .0 molL −1 ⇌ 2N H 3 ( g ) 0 .5 molL −1 Now,we know that, Q c = [ N H 3 ] 2 [ N 2 ] [ H 2 ] 3 = ( 0.5 ) 2 ( 3.0 ) ( 2.0 ) 3 =0.0104 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@B942@

As given, Kc = 0.061

Since Qc ≠ Kc the reaction is not at equilibrium.

and Qc < Kc , reaction will proceed in the forward direction to reach equilibrium.

Q.4 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl( g ) ⇌ B r 2 ( g )+C l 2 ( g )MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaaGOmaiaadkeacaWGYbGaam4qaiaadYgadaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaamOqaiaadkhadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGdbGaamiBamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaaa@54DA@

for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?

Ans.

Let the amount of Br2 and Cl2 produced at equilibrium be x mol L-1. The given reaction is:

2BrCl( g ) ⇌ B r 2 ( g ) + C l 2 ( g ) Initial Conc. 3 .3×10 −3 0 0 At equilibrium( 3.3× 10 −3 −2x ) x x Thus, ⇒ [ B r 2 ][ C l 2 ] [ BrC l 2 ] = K c ⇒ x⋅x ( 3.3× 10 −3 −2x ) 2 =32 ⇒ x ( 3.3× 10 −3 −2x ) =5.66 ⇒x=18.678× 10 −3 −11.32x ⇒12.32x=18.678× 10 −3 ⇒x=1.5× 10 −3 Therefore, at equilibrium, [ BrCl ]=3.3× 10 −3 −( 2×1.5× 10 −3 ) =3.3× 10 −3 −3.0× 10 −3 =0.3× 10 −3 =3.0× 10 −4 mol L −1MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@5196@

Q.5 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass

C( s )+C O 2 ( g ) ⇌ 2CO( g )MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9LqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaam4qamaabmaabaGaam4CaaGaayjkaiaawMcaaiabgUcaRiaadoeacaWGpbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaaikdacaWGdbGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaaaaa@5106@

Calculate Kc for this reaction at the above temperature.

Ans.

Let us assume that the total mass of the gaseous mixture = 100 g
Mass of CO = 90.55 g and
Mass of CO2 = (100 – 90.55) g = 9.45 g
Number of moles of CO (nCO) = 90.55/28 = 3.234 mol
[Molecular mass of CO = (12+16) = 28]
Number of moles of CO2 (nCO2) = 9.45/44 = 0.215 mol
[Molecular mass of CO2 = (12+32) = 44]
Partial pressure of CO,

Q.6 Calculate a) DG° and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K

NO( g )+ 1 2 O 2 ( g ) ⇌ N O 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaamOtaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYaaaaiaad+eadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaamOtaiaad+eadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@52CA@

Ans.

Where ΔfG° (NO2) = 52.0 kJ/mol

ΔfG° (NO) = 87.0 kJ/mol

ΔfG° (O2) = 0 kJ/mol

Ans. (a) For the above reaction,

ΔG°= ΔG° (products) – ΔG° (reactants)

= 52.0 – (87.0 + 0)

= -35.0 KJ mol-1

(b) -ΔG° = RT ln Kc

or – ΔG° = 2.303 RT log Kc

or log Kc = -ΔG°/2.303 RT

⇒log K c = −( −35× 10 −3 ) 2.303×8.314×298 ⇒ K c =antilog( 6.134 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@71CC@

= 1.36 x 106

Thus, the equilibrium constant for the given reaction Kc is 1.36 × 106.

Q.7 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

a) PC l 5 ( g ) ⇌ PC l 3 ( g )+C l 2 ( g ) b) CaO( s )+C O 2 ( g ) ⇌ CaC O 3 ( s ) c) 3 Fe( s )+4 H 2 O( I ) ⇌ F e 3 O 4 ( s )+4 H 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8C65@

Ans.

According to Le-Chatelier principle, the stress of a decrease in pressure is relieved by the net reaction in the direction that increases the number of moles of gas.

  1. Since the number of moles of gases is more on the products side. The reaction will proceed in the forward direction. As a result, the number of moles of the reaction products increases.
  2. Since the number of moles of gases is more on the reactants side. The reaction will proceed in the backward direction. As a result, the number of moles of the reaction products decreases.
  3. In the 3rd reaction, the number of moles of gases is same on both sides (reactants as well as products) of the chemical equation. It depicts that equilibrium is not affected by the pressure change. Thus, the number of moles of reaction products remains the same.

Q.8 Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

(i) COC l 2 ( g ) ⇌ CO( g )+C l 2 ( g ) (ii) C H 4 ( g )+2 S 2 ( g ) ⇌ C S 2 ( g )+2 H 2 S( g ) (iii) C O 2 ( g )+C( s ) ⇌ 2CO( g ) (iv) 2H 2 ( g )+CO( g ) ⇌ C H 3 OH( g ) (v) CaC O 3 ( s ) ⇌ CaO( s )+C O 2 ( g ) (vi) 4N H 3 ( g )+5 O 2 ( g ) ⇌ 4NO( g )+6 H 2 O( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@D790@

Ans.

Only those reactions will be affected in which (np ≠ nr)
Here, nr = the number of moles of gaseous reactant
np = the number of moles of gaseous product
So, (i), (iii), (iv), (v) and (vi) reactions will get affected by increasing the pressure.
By applying Le Chatelier’s principle, we can predict the direction of the reaction. The stress of increase in pressure is relieved by the net reaction in the direction that decreases the number of moles of gas.
Reaction (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.
Reactions (i), (iii), (v) and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

Q.9 The equilibrium constant for the following reaction is 1.6 ×105 at 1024 K.

H 2 ( g )+B r 2 ( g ) ⇌ 2HBr( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaamisamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaadkeacaWGYbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaaikdacaWGibGaamOqaiaadkhadaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@52F3@

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Ans.

Equilibrium constant for the forward reaction,

Kp = 1.6 x 105

Let us calculate equilibrium constant for the backward reaction.

Kp’ = 1/Kp

= 1 1.6× 10 5 =6.25× 10 −6 The backward reaction can be written as 2HBr( g ) ⇌ H 2 ( g )+B r 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0de9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8722@

Let us assume that the pressure of gaseous H2 or gaseous Br2 (at equilibrium) is p.

2HBr( g ) Initial conc. 10 At Equilibrium( 10−2p ) ⇌ H 2 ( g ) 0 p + B r 2 ( g ) 0 p MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xf9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7D47@

Thus,

p H 2 × p B r 2 p 2 HBr =K p ⇒ p×p ( 10−2p ) 2 =6.25× 10 −6 ⇒ p×p ( 10−2p ) =2.5× 10 −3 ⇒p=2.5× 10 −2 −( 5.0× 10 −3 )p ⇒p+( 5.0× 10 −3 )p=2.5× 10 −2 ⇒p=2.49× 10 −2 bar MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xf9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@BA0A@

Therefore, at equilibrium,

[H2] = [Br2] = 2.49 x 10-2 bar

[HBr] = 10 – 2 x (2.49 x 10-2) bar

= 10 bar (approx)

Q.10 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

C H 4 ( g )+ H 2 O( g ) ⇌ CO( g )+3 H 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaam4qaiaadIeadaWgaaWcbaGaaGinaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGibWaaSbaaSqaaiaaikdaaeqaaOGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGdbGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaaiodacaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@57BD@

(a) Write as expression for Kp for the above reaction.

(b) How will the values of Kp and composition of equilibrium mixture be affected by

(i) Increasing the pressure

(ii) Increasing the temperature

(iii) Using a catalyst

Ans.

(a) The expression of Kp can be written as follows

K p = p CO × p H 2 3 p C H 4 × p H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaam4samaaBaaaleaacaWGWbaabeaakiabg2da9maalaaabaGaamiCamaaBaaaleaacaWGdbGaam4taaqabaGccqGHxdaTcaWGWbWaa0baaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaaaleaacaaIZaaaaaGcbaGaamiCamaaBaaaleaacaWGdbGaamisamaaBaaameaacaaI0aaabeaaaSqabaGccqGHxdaTcaWGWbWaaSbaaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaWccaWGpbaabeaaaaaaaa@54EB@

(b)

  1. As nr < np, according to Le Chatelier’s principle the equilibrium will shift in the backward direction.
  2. As given, the reaction is endothermic. By Le Chatelier’s principle, the equilibrium will shift in the forward direction.
  3. The equilibrium constant of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Q.11 Describe the effect of:

a) Addition of H2

b) Addition of CH3OH

c) Removal of CO

d) Removal of CH3OH

On the equilibrium of the reaction:

2 H 2 ( g )+CO( g ) ⇌ C H 3 OH( g )MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacqaaamaayzcaapdaiIJaaGOmaiaadIeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGdbGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGdbGaamisamaaBaaaleaacaaIZaaabeaakiaad+eacaWGibWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@5372@

Ans.

a) According to Le Chatelier’s principle, if H2 is added to the reaction mixture, the equilibrium of the given reaction will shift in the forward direction. On adding H2 to reaction, to maintain the equilibrium constant some amount of CO reacts with H2 to form alcohol. Thus, the reaction will shift towards the forward direction.

b) On addition of CH3OH, the rate of backward reaction increases than the rate of forward reaction. Hence, the equilibrium will shift in the backward direction. . To maintain, the equilibrium constant some amount of CH3OH breaks into CO and H2.

c) On removing CO, the rate of forward reaction decreases than the rate of backward reaction. Hence, the equilibrium will shift in the backward direction. To maintain the equilibrium constant, some amount of CH3OH breaks into CO and H2.

d) On removing CH3OH, the equilibrium will shift in the forward direction. To maintain the equilibrium constant some amount of CO reacts with H2 to form alcohol. Thus, the reaction will shift towards the forward direction.

Q.12 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 × 10-3. If decomposition is depicted as,

PC l 5 ( g ) ⇌ P C 3 ( g )+C l 2 ( g ), Δ f H º=124.0 kJmol −1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamiuaiaadoeacaWGSbWaaSbaaSqaaiaaiwdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaadcfacaWGdbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaey4kaSIaam4qaiaadYgadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacaGGSaGaaeiiaiabfs5aenaaBaaaleaacaWGMbaabeaakiaadIeacaWLjaGaaiOUaiabg2da9iaaigdacaaIYaGaaGinaiaac6cacaaIWaGaaeiiaiaabUgacaqGkbGaaeyBaiaab+gacaqGSbWaaWbaaSqabeaacqGHsislcaaIXaaaaaaa@66EC@

a) Write an expression for Kc for the reaction.

b) What is the value of Kc for the reverse reaction at the same temperature?

c) What would be the effect on Kc

If (i) more PCl5 is added (ii) pressure is increased? (iii) The temperature is increased?

Ans.

a) The expression for Kc for the given reaction

K c = [ PC l 3 ( g ) ][ C l 2 ( g ) ] [ PC l 5 ( g ) ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWGJbaabeaakiabg2da9maalaaabaWaamWaaeaacaWGqbGaam4qaiaadYgadaWgaaWcbaGaaG4maaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaiaawUfacaGLDbaadaWadaqaaiaadoeacaWGSbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaacaGLBbGaayzxaaaabaWaamWaaeaacaWGqbGaam4qaiaadYgadaWgaaWcbaGaaGynaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaiaawUfacaGLDbaaaaaaaa@5A81@

b) The value of Kc for the backward reaction:

K’c= 1/ Kc

=1 / (8.3 x 10-3)

= 1.2048 x 102

= 120.48

c) i) Kc is constant at constant temperature, so no effect on adding PCl5.

ii) The terms in Kc is independent of pressure, so no effect will occur if pressure is increased.

iii) The given reaction is endothermic, on increasing in temperature kf will increase.

As Kc= kf/kb

Thus, the value of Kc also increases.

Q.13 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO( g )+ H 2 O( g ) ⇌ C O 2 ( g )+ H 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4qaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGibWaaSbaaSqaaiaaikdaaeqaaOGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGdbGaam4tamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaadIeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@57CF@

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that

p CO = p H 2 O =4.0 bar, MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamiCamaaBaaaleaacaWGdbGaam4taaqabaGccqGH9aqpcaWGWbWaaSbaaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaWccaWGpbaabeaakiabg2da9iaaisdacaGGUaGaaGimaiaabccacaWGIbGaamyyaiaadkhaaaa@4E

what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C

Ans.

Let the partial pressure of both CO2 (gas) and H2 (gas) be p atm.

Co( g ) Initial conc. 4.0 bar At Equilibrium ( 4.0−p ) + H 2 O( g ) 4.0 bar ( 4.0−p ) ⇌ C O 2 ( g ) 0 p + H 2 ( g ) 0 p MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9460@

And Kp = 10.1

Thus,

p C O 2 × p H 2 p CO × p H 2 O = K p ⇒ p×p ( 4.0−p ) 2 =10.1 ⇒ p×p ( 4.0−p ) =3.178 ⇒p=12.712−3.178 p 4.178 p=12.712 ⇒p=3.04 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@975B@

Thus, at equilibrium, the partial pressure of H2 will be 3.04 bar.

Q.14 Predict which of the following reaction will have appreciable concentration of reactants and products:

a) C l 2 ( g ) ⇌ 2Cl( g ), K c =5× 10 −39 b) C l 2 ( g )+2NO( g ) ⇌ NOCl( g ), K c =3.7× 10 8 c) C l 2 ( g )+2N O 2 ( g ) ⇌ 2N O 2 Cl( g ), K c =1.8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9B59@

Ans.

For the reaction (c), Kc is neither high nor very low. Therefore, in this reaction, reactants and products will be present in appreciable concentration.

Q.15 The value of Kc for the reaction

3 O 2 ( g ) ⇌ 2 O 3 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaaG4maiaad+eadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaaGOmaiaad+eadaWgaaWcbaGaaG4maaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@4DAB@

is 2.0 × 10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 × 10–2, what is the concentration of O3?

Ans.

The given reaction is:

3 O 2 ( g ) ⇌ 2 O 3 ( g ) K c = [ O 3 ( g ) ] 2 [ O 2 ( g ) ] 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@5EEE@

It is given that Kc= 2.0 ×10–50 and [O2] = 1.6 ×10–2

⇒2.0× 10 −50 = [ O 3 ( g ) ] 2 ( 1.6× 10 −2 ) 3 ⇒ [ O 3 ( g ) ] 2 =( 2.0× 10 −50 ) ( 1.6× 10 −2 ) 3 ⇒ [ O 3 ( g ) ] 2 =8.192× 10 −56 ⇒[ O 3 ( g ) ]=2.86× 10 28 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacqGHshI3caaIYaGaaiOlaiaaicdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaiaaicdaaaGccqGH9aqpdaWcaaqaamaadmaabaGaam4tamaaBaaaleaacaaIZaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaGaay5waiaaw2faamaaCaaaleqabaGaaGOmaaaaaOqaamaabmaabaGaaGymaiaac6cacaaI2aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaaaaaGcbaGaeyO0H49aamWaaeaacaWGpbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaacaGLBbGaayzxaaWaaWbaaSqabeaacaaIYaaaaOGaeyypa0ZaaeWaaeaacaaIYaGaaiOlaiaaicdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaiaaicdaaaaakiaawIcacaGLPaaadaqadaqaaiaaigdacaGGUaGaaGOnaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaaaaaGcbaGaeyO0H49aamWaaeaacaWGpbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaacaGLBbGaayzxaaWaaWbaaSqabeaacaaIYaaaaOGaeyypa0JaaGioaiaac6cacaaIXaGaaGyoaiaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaiaaiAdaaaaakeaacqGHshI3daWadaqaaiaad+eadaWgaaWcbaGaaG4maaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaiaawUfacaGLDbaacqGH9aqpcaaIYaGaaiOlaiaaiIdacaaI2aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiaaikdacaaI4aaaaOGaaeiiaiaab2eaaaaa@A3A7@

Hence, the concentration O3 is 2.86 x 10-28 M.

Q.16 The reaction,

CO( g )+3 H 2 ( g ) ⇌ C H 4 ( g )+ H 2 O( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4qaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaaIZaGaamisamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGdbGaamisamaaBaaaleaacaaI0aaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaadIeadaWgaaWcbaGaaGOmaaqabaGccaWGpbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@5887@

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Ans.

Equilibrium

Conc. 0.30 0.10 x 0.02

CO( g )+3 H 2 ( g ) ⇌ C H 4 ( g )+ H 2 O( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4qaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaaIZaGaamisamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGdbGaamisamaaBaaaleaacaaI0aaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaadIeadaWgaaWcbaGaaGOmaaqabaGccaWGpbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@5887@

The equilibrium constant for the reaction,

K c = [ C H 4 ( g ) ][ H 2 O( g ) ] [ CH( g ) ] [ H 2 O( g ) ] 3 ⇒ x×0.02 0.3× ( 0.1 ) 3 =3.90 ⇒x= 3.90×0.3× ( 0.1 ) 3 0.02 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@89EB@

or, x = 0.0585 M = 5.85 x 10-2 M

Hence, the concentration of CH4 at equilibrium is 5.85 × 10–2 M.

Q.17 What is meant by the conjugate acid/base pair? Find the conjugate acid/base for the following species:

HNO2, CN, HClO4, F, OH, CO32- and S2-

Ans.

The acid-base pair which differs by a proton is called conjugate acid/base pair.

Species Conjugate acid/base
HNO2 NO2– (base)
CN– HCN (acid)
HClO4– ClO4– (base)
F– HF (acid)
OH– H2O(acid)/ O2– (base)
CO32– HCO3– (acid)
S2– HS– (acid)

Q.18 Which of the followings are Lewis acids? H2O, BF3, H+ and NH4+.

Ans.

Lewis acid can accept a pair of electrons. All cations are Lewis acid. Thus, BF3, H+ and NH4+ are considered as Lewis acids.

Q.19 What will be the conjugate bases for the Bronsted acids: HF, H2SO4 and HCO3–?

Ans.

The table below lists the conjugate bases for the given Bronsted acids.

Bronsted acids Conjugate bases
HF F–
H2SO4 HSO4–
HCO3– CO32–

Q.20 Write the conjugate acids for the following Bronsted bases: NH2–, NH3 and HCOO–.

Ans.

The table below lists the conjugate acids for the given Bronsted bases:

Bronsted bases Conjugate acids
NH2– NH3
NH3 NH4+
HCOO– HCOOH

Q.21 The species: H2O, HCO3, HSO4 and NH3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.

Ans.

The table below lists the conjugate acids and conjugate bases for the given species.

Species Conjugate acids Conjugate bases
H2O H3O+ OH–
HCO3– H2CO3 CO32–
HSO4– H2SO4 SO42–
NH3 NH4+ NH2–

Q.22 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

  1. OH– (b) F– (c) H+ and (d) BCl3

Ans.

  1. OH acts as Lewis base as it can donate its lone pair of electrons.
  2. F– is Lewis base as it can donate one of its lone pair of electrons.
  3. H+ acts as Lewis acid since it can accept a lone pair of electrons.
  4. BCl3 is Lewis acid since it can accept a lone pair of electrons in vacant p-orbital of Boron.

Q.23 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH?

Ans.

[ H + ]= 3.8510 −3 M pH=−log[ H + ] ⇒pH=−log( 3.8× 10 −3 M ) ⇒pH=−log3.8−log 10 −3 ⇒pH=−log3.8+3 =−0.58+3 =2.42MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaadaWadaqaaiaadIeadaahaaWcbeqaaiabgUcaRaaaaOGaay5waiaaw2faaiabg2da9iaaiodacaGGUaGaaGioaiaaiwdacaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaG4maaaakiaad2eaaeaacaWGWbGaamisaiabg2da9iabgkHiTiGacYgacaGGVbGaai4zamaadmaabaGaamisamaaCaaaleqabaGaey4kaScaaaGccaGLBbGaayzxaaaabaGaeyO0H4TaamiCaiaadIeacqGH9aqpcqGHsislciGGSbGaai4BaiaacEgadaqadaqaaiaaiodacaGGUaGaaGioaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIZaaaaOGaamytaaGaayjkaiaawMcaaaqaaiabgkDiElaadchacaWGibGaeyypa0JaeyOeI0IaciiBaiaac+gacaGGNbGaaG4maiaac6cacaaI4aGaeyOeI0IaciiBaiaac+gacaGGNbGaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiodaaaaakeaacqGHshI3caWGWbGaamisaiabg2da9iabgkHiTiGacYgacaGGVbGaai4zaiaaiodacaGGUaGaaGioaiabgUcaRiaaiodaaeaacqGH9aqpcqGHsislcaaIWaGaaiOlaiaaiwdacaaI4aGaey4kaSIaaG4maaqaaiabg2da9iaaikdacaGGUaGaaGinaiaaikdaaaaa@9318@

Q.24 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Ans.

Given pH= 3.76

From the relation,

pH=−log[ H + ] ⇒log[ H + ]=−pH ⇒[ H + ]=antilog( −pH ) ⇒[ H + ]=antilog( −3.76 ) =1.74× 10 −4 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8720@

Hence, the concentration of hydrogen ion in vinegar is 1.74 × 10–4 M.

Q.25 The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Ans.

From the relation,

K w = K a × K b MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWG3baabeaakiabg2da9iaadUeadaWgaaWcbaGaamyyaaqabaGccqGHxdaTcaWGlbWaaSbaaSqaaiaadkgaaeqaaaaa@4987@

Here,

Kw= ionic product of water

Kb = ionization constant of conjugate base

Ka = ionization constant of conjugate acid

K b = K w K a MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWGIbaabeaakiabg2da9maalaaabaGaam4samaaBaaaleaacaWG3baabeaaaOqaaiaadUeadaWgaaWcbaGaamyyaaqabaaaaaaa@4780@

Ka of HF = 6.8 × 10–4

Kw of water = 1 x 10-14

Hence, Kb of its conjugate base F–

= K w K a = 1× 10 −14 6.8× 10 −4 =1.5× 10 −11 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacqGH9aqpdaWcaaqaaiaadUeadaWgaaWcbaGaam4DaaqabaaakeaacaWGlbWaaSbaaSqaaiaadggaaeqaaaaaaOqaaiabg2da9maalaaabaGaaGymaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIXaGaaGinaaaaaOqaaiaaiAdacaGGUaGaaGioaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaaaakiabg2da9iaaigdacaGGUaGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIXaGaaGymaaaaaaaa@5E8E@

Again, Ka of HCOOH = 1.8 × 10–4

So, Kb of its conjugate base HCOO–

= K w K a = 1× 10 −14 1.8× 10 −4 =5.6× 10 −11 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaeyypa0ZaaSaaaeaacaWGlbWaaSbaaSqaaiaadEhaaeqaaaGcbaGaam4samaaBaaaleaacaWGHbaabeaaaaGccqGH9aqpdaWcaaqaaiaaigdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaisdaaaaakeaacaaIXaGaaiOlaiaaiIdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGinaaaaaaGccqGH9aqpcaaI1aGaaiOlaiaaiAdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaigdaaaaaaa@5E87@

Given, Ka of HCN = 4.8 × 10–9

= K w K a = 1× 10 −14 4.8× 10 −9 =2.08× 10 −6 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaeyypa0ZaaSaaaeaacaWGlbWaaSbaaSqaaiaadEhaaeqaaaGcbaGaam4samaaBaaaleaacaWGHbaabeaaaaGccqGH9aqpdaWcaaqaaiaaigdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaisdaaaaakeaacaaI0aGaaiOlaiaaiIdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGyoaaaaaaGccqGH9aqpcaaIYaGaaiOlaiaaicdacaaI4aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaa@5E92@

Q.26 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

Ans.

Ionization of phenol involves the following reaction:

C 6 H 5 OH Initial conc. 0.05 At equilibrium ( 0.05−x ) + H 2 O ⇌ C 6 H 5 O 0 x − + H 3 O + 0 x K a = [ C 6 H 5 O − ][ H 3 O + ] [ C 6 H 5 OH ] ⇒ K a x⋅x ( 0.05−x ) The value of the ionization constant is very less, x will be very small. We can ignore x in the denominator. x= ( 1× 10 −10 ×0.05 ) x= ( 5× 10 −12 ) =2.2× 10 −6 M=[ H 3 O + ] Since[ H 3 O + ]=[ C 6 H 5 O − ] [ C 6 H 5 O − ]=2.2× 10 −6 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@617E@

Let the degree of ionization of phenol be α.

C 6 H 5 OH Conc.( 0.05−0.05α ) + H 2 O ⇌ C 6 H 5 O − 0.05α + H 3 O + 0.05α C 6 H 5 ONa Conc. ⇌ C 6 H 5 O − + N a + 0.01 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9D83@

[C6H5OH]= 0.05 – 0.05α; 0.05 M

[C6H5O] = 0.01 + 0.05α; 0.01 M

[H3O+] = 0.05α

K a = [ C 6 H 5 O − ][ H 3 O + ] [ C 6 H 5 OH ] K a = ( 0.01 )( 0.05α ) ( 0.05 ) 1× 10 −10 =0.01α α=1× 10 −8 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7ED5@

Q.27 The first ionization constant of H2S is 9.1 × 10–8.

(i) Calculate the concentration of HS– ion in its 0.1 M solution.

(ii)How will this concentration be affected if the solution is 0.1 M in HCl?

(iii)Also if the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.

Ans.

(i) Let us calculate the concentration of HS– ion:

K a = [ H + ][ H S − ] [ H 2 S ] ⇒9.1× 10 −8 = x⋅x ( 0.1−x ) ⇒( 9.1× 10 −8 )( 0.1−x )= x 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@79C0@

Since value of x is very small, so we can consider

0.1 – x = 0.1

⇒9.1× 10 −8 = x 2 0.1 ⇒ x 2 =9.1× 10 −9 [ H S − ]=x=9.54× 10 −5 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6DBA@

(ii) In the presence of 0.1 M HCl, suppose H2S dissociates y M.

[ H 2 S ] Initial conc. 0.1 At equilibrium 0.1−y ⇌ H S − 0 y + H + 0 y MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8BEE@

Thus, at equilibrium,

[H2S] = 0.1 – y» 0.1,

[H+]= 0.1 + y » 0.1

[HS]= y M

Ka = 0.1×y 0.1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWGHbaabeaakiabg2da9maalaaabaGaaGimaiaac6cacaaIXaGaey41aqRaamyEaaqaaiaaicdacaGGUaGaaGymaaaaaaa@4AFE@

= 9.1 x 10-8

Thus, y = 9.1 x 10-8 M

(iii)The ionization of H2S is as follows –

H 2 S Initial conc. (M) 9 .4×10 −5 final conc. (M) 9 .4×10 −5 −X ⇌ 2 H + 9 .4×10 −5 X + S 2− 0 X MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9882@

Therefore,

K a 2 = [ H + ][ S 2− ] [ H S − ] K a 2 = ( 9.54× 10 −5 )( X ) 9.54× 10 −5 X=1.2× 10 −13 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGlbWaaSbaaSqaaiaadggadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0ZaaSaaaeaadaWadaqaaiaadIeadaahaaWcbeqaaiabgUcaRaaaaOGaay5waiaaw2faamaadmaabaGaam4uamaaCaaaleqabaGaaGOmaiabgkHiTaaaaOGaay5waiaaw2faaaqaamaadmaabaGaamisaiaadofadaahaaWcbeqaaiabgkHiTaaaaOGaay5waiaaw2faaaaaaeaacaWGlbWaaSbaaSqaaiaadggadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0ZaaSaaaeaadaqadaqaaiaaiMdacaGGUaGaaGynaiaaisdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaOGaayjkaiaawMcaamaabmaabaGaamiwaaGaayjkaiaawMcaaaqaaiaaiMdacaGGUaGaaGynaiaaisdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaaaakeaacaWGybGaeyypa0JaaGymaiaac6cacaaIYaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaigdacaaIZaaaaaaaaa@74A5@

Hence,

[ S 2− ]=1.2× 10 −13 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIZaamWaaeaacaWGtbWaaWbaaSqabeaacaaIYaGaeyOeI0caaaGccaGLBbGaayzxaaGaeyypa0JaaGymaiaac6cacaaIYaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaigdacaaIZaaaaOGaamytaaaa@4F6C@

In the presence of 0.1 M of HCl, let the concentration of S2- be Y M.

[ H 2 S ] Initial conc. 0.1 final conc. 0.1−y ⇌ H S − 0 y + H + 0 y MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@828F@

And

HCl ⇌ H + 0.1M + C l − 0.1M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamisaiaadoeacaWGSbWaa4GcaSqabeaaaOGaayjWHiaaw2BiamaaxababaGaamisamaaCaaaleqabaGaey4kaScaaaabaeqabaaabaqcLbEacaqGWaGaaeOlaiaabgdacaqGnbaaaSqabaGccqGHRaWkdaWfqaqaaiaadoeacaWGSbWaaWbaaSqabeaacqGHsislaaaaeaqabeaaaeaajug4biaaicdacaGGUaGaaGymaiaad2eaaaWcbeaaaaa@54DB@ K a 2 = [ H + ][ S 2− ] [ H S − ] 1.2× 10 −13 = ( 0.1 )( Y ) 9.1× 10 −8 10.92× 10 −21 =0.1Y 10.92× 10 −21 0.1 =Y Y=1.092× 10 −19 M [ S 2− ]=1.092× 10 −19 MMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@A197@

Q.28 The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the conc. of acetate ion in the solution and its pH.

Ans.

The given reaction is

C H 3 COOH ⇌ C H 3 CO O − + H + Degree of dissociation, α= K a /C MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6CEF@

Ka= ionization constant

C = concentration

C = 0.05 M Ka= 1.74 × 10–5

α= 1.74× 10 −5 5× 10 −2 = ( 34.8× 10 −5 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacqaHXoqycqGH9aqpdaGcaaqaamaalaaabaGaaGymaiaac6cacaaI3aGaaGinaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI1aaaaaGcbaGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaaaaaeqaaaGcbaGaeyypa0ZaaOaaaeaadaqadaqaaiaaiodacaaI0aGaaiOlaiaaiIdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaOGaayjkaiaawMcaaaWcbeaaaaaa@5CF3@

= 1.86 x 10-2

[CH3COO]= = 0.05 x 1.86 x 10-2

= 0.093 x 10-2 M

[CH3COO] = [H+] = 0.093 x 10-2 M

pH = – log [H+]= – log (0.093 x 10-2)

or, pH = 3.03

Q.29 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

Ans.

Let the organic acid be HA.

HA ⇌ H + + A − MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamisaiaadgeadaGdkaWcbeqaaaGccaGLahIaayzVHaGaamisamaaCaaaleqabaGaey4kaScaaOGaey4kaSIaamyqamaaCaaaleqabaGaeyOeI0caaaaa@49F6@

Concentration of HA = 0.01 M

pH = 4.15

⇒−log[ H + ]=4.15 ⇒[ H + ]=antilog( −4.15 ) ⇒[ H + ]=7.08× 10 −5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@713D@

Let us calculate ionization constant by the expression mentioned here.

K a = [ H + ][ A − ] [ HA ] [ H + ]=[ A −1 ]=7.08× 10 −5 [ H + ]=0.01 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6974@

Then,

K a = ( 7.08× 10 −5 )( 7.08× 10 −5 ) ( 0.01 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWGHbaabeaakiabg2da9maalaaabaWaaeWaaeaacaaI3aGaaiOlaiaaicdacaaI4aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiwdaaaaakiaawIcacaGLPaaadaqadaqaaiaaiEdacaGGUaGaaGimaiaaiIdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaOGaayjkaiaawMcaaaqaamaabmaabaGaaGimaiaac6cacaaIWaGaaGymaaGaayjkaiaawMcaaaaaaaa@5BD3@

⇒Ka= 5.01 x 10-7

⇒pKa = – log Ka

= – log (5.01 x 10-7)

= 6.3001

Q.30 Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

Ans.

(a) 0.003 M HCl:

Considering the equilibrium: <

HCl( aq ) ⇌ H + ( aq )+C l − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamisaiaadoeacaWGSbWaaeWaaeaacaWGHbGaamyCaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGibWaaWbaaSqabeaacqGHRaWkaaGcdaqadaqaaiaadggacaWGXbaacaGLOaGaayzkaaGaey4kaSIaam4qaiaadYgadaahaaWcbeqaaiabgkHiTaaakmaabmaabaGaamyyaiaadghaaiaawIcacaGLPaaaaaa@5615@

Thus, [H+] = [HCl] = 3 x 10–3

pH= – log [H+]= –log (3 x 10–3)= 2.52

(b) 0.005 M NaOH:

Considering the equilibrium:

NaOH( aq ) ⇌ N a + ( aq )+O H − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamOtaiaadggacaWGpbGaamisamaabmaabaGaamyyaiaadghaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaamOtaiaadggadaahaaWcbeqaaiabgUcaRaaakmaabmaabaGaamyyaiaadghaaiaawIcacaGLPaaacqGHRaWkcaWGpbGaamisamaaCaaaleqabaGaeyOeI0caaOWaaeWaaeaacaWGHbGaamyCaaGaayjkaiaawMcaaaaa@57BD@

Thus, [OH–] = 5 x 10–3 M

From the relation,

[H+] [OH–] = 1 x 10-14

Or, [H+] = (1 x 10–14) / (5 x 10–3)

= 2 x 10–12

pH= – log [H+] = – log (2 x 10–12) = 11.70

(c) 0.002 M HBr:

Considering the equilibrium:

HBr( aq ) ⇌ H + ( aq )+B r − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamisaiaadkeacaWGYbWaaeWaaeaacaWGHbGaamyCaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGibWaaWbaaSqabeaacqGHRaWkaaGcdaqadaqaaiaadggacaWGXbaacaGLOaGaayzkaaGaey4kaSIaamOqaiaadkhadaahaaWcbeqaaiabgkHiTaaakmaabmaabaGaamyyaiaadghaaiaawIcacaGLPaaaaaa@561F@

[H+] = 2 x10–3 M

pH = – log [H+] = – log (2 x 10–3) = 2.70

(d) 0.002 M KOH:

Considering the equilibrium:

KOH( aq ) ⇌ K + ( aq )+O H − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4saiaad+eacaWGibWaaeWaaeaacaWGHbGaamyCaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGlbWaaWbaaSqabeaacqGHRaWkaaGcdaqadaqaaiaadggacaWGXbaacaGLOaGaayzkaaGaey4kaSIaam4taiaadIeadaahaaWcbeqaaiabgkHiTaaakmaabmaabaGaamyyaiaadghaaiaawIcacaGLPaaaaaa@55EB@

Thus, [OH–] = 2 x 10–3 M

From the relation,

[H+] [OH–] = 1 x 10-14

or [H+] = (1 x 10–14) / (2 x 10–3) = 5 x 10–12

pH = – log [H+] = – log (5 x 10–12) = 11.30

Q.31 Calculate the pH of the following solutions:

  1. 2 g of TlOH dissolved in water to give 2 litre of the solution.
  2. 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution.
  3. 0.3 g of NaOH dissolved in water to give 200 mL of the solution
  4. 1 mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.

Ans.

(a)

Molar conc. of TIOH= 2g 221gmo l −1 × 1 2L =4.52× 10 −3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGnbGaam4BaiaadYgacaWGHbGaamOCaiaabccacaqGJbGaae4Baiaab6gacaqGJbGaaeOlaiaabccacaqGVbGaaeOzaiaabccacaqGubGaaeysaiaab+eacaqGibGaaeypamaalaaabaGaaGOmaiaadEgaaeaacaaIYaGaaGOmaiaaigdacaWGNbGaamyBaiaad+gacaWGSbWaaWbaaSqabeaacqGHsislcaaIXaaaaaaakiabgEna0oaalaaabaGaaGymaaqaaiaaikdacaWGmbaaaaqaaiabg2da9iaaisdacaGGUaGaaGynaiaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaG4maaaaaaaa@68A8@

[OH–]= [TlOH] = 4.52 x 10–3 M

[H+] [OH–] = 1 x 10-14

[H+]= (1 x 10–14)/ (4.52 x 10–3) = 2.21 x 10–12 M

pH = – log (2.21 x 10–12) = 12 – (0.3424) = 11.66

(b)

Molar conc. of Ca ( OH ) 2 = 0.3g 74gmo l −1 × 1 0.5 L =8.11× 10 −3 Ca ( OH ) 2 ( aq ) ⇌ C a 2+ ( aq )+2O H − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8904@

[OH–] = 2 [Ca(OH)2]

= 2 x (8.11 x 10–3) M

= 16.22 x 10–3 M

pOH = – log (16.22 x 10–3)

= 3 – 1.2101

= 1.79

pH = 14 – 1.79

(c)

Molar conc. of NaOH= 0.3g 40 gmo l −1 × 1 0.2 L =3.75× 10 −2 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGnbGaam4BaiaadYgacaWGHbGaamOCaiaabccacaqGJbGaae4Baiaab6gacaqGJbGaaeOlaiaabccacaqGVbGaaeOzaiaabccacaqGobGaaeyyaiaab+eacaqGibGaaeypamaalaaabaGaaGimaiaac6cacaaIZaGaam4zaaqaaiaaisdacaaIWaGaaeiiaiaadEgacaWGTbGaam4BaiaadYgadaahaaWcbeqaaiabgkHiTiaaigdaaaaaaOGaey41aq7aaSaaaeaacaaIXaaabaGaaGimaiaac6cacaaIYaGaaeiiaiaadYeaaaaabaGaeyypa0JaaG4maiaac6cacaaI3aGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaOGaamytaaaaaa@6CFD@

[OH–]= [NaOH] = 3.75 x 10–2 M

pOH = – log (3.75 x 10–2 M) = 2 – 0.0574 = 1.43

pH = 14 – 1.43 = 12.57

(d) From the relation,

M1V1 = M2V2

or 13.6 M x 1 mL = M2 x 1000 mL

or M2 = 1.36 x 10–2 M

[H+] = [OH–] = 1.36 x 10–2 M

pH = – log (1.36 x 10–2) = 2 – 0.1335 = 1.87

Q.32 The degree of ionization of a 0.1 M bromo acetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Ans.

Degree of ionization, a = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H3O+= c.a = 0.1 × 0.132

= 0.0132

pH=−log[ H + ] = – log ( 0.0132 ) = 1.88 K a =C α 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGWbGaamisaiabg2da9iabgkHiTiGacYgacaGGVbGaai4zamaadmaabaGaamisamaaCaaaleqabaGaey4kaScaaaGccaGLBbGaayzxaaaabaGaaeiiaiaacckacaGGGcGaaiiOaiaacckacqGH9aqpcaqGGaGaai4eGiaabccacaqGSbGaae4BaiaabEgacaqGGaWaaeWaaeaacaaIWaGaaiOlaiaaicdacaqGXaGaae4maiaabkdaaiaawIcacaGLPaaacaqGGaGaeyypa0JaaeiiaiaabgdacaGGUaGaaeioaiaabIdaaeaacaqGlbWaaSbaaSqaaiaadggaaeqaaOGaeyypa0Jaam4qaiabeg7aHnaaCaaaleqabaGaaGOmaaaaaaaa@67D8@

= 0.1 x (0.132)2

Ka = 0.0017

and pKa = – log (0.0017)

= 2.77

Q.33 The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate the ionization constant and pKb.

Ans.

Considering the reaction

Cod+ H 2 O ⇌ Cod H + +O H − MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4qaiaad+gacaWGKbGaey4kaSIaamisamaaBaaaleaacaaIYaaabeaakiaad+eadaGdkaWcbeqaaaGccaGLahIaayzVHaGaam4qaiaad+gacaWGKbGaamisamaaCaaaleqabaGaey4kaScaaOGaey4kaSIaam4taiaadIeadaahaaWcbeqaaiabgkHiTaaaaaa@51FD@

pH = 9.95, pOH = 14 – pH

= 14 – 9.95

= 4.05

or – log [OH–] = 4.05

or log[OH–] = – 4.05

or [OH–] = antilog (– 4.05) = 8.913 x 10-5

cα=8.91× 10 −5 α= 8.91× 10 −5 5× 10 −3 =1.782× 10 −2 K b =c α 2 K b =0.005× ( 1.782 ) 2 × 10 −4 =0.005×3.1755× 10 −4 =0.005× 10 −4 =1.58× 10 −6 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGJbGaeqySdeMaeyypa0JaaGioaiaac6cacaaI5aGaaGymaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI1aaaaaGcbaGaeqySdeMaeyypa0ZaaSaaaeaacaaI4aGaaiOlaiaaiMdacaaIXaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiwdaaaaakeaacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiodaaaaaaOGaeyypa0JaaGymaiaac6cacaaI3aGaaGioaiaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaOqaaiaadUeadaWgaaWcbaGaamOyaaqabaGccqGH9aqpcaWGJbGaeqySde2aaWbaaSqabeaacaaIYaaaaaGcbaGaam4samaaBaaaleaacaWGIbaabeaakiabg2da9iaaicdacaGGUaGaaGimaiaaicdacaaI1aGaey41aq7aaeWaaeaacaaIXaGaaiOlaiaaiEdacaaI4aGaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaaGcbaGaaCzcaiabg2da9iaaicdacaGGUaGaaGimaiaaicdacaaI1aGaey41aqRaaG4maiaac6cacaaIXaGaaG4naiaaiwdacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaisdaaaaakeaacaWLjaGaeyypa0JaaGimaiaac6cacaaIWaGaaGimaiaaiwdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGinaaaaaOqaaiaaxMaacqGH9aqpcaaIXaGaaiOlaiaaiwdacaaI4aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaaa@A885@

pKb = – log (1.588 x 10–6) = 6 – 0.1987 = 5.8

Q.34 What is the pH of 0.001 M aniline solution? The ionization constant of aniline = 4.27 x 10-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Ans.

Kb= 4.27 × 10–10 and c = 0.001 M

Kb= cα2

⇒4.27× 10 −10 =0.001× α 2 ⇒4270× 10 −10 = α 2 ⇒65.34× 10 −5 =α=6.53× 10 −4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7BDE@

Thus, [OH–] = c . α = 0.001 x 65.34 x 10–5

= 0.065 x 10–5

pOH = – log [OH]

= – log (0.065 x 10–5)

= – log (65 x 10–8)

= 6.187

We know,

pH + pOH = 14

or pH = 14 – 6.187 = 7.813

Again,

Ka x Kb = Kw

or Ka x (4.27 × 10–10) = 10–14

Ka = 10 −14 4.27× 10 −10 =2.34× 10 −5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWGHbaabeaakiabg2da9maalaaabaGaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaigdacaaI0aaaaaGcbaGaaGinaiaac6cacaaIYaGaaG4naiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIXaGaaGimaaaaaaGccqGH9aqpcaaIYaGaaiOlaiaaiodacaaI0aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiwdaaaaaaa@5A13@

The ionization constant of the conjugate acid of aniline is 2.34 × 10–5.

Q.35 Calculate the degree of ionization of 0.05 M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?

Ans.

Given values are as follows-

C = 0.05 M = 5 x 10-2

pKa = 4.74

pKa = – log Ka = 4.74

or Ka = antilog (– 4.74) = 1.82 x 10–5

From the relation,

Ka = cα2

⇒α= ( K a /c ) ⇒α= 1.82× 10 −5 5× 10 −2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacqGHshI3cqaHXoqycqGH9aqpdaGcaaqaamaabmaabaGaam4samaaBaaaleaacaWGHbaabeaakiaac+cacaWGJbaacaGLOaGaayzkaaaaleqaaaGcbaGaeyO0H4TaeqySdeMaeyypa0ZaaOaaaeaadaWcaaqaaiaaigdacaGGUaGaaGioaiaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaOqaaiaaiwdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaaaabeaaaaaa@5E74@

= 1.908 x 10–2

When HCl is added to the solution, the concentration of H+ ions increases. Thus the equilibrium will shift in the backward direction. As a result, the dissociation of acetic acid will decrease.

  1. When 0.01 M HCl is taken:

Let x be the amount of acetic acid dissociated after the addition of HCl.

C H 3 COOH Initial conc. 0.05 M At equilibrium ( 0.05−x ) ⇌ H + 0 ( 0.01+x ) + C H 3 COO− 0 X MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9952@

As a very small amount of acetic acid is dissociated, the values of (0.05 – x) and (0.01 + x) can be taken as 0.05 M and 0.01 M respectively.

K a = [ C H 3 CO O − ][ H + ] [ C H 3 COOH ] Putting the values in the above equation: K a = ( 0.01 )x 0.05 ⇒x= 1.82× 10 −5 ×0.05 0.01 ⇒x=1.82× 10 −3 ×0.05 M By definition, α= Amount of aicd dissociated Amount of acid taken ⇒α= 1.82× 10 −3 ×0.05 0.05 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@FBB0@

=1.82 x 10–3
1. When 0.1 M HCl is taken.
Let us assume that the amount of acetic acid dissociated in this case is X. The concentrations of various species involved in the reaction are:
[CH3COOH]= 0.05 – X ~ 0.05M
[CH3COO–] = X
[H+]= 0.1 + X ~ 0.1

K a = [ C H 3 CO O − ][ H + ] [ C H 3 COOH ] K a = ( 0.01 )x 0.05 ⇒x= 1.82× 10 −5 ×0.05 0.01 ⇒x=1.82× 10 −3 ×0.05 M Now, α= Amount of aicd dissociated Amount of acid taken ⇒α= 1.82× 10 −4 ×0.05 0.05 = 1.82 x 1 0 –4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@D855@

Q.36 The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Ans.

c = 0.02 M and Kb = 5.4 × 10–4

α= ( K b /c ) = 5.4× 10 −4 0.02 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacqaHXoqycqGH9aqpdaGcaaqaamaabmaabaGaam4samaaBaaaleaacaWGIbaabeaakiaac+cacaWGJbaacaGLOaGaayzkaaaaleqaaaGcbaGaeyypa0ZaaOaaaeaadaWcaaqaaiaaiwdacaGGUaGaaGinaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaaGcbaGaaGimaiaac6cacaaIWaGaaGOmaaaaaSqabaaaaaa@542B@

=0.1643

When 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

NaOH( aq ) ⇌ N a + ( aq ) 0.1 M + O H − ( aq ) 0.1 M ( C H 3 ) 2 NH ( 0.02−x )~0.02 + H 2 O ⇌ ( C H 3 ) 2 N H 2 + x + O H − ( x+0.1 )~0.1 [ ( C H 3 ) 2 N H 2 + ]=x [ OH – ] = x + 0.1 ~ 0.1 K b = [ ( C H 3 ) 2 N H 2 + ][ O H − ] ( C H 3 ) 2 NH ⇒5.4× 10 −4 = x×0.1 0.02 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@DECB@

⇒x = 0.0054

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Q.37 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(a) Human muscle-fluid, 6.83

(b) Human stomach-fluid, 1.2

(c) Human blood, 7.38

(d) Human saliva, 6.4.

Ans.

(a) For Human muscle fluid,

pH = 6.83

or pH = – log [H+]

or 6.83 = – log [H+]

or [H+] = antilog (-6.83)

or [H+] = 1.48 × 10–7 M

(b) For Human stomach fluid,

pH = 1.2 = – log [H+]

or [H+] = antilog (-1.2)

or [H+] = 0.063 M

(c) For Human blood,

pH = 7.38 = – log [H+]

or [H+] = antilog (-7.38)

or [H+] = 4.17 × 10–8 M

(d) For Human saliva,

pH = 6.4 = – log [H+]

or [H+] = antilog (-6.4)

or [H+] = 3.98 × 10–7 M

Q.38 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Ans.

The hydrogen ion concentration can be easily calculated by the following relation

pH=−log[ H + ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamiCaiaadIeacqGH9aqpcqGHsislciGGSbGaai4BaiaacEgadaWadaqaaiaadIeadaahaaWcbeqaaiabgUcaRaaaaOGaay5waiaaw2faaaaa@4AF6@

  1. pH of milk = 6.8

Using the above relation,

pH = – log [H+]

⇒ log [H+] = –6.8

⇒ [H+] = antilog (–6.8)

= 1.5 x 10–7 M

  1. pH of black coffee = 5.0

Since, pH = – log [H+]

⇒ 5.0 = – log [H+]

⇒ log [H+] = –5.0

⇒ [H+] = antilog(–5.0) = 10–5 M

  1. pH of tomato juice = 4.2

pH = – log [H+]

⇒ 4.2 = – log [H+]

⇒ log [H+] = –4.2

⇒ [H+] = antilog(–4.2)

=6.31 x10–5 M

  1. pH of lemon juice = 2.2

pH = – log [H+]

⇒ 2.2 = – log [H+]

⇒ log [H+] = –2.2

⇒[H+] = antilog(–2.2)

=6.31 x10–3 M

  1. pH of egg white = 7.

pH = – log [H+]

⇒7.8 = – log [H+]

⇒log [H+] = –7.8

⇒ [H+] = antilog(–7.8)

=1.58 x 10–8 M

Q.39 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Ans.

Let us calculate the concentration of KOH in the solution:

[ KOH ] ( aq ) = 0.561 200/1000 g/L =2.805g/L = 2.805 56.11 M =0.05 M The ionic equilibrium for the aqueous KOH is given below: KOH( aq ) ⇌ K + ( aq )+O H − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaadaWadaqaaiaadUeacaWGpbGaamisaaGaay5waiaaw2faamaaBaaaleaadaqadaqaaiaadggacaWGXbaacaGLOaGaayzkaaaabeaakiabg2da9maalaaabaGaaGimaiaac6cacaaI1aGaaGOnaiaaigdaaeaacaaIYaGaaGimaiaaicdacaGGVaGaaGymaiaaicdacaaIWaGaaGimaaaacaWGNbGaai4laiaadYeaaeaacqGH9aqpcaaIYaGaaiOlaiaaiIdacaaIWaGaaGynaiaadEgacaGGVaGaamitaaqaaiabg2da9maalaaabaGaaGOmaiaac6cacaaI4aGaaGimaiaaiwdaaeaacaaI1aGaaGOnaiaac6cacaaIXaGaaGymaaaacaWGnbaabaGaeyypa0JaaGimaiaac6cacaaIWaGaaGynaiaabccacaqGnbaabaGaaeivaiaabIgacaqGLbGaaeiiaiaabMgacaqGVbGaaeOBaiaabMgacaqGJbGaaeiiaiaabwgacaqGXbGaaeyDaiaabMgacaqGSbGaaeyAaiaabkgacaqGYbGaaeyAaiaabwhacaqGTbGaaeiiaiaabAgacaqGVbGaaeOCaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyyaiaabghacaqG1bGaaeyzaiaab+gacaqG1bGaae4CaiaabccacaqGlbGaae4taiaabIeacaqGGaGaaeyAaiaabohacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeOyaiaabwgacaqGSbGaae4BaiaabEhacaGG6aaabaGaam4saiaad+eacaWGibWaaeWaaeaacaWGHbGaamyCaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGlbWaaWbaaSqabeaacqGHRaWkaaGcdaqadaqaaiaadggacaWGXbaacaGLOaGaayzkaaGaey4kaSIaam4taiaadIeadaahaaWcbeqaaiabgkHiTaaakmaabmaabaGaamyyaiaadghaaiaawIcacaGLPaaaaaaa@B27D@

Thus, [K+] = [OH–] = 0.05M
[H+] x [OH–] = Kw
or [H+] = Kw/ [OH–]

= 10 −14 0.05 =2× 10 −13 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaeyypa0ZaaSaaaeaacaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaisdaaaaakeaacaaIWaGaaiOlaiaaicdacaaI1aaaaiabg2da9iaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaiodaaaGccaqGGaGaaeytaaaa@5203@

pH = – log [H+]
⇒ pH = – log (2 x 10–13) = 12.70

Q.40 The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Ans.

Molecular mass of Sr(OH)2 = 87.6 + 34 = 121.6 gmol-1

Concentration of Sr(OH)2

= 19.23g L −1 121.6gmo l −1 =0.1581 M Assuming complete dissociation of Sr ( OH ) 2 Sr ( OH ) 2 ⇌ S r 2+ +2O H − MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8E69@

[Sr2+] = 0.1581 M

[OH–] = 2 x 0.1581 M = 0.3162 M

pOH = – log (0.3162) = 0.5

pH = 14 – 0.5 = 13.5

Q.41 The ionization constant of propanoic acid is 1.32 ×10–5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?

Ans.

Let the degree of ionization of propanoic acid be α.

Assume propionic acid as HA, we have:

HA+ H 2 O 0.05−0.05α ⇌ H 3 O + 0.05α + A − 0.05α Let us assumeas 0.05−0.05α as 0.05. K a = [ H 3 O + ][ A − ] [ HA ] = ( 0.05α )( 0.05α ) 0.05 =0.05 α 2 α= K a 0.05 = 1.32× 10 −5 .05 =1.63× 10 −2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@C8A7@

[H3O+] = 0.05 α = 0.05 x 1.63 x 10–2 = 8.15 x 10–4 M

pH = – log [H+]

= – log [H3O+]

= – log (8.15 x 10–4)

= 3.09

In the presence of 0.1 M of HCl, let α’ be the degree of ionization.

[H3O+] = 0.01

[A–] = 0.05 α’

[HA]= 0.05

K a = ( 0.01 )( 0.05α ) 0.05 ⇒1.32× 10 −5 =0.01×α ⇒α=1.32× 10 −3MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@729A@

Q.42 The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Ans.

Concentration c = 0.1M

pH = 2.34

– log [H+] = 2.34

or [H+] = antilog (-2.34)

or [H+] = 4.5 x 10–3

Also [H+] = c . α

⇒ 4.5 x 10–3 = 0.1 x α

⇒ α = 45 x 10–3= 0.045

Ka = c . α2

= 0.1 (45 x 10–3)2

= 2.02 x 10–4

Q.43 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Ans.

NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2).

N O 2 − + H 2 O 0.04−x ⇌ HN O 2 x + O H − x MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIZaaCbeaeaacaWGobGaam4tamaaDaaaleaacaaIYaaabaGaeyOeI0caaOGaey4kaSIaamisamaaBaaaleaacaaIYaaabeaakiaad+eaaSabaeqabaaabaqcLbEacaaIWaGaaiOlaiaaicdacaaI0aGaeyOeI0IaamiEaaaaleqaaOWaa4GcaSqabeaaaOGaayjWHiaaw2BiamaaxababaGaamisaiaad6eacaWGpbWaaSbaaSqaaiaaikdaaeqaaaabaeqabaaabaqcLbEacaWG4baaaSqabaGccqGHRaWkdaWfqaqaaiaad+eacaWGibWaaWbaaSqabeaacqGHsislaaaaeaqabeaaaeaajug4biaadIhaaaWcbeaaaaa@5D44@ K h = [ HN O 2 ][ O H − ] [ N O 2 − ] ⇒ K w K a = 10 −14 4.5× 10 −4 =0.22× 10 −10 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7056@

Suppose x mole of salt undergoes hydrolysis, the concentration of various species present in the solution will be:

[NO2–]=0.04 – x ~ 0.04

[HNO2] = x

[OH–] = x

K h = [ HN O 2 ][ O H − ] [ N O 2 − ] K h = x 2 0.04 =0.22× 10 −10 ⇒ x 2 =0.0088× 10 −10 ⇒x=0.093× 10 −5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8266@

[OH–] = 0.093 x 10–5 M

[H+]= Kw /(0.093 x 10–5)

= 10–14 / (0.093 x 10–5)

= 10.75 x 10–9 M

pH = – log (10.75 x 10–9) = 7.96

Therefore, degree of hydrolysis h

h= x 0.04 = 0.093× 10 −5 0.04 =2.325× 10 −5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGObGaeyypa0ZaaSaaaeaacaWG4baabaGaaGimaiaac6cacaaIWaGaaGinaaaacqGH9aqpdaWcaaqaaiaaicdacaGGUaGaaGimaiaaiMdacaaIZaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiwdaaaaakeaacaaIWaGaaiOlaiaaicdacaaI0aaaaaqaaiabg2da9iaaikdacaGGUaGaaG4maiaaikdacaaI1aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiwdaaaaaaaa@5DA4@

Q.44 A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Ans.

pH = – log [H+]

⇒3.44 = – log [H+]

⇒[H+] = – antilog pH

⇒[H+] = – antilog (3.44)

⇒[H+] = 3.63 x 10–4

K h = [ pyridinium chloride ][ H + ] [ pyridinium hydrochloride ] ⇒ K h ( 3.63× 10 −4 ) 2 0.02 =6.6× 10 −6 ⇒ K h = K w K a = 10 −14 6.6× 10 −6 =1.51× 10 −9MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGlbWaaSbaaSqaaiaadIgaaeqaaOGaeyypa0ZaaSaaaeaadaWadaqaaiaadchacaWG5bGaamOCaiaadMgacaWGKbGaamyAaiaad6gacaWGPbGaamyDaiaad2gacaqGGaGaae4yaiaabIgacaqGSbGaae4BaiaabkhacaqGPbGaaeizaiaabwgaaiaawUfacaGLDbaadaWadaqaaiaadIeadaahaaWcbeqaaiabgUcaRaaaaOGaay5waiaaw2faaaqaamaadmaabaGaamiCaiaadMhacaWGYbGaamyAaiaadsgacaWGPbGaamOBaiaadMgacaWG1bGaamyBaiaabccacaqGObGaaeyEaiaabsgacaqGYbGaae4BaiaabogacaqGObGaaeiBaiaab+gacaqGYbGaaeyAaiaabsgacaqGLbaacaGLBbGaayzxaaaaaaqaaiabgkDiElaadUeadaWgaaWcbaGaamiAaaqabaGcdaWcaaqaamaabmaabaGaaG4maiaac6cacaaI2aGaaG4maiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGimaiaac6cacaaIWaGaaGOmaaaacqGH9aqpcaaI2aGaaiOlaiaaiAdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGOnaaaaaOqaaiabgkDiElaadUeadaWgaaWcbaGaamiAaaqabaGccqGH9aqpdaWcaaqaaiaadUeadaWgaaWcbaGaam4DaaqabaaakeaacaWGlbWaaSbaaSqaaiaadggaaeqaaaaakiabg2da9maalaaabaGaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaigdacaaI0aaaaaGcbaGaaGOnaiaac6cacaaI2aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiAdaaaaaaOGaeyypa0JaaGymaiaac6cacaaI1aGaaGymaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI5aaaaaaaaa@AD51@

Q.45 Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF.

Ans.

NaCN, NaNO2, KF solutions are basic, as they are salts of strong base and weak acid.

NaCl, KBr solutions are neutral, as they are salts of strong base and strong acid.

NH4NO3 solution is acidic, as it is a salt of strong acid and weak base.

Q.46 The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Ans.

It is given that Ka for ClCH2COOH is 1.35 × 10–3

K a =c α 2 α= K a C Let us put the values in the above expression α= 1.35× 10 −3 0.1 ⇒α= 1.35× 10 −2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9028@

= 0.116

[H+] = c . a

= 0.1 x 0.116

= 0.0116

⇒pH=−log[ H + ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaeyO0H4TaamiCaiaadIeacqGH9aqpcqGHsislciGGSbGaai4BaiaacEgadaWadaqaaiaadIeadaahaaWcbeqaaiabgUcaRaaaaOGaay5waiaaw2faaaaa@4D53@

= – log (.0116)

=1.94

ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.

pH=− 1 2 [ log K w +log K a −logc ] ⇒ph=− 1 2 [ log 10 −14 +log( 1.35× 10 −3 )−log0.1 ] ⇒pH=− 1 2 [ −14+( −3+0.1303 )−( −1 ) ] =7.94MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9842@

Q.47 Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

Ans.

Ionic product of water

K w =[ H + ][ O H − ] Since, [ H + ]=[ O H − ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGlbWaaSbaaSqaaiaadEhaaeqaaOGaeyypa0ZaamWaaeaacaWGibWaaWbaaSqabeaacqGHRaWkaaaakiaawUfacaGLDbaadaWadaqaaiaad+eacaWGibWaaWbaaSqabeaacqGHsislaaaakiaawUfacaGLDbaaaeaacaWGtbGaamyAaiaad6gacaWGJbGaamyzaiaacYcaaeaadaWadaqaaiaadIeadaahaaWcbeqaaiabgUcaRaaaaOGaay5waiaaw2faaiabg2da9maadmaabaGaam4taiaadIeadaahaaWcbeqaaiabgkHiTaaaaOGaay5waiaaw2faaaaaaa@5B08@

Thus, Kw = [H+]2 and Kw at 310 K is 2.7 × 10–14

⇒2.7 × 10–14 = [H+]2

⇒ [H+] = 1.64 x 10–7

pH = – log [H+]

= – log (1.64 x 10–7)

= 6.78

Hence, the pH of neutral water is 6.78.

Q.48 Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2 M Ca(OH)2+ 25 mL of 0.1 M HCl

b) 10 mL of 0.01 M H2SO4+ 10 mL of 0.01 M Ca(OH)2

c) 10 mL of 0.1 M H2SO4+ 10 mL of 0.1 M KOH

Ans.

(a)

10 mL of 0.2 M Ca(OH)2 = 10 x 0.2 = 2 millimoles of Ca(OH)2

25 mL of 0.1M HCl = 25 x 0.1 millimoles = 2.5 millimoles of HCl

Ca ( OH ) 2 2HCl → CaC l 2 +2 H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4qaiaadggadaqadaqaaiaad+eacaWGibaacaGLOaGaayzkaaWaaSbaaSqaaiaaikdaaeqaaOGaaGOmaiaadIeacaWGdbGaamiBamaaoqcaleaaaeqakiaawkziaiaadoeacaWGHbGaam4qaiaadYgadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaaIYaGaamisamaaBaaaleaacaaIYaaabeaakiaad+eaaaa@53C7@

2 millimoles of HCl reacts with = 1 millimoles of Ca(OH)2

1 millimoles of HCl reacts with = 1/2 millimoles of Ca(OH)2

2.5 millimoles of HCl reacts with = ½ x 2.5 = 1.25 millimoles of Ca(OH)2

Amount of Ca(OH)2 left = 2 – 1.25 = 0.75 millimoles

Total volume of the solution (10 + 25) mL = 35 mL

Molarity of Ca(OH)2 in the solution = (0.75/35)M = 0.0214 M

[OH–] = 2 x 0.0214 M = 0.0428 M = 4.28 x 10–2 M

pOH = – log (4.28 x 10–2) = 2 – 0.6314 » 1.37

pH = 14 – 1.37 = 12.63

(b) 10 mL 0.01 M H2SO4 = 10 x 0.01 millimoles of H2SO4 = 0.1 millimole

10 mL of 0.01 M Ca(OH)2 = 10 x 0.01 millimoles of Ca(OH)2 = 0.1 millimole

Ca ( OH ) 2 + H 2 S O 4 → CaS O 4 +2 H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4qaiaadggadaqadaqaaiaad+eacaWGibaacaGLOaGaayzkaaWaaSbaaSqaaiaaikdaaeqaaOGaey4kaSIaamisamaaBaaaleaacaaIYaaabeaakiaadofacaWGpbWaaSbaaSqaaiaaisdaaeqaaOWaa4ajaSqaaaqabOGaayPKHaGaam4qaiaadggacaWGtbGaam4tamaaBaaaleaacaaI0aaabeaakiabgUcaRiaaikdacaWGibWaaSbaaSqaaiaaikdaaeqaaOGaam4taaaa@55BB@

1 mole of Ca(OH)2 reacts with 1 mole of H2SO4.

(c) Thus, 0.1 millimole of Ca(OH)2 reacts completely with 0.1 millimole of H2SO4. The solution will be neutral with pH value as 7.0.10 mL 0.1 M H2SO4 = 1 millimole of H2SO4

10 mL 0.1 M KOH = 1 millimole of KOH

2KOH+ H 2 S O 4 → K 2 S O 4 +2 H 2 O MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaaGOmaiaadUeacaWGpbGaamisaiabgUcaRiaadIeadaWgaaWcbaGaaGOmaaqabaGccaWGtbGaam4tamaaBaaaleaacaaI0aaabeaakmaaoqcaleaaaeqakiaawkziaiaadUeadaWgaaWcbaGaaGOmaaqabaGccaWGtbGaam4tamaaBaaaleaacaaI0aaabeaakiabgUcaRiaaikdacaWGibWaaSbaaSqaaiaaikdaaeqaaOGaam4taaaa@5332@

2 millimole of KOH reacts with = 1 millimole of H2SO4

1 millimole of KOH reacts with = ½ = 0.5 millimole of H2SO4

Amount of H2SO4 left = (1 – .05) = 0.5 millimole

Volume of the reaction mixture = 10 + 10 = 20 mL

Molarity of H2SO4 in the solution = 0.5/20 = 2.5 x 10–2 M

[H+] = 2 x (2.5 x 102) = 5 x 10–2

pH = – log (5 x 10–2) = 2 – 0.699 = 1.3

Q.49 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given: Ag2CrO4 = 1.1 x 10–12,

BaCrO4 = 1.2 x 10–10, Fe(OH)3 = 1.0 x 10–38,

PbCl2 = 1.6 x 10–5, Hg2I2 = 4.5 x 10–29

Determine also the molarities of individual ions.

Ans.

  1. Silver chromate:

A g 2 Cr O 4 ⇌ 2A g + +Cr O 4 2− K sp = [ A g + ] 2 [ Cr O 4 2− ]MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6453@

Let the solubility of Ag2CrO4 be s.

For [Ag+] = 2s

[CrO42–] = s

Then,

Ksp = (2s) 2.s = 4 s3

⇒ 1.1 x 10–12 = 4 s3

⇒ s = 0.65 x 10–4 M

Molarity of Ag+ = 2 s

= 2 x (0.65 x 10–4)

= 1.3 x 10–4 M

Molarity of CrO42– = s = 0.65 x 10–4 M

  1. Barium chromate:

BaCr O 4 ⇌ B a 2+ +Cr O 4 2− K sp =[ B a + ][ Cr O 4 2− ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGcbGaamyyaiaadoeacaWGYbGaam4tamaaBaaaleaacaaI0aaabeaakmaaoOaaleqabaaakiaawcCicaGL9gcacaWGcbGaamyyamaaCaaaleqabaGaaGOmaiabgUcaRaaakiabgUcaRiaadoeacaWGYbGaam4tamaaDaaaleaacaaI0aaabaGaaGOmaiabgkHiTaaaaOqaaiaadUeadaWgaaWcbaGaam4CaiaadchaaeqaaOGaeyypa0ZaamWaaeaacaWGcbGaamyyamaaCaaaleqabaGaey4kaScaaaGccaGLBbGaayzxaaWaamWaaeaacaWGdbGaamOCaiaad+eadaqhaaWcbaGaaGinaaqaaiaaikdacqGHsislaaaakiaawUfacaGLDbaaaaaa@625F@ K sp =[ B a + ][ Cr O 4 2− ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaam4samaaBaaaleaacaWGZbGaamiCaaqabaGccqGH9aqpdaWadaqaaiaadkeacaWGHbWaaWbaaSqabeaacqGHRaWkaaaakiaawUfacaGLDbaadaWadaqaaiaadoeacaWGYbGaam4tamaaDaaaleaacaaI0aaabaGaaGOmaiabgkHiTaaaaOGaay5waiaaw2faaaaa@506D@

The solubility of BaCrO4 is s.

Thus, [Ba2+] = s

and [CrO42–] = s

Ksp = s . s = s2

⇒ s2 = 1.2 x 10–10

⇒ s = 1.09 x 10–5 M

Molarity of Ba2+ = Molarity of CrO42– = 1.09 x 10–5 M

  1. Ferric hydroxide:

Fe ( OH ) 3 ⇌ F e 3+ +3O H − K sp =[ F e 3+ ] [ O H − ] 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGgbGaamyzamaabmaabaGaam4taiaadIeaaiaawIcacaGLPaaadaWgaaWcbaGaaG4maaqabaGcdaGdkaWcbeqaaaGccaGLahIaayzVHaGaamOraiaadwgadaahaaWcbeqaaiaaiodacqGHRaWkaaGccqGHRaWkcaaIZaGaam4taiaadIeadaahaaWcbeqaaiabgkHiTaaaaOqaaiaadUeadaWgaaWcbaGaam4CaiaadchaaeqaaOGaeyypa0ZaamWaaeaacaWGgbGaamyzamaaCaaaleqabaGaaG4maiabgUcaRaaaaOGaay5waiaaw2faamaadmaabaGaam4taiaadIeadaahaaWcbeqaaiabgkHiTaaaaOGaay5waiaaw2faamaaCaaaleqabaGaaG4maaaaaaaa@609A@

Let the solubility product of Fe(OH)3 be s.

Thus, [Fe3+] = s

and [OH–] = 3s

∴ Ksp = s. (3s)3 = 27 s4

⇒ Ksp = 27 s4

⇒ 1.0 x 10–38 = 27 s4

⇒ 0.037 x 10–38 = s4

⇒ s = 1.39 x 10–10 M

Molarity of Fe3+ = s = 1.39 x 10–10 M

Molarity of OH– = 3s = 3 x 1.39 x 10–10 M

= 4.17 x 10–10 M

  1. Lead Chloride:

PbC l 2 ⇌ P b 2+ +2C l − K sp =[ P b 2+ ] [ C l − ] 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGqbGaamOyaiaadoeacaWGSbWaaSbaaSqaaiaaikdaaeqaaOWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaadcfacaWGIbWaaWbaaSqabeaacaaIYaGaey4kaScaaOGaey4kaSIaaGOmaiaadoeacaWGSbWaaWbaaSqabeaacqGHsislaaaakeaacaWGlbWaaSbaaSqaaiaadohacaWGWbaabeaakiabg2da9maadmaabaGaamiuaiaadkgadaahaaWcbeqaaiaaikdacqGHRaWkaaaakiaawUfacaGLDbaadaWadaqaaiaadoeacaWGSbWaaWbaaSqabeaacqGHsislaaaakiaawUfacaGLDbaadaahaaWcbeqaaiaaikdaaaaaaaa@5F69@

Let the solubility product of PbCl2 be s.

Thus, [Pb2+] = s

and [Cl–] = 2s

∴ Ksp = s. (2s)2

⇒ Ksp = 4 s3

⇒ 1.6 x 10–5 = 4 s3

⇒ s3 = 4 x 10–6

⇒ s = 1.58 x 10–2 M

Molarity of Pb2+ = s = 1.58 x 10–2 M

Molarity of Cl– = 2s = 2 x 1.58 x 10–2 M

= 3.16 x 10–2 M

5. Mercurous iodide:

H g 2 I 2 ⇌ H g 2 2+ +2 I − K sp =[ H g 2 2+ ] [ I − ] 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqafaaadaaQHaawMaaWZaaI4eaacaWGibGaam4zamaaBaaaleaacaaIYaaabeaakiaadMeadaWgaaWcbaGaaGOmaaqabaGcdaGdkaWcbeqaaaGccaGLahIaayzVHaGaamisaiaadEgadaqhaaWcbaGaaGOmaaqaaiaaikdacqGHRaWkaaGccqGHRaWkcaaIYaGaamysamaaCaaaleqabaGaeyOeI0caaaGcbaGaam4samaaBaaaleaacaWGZbGaamiCaaqabaGccqGH9aqpdaWadaqaaiaadIeacaWGNbWaa0baaSqaaiaaikdaaeaacaaIYaGaey4kaScaaaGccaGLBbGaayzxaaWaamWaaeaacaWGjbWaaWbaaSqabeaacqGHsislaaaakiaawUfacaGLDbaadaahaaWcbeqaaiaaikdaaaaaaaa@5F09@

Let the solubility product of Hg2I2 be s.

[Hg22+] = s

and [I–] = 2s

∴ Ksp= s . (2s)2

⇒ Ksp= 4 s3

⇒4.5 x 10–29 = 4 s3

⇒ s = 2.24 x 10–10 M

Molarity of [Hg22+] = s = 2.24 x 10–10 M

Molarity of [I–] = 2s = 2 x 2.24 x 10–10 M

= 4.48 x 10–10 M

Q.50 The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Ans.

Let the solubility of Ag2CrO4 = s

A g 2 Cr O 4 ⇌ 2A g + +Cr O 4 2− K sp = [ A g + ] 2 [ Cr O 4 2− ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6453@

Let the solubility of Ag2CrO4 be s.

For [Ag+] = 2s

[CrO42–] = s

Then,

Ksp= (2s2). s = 4 s3

⇒ 1.1 x 10–12 = 4 s3

⇒ s = 0.65 x 10–4 M = 6.5 x 10–5 M

Let s’ be the solubility of AgBr.

AgBr( s ) ⇌ A g + +B r − MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIJaamyqaiaadEgacaWGcbGaamOCamaabmaabaGaam4CaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGbbGaam4zamaaCaaaleqabaGaey4kaScaaOGaey4kaSIaamOqaiaadkhadaahaaWcbeqaaiabgkHiTaaaaaa@5031@

Thus, [Ag+] = [Br–] = s

⇒Ksp = (s’)2 = 5.0 x 10–13

⇒ s’ = 7.07 x 10–7 M

Therefore, the ratio of the molarities of their saturated solution is

s s = 6.5× 10 −5 M 7.07× 10 −7 M =91.9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaamaaOgcayzcaapdaiIZaaSaaaeaacaWGZbaabaGaam4CaiaacEcaaaGaeyypa0ZaaSaaaeaacaaI2aGaaiOlaiaaiwdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaakiaad2eaaeaacaaI3aGaaiOlaiaaicdacaaI3aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaaiEdaaaGccaWGnbaaaiabg2da9iaaiMdacaaIXaGaaiOlaiaaiMdaaaa@59F6@

Q.51 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp= 7.4 × 10–8).

Ans.

When we mix equal volumes of sodium iodate and cupric chlorate solutions, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Considering the ionization of both compounds:

NaI O 3 0.001 M → N a + + I O 3 − 0.001 M Cu ( Cl O 3 ) 2 0.001 M → C u 2+ + 2C l 3 − 0.001 M The solubility equilibrium for copper iodate can be written as: Cu ( I O 3 ) 2 ( aq ) ⇌ C u 2+ ( aq )+2I O 3 − ( aq ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@CF03@

Ionic product of copper iodate:

= [Cu2+] [IO3–]2

= (0.001) (0.001)2

= 1 x 10–9

The Ksp value for copper iodate, Ksp = 7.4 × 10–8

Since the ionic product (1 × 10–9) is less than Ksp (7.4 × 10–8), precipitation will not occur.

Q.52 The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Ans.

Given,

pH = – log [H+] = 3.19

⇒[H+] = antilog (–3.19)

= 6.457 x 10–4 M

C 6 H 5 COOH( aq ) ⇌ C 6 H 5 CO O − ( aq )+ H + ( aq ) K a = [ C 6 H 5 CO O − ( aq ) ][ H + ( aq ) ] [ C 6 H 5 COOH( aq ) ] ⇒ [ C 6 H 5 COOH( aq ) ] [ C 6 H 5 CO O − ( aq ) ] = [ H + ( aq ) ] K a = 6.46× 10 −4 6.46× 10 −5 =10 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@B7BE@

Let the solubility of C6H5COOAg be x mol/L

Then,

[Ag+] = x

⇒ [C6H5COOH] + [C6H5COO–] = x

⇒ 10 [C6H5COO–] + [C6H5COO–] = x

⇒11 [C6H5COO–] = x

⇒ [C6H5COO–] = x/11

K sp =[ A g + ][ C 6 H 5 CO O − ] ⇒2.5× 10 −13 =x.( x/11 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabaeqabaGaam4samaaBaaaleaacaWGZbGaamiCaaqabaGccqGH9aqpdaWadaqaaiaadgeacaWGNbWaaWbaaSqabeaacqGHRaWkaaaakiaawUfacaGLDbaadaWadaqaaiaadoeadaWgaaWcbaGaaGOnaaqabaGccaWGibWaaSbaaSqaaiaaiwdaaeqaaOGaam4qaiaad+eacaWGpbWaaWbaaSqabeaacqGHsislaaaakiaawUfacaGLDbaaaeaacqGHshI3caaIYaGaaiOlaiaaiwdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGymaiaaiodaaaGccqGH9aqpcaWG4bGaaiOlamaabmaabaGaamiEaiaac+cacaaIXaGaaGymaaGaayjkaiaawMcaaaaaaa@616E@

⇒ x = 1.66 x 10–6 mol/L

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10–6 mol/L.

Let the solubility of C6H5COOAg be y mol/L in pure water.

[Ag+] =y M

and [CH3COO–]= y M

K sp =[ A g + ][ C 6 H 5 CO O − ] ⇒y= Ksp = ( 2.5× 10 −13 ) =5× 10 −7 mol/L x y = 1.66× 10 −6 5× 10 −7 =3.32 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@82D2@

Hence, C6H5COOAg is approximately 3.32 times more soluble in a low pH solution.

Q.53 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp= 6.3 × 10–18).

Ans.

Let the maximum concentration of each solution be x mol/L. After mixing, the concentrations of each solution will be reduced to half i.e., (x/2) mol/L.

∴[ FeS O 4 ]=[ N a 2 S ]= x 2 M ∴[ F e 2+ ]=∴[ FeS O 4 ]= x 2 M ∴[ S 2− ]=[ N a 2 S ]= x 2 M FeS( s ) ⇌ F e 2+ ( aq )+ S 2− ( aq ) K sp =[ F e 2+ ][ S 2− ] ⇒6.3× 10 −18 = x 2 × x 2 ⇒6.3× 10 −18 = x 2 4 ⇒x=5.02× 10 −9 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@C2FF@

If the concentrations of both solutions are equal to or less than 5.02 × 10–9 M, then there will be no precipitation of iron Sulphide.

Q.54 What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).

Ans.

CaS O 4 ( s ) ⇌ C a 2+ ( aq )+S O 4 2− ( aq ) K sp =[ C a 2+ ][ S O 4 2− ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@69B4@

Let the solubility of CaSO4 be s.

Then, Ksp = s2

⇒ 9.1 x 10–6 = s2

⇒ s = 3.02 x 10–3 mol/L

Molecular mass of CaSO4 = 136 g/mol

Solubility of CaSO4 in gram/L = 3.02 x 10–3 x 136 g/L

= 0.411 g/L

Thus, to dissolve 0.411 g of CaSO4, we need water = 1 L

And to dissolve 1 g of CaSO4, we need water = (1/0.411) L

= 2.43 L

Q.55 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?

Given, Ksp for FeS = 6.3 x 10–18, MnS = 2.5 x 10–13, ZnS = 1.6 x 10–24 and CdS = 8.0 x 10–27

Ans.

For precipitation to take place, it is required that the calculated ionic product exceeds the Ksp value.

10mL of solution containing S2- ion is mixed with 5 mL of metal salt solution.

Before mixing,

[ S 2− ]=1.0× 10 −19 M ( Volume= 10 mL ) [ M 2+ ]=0.04M ( Volume= 5 mL ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7272@

After mixing,

[S2–] = 1.0 x 10–19 x (10/15) = 6.67 x 10–20

[M2+]= [Fe3+] = [Mn2+] = [Zn2+] = [Cd2+]

= 0.04 x (5/15) = 1.33 x 10–2 M

∴ Ionic product of each will be=

[M2+][S2–] = (1.33 x 10–2) (6.67 x 10–20)

= 8.87 x 10–22

This ionic product exceeds the Ksp of ZnS and CdS. Therefore, precipitation will occur in ZnCl2 and CdCl2 solutions.

Q.56 A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially?

c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Ans.

(a) If the volume of the container is increased suddenly, the vapour pressure decreases initially. It is so because the volume of the vapour remains the same, but it is distributed in a larger volume.

(b) Rate of evaporation is directly proportional to the temperature. Since the temperature remains constant, the rate of evaporation also remains unchanged. When the volume of the container is increased, the density of the vapour phase decreases. At the same time, the rate of collisions of the vapour particles also decreases. As a result, the rate of condensation decreases initially.

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. Only the volume changes while the temperature remains constant. The vapour pressure depends on the temperature and not on the volume. The final vapour pressure will be equal to original vapour pressure of the system.

Q.57 What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] = 0.60 M, [O2] = 0.82 M and [SO3] =1.90 M?

2S O 2 ( g )+ O 2 ⇌ 2S O 3 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaaiaaikdacaWGtbGaam4tamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaad+eadaWgaaWcbaGaaGOmaaqabaGcdaGdkaWcbeqaaaGccaGLahIaayzVHaGaaGOmaiaadofacaWGpbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@4EFE@

Ans.

The expression for the equilibrium constant (Kc) for the give reaction is:

K c = [ S O 3 ] 2 [ S O 2 ] 2 [ O 2 ] K c = ( 1.90 ) 2 M 2 ( 0.60 ) 2 ( 0.821 ) M 3 =12.239 M −1 ( approx ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@75E2@

Hence, Kc for the equilibrium Kc is 12.239 M-1.

Q.58 At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms

I 2 ( g ) ⇌ 2I( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaaiaadMeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaaGOmaiaadMeadaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@48EB@

Calculate Kp for the equilibrium.

Ans.

Partial pressure of I atoms,

p I = 40 100 × p total = 40 100 × 10 5 =4× 10 4 Pa Partial pressure of I 2 molecules p I 2 = 60 100 × p total = 60 100 × 10 5 =6× 10 4 Pa Let us write the expression of K p K p = ( pI ) 2 p I 2 Place the value of partial pressure of iodine atom and iodine molecule in this equation. We get K p = ( 4× 10 4 ) 2 P a 2 6× 10 6 Pa =2.67× 10 4 Pa MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@3535@

Q.59 Write the expression for the equilibrium constant, Kc for each of the following reactions:

(i) 2NOCl( g ) ⇌ 2NO( g )+C l 2 ( g ) (ii) 2Cu ( N O 3 ) 2 ( s ) ⇌ 2CUO( s )+4N O 2 ( g )+ O 2 ( g ) (iii) C H 3 COO C 2 H 5 ( aq )+ H 2 O( I ) ⇌ C H 3 COOH( aq )+ C 2 H 5 OH( aq ) (iv) F e 3+ ( aq )+3O H − ( aq ) ⇌ Fe ( OH ) 3 ( s ) (v) I 2 ( s )+5 F 2 ⇌ 2I F 5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@D2E5@

Ans.

(i) K c = [ N O ( g ) ] 2 [ C l 2( g ) ] 2 [ NOC l ( g ) ] 2 (ii) K c = [ N O 2( g ) ] 4 [ O 2( g ) ] ( The concentration of a solid is considered as constant. ) (iii) K c = [ C H 3 COOH( aq ) ][ C 2 H 5 OH( aq ) ] [ C H 3 COO C 2 H 5 ( aq ) ] ( When water is used as solvent, its concentration remains constant. ) (iv) K c = 1 [ F e 3+ ( aq ) ] [ O H − ( aq ) ] 3 ( The concentration of a solid is considered as constant. ) (v) K c = [ I F 5 ] [ F 2 ] 5 2 ( The concentration of a solid is considered as constant. ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@9822@

Q.60 Find out the value of Kc for each of the following equilibrium from the value of Kp:

(i) 2NOCl( g ) ⇌ 2NO( g )+C l 2 ( g ); K p =1.8× 10 −2 at 500 K (ii) CaC O 3 ( s ) ⇌ CaO( s )+C O 2 ( g ); K p =167 at 1073 K MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8E97@

Ans.

The relationship between Kp and Kc is written as:

K p = K c ( RT ) Δn MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaam4samaaBaaaleaacaWGWbaabeaakiabg2da9iaadUeadaWgaaWcbaGaam4yaaqabaGcdaqadaqaaiaadkfacaWGubaacaGLOaGaayzkaaWaaWbaaSqabeaacqqHuoarcaWGUbaaaaaa@4910@

(i) For the given reaction, the required values of these terms are –

Δn= 3 – 2 = 1

R= 0.0831 bar L mol–1 K–1

T= 500 K

Kp= 1.8 × 10–2

Let us place these values in the following equation,

K p = K c ( RT ) Δn ⇒1.8× 10 −2 = K c ( 0.0831×500 ) 1 ⇒ K c = 1.8× 10 −2 0.0831×500 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@76D9@

= 4.33 x 10-4 (approx)

ii) Δn = 2 – 1 = 1

R = 0.0831 bar L mol–1 K–1

T = 1073 K

Kp= 167

Now,

K p = K c ( RT ) Δn MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaam4samaaBaaaleaacaWGWbaabeaakiabg2da9iaadUeadaWgaaWcbaGaam4yaaqabaGcdaqadaqaaiaadkfacaWGubaacaGLOaGaayzkaaWaaWbaaSqabeaacqqHuoarcaWGUbaaaaaa@4910@ ⇒167= K c ( 0.0831×1073 ) 1 ⇒ K c = 167 0.0831×1073 =1.87( approx )MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6E1A@

Q.61 For the following equilibrium Kc= 6.3 x 1014 at 1000K,

NO( g )+ O 3 ( g ) ⇌ N O 2 ( g )+ O 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaamOtaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGpbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2Biaiaad6eacaWGpbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaey4kaSIaam4tamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaaa@54E9@

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

Ans.

For the forward reaction, Kc = 6.3 x 1014

Kc for the backward (reverse) reaction, Kc’ = 1/Kc

= 1 6.3× 10 14 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOnaiaac6cacaaIZaGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiaaigdacaaI0aaaaaaaaaa@4794@

= 1.59 x 10-15

Q.62 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Ans.

For a pure substance (both solids and liquids), the concentration term can be represented as-

Pure substance= Number of moles Volume = Mass/Molecular mass Volume = Mass Volume×Molecular mass = Density Molecular mass MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@A427@

At a particular temperature, the molecular mass and the density of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances can be ignored in the equilibrium constant expression.

Q.63 Reaction between N2 and O2 takes place as follows:

2 N 2 ( g )+ O 2 ( g ) ⇌ 2 N 2 O( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaaGOmaiaad6eadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGpbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaaikdacaWGobWaaSbaaSqaaiaaikdaaeqaaOGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaaaaa@5161@

If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which

Kc= 2.0 × 10–37, determine the composition of equilibrium mixture.

Ans.

Let us assume that the concentration Of N2O at equilibrium be x.

Let us assume that the concentration Of N 2 O at equilibrium be x. 2 N 2 ( g ) Initial conc. 0.482mol At equilibrium ( 0.482−x )mol + O 2 ( g ) 0.933 mol ( 0.933−x )mol ⇌ 2 N 2 O( g ) 0 x mol The reaction is carried out in 10 L vessel, at equilibrium [ N 2 ]= 0.482−x 10 [ O 2 ]= 0.933−x 10 [ N 2 O ]=x/10 The value of equilibrium constant is very small ( as given ). Therefore, the amount of N 2 and O 2 reacted is also very small. Thus x can be neglected from the molar concentration terms of N 2 and O 2 . [ N 2 ]= 0.482 10 =0.0482 mol L −1 [ O 2 ]= 0.933 10 =0.0933 mol L −1 K c = [ N 2 O ( g ) ] 2 [ N 2( g ) ] 2 [ O 2( g ) ] ⇒2.0× 10 −37 = ( x 10 ) 2 ( 0.0482 ) 2 ( 0.0933 ) ⇒ x 2 100 =2.0× 10 −37 × ( 0.0482 ) 2 ×( 0.0933 ) ⇒ x 2 =43.35× 10 −40 ⇒x=6.6× 10 −20 [ N 2 O ]= x 10 = 6.6× 10 −20 10 =6.6× 10 −21 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@810E@

Q.64 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO( g )+B r 2 ( g ) ⇌ 2NOBr( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaaGOmaiaad6eacaWGpbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaey4kaSIaamOqaiaadkhadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaaGOmaiaad6eacaWGpbGaamOqaiaadkhadaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@52F9@

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

Ans.

The given reaction is:

2NO( g ) 2 mol + B r 2 ( g ) 1 mol ⇌ 2NOBr( g ) 2 mol MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyYaaCbeaeaacaaIYaGaamOtaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaaaSabaeqabaaabaqcLbEacaaIYaGaaeiiaiaab2gacaqGVbGaaeiBaaaaleqaaOGaey4kaSYaaCbeaeaacaWGcbGaamOCamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaWceaqabeaaaeaajug4biaaigdacaqGGaGaaeyBaiaab+gacaqGSbaaaSqabaGcdaGdkaWcbeqaaaGccaGLahIaayzVHaWaaCbeaeaacaaIYaGaamOtaiaad+eacaWGcbGaamOCamaabmaabaGaam4zaaGaayjkaiaawMcaaaWceaqabeaaaeaajug4biaaikdacaqGGaGaaeyBaiaab+gacaqGSbaaaSqabaaaaa@646A@

2 moles of NOBr are formed from 1 mole of Br.

Thus, 0.0518 mole of NOBr are formed from

= (0.0518/2) mole of Br or NO

= 0.0259 mole of Br or 0.0259 mole of NO.

The initial amount of NO and Br present is as follows:

[NO] = 0.087 mol

[Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518 = 0.0352 mol

And the amount of Br present at equilibrium is:

[Br2] = 0.0437 – 0.0259 = 0.0178 mol

Q.65 At 450 K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.

2S O 2 ( g )+ O 2 ( g ) ⇌ 2S o 3 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaaGOmaiaadofacaWGpbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaey4kaSIaam4tamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaaIYaGaam4uaiaad+gadaWgaaWcbaGaaG4maaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@5260@

What is Kc at this temperature?

Ans.

For the given reaction,

Δn = 2 – 3 = -1

T= 450 K

Kp = 2.0 x 1010 bar-1

R= 0.0831 L bar K–1 mol–1

From the relation,

K p = K c ( RT ) Δn ⇒2.0× 10 10 ba r −1 = K c ( 0.0831 L bar K −1 mo l −1 ×450K ) −1 K c = ( 2.0× 10 10 ba r −1 ) ( 0.0831 L bar K −1 mo l −1 ×450 K ) −1 =7.48× 10 11 L mol −1 =7.48× 10 11 M −1

Q.66 A sample of HI (g) is placed in flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?

2HI( g ) ⇌ H 2 ( g )+ I 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaaGOmaiaadIeacaWGjbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaadIeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGjbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@4F9B@

Ans.

Let us denote concentrations in terms of pressure. The initial concentration of HI is 0.2 atm. At equilibrium, partial pressure of HI is 0.04 atm. Therefore, a decrease in the pressure of HI is (0.2 – 0.04) = 0.16.

2HI( g ) Initial conc. 0.2 atm At equilibrium 0.04 atm ⇌ H 2 ( g ) 0 0.16/2 atm =0.08 atm + I 2 ( g ) 0 0.16/2 atm =0.08 atm K p = p H 2 × p I 2 p HI 2 = ( 0.08 atm×0.08 atm ) ( 0.04 atm ) 2 = .0064 .0016 =4.0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@D490@

Q.67 A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction is 1.7 x 102.

N 2 ( g )+3 H 2 ( g ) ⇌ 2N H 3 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaamOtamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaaiodacaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaaikdacaWGobGaamisamaaBaaaleaacaaIZaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaaa@5155@

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans.

For the given reaction

N 2 ( g )+3 H 2 ( g ) ⇌ 2N H 3 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaamOtamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaaiodacaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaaikdacaWGobGaamisamaaBaaaleaacaaIZaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaaa@5155@

The given reaction, the concentration for various species is as follows –

[N2]= (1.57/20) mol L-1

[H2]= (1.92/20) mol L-1

[NH3]= (8.13/20) mol L-1

Now, let us calculate reaction quotient Qc.

Q c = [ N H 3 ] 2 [ N 2 ] [ H 2 ] 3 = [ 8.13 20 mol L −1 ] 2 [ 1.57 20 mol L −1 ] [ 1.92 20 mol L −1 ] 3 =2.38× 10 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8053@

As Qc ≠ Kc, the reaction mixture is not at equilibrium.

And Qc >Kc, the net reaction will proceed in the backward (reverse) direction.

Q.68

The equilibrium constant expression for a gas reaction is, K c = [ NH 3 ] 4 [ O 2 ] 5 [ NO ] 4 [ H 2 O ] 6 Write the balanced chemical equation corresponding to this expression. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@C915@

Ans.

The balanced chemical equation can be written as: 4NO( g )+6 H 2 O( g ) ⇌ 4N H 3 ( g )+5 O 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@84C4@

Q.69 One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,

H 2 O( g )+CO( g ) ⇌ H 2 ( g )+C O 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaamisamaaBaaaleaacaaIYaaabeaakiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGdbGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaey4kaSIaam4qaiaad+eadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaaa@5598@

Calculate the equilibrium constant for the reaction.

Ans.

H 2 O( g ) Initial conc. 1 10 M At equilibrium 1−0.4 10 M =0.06 M + CO( g ) 1 10 M 1−0.4 10 M =0.06 M ⇌ H 2 ( g ) 0 0.4 10 M =0.04 M + C O 2 ( g ) 0 0.4 10 M =0.04 M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyYaaCbeaeaacaWGibWaaSbaaSqaaiaaikdaaeqaaOGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaaaWceaqabeaaaeaajug4biaadMeacaWGUbGaamyAaiaadshacaWGPbGaamyyaiaadYgacaqGGaGaae4yaiaab+gacaqGUbGaae4yaiaab6cacaqGGaGcdaWcaaqaaiaaigdaaeaacaaIXaGaaGimaaaacaWGnbaaleaaaeaajug4biaadgeacaWG0bGaaeiiaiaabwgacaqGXbGaaeyDaiaabMgacaqGSbGaaeyAaiaabkgacaqGYbGaaeyAaiaabwhacaqGTbGaaeiiaOWaaSaaaeaacaaIXaGaeyOeI0IaaGimaiaac6cacaaI0aaabaGaaGymaiaaicdaaaGaamytaaWcbaaabaGccqGH9aqpcaaIWaGaaiOlaiaaicdacaaI2aGaaeiiaiaab2eaaaWcbeaakiabgUcaRmaaxababaGaam4qaiaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaaaSabaeqabaaabaGcdaWcaaqaaiaaigdaaeaacaaIXaGaaGimaaaacaWGnbaaleaaaeaakmaalaaabaGaaGymaiabgkHiTiaaicdacaGGUaGaaGinaaqaaiaaigdacaaIWaaaaiaad2eaaSqaaaqaaOGaeyypa0JaaGimaiaac6cacaaIWaGaaGOnaiaabccacaqGnbaaaSqabaGcdaGdkaWcbeqaaaGccaGLahIaayzVHaWaaCbeaeaacaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaalqaabeqaaaqaaOGaaGimaaWcbaaabaqcLbEadaWcaaWcbaqcLbEacaaIWaGaaiOlaiaaisdaaSqaaKqzGhGaaGymaiaaicdaaaGaamytaaWcbaaabaGccqGH9aqpcaaIWaGaaiOlaiaaicdacaaI0aGaaeiiaiaab2eaaaWcbeaakiabgUcaRmaaxababaGaam4qaiaad+eadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaaaSabaeqabaaabaGccaaIWaaaleaaaeaakmaalaaaleaajug4biaaicdacaGGUaGaaGinaaWcbaqcLbEacaaIXaGaaGimaaaacaWGnbaaleaaaeaakiabg2da9iaaicdacaGGUaGaaGimaiaaisdacaqGGaGaaeytaaaaleqaaaaa@ACD2@

At equilibrium, the given concentrations are:

[H2O]= (1 – 0.40)/10 mol L-1 = 0.06 mol L-1

[CO]= 0.06 mol L-1

[H2]=0.4/10 mol L-1 = 0.04 mol L-1

[CO2]= 0.04 mol L-1

K c = [ H 2 ][ C O 2 ] [ CO ][ H 2 O ] = 0.04×0.04 0.06×0.06 =0.444MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@6735@

Q.70 At 700 K, equilibrium constant for the reaction

H 2 ( g )+ I 2 ( g ) ⇌ 2HI( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaamisamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaadMeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaaGOmaiaadIeacaWGjbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@4F9B@

is 54.8. If 0.5 mol L–1of HI (g) is present at equilibrium at 700 K, what are the concentration of H2 (g) and I2 (g) assuming that we initially started with HI (g) and allowed it to reach equilibrium at 700 K?

Ans.

The equilibrium constant of the forward reaction Kc = 54.8
Thus, the equilibrium constant of the backward reaction K’c = 1/54.8
The backward reaction can be written as:

2HI( g ) ⇌ H 2 ( g )+ I 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaciaa4naaOyIaaGOmaiaadIeacaWGjbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaWaa4GcaSqabeaaaOGaayjWHiaaw2BiaiaadIeadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGjbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@4F9B@

[HI]= 0.5 mol L-1
Let the concentrations of hydrogen and iodine at equilibrium be x mol L-1
[H2] = [I2] = x mol L-1

[ H 2 ][ I 2 ] [ HI ] 2 =K c ⇒ x⋅x ( 0.5 ) 2 = 1 54.8 ⇒ x 2 = 0.25 54.8 Or, x=0.068 mol L −1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@78D6@

Hence, at equilibrium, [H2] = [I2] = 0.068 mol L-1

Q.71 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl( g ) ⇌ I 2 ( g )+C l 2 ( g ); K c =0.14 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacmaa4naaG2baOyIaaGOmaiaadMeacaWGdbGaamiBamaabmaabaGaam4zaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGjbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaey4kaSIaam4qaiaadYgadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacaGG7aGaaCzcaiaadUeadaWgaaWcbaGaam4yaaqabaGccqGH9aqpcaaIWaGaaiOlaiaaigdacaaI0aaaaa@596B@

Ans.

For the given reaction,

2ICl( g ) Initial conc. 0.78 M At equilibrium ( 0.78-2x ) M ⇌ I 2 ( g ) 0 x M + C l 2 ( g ) 0 x M MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@8955@

Now we can write from the expression,

Q.72 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C 2 H 6 ( g ) ⇌ C 2 H 4 ( g )+ H 2 ( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacmaa4naaG2baOyIaam4qamaaBaaaleaacaaIYaaabeaakiaadIeadaWgaaWcbaGaaGOnaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaadaGdkaWcbeqaaaGccaGLahIaayzVHaGaam4qamaaBaaaleaacaaIYaaabeaakiaadIeadaWgaaWcbaGaaGinaaqabaGcdaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWGibWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaaaa@533A@

Ans.

Let us assume that p is the pressure exerted by ethane as well as hydrogen gas at equilibrium. Note down the chemical reaction,

C 2 H 6 ( g ) Initial conc. 4.0 atm At equilibrium ( 4.0−p )atm ⇌ C 2 H 4 ( g ) 0 p atm + H 2 ( g ) 0 p atm p C 2 H 4 × p H 2 p C 2 H 6 = K p ⇒ p×p ( 40−p ) =0.04 ⇒ p 2 =0.16−0.04p ⇒ p 2 +0.04p−0.16=0 p= −0.04± ( 0.04 ) 2 −4×1×( 0.16 ) 2×1 = −0.04±0.80 2 =0.76/2 p=0.38 Hence, at equilibrium, [ C 2 H 6 ]=4−p=4−0.38=3.62 atm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabceqadaaEdaaODaakMeaadaWfqaqaaiaadoeadaWgaaWcbaGaaGOmaaqabaGccaWGibWaaSbaaSqaaiaaiAdaaeqaaOWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaalqaabeqaaaqaaKqzGhGaamysaiaad6gacaWGPbGaamiDaiaadMgacaWGHbGaamiBaiaabccacaqGJbGaae4Baiaab6gacaqGJbGaaeOlaiaabccacaqG0aGaaeOlaiaabcdacaqGGaGaaeyyaiaabshacaqGTbaaleaaaeaajug4biaabgeacaqG0bGaaeiiaiaabwgacaqGXbGaaeyDaiaabMgacaqGSbGaaeyAaiaabkgacaqGYbGaaeyAaiaabwhacaqGTbGaaeiiaOWaaeWaaeaacaaI0aGaaiOlaiaaicdacqGHsislcaWGWbaacaGLOaGaayzkaaGaamyyaiaadshacaWGTbaaaSqabaGcdaGdkaWcbeqaaaGccaGLahIaayzVHaWaaCbeaeaacaWGdbWaaSbaaSqaaiaaikdaaeqaaOGaamisamaaBaaaleaacaaI0aaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaWceaqabeaaaeaajug4biaaicdaaSqaaaqaaKqzGhGaamiCaiaabccacaqGHbGaaeiDaiaab2gaaaWcbeaakiabgUcaRmaaxababaGaamisamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaaWceaqabeaaaeaajug4biaaicdaaSqaaaqaaKqzGhGaamiCaiaabccacaqGHbGaaeiDaiaab2gaaaWcbeaaaOqaamaalaaabaGaamiCamaaBaaaleaacaWGdbWaaSbaaWqaaiaaikdaaeqaaSGaamisamaaBaaameaacaaI0aaabeaaaSqabaGccqGHxdaTcaWGWbWaaSbaaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaaaleqaaaGcbaGaamiCamaaBaaaleaacaWGdbWaaSbaaWqaaiaaikdaaeqaaSGaamisamaaBaaameaacaaI2aaabeaaaSqabaaaaOGaeyypa0Jaam4samaaBaaaleaacaWGWbaabeaaaOqaaiabgkDiEpaalaaabaGaamiCaiabgEna0kaadchaaeaadaqadaqaaiaaisdacaaIWaGaeyOeI0IaamiCaaGaayjkaiaawMcaaaaacqGH9aqpcaaIWaGaaiOlaiaaicdacaaI0aaabaGaeyO0H4TaamiCamaaCaaaleqabaGaaGOmaaaakiabg2da9iaaicdacaGGUaGaaGymaiaaiAdacqGHsislcaaIWaGaaiOlaiaaicdacaaI0aGaamiCaaqaaiabgkDiElaadchadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIWaGaaiOlaiaaicdacaaI0aGaamiCaiabgkHiTiaaicdacaGGUaGaaGymaiaaiAdacqGH9aqpcaaIWaaabaGaamiCaiabg2da9maalaaabaGaeyOeI0IaaGimaiaac6cacaaIWaGaaGinaiabgglaXoaakaaabaWaaeWaaeaacaaIWaGaaiOlaiaaicdacaaI0aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGinaiabgEna0kaaigdacqGHxdaTdaqadaqaaiaaicdacaGGUaGaaGymaiaaiAdaaiaawIcacaGLPaaaaSqabaaakeaacaaIYaGaey41aqRaaGymaaaaaeaacqGH9aqpdaWcaaqaaiabgkHiTiaaicdacaGGUaGaaGimaiaaisdacqGHXcqScaaIWaGaaiOlaiaaiIdacaaIWaaabaGaaGOmaaaaaeaacqGH9aqpcaaIWaGaaiOlaiaaiEdacaaI2aGaai4laiaaikdaaeaacaWGWbGaeyypa0JaaGimaiaac6cacaaIZaGaaGioaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaiilaiaabccacaqGHbGaaeiDaiaabccacaqGLbGaaeyCaiaabwhacaqGPbGaaeiBaiaabMgacaqGIbGaaeOCaiaabMgacaqG1bGaaeyBaiaacYcaaeaadaWadaqaaiaadoeadaWgaaWcbaGaaGOmaaqabaGccaWGibWaaSbaaSqaaiaaiAdaaeqaaaGccaGLBbGaayzxaaGaeyypa0JaaGinaiabgkHiTiaadchacqGH9aqpcaaI0aGaeyOeI0IaaGimaiaac6cacaaIZaGaaGioaiabg2da9iaaiodacaGGUaGaaGOnaiaaikdacaqGGaGaaeyyaiaabshacaqGTbaaaaa@23F9@

Q.73 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

C H 3 COOH( l )+ C 2 H 5 OH( l ) ⇌ C H 3 COO C 2 H 5 ( l )+ H 2 O( l ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacmaa4naaG2baOyIaam4qaiaadIeadaWgaaWcbaGaaG4maaqabaGccaWGdbGaam4taiaad+eacaWGibWaaeWaaeaacaWGSbaacaGLOaGaayzkaaGaey4kaSIaam4qamaaBaaaleaacaaIYaaabeaakiaadIeadaWgaaWcbaGaaGynaaqabaGccaWGpbGaamisamaabmaabaGaamiBaaGaayjkaiaawMcaamaaoOaaleqabaaakiaawcCicaGL9gcacaWGdbGaamisamaaBaaaleaacaaIZaaabeaakiaadoeacaWGpbGaam4taiaadoeadaWgaaWcbaGaaGOmaaqabaGccaWGibWaaSbaaSqaaiaaiwdaaeqaaOWaaeWaaeaacaWGSbaacaGLOaGaayzkaaGaey4kaSIaamisamaaBaaaleaacaaIYaaabeaakiaad+eadaqadaqaaiaadYgaaiaawIcacaGLPaaaaaa@63D7@

  1. Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
  2. At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
  3. Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Ans.

Reaction quotient,

Q c = [ C H 3 COO C 2 H 5 ][ H 2 O ] [ C H 3 COOH ][ C 2 H 5 OH ] ( ii ) Let us assume that the volume of the reaction mixture is V. Also, we will consider that water is a solvent and is present in excess. The given chemical reaction is: C H 3 COOH( l ) Initial conc. 1/V M At equilibrium ( 1−0.171 )/V =0.829/V + C 2 H 5 OH( l ) 0.18/V M ( 0.18-0.171 )/V =0.009/V ⇌ C H 3 COO C 2 H 5 ( l ) 0 0.171/V + H 2 O( l ) 0 0.171/V Therefore, equilibrium constant for the given reaction is: K c = [ C H 3 COO C 2 H 5 ][ H 2 O ] [ C H 3 COOH ][ C 2 H 5 OH ] = 0.171 V × 0.171 V 0.829 V × 0.829 V =3.92( approx ) (iii) Let us assume again that the volume of the reaction mixture as V. C H 3 COOH( l ) Initial conc. 1.0/V M At equilibrium ( 1−2.14 )/V =0.786/V M + C 2 H 5 OH( l ) 0.5/V M ( 0.5-0.214 )/V =0.286/V ⇌ C H 3 COO C 2 H 5 ( l ) 0 0.214/V M + H 2 O( l ) 0 0.214/V M The expression for the quotient is, Q c = [ C H 3 COO C 2 H 5 ][ H 2 O ] [ C H 3 COOH ][ C 2 H 5 OH ] = 0.214 V × 0.214 V 0.786 V × 0.786 V =0.2037 =0.204( approx ) As Q c < K c ,equilibrium has not been reached. 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FAQs (Frequently Asked Questions)

1. What is the CBSE Chemistry NCERT Class 11 Chapter 7 about?

Chapter 7 provides a brief overview of the various concepts of equilibrium in chemical and physical processes, as well as details on how equilibrium is dynamic. The law of mass action, various factors affecting equilibrium, and the equilibrium constant based on Le Chatelier’s principle are also discussed in this chapter. Equilibrium is the most important part of chemistry because it explains how objects behave.. Chemical theories and models are used to explain equilibrium in this chapter. 

2. What does buffer solution mean?

A buffer solution is a water-solvent solution which is a mix that is either made of a weak base and the conjugate acid, or a weak acid and the conjugate base. Dilution or the addition of small amounts of acid or alkali to it does not change the pH. 

3. How can the NCERT Solutions for Class 11 Chemistry 7 help you to prepare for your board exam?

Students who refer to NCERT Solutions for Class 11 Chemistry Chapter 7 improve their chances of passing their final exams with high scores and acing the topic. The solutions are created as per the updated syllabus, and cover all of the important topics of the chapter. As a result, answering these questions will boost students’ confidence when they prepare for board exams. 

4. Make a list of the key points covered in Chapter 7 of the NCERT Solutions for Class 11 Chemistry.

The following are some of the important topics covered in the NCERT Solutions for Class 11 Chemistry from Chapter 7: 

– The factors that influence equilibrium 

– Equilibrium in homogeneous and heterogeneous systems 

– Chemical and physical processes that are in equilibrium 

– The relationship between the equilibrium constant K and the rate of change 

– The use of an equilibrium constant 

5. What are the characteristics of chemical equilibrium?

Chemical equilibrium can be achieved from any side of the reaction. 

-Even after achieving this state, the reaction continues. 

-Both reactants and products have the same concentration. 

-Catalyst aids in the acceleration of reactions.Â