# 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 PCl _{5} was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl_{5} was found to be 0.5 × 10^{–1 }mol L^{–1}. If value of K_{c} is 8.3 × 10^{–3}, what are the concentrations of PCl_{3} and Cl_{2} at equilibrium?**

PC{l}_{5}(g)\underset{}{\rightleftharpoons}PC{l}_{3}(g)+C{l}_{2}(g)

**Ans.**

At equilibrium, let us assume that the concentration of PCl_{3} and Cl_{2} as x mol L^{-1 }

\underset{At\text{equilibrium 0}{\text{.5\xd710}}^{-1}mol{L}^{-1}}{PC{l}_{5}(g)}\underset{}{\rightleftharpoons}\underset{x{\text{molL}}^{-1}}{PC{l}_{3}(g)}+\underset{x{\text{molL}}^{-1}}{C{l}_{2}(g)}

The value of equilibrium constant K_{c }= 8.3 x 10^{-3 }

Now we can write the expression for equilibrium as:

\begin{array}{l}\Rightarrow \text{}\frac{\left[PC{l}_{3}\right]\left[C{l}_{2}\right]}{\left[PC{l}_{5}\right]}={K}_{c}\\ \Rightarrow \text{}\frac{x\cdot +x}{\left(0.5\times {10}^{-1}\right)}=8.3\times {10}^{-3}\end{array}

⇒ x^{2}= 4.15 x 10^{-4}

⇒ x = 2.04 x 10^{-2}

= 0.0204

= 0.02 (approx)

At equilibrium, we can write as-

[PCl_{3}] = [Cl_{2}] = 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 CO _{2}.**

FeO\left(s\right)+CO\left(g\right)\underset{}{\rightleftharpoons}Fe\left(s\right)+C{O}_{2}\left(g\right),\text{}{K}_{p}=0.265\text{}at\text{}1050K

What are the equilibrium partial pressures of CO and CO_{2} at 1050 K if the initial partial pressures are: Pco=1.4atm and Pco_{2} = 0.80 atm?

**Ans.**

Pco=1.4atm and Pco_{2} = 0.80 atm ?

\begin{array}{l}\underset{Initially}{FeO\left(s\right)}+\underset{1.4\text{atm}}{CO\left(g\right)}\underset{}{\rightleftharpoons}Fe\left(s\right)+\underset{0.80\text{atm}}{C{O}_{2}\left(g\right)}\\ {Q}_{p}=\frac{{p}_{C{O}_{2}}}{{p}_{CO}}=\frac{0.80}{1.4}\end{array}

= 0.571

Given K_{p} = 0.265 at 1050 K

Q_{p }> K_{p}

Thus, the reaction will proceed in the backward direction.

To achieve the equilibrium, the pressure of CO will increase and the pressure of CO_{2} will decrease.

Let us assume that the increase in pressure of CO = decrease in pressure of CO_{2} = p.

\begin{array}{l}{K}_{p}=\frac{{p}_{C{O}_{2}}}{{p}_{CO}}\\ \Rightarrow \text{}0.265=\frac{0.80-p}{1.4+p}\\ \Rightarrow \text{}0.371+0.265p=0.80-p\\ \Rightarrow \text{}1.265p=0.429\end{array}

⇒ p=0.339 atm

At equilibrium, partial pressure of CO_{2},

{p}_{C{O}_{2}}=0.80-0.339=0.461\text{atm}

And at equilibrium, partial pressure of CO,

p_{CO}= 1.4 + 0.339 = 1.739 atm

**Q.3** **Equilibrium constant, K _{c} for the reaction**

{N}_{2}\left(g\right)+3{H}_{2}\left(g\right)\underset{}{\rightleftharpoons}2N{H}_{3}\left(g\right)\text{at}500K\text{is 0}\text{.061}

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^{-1 }N_{2}, 2.0 mol L^{–1 }H_{2} and 0.5 mol L^{–1 }NH_{3}. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

**Ans.**

\begin{array}{l}\underset{At\text{a particular time: 3}{\text{.0 molL}}^{-1}}{{N}_{2}\left(g\right)}+\underset{\text{2}{\text{.0 molL}}^{-1}}{3{H}_{2}\left(g\right)}\underset{}{\rightleftharpoons}\underset{\text{0}{\text{.5 molL}}^{-1}}{2N{H}_{3}\left(g\right)}\\ Now,we\text{know that,}\\ {Q}_{c}=\frac{{\left[N{H}_{3}\right]}^{2}}{\left[{N}_{2}\right]{\left[{H}_{2}\right]}^{3}}\\ =\frac{{\left(0.5\right)}^{2}}{\left(3.0\right){\left(2.0\right)}^{3}}=0.0104\end{array}

As given, K_{c }= 0.061

Since Q_{c} ≠ K_{c} the reaction is not at equilibrium.

and Q_{c }< K_{c }, 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\left(g\right)\underset{}{\rightleftharpoons}B{r}_{2}\left(g\right)+C{l}_{2}\left(g\right)

for which K_{c }= 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 Br_{2} and Cl_{2} produced at equilibrium be x mol L^{-1}. The given reaction is:

\begin{array}{l}\begin{array}{ccccc}2BrCl\left(g\right)& \underset{}{\rightleftharpoons}& B{r}_{2}\left(g\right)& +& C{l}_{2}\left(g\right)\\ Initial\text{Conc}\text{. 3}{\text{.3\xd710}}^{-3}& & 0& & 0\\ At\text{equilibrium}\left(3.3\times {10}^{-3}-2x\right)& & x& & x\end{array}\\ Thus,\\ \Rightarrow \frac{\left[B{r}_{2}\right]\left[C{l}_{2}\right]}{\left[BrC{l}^{2}\right]}={K}_{c}\\ \Rightarrow \frac{x\cdot x}{{\left(3.3\times {10}^{-3}-2x\right)}^{2}}=32\\ \Rightarrow \frac{x}{\left(3.3\times {10}^{-3}-2x\right)}=5.66\\ \Rightarrow x=18.678\times {10}^{-3}-11.32x\\ \Rightarrow 12.32x=18.678\times {10}^{-3}\\ \Rightarrow x=1.5\times {10}^{-3}\\ Therefore,\text{at equilibrium,}\\ \left[BrCl\right]=3.3\times {10}^{-3}-\left(2\times 1.5\times {10}^{-3}\right)\\ =3.3\times {10}^{-3}-3.0\times {10}^{-3}\\ =0.3\times {10}^{-3}\\ =3.0\times {10}^{-4}mol{L}^{-1}\end{array}

**Q.5** **At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO _{2 }in equilibrium with solid carbon has 90.55% CO by mass**

C\left(s\right)+C{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}2CO\left(g\right)

Calculate K_{c} 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 CO_{2 }= (100 – 90.55) g = 9.45 g

Number of moles of CO (n_{CO}) = 90.55/28 = 3.234 mol

[Molecular mass of CO = (12+16) = 28]

Number of moles of CO_{2 }(n_{CO2}) = 9.45/44 = 0.215 mol

[Molecular mass of CO_{2 }= (12+32) = 44]

Partial pressure of CO,

**Q.6** **Calculate a) DG° and b) the equilibrium constant for the formation of NO _{2 }from NO and O_{2 }at 298 K**

NO\left(g\right)+\frac{1}{2}{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}N{O}_{2}\left(g\right)

**Ans.**

Where Δ_{f}G° (NO_{2}) = 52.0 kJ/mol

Δ_{f}G° (NO) = 87.0 kJ/mol

Δ_{f}G° (O_{2}) = 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 K_{c}

or – ΔG° = 2.303 RT log K_{c}

or log K_{c} = -ΔG°/2.303 RT

\begin{array}{l}\Rightarrow \mathrm{log}{\text{K}}_{c}=\frac{-\left(-35\times {10}^{-3}\right)}{2.303\times 8.314\times 298}\\ \Rightarrow {K}_{c}=anti\mathrm{log}\left(6.134\right)\end{array}

= 1.36 x 10^{6}

Thus, the equilibrium constant for the given reaction K_{c }is 1.36 × 10^{6}.

**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?**

\begin{array}{l}a)\text{}PC{l}_{5}\left(g\right)\underset{}{\rightleftharpoons}PC{l}_{3}\left(g\right)+C{l}_{2}\left(g\right)\\ b)\text{}CaO\left(s\right)+C{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}CaC{O}_{3}\left(s\right)\\ c)\text{}3\text{}Fe\left(s\right)+4{H}_{2}O\left(I\right)\underset{}{\rightleftharpoons}F{e}_{3}{O}_{4}\left(s\right)+4{H}_{2}\left(g\right)\end{array}

**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.

- 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.
- 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.
- In the 3
^{rd}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.**

\begin{array}{l}(i)\text{}COC{l}_{2}\left(g\right)\underset{}{\rightleftharpoons}CO\left(g\right)+C{l}_{2}\left(g\right)\\ (ii)\text{}C{H}_{4}\left(g\right)+2{S}_{2}\left(g\right)\underset{}{\rightleftharpoons}C{S}_{2}\left(g\right)+2{H}_{2}S\left(g\right)\\ (iii)\text{}C{O}_{2}\left(g\right)+C\left(s\right)\underset{}{\rightleftharpoons}2CO\left(g\right)\\ (iv){\text{2H}}_{2}\left(g\right)+CO\left(g\right)\underset{}{\rightleftharpoons}C{H}_{3}OH\left(g\right)\\ (v)\text{}CaC{O}_{3}\left(s\right)\underset{}{\rightleftharpoons}CaO\left(s\right)+C{O}_{2}\left(g\right)\\ (vi)\text{}4N{H}_{3}\left(g\right)+5{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}4NO\left(g\right)+6{H}_{2}O\left(g\right)\end{array}

**Ans.**

Only those reactions will be affected in which (n_{p} ≠ n_{r})

Here, n_{r} = the number of moles of gaseous reactant

n_{p} = 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 ×10 ^{5 }at 1024 K.**

{H}_{2}\left(g\right)+B{r}_{2}\left(g\right)\underset{}{\rightleftharpoons}2HBr\left(g\right)

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,

K_{p} = 1.6 x 10^{5}

Let us calculate equilibrium constant for the backward reaction.

K_{p}^{’} = 1/K_{p}

\begin{array}{l}=\frac{1}{1.6\times {10}^{5}}=6.25\times {10}^{-6}\\ \text{The backward reaction can be written as}\\ 2HBr\left(g\right)\underset{}{\rightleftharpoons}{H}_{2}\left(g\right)+B{r}_{2}\left(g\right)\end{array}

Let us assume that the pressure of gaseous H_{2} or gaseous Br_{2} (at equilibrium) is p.

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 10}\\ \\ \text{At Equilibrium}\left(10-2p\right)\end{array}}{2HBr\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ p\end{array}}{{H}_{2}\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ p\end{array}}{B{r}_{2}\left(g\right)}

Thus,

\begin{array}{l}\frac{{p}_{{H}_{2}}\times {p}_{B{r}_{2}}}{{p}^{2}HBr}=K{\u2018}_{p}\\ \Rightarrow \frac{p\times p}{{\left(10-2p\right)}^{2}}=6.25\times {10}^{-6}\\ \Rightarrow \frac{p\times p}{\left(10-2p\right)}=2.5\times {10}^{-3}\\ \Rightarrow p=2.5\times {10}^{-2}-\left(5.0\times {10}^{-3}\right)p\\ \Rightarrow p+\left(5.0\times {10}^{-3}\right)p=2.5\times {10}^{-2}\\ \Rightarrow p=2.49\times {10}^{-2}\text{}bar\end{array}

Therefore, at equilibrium,

[H_{2}] = [Br_{2}] = 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}\left(g\right)+{H}_{2}O\left(g\right)\underset{}{\rightleftharpoons}CO\left(g\right)+3{H}_{2}\left(g\right)

(a) Write as expression for K_{p} for the above reaction.

(b) How will the values of K_{p} 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 K_{p} can be written as follows

{K}_{p}=\frac{{p}_{CO}\times {p}_{{H}_{2}}^{3}}{{p}_{C{H}_{4}}\times {p}_{{H}_{2}O}}

(b)

- As n
_{r}< n_{p}, according to Le Chatelier’s principle the equilibrium will shift in the backward direction. - As given, the reaction is endothermic. By Le Chatelier’s principle, the equilibrium will shift in the forward direction.
- 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 H_{2}

b) Addition of CH_{3}OH

c) Removal of CO

d) Removal of CH_{3}OH

On the equilibrium of the reaction:

2{H}_{2}\left(g\right)+CO\left(g\right)\underset{}{\rightleftharpoons}C{H}_{3}OH\left(g\right)

**Ans.**

a) According to Le Chatelier’s principle, if H_{2} is added to the reaction mixture, the equilibrium of the given reaction will shift in the forward direction. On adding H_{2} to reaction, to maintain the equilibrium constant some amount of CO reacts with H_{2} to form alcohol. Thus, the reaction will shift towards the forward direction.

b) On addition of CH_{3}OH, 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 CH_{3}OH breaks into CO and H_{2. }

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 CH_{3}OH breaks into CO and H_{2.}

d) On removing CH_{3}OH, the equilibrium will shift in the forward direction. To maintain the equilibrium constant some amount of CO reacts with H_{2} to form alcohol. Thus, the reaction will shift towards the forward direction.

**Q.12** **At 473 K, equilibrium constant K _{c} for decomposition of phosphorus pentachloride, PCl_{5 }is 8.3 × 10^{-3}. If decomposition is depicted as,**

PC{l}_{5}\left(g\right)\underset{}{\rightleftharpoons}P{C}_{3}\left(g\right)+C{l}_{2}\left(g\right),\text{}{\Delta}_{f}H\text{}\xba=124.0{\text{kJmol}}^{-1}

a) Write an expression for K_{c} for the reaction.

b) What is the value of K_{c} for the reverse reaction at the same temperature?

c) What would be the effect on K_{c}

If (i) more PCl_{5 }is added (ii) pressure is increased? (iii) The temperature is increased?

**Ans.**

a) The expression for K_{c} for the given reaction

{K}_{c}=\frac{\left[PC{l}_{3}\left(g\right)\right]\left[C{l}_{2}\left(g\right)\right]}{\left[PC{l}_{5}\left(g\right)\right]}

b) The value of K_{c} for the backward reaction:

K’_{c}= 1/ K_{c}

=1 / (8.3 x 10^{-3})

= 1.2048 x 10^{2}

= 120.48

c) i) K_{c} is constant at constant temperature, so no effect on adding PCl_{5}.

ii) The terms in K_{c} is independent of pressure, so no effect will occur if pressure is increased.

iii) The given reaction is endothermic, on increasing in temperature k_{f} will increase.

As K_{c}= k_{f}/k_{b }

Thus, the value of K_{c} 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 H _{2}. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,**

CO\left(g\right)+{H}_{2}O\left(g\right)\underset{}{\rightleftharpoons}C{O}_{2}\left(g\right)+{H}_{2}\left(g\right)

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\text{}ba\mathrm{r,}

what will be the partial pressure of H_{2} at equilibrium? K_{p}= 10.1 at 400°C

**Ans.**

Let the partial pressure of both CO_{2} (gas) and H_{2 }(gas) be p atm.

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 4}\text{.0 bar}\\ \\ \text{At Equilibrium}\left(4.0-p\right)\end{array}}{Co\left(g\right)}+\underset{\begin{array}{l}\\ \text{4}\text{.0 bar}\\ \\ \left(4.0-p\right)\end{array}}{{H}_{2}O\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{p}\end{array}}{C{O}_{2}\left(g\right)}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{p}\end{array}}{{H}_{2}\left(g\right)}

And K_{p} = 10.1

Thus,

\begin{array}{l}\frac{{p}_{C{O}_{2}}\times {p}_{{H}_{2}}}{{p}_{CO}\times {p}_{{H}_{2}O}}={K}_{p}\\ \Rightarrow \frac{p\times p}{{\left(4.0-p\right)}^{2}}=10.1\\ \Rightarrow \frac{p\times p}{\left(4.0-p\right)}=3.178\\ \Rightarrow p=12.712-3.178\text{p}\\ \text{4}\text{.178 p}=12.712\\ \Rightarrow p=3.04\end{array}

Thus, at equilibrium, the partial pressure of H_{2} will be 3.04 bar.

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

\begin{array}{l}a)\text{}C{l}_{2}\left(g\right)\underset{}{\rightleftharpoons}2Cl\left(g\right),\text{}{K}_{c}=5\times {10}^{-39}\\ b)\text{}C{l}_{2}\left(g\right)+2NO\left(g\right)\underset{}{\rightleftharpoons}NOCl\left(g\right),\text{}{K}_{c}=3.7\times {10}^{8}\\ c)\text{}C{l}_{2}\left(g\right)+2N{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}2N{O}_{2}Cl\left(g\right),\text{}{K}_{c}=1.8\end{array}

**Ans.**

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

**Q.15** **The value of K _{c} for the reaction**

3{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}2{O}_{3}\left(g\right)

is 2.0 × 10^{–50 }at 25°C. If the equilibrium concentration of O_{2 }in air at 25°C is 1.6 × 10^{–2}, what is the concentration of O_{3}?

**Ans.**

The given reaction is:

\begin{array}{l}3{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}2{O}_{3}\left(g\right)\\ {K}_{c}=\frac{{\left[{O}_{3}\left(g\right)\right]}^{2}}{{\left[{O}_{2}\left(g\right)\right]}^{3}}\end{array}

It is given that K_{c}= 2.0 ×10^{–50} and [O_{2}] = 1.6 ×10^{–2}

\begin{array}{l}\Rightarrow 2.0\times {10}^{-50}=\frac{{\left[{O}_{3}\left(g\right)\right]}^{2}}{{\left(1.6\times {10}^{-2}\right)}^{3}}\\ \Rightarrow {\left[{O}_{3}\left(g\right)\right]}^{2}=\left(2.0\times {10}^{-50}\right){\left(1.6\times {10}^{-2}\right)}^{3}\\ \Rightarrow {\left[{O}_{3}\left(g\right)\right]}^{2}=8.192\times {10}^{-56}\\ \Rightarrow \left[{O}_{3}\left(g\right)\right]=2.86\times {10}^{28}\text{M}\end{array}

Hence, the concentration O_{3} is 2.86 x 10^{-28 }M.

**Q.16** **The reaction,**

CO\left(g\right)+3{H}_{2}\left(g\right)\underset{}{\rightleftharpoons}C{H}_{4}\left(g\right)+{H}_{2}O\left(g\right)

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H_{2 }and 0.02 mol of H_{2}O and an unknown amount of CH_{4 }in the flask. Determine the concentration of CH_{4 }in the mixture. The equilibrium constant, K_{c} for the reaction at the given temperature is 3.90.

**Ans.**

Equilibrium

Conc. 0.30 0.10 x 0.02

CO\left(g\right)+3{H}_{2}\left(g\right)\underset{}{\rightleftharpoons}C{H}_{4}\left(g\right)+{H}_{2}O\left(g\right)

The equilibrium constant for the reaction,

\begin{array}{l}{K}_{c}=\frac{\left[C{H}_{4}\left(g\right)\right]\left[{H}_{2}O\left(g\right)\right]}{\left[CH\left(g\right)\right]{\left[{H}_{2}O\left(g\right)\right]}^{3}}\\ \Rightarrow \frac{x\times 0.02}{0.3\times {\left(0.1\right)}^{3}}=3.90\\ \Rightarrow x=\frac{3.90\times 0.3\times {\left(0.1\right)}^{3}}{0.02}\end{array}

or, x = 0.0585 M = 5.85 x 10^{-2} M

Hence, the concentration of CH_{4 }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:**

HNO_{2}, CN^{–}, HClO_{4}^{–}, F^{–}, OH^{–}, CO_{3}^{2-} and S^{2-}

**Ans.**

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

Species | Conjugate acid/base |

HNO_{2} |
NO_{2}^{– }(base) |

CN^{–} |
HCN (acid) |

HClO_{4}^{–} |
ClO_{4}^{– }(base) |

F^{–} |
HF (acid) |

OH^{–} |
H_{2}O(acid)/ O^{2– }(base) |

CO_{3}^{2–} |
HCO_{3}^{– }(acid) |

S^{2–} |
HS^{– }(acid) |

**Q.18** **Which of the followings are Lewis acids? H _{2}O, BF_{3}, H^{+ }and NH_{4}^{+}.**

**Ans.**

Lewis acid can accept a pair of electrons. All cations are Lewis acid. Thus, BF_{3}, H^{+ }and NH_{4}^{+} are considered as Lewis acids.

**Q.19** **What will be the conjugate bases for the Bronsted acids: HF, H _{2}SO_{4 }and HCO_{3}^{–}?**

**Ans.**

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

Bronsted acids | Conjugate bases |

HF | F^{–} |

H_{2}SO_{4} |
HSO_{4}^{–} |

HCO_{3}^{–} |
CO_{3}^{2–} |

**Q.20** **Write the conjugate acids for the following Bronsted bases: NH _{2}^{–}, NH_{3} and HCOO^{–}.**

**Ans.**

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

Bronsted bases | Conjugate acids |

NH_{2}^{–} |
NH_{3} |

NH_{3} |
NH_{4}^{+} |

HCOO^{–} |
HCOOH |

**Q.21** **The species: H _{2}O, HCO_{3}^{–}, HSO_{4}^{–} and NH_{3} 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 |

H_{2}O |
H_{3}O^{+} |
OH^{–} |

HCO_{3}^{–} |
H_{2}CO_{3} |
CO_{3}^{2–} |

HSO_{4}^{–} |
H_{2}SO_{4} |
SO_{4}^{2–} |

NH_{3} |
NH_{4}^{+} |
NH_{2}^{–} |

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

**OH**^{–}(b) F^{–}(c) H^{+}and (d) BCl_{3}

**Ans.**

- OH
^{–}acts as Lewis base as it can donate its lone pair of electrons. - F
^{– }is Lewis base as it can donate one of its lone pair of electrons. - H
^{+}acts as Lewis acid since it can accept a lone pair of electrons. - BCl
_{3 }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.**

\begin{array}{l}\left[{H}^{+}\right]={3.8510}^{-3}M\\ pH=-\mathrm{log}\left[{H}^{+}\right]\\ \Rightarrow pH=-\mathrm{log}\left(3.8\times {10}^{-3}M\right)\\ \Rightarrow pH=-\mathrm{log}3.8-\mathrm{log}{10}^{-3}\\ \Rightarrow pH=-\mathrm{log}3.8+3\\ =-0.58+3\\ =2.42\end{array}

**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,

\begin{array}{l}pH=-\mathrm{log}\left[{H}^{+}\right]\\ \Rightarrow \mathrm{log}\left[{H}^{+}\right]=-pH\\ \Rightarrow \left[{H}^{+}\right]=anti\mathrm{log}\left(-pH\right)\\ \Rightarrow \left[{H}^{+}\right]=anti\mathrm{log}\left(-3.76\right)\\ =1.74\times {10}^{-4}M\end{array}

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}\times {K}_{b}

Here,

K_{w}= ionic product of water

K_{b} = ionization constant of conjugate base

K_{a} = ionization constant of conjugate acid

{K}_{b}=\frac{{K}_{w}}{{K}_{a}}

K_{a} of HF = 6.8 × 10^{–4}

K_{w }of water = 1 x 10^{-14}

Hence, K_{b }of its conjugate base F^{–}

\begin{array}{l}=\frac{{K}_{w}}{{K}_{a}}\\ =\frac{1\times {10}^{-14}}{6.8\times {10}^{-4}}=1.5\times {10}^{-11}\end{array}

Again, K_{a} of HCOOH = 1.8 × 10^{–4}

So, K_{b} of its conjugate base HCOO^{–}

=\frac{{K}_{w}}{{K}_{a}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-4}}=5.6\times {10}^{-11}

Given, K_{a} of HCN = 4.8 × 10^{–9}

=\frac{{K}_{w}}{{K}_{a}}=\frac{1\times {10}^{-14}}{4.8\times {10}^{-9}}=2.08\times {10}^{-6}

**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:

\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.05}\\ \\ \text{At equilibrium}\left(0.05-x\right)\end{array}}{{C}_{6}{H}_{5}OH}+{H}_{2}O\underset{}{\rightleftharpoons}{\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{x}\end{array}}{{C}_{6}{H}_{5}O}}^{-}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{x}\end{array}}{{H}_{3}{O}^{+}}\\ {K}_{a}=\frac{\left[{C}_{6}{H}_{5}{O}^{-}\right]\left[{H}_{3}{O}^{+}\right]}{\left[{C}_{6}{H}_{5}OH\right]}\\ \Rightarrow {K}_{a}\frac{x\cdot x}{\left(0.05-x\right)}\\ \text{The value of the ionization constant is very less},\text{x will be very small}.\text{We can ignore x in the denominator}.\\ x=\sqrt{\left(1\times {10}^{-10}\times 0.05\right)}\\ x=\sqrt{\left(5\times {10}^{-12}\right)}\\ =2.2\times {10}^{-6}M=\left[{H}_{3}{O}^{+}\right]\\ Since\left[{H}_{3}{O}^{+}\right]=\left[{C}_{6}{H}_{5}{O}^{-}\right]\\ \left[{C}_{6}{H}_{5}{O}^{-}\right]=2.2\times {10}^{-6}M\end{array}

Let the degree of ionization of phenol be α.

\begin{array}{l}\underset{\begin{array}{l}\\ Conc.\left(0.05-0.05\alpha \right)\text{}\end{array}}{{C}_{6}{H}_{5}OH}+{H}_{2}O\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{C}_{6}{H}_{5}{O}^{-}}+\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{H}_{3}{O}^{+}}\\ \underset{\begin{array}{l}\\ Conc.\text{}\end{array}}{{C}_{6}{H}_{5}ONa}\underset{}{\rightleftharpoons}{C}_{6}{H}_{5}{O}^{-}+\underset{\begin{array}{l}\\ 0.01\end{array}}{N{a}^{+}}\end{array}

[C_{6}H_{5}OH]= 0.05 – 0.05α; 0.05 M

[C_{6}H_{5}O^{–}] = 0.01 + 0.05α; 0.01 M

[H_{3}O^{+}] = 0.05α

\begin{array}{l}{K}_{a}=\frac{\left[{C}_{6}{H}_{5}{O}^{-}\right]\left[{H}_{3}{O}^{+}\right]}{\left[{C}_{6}{H}_{5}OH\right]}\\ {K}_{a}=\frac{\left(0.01\right)\left(0.05\alpha \right)}{\left(0.05\right)}\\ 1\times {10}^{-10}=0.01\alpha \\ \alpha =1\times {10}^{-8}\end{array}

**Q.27** **The first ionization constant of H _{2}S 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 H_{2}S is 1.2 × 10^{–13}, calculate the concentration of S^{2–} under both conditions.

**Ans.**

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

\begin{array}{l}{K}_{a}=\frac{\left[{H}^{+}\right]\left[H{S}^{-}\right]}{\left[{H}_{2}S\right]}\\ \Rightarrow 9.1\times {10}^{-8}=\frac{x\cdot x}{\left(0.1-x\right)}\\ \Rightarrow \left(9.1\times {10}^{-8}\right)\left(0.1-x\right)={x}^{2}\end{array}

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

0.1 – x = 0.1

\begin{array}{l}\Rightarrow 9.1\times {10}^{-8}=\frac{{x}^{2}}{0.1}\\ \Rightarrow {x}^{2}=9.1\times {10}^{-9}\\ \left[H{S}^{-}\right]=x=9.54\times {10}^{-5}M\end{array}

(ii) In the presence of 0.1 M HCl, suppose H_{2}S dissociates y M.

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.1}\\ \\ \text{At equilibrium 0}\text{.1}-y\text{}\end{array}}{\left[{H}_{2}S\right]}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{H{S}^{-}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{{H}^{+}}

Thus, at equilibrium,

[H_{2}S] = 0.1 – y» 0.1,

[H^{+}]= 0.1 + y » 0.1

[HS^{–}]= y M

{K}_{a}=\frac{0.1\times y}{0.1}

= 9.1 x 10^{-8}

Thus, y = 9.1 x 10^{-8} M

(iii)The ionization of H_{2}S is as follows –

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. (M) 9}{\text{.4\xd710}}^{-5}\text{}\\ \\ final\text{conc}\text{. (M) 9}{\text{.4\xd710}}^{-5}-X\end{array}}{{H}_{2}S}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ \text{9}{\text{.4\xd710}}^{-5}\\ \\ \text{X}\end{array}}{2{H}^{+}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{X}\end{array}}{{S}^{2-}}

Therefore,

\begin{array}{l}{K}_{{a}_{2}}=\frac{\left[{H}^{+}\right]\left[{S}^{2-}\right]}{\left[H{S}^{-}\right]}\\ {K}_{{a}_{2}}=\frac{\left(9.54\times {10}^{-5}\right)\left(X\right)}{9.54\times {10}^{-5}}\\ X=1.2\times {10}^{-13}\end{array}

Hence,

\left[{S}^{2-}\right]=1.2\times {10}^{-13}M

In the presence of 0.1 M of HCl, let the concentration of S^{2-} be Y M.

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.1}\\ \\ final\text{conc}\text{. 0}\text{.1}-y\end{array}}{\left[{H}_{2}S\right]}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{H{S}^{-}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{y}\end{array}}{{H}^{+}}

And

HCl\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ \text{0}\text{.1M}\end{array}}{{H}^{+}}+\underset{\begin{array}{l}\\ 0.1M\end{array}}{C{l}^{-}} \begin{array}{l}{K}_{{a}_{2}}=\frac{\left[{H}^{+}\right]\left[{S}^{2-}\right]}{\left[H{S}^{-}\right]}\\ 1.2\times {10}^{-13}=\frac{\left(0.1\right)\left(Y\right)}{9.1\times {10}^{-8}}\\ 10.92\times {10}^{-21}=0.1Y\\ \frac{10.92\times {10}^{-21}}{0.1}=Y\\ Y=1.092\times {10}^{-19}M\\ \left[{S}^{2-}\right]=1.092\times {10}^{-19}M\end{array}

**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

\begin{array}{l}C{H}_{3}COOH\underset{}{\rightleftharpoons}C{H}_{3}CO{O}^{-}+{H}^{+}\\ \text{Degree of dissociation},\\ \alpha =\sqrt{{K}_{a}/C}\end{array}

K_{a}= ionization constant

C = concentration

C = 0.05 M K_{a}= 1.74 × 10^{–5}

\begin{array}{l}\alpha =\sqrt{\frac{1.74\times {10}^{-5}}{5\times {10}^{-2}}}\\ =\sqrt{\left(34.8\times {10}^{-5}\right)}\end{array}

= 1.86 x 10^{-2}

[CH_{3}COO^{–}]= = 0.05 x 1.86 x 10^{-2}

= 0.093 x 10^{-2} M

[CH_{3}COO^{–}] = [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 pK _{a}.**

**Ans.**

Let the organic acid be HA.

HA\underset{}{\rightleftharpoons}{H}^{+}+{A}^{-}

Concentration of HA = 0.01 M

pH = 4.15

\begin{array}{l}\Rightarrow -\mathrm{log}\left[{H}^{+}\right]=4.15\\ \Rightarrow \left[{H}^{+}\right]=anti\mathrm{log}\left(-4.15\right)\\ \Rightarrow \left[{H}^{+}\right]=7.08\times {10}^{-5}\end{array}

Let us calculate ionization constant by the expression mentioned here.

\begin{array}{l}{K}_{a}=\frac{\left[{H}^{+}\right]\left[{A}^{-}\right]}{\left[HA\right]}\\ \left[{H}^{+}\right]=\left[{A}^{-1}\right]=7.08\times {10}^{-5}\\ \left[{H}^{+}\right]=0.01\end{array}

Then,

{K}_{a}=\frac{\left(7.08\times {10}^{-5}\right)\left(7.08\times {10}^{-5}\right)}{\left(0.01\right)}

⇒K_{a}= 5.01 x 10^{-7}

⇒pK_{a }= – log K_{a}

= – 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\left(aq\right)\underset{}{\rightleftharpoons}{H}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)

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\left(aq\right)\underset{}{\rightleftharpoons}N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)

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\left(aq\right)\underset{}{\rightleftharpoons}{H}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)

[H^{+}] = 2 x10^{–3 }M

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

(d) 0.002 M KOH:

Considering the equilibrium:

KOH\left(aq\right)\underset{}{\rightleftharpoons}{K}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)

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:**

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

**Ans.**

(a)

\begin{array}{l}Molar\text{conc}\text{. of TIOH=}\frac{2g}{221gmo{l}^{-1}}\times \frac{1}{2L}\\ =4.52\times {10}^{-3}\end{array}

[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)

\begin{array}{l}Molar\text{conc}\text{. of Ca}{\left(OH\right)}_{2}\text{=}\frac{0.3g}{74gmo{l}^{-1}}\times \frac{1}{0.5\text{}L}\\ =8.11\times {10}^{-3}\\ Ca{\left(OH\right)}_{2}\left(aq\right)\underset{}{\rightleftharpoons}C{a}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\end{array}

[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)

\begin{array}{l}Molar\text{conc}\text{. of NaOH=}\frac{0.3g}{40\text{}gmo{l}^{-1}}\times \frac{1}{0.2\text{}L}\\ =3.75\times {10}^{-2}M\end{array}

[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,

M_{1}V_{1} = M_{2}V_{2}

or 13.6 M x 1 mL = M_{2} x 1000 mL

or M_{2} = 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 H_{3}O^{+}= c.a = 0.1 × 0.132

= 0.0132

\begin{array}{l}pH=-\mathrm{log}\left[{H}^{+}\right]\\ \text{}=\text{}\u2013\text{log}\left(0.0\text{132}\right)\text{}=\text{1}.\text{88}\\ {\text{K}}_{a}=C{\alpha}^{2}\end{array}

= 0.1 x (0.132)^{2}

K_{a }= 0.0017

and pK_{a }= – log (0.0017)

= 2.77

**Q.33** **The pH of 0.005M codeine (C _{18}H_{21}NO_{3}) solution is 9.95. Calculate the ionization constant and pK_{b}.**

**Ans.**

Considering the reaction

Cod+{H}_{2}O\underset{}{\rightleftharpoons}Cod{H}^{+}+O{H}^{-}

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}

\begin{array}{l}c\alpha =8.91\times {10}^{-5}\\ \alpha =\frac{8.91\times {10}^{-5}}{5\times {10}^{-3}}=1.782\times {10}^{-2}\\ {K}_{b}=c{\alpha}^{2}\\ {K}_{b}=0.005\times {\left(1.782\right)}^{2}\times {10}^{-4}\\ \text{}=0.005\times 3.1755\times {10}^{-4}\\ \text{}=0.005\times {10}^{-4}\\ \text{}=1.58\times {10}^{-6}\end{array}

pK_{b }= – 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.**

K_{b}= 4.27 × 10^{–10} and c = 0.001 M

K_{b}= cα^{2}

\begin{array}{l}\Rightarrow 4.27\times {10}^{-10}=0.001\times {\alpha}^{2}\\ \Rightarrow 4270\times {10}^{-10}={\alpha}^{2}\\ \Rightarrow 65.34\times {10}^{-5}=\alpha =6.53\times {10}^{-4}\end{array}

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,

K_{a} x K_{b} = K_{w }

or K_{a }x (4.27 × 10^{–10}) = 10^{–14 }

{K}_{a}=\frac{{10}^{-14}}{4.27\times {10}^{-10}}=2.34\times {10}^{-5}

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 pK _{a} 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}

pK_{a }= 4.74

pK_{a }= – log K_{a }= 4.74

or K_{a} = antilog (– 4.74) = 1.82 x 10^{–5}

From the relation,

K_{a }= cα^{2}

\begin{array}{l}\Rightarrow \alpha =\sqrt{\left({K}_{a}/c\right)}\\ \Rightarrow \alpha =\sqrt{\frac{1.82\times {10}^{-5}}{5\times {10}^{-2}}}\end{array}

= 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.

- When 0.01 M HCl is taken:

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

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.05 M}\\ \\ \text{At equilibrium}\left(0.05-x\right)\end{array}}{C{H}_{3}COOH}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ \left(0.01+x\right)\end{array}}{{H}^{+}}+\underset{\begin{array}{l}\\ 0\\ \\ X\end{array}}{C{H}_{3}COO-}

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.

\begin{array}{l}{K}_{a}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[{H}^{+}\right]}{\left[C{H}_{3}COOH\right]}\\ \text{Putting the values in the above equation}:\\ {K}_{a}=\frac{\left(0.01\right)x}{0.05}\\ \Rightarrow x=\frac{1.82\times {10}^{-5}\times 0.05}{0.01}\\ \Rightarrow x=1.82\times {10}^{-3}\times 0.05\text{}M\\ \text{By definition},\\ \alpha =\frac{Amount\text{of aicd dissociated}}{Amount\text{of acid taken}}\\ \Rightarrow \alpha =\frac{1.82\times {10}^{-3}\times 0.05}{0.05}\end{array}

=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:

[CH_{3}COOH]= 0.05 – X ~ 0.05M

[CH_{3}COO^{–}] = X

[H^{+}]= 0.1 + X ~ 0.1

\begin{array}{l}{K}_{a}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[{H}^{+}\right]}{\left[C{H}_{3}COOH\right]}\\ {K}_{a}=\frac{\left(0.01\right)x}{0.05}\\ \Rightarrow x=\frac{1.82\times {10}^{-5}\times 0.05}{0.01}\\ \Rightarrow x=1.82\times {10}^{-3}\times 0.05\text{}M\\ Now,\\ \alpha =\frac{Amount\text{of aicd dissociated}}{Amount\text{of acid taken}}\\ \Rightarrow \alpha =\frac{1.82\times {10}^{-4}\times 0.05}{0.05}\\ =\text{1}.\text{82 x 1}{0}^{\u2013\text{4}}\end{array}

**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 K_{b }= 5.4 × 10^{–4 }

\begin{array}{l}\alpha =\sqrt{\left({K}_{b}/c\right)}\\ =\sqrt{\frac{5.4\times {10}^{-4}}{0.02}}\end{array}

=0.1643

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

\begin{array}{l}NaOH\left(aq\right)\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0.1\text{M}\end{array}}{N{a}^{+}\left(aq\right)}+\underset{\begin{array}{l}\\ 0.1\text{M}\end{array}}{O{H}^{-}\left(aq\right)}\\ \underset{\left(0.02-x\right)~0.02}{\underset{}{{\left(C{H}_{3}\right)}_{2}NH}}+{H}_{2}O\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ x\end{array}}{{\left(C{H}_{3}\right)}_{2}N{H}_{2}^{+}}+\underset{\begin{array}{l}\\ \left(x+0.1\right)~0.1\end{array}}{O{H}^{-}}\\ \left[{\left(C{H}_{3}\right)}_{2}N{H}_{2}^{+}\right]=x\\ \left[{\text{OH}}^{\u2013}\right]\text{}=\text{x}+\text{}0.\text{1}~\text{}0.\text{1}\\ {\text{K}}_{b}=\frac{\left[{\left(C{H}_{3}\right)}_{2}N{H}_{2}^{+}\right]\left[O{H}^{-}\right]}{{\left(C{H}_{3}\right)}_{2}NH}\\ \Rightarrow 5.4\times {10}^{-4}=\frac{x\times 0.1}{0.02}\end{array}

⇒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=-\mathrm{log}\left[{H}^{+}\right]

- 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

- 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

- 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

- 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

- 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:

\begin{array}{l}{\left[KOH\right]}_{\left(aq\right)}=\frac{0.561}{200/1000}g/L\\ =2.805g/L\\ =\frac{2.805}{56.11}M\\ =0.05\text{M}\\ \text{The ionic equilibrium for the aqueous KOH is given below}:\\ KOH\left(aq\right)\underset{}{\rightleftharpoons}{K}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\end{array}

Thus, [K^{+}] = [OH^{–}] = 0.05M

[H^{+}] x [OH^{–}] = K_{w}

or [H^{+}] = K_{w}/ [OH^{–}]

=\frac{{10}^{-14}}{0.05}=2\times {10}^{-13}\text{M}

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}

\begin{array}{l}=\frac{19.23g{L}^{-1}}{121.6gmo{l}^{-1}}\\ =0.1581\text{M}\\ \text{Assuming complete dissociation of Sr}{\left(\text{OH}\right)}_{\text{2}}\\ Sr{\left(OH\right)}_{2}\underset{}{\rightleftharpoons}S{r}^{2+}+2O{H}^{-}\end{array}

[Sr^{2+}] = 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:

\begin{array}{l}\underset{\begin{array}{l}\\ 0.05-0.05\alpha \end{array}}{HA+{H}_{2}O}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{H}_{3}{O}^{+}}+\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{{A}^{-}}\\ \text{Let us assumeas 0}\text{.05}-0.05\alpha \text{as}0.0\text{5}.\\ {K}_{a}=\frac{\left[{H}_{3}{O}^{+}\right]\left[{A}^{-}\right]}{\left[HA\right]}\\ =\frac{\left(0.05\alpha \right)\left(0.05\alpha \right)}{0.05}=0.05{\alpha}^{2}\\ \alpha =\sqrt{\frac{{K}_{a}}{0.05}}\\ =\sqrt{\frac{1.32\times {10}^{-5}}{.05}}\\ =1.63\times {10}^{-2}\end{array}

[H_{3}O^{+}] = 0.05 α = 0.05 x 1.63 x 10^{–2} = 8.15 x 10^{–4} M

pH = – log [H^{+}]

= – log [H_{3}O^{+}]

= – log (8.15 x 10^{–4})

= 3.09

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

[H_{3}O^{+}] = 0.01

[A^{–}] = 0.05 α’

[HA]= 0.05

\begin{array}{l}{K}_{a}=\frac{\left(0.01\right)\left(0.05\alpha \u2018\right)}{0.05}\\ \Rightarrow 1.32\times {10}^{-5}=0.01\times \alpha \u2018\\ \Rightarrow \alpha \u2018=1.32\times {10}^{-3}\end{array}

**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

K_{a} = c . α^{2 }

= 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.**

NaNO_{2 }is the salt of a strong base (NaOH) and a weak acid (HNO_{2}).

\underset{\begin{array}{l}\\ 0.04-x\end{array}}{N{O}_{2}^{-}+{H}_{2}O}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ x\end{array}}{HN{O}_{2}}+\underset{\begin{array}{l}\\ x\end{array}}{O{H}^{-}} \begin{array}{l}{K}_{h}=\frac{\left[HN{O}_{2}\right]\left[O{H}^{-}\right]}{\left[N{O}_{2}^{-}\right]}\\ \Rightarrow \frac{{K}_{w}}{{K}_{a}}=\frac{{10}^{-14}}{4.5\times {10}^{-4}}=0.22\times {10}^{-10}\end{array}

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

[NO_{2}^{–}]=0.04 – x ~ 0.04

[HNO_{2}] = x

[OH^{–}] = x

\begin{array}{l}{K}_{h}=\frac{\left[HN{O}_{2}\right]\left[O{H}^{-}\right]}{\left[N{O}_{2}^{-}\right]}\\ {K}_{h}=\frac{{x}^{2}}{0.04}=0.22\times {10}^{-10}\\ \Rightarrow {x}^{2}=0.0088\times {10}^{-10}\\ \Rightarrow x=0.093\times {10}^{-5}\end{array}

[OH^{–}] = 0.093 x 10^{–5} M

[H^{+}]= K_{w }/(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

\begin{array}{l}h=\frac{x}{0.04}=\frac{0.093\times {10}^{-5}}{0.04}\\ =2.325\times {10}^{-5}\end{array}

**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}

\begin{array}{l}{K}_{h}=\frac{\left[pyridinium\text{chloride}\right]\left[{H}^{+}\right]}{\left[pyridinium\text{hydrochloride}\right]}\\ \Rightarrow {K}_{h}\frac{{\left(3.63\times {10}^{-4}\right)}^{2}}{0.02}=6.6\times {10}^{-6}\\ \Rightarrow {K}_{h}=\frac{{K}_{w}}{{K}_{a}}=\frac{{10}^{-14}}{6.6\times {10}^{-6}}=1.51\times {10}^{-9}\end{array}

**Q.45** **Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH _{4}NO_{3}, NaNO_{2} and KF.**

**Ans.**

NaCN, NaNO_{2}, 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.

NH_{4}NO_{3} 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 K_{a} for ClCH_{2}COOH is 1.35 × 10^{–3}

\begin{array}{l}{K}_{a}=c{\alpha}^{2}\\ \alpha =\sqrt{\frac{{K}_{a}}{C}}\\ \text{Let us put}\text{the values in the above expression}\\ \alpha =\sqrt{\frac{1.35\times {10}^{-3}}{0.1}}\\ \Rightarrow \alpha =\sqrt{1.35\times {10}^{-2}}\end{array}

= 0.116

[H^{+}] = c . a

= 0.1 x 0.116

= 0.0116

\Rightarrow pH=-\mathrm{log}\left[{H}^{+}\right]

= – log (.0116)

=1.94

ClCH_{2}COONa is the salt of a weak acid i.e., ClCH_{2}COOH and a strong base i.e., NaOH.

\begin{array}{l}pH=-\frac{1}{2}\left[\mathrm{log}{K}_{w}+\mathrm{log}{K}_{a}-\mathrm{log}c\right]\\ \Rightarrow ph=-\frac{1}{2}\left[\mathrm{log}{10}^{-14}+\mathrm{log}\left(1.35\times {10}^{-3}\right)-\mathrm{log}0.1\right]\\ \Rightarrow pH=-\frac{1}{2}\left[-14+\left(-3+0.1303\right)-\left(-1\right)\right]\\ =7.94\end{array}

**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

\begin{array}{l}{K}_{w}=\left[{H}^{+}\right]\left[O{H}^{-}\right]\\ Since,\\ \left[{H}^{+}\right]=\left[O{H}^{-}\right]\end{array}

Thus, K_{w }= [H^{+}]^{2 } and K_{w} 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 H_{2}SO_{4}+ 10 mL of 0.01 M Ca(OH)_{2}

c) 10 mL of 0.1 M H_{2}SO_{4}+ 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{\left(OH\right)}_{2}2HCl\stackrel{}{\to}CaC{l}_{2}+2{H}_{2}O

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 H_{2}SO_{4} = 10 x 0.01 millimoles of H_{2}SO_{4} = 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{\left(OH\right)}_{2}+{H}_{2}S{O}_{4}\stackrel{}{\to}CaS{O}_{4}+2{H}_{2}O

1 mole of Ca(OH)_{2} reacts with 1 mole of H_{2}SO_{4}.

(c) Thus, 0.1 millimole of Ca(OH)_{2} reacts completely with 0.1 millimole of H_{2}SO_{4}. The solution will be neutral with pH value as 7.0.10 mL 0.1 M H_{2}SO_{4} = 1 millimole of H_{2}SO_{4}

10 mL 0.1 M KOH = 1 millimole of KOH

2KOH+{H}_{2}S{O}_{4}\stackrel{}{\to}{K}_{2}S{O}_{4}+2{H}_{2}O

2 millimole of KOH reacts with = 1 millimole of H_{2}SO_{4}

1 millimole of KOH reacts with = ½ = 0.5 millimole of H_{2}SO_{4}

Amount of H_{2}SO_{4 }left = (1 – .05) = 0.5 millimole

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

Molarity of H_{2}SO_{4} in the solution = 0.5/20 = 2.5 x 10^{–2} M

[H^{+}] = 2 x (2.5 x 10^{–}^{2}) = 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: Ag _{2}CrO_{4} = 1.1 x 10^{–12},**

**BaCrO _{4 }= 1.2 x 10^{–10}, Fe(OH)_{3} = 1.0 x 10^{–38},**

**PbCl _{2 }= 1.6 x 10^{–5}, Hg_{2}I_{2 }= 4.5 x 10^{–29}**

**Determine also the molarities of individual ions.**

**Ans.**

- Silver chromate:

\begin{array}{l}A{g}_{2}Cr{O}_{4}\underset{}{\rightleftharpoons}2A{g}^{+}+Cr{O}_{4}^{2-}\\ {K}_{sp}={\left[A{g}^{+}\right]}^{2}\left[Cr{O}_{4}^{2-}\right]\end{array}

Let the solubility of Ag_{2}CrO_{4} be s.

For [Ag^{+}] = 2s

[CrO_{4}^{2–}] = s

Then,

K_{sp }= (2s)^{ 2}.s = 4 s^{3}

⇒ 1.1 x 10^{–12 }= 4 s^{3}

⇒ 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 CrO_{4}^{2–} = s = 0.65 x 10^{–4} M

- Barium chromate:

\begin{array}{l}BaCr{O}_{4}\underset{}{\rightleftharpoons}B{a}^{2+}+Cr{O}_{4}^{2-}\\ {K}_{sp}=\left[B{a}^{+}\right]\left[Cr{O}_{4}^{2-}\right]\end{array} {K}_{sp}=\left[B{a}^{+}\right]\left[Cr{O}_{4}^{2-}\right]

The solubility of BaCrO_{4 }is s.

Thus, [Ba^{2+}] = s

and [CrO_{4}^{2–}] = s

K_{sp }= s . s = s^{2}

⇒ s^{2 }= 1.2 x 10^{–10 }

⇒ s = 1.09 x 10^{–5} M

Molarity of Ba^{2+} = Molarity of CrO_{4}^{2–} = 1.09 x 10^{–5} M

- Ferric hydroxide:

\begin{array}{l}Fe{\left(OH\right)}_{3}\underset{}{\rightleftharpoons}F{e}^{3+}+3O{H}^{-}\\ {K}_{sp}=\left[F{e}^{3+}\right]{\left[O{H}^{-}\right]}^{3}\end{array}

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

Thus, [Fe^{3+}] = s

and [OH^{–}] = 3s

∴ K_{sp }= s. (3s)^{3 } = 27 s^{4 }

⇒ K_{sp} = 27 s^{4 }

⇒ 1.0 x 10^{–38} = 27 s^{4 }

⇒ 0.037 x 10^{–38} = s^{4}

⇒ s = 1.39 x 10^{–10} M

Molarity of Fe^{3+} = 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

- Lead Chloride:

\begin{array}{l}PbC{l}_{2}\underset{}{\rightleftharpoons}P{b}^{2+}+2C{l}^{-}\\ {K}_{sp}=\left[P{b}^{2+}\right]{\left[C{l}^{-}\right]}^{2}\end{array}

Let the solubility product of PbCl_{2} be s.

Thus, [Pb^{2+}] = s

and [Cl^{–}] = 2s

∴ K_{sp }= s. (2s)^{2}

⇒ K_{sp }= 4 s^{3}

⇒ 1.6 x 10^{–5} = 4 s^{3}

⇒ s^{3} = 4 x 10^{–6}

⇒ s = 1.58 x 10^{–2} M

Molarity of Pb^{2+} = 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:

\begin{array}{l}H{g}_{2}{I}_{2}\underset{}{\rightleftharpoons}H{g}_{2}^{2+}+2{I}^{-}\\ {K}_{sp}=\left[H{g}_{2}^{2+}\right]{\left[{I}^{-}\right]}^{2}\end{array}

Let the solubility product of Hg_{2}I_{2} be s.

[Hg_{2}^{2+}] = s

and [I^{–}] = 2s

∴ K_{sp}= s . (2s)^{2}

⇒ K_{sp}= 4 s^{3}

⇒4.5 x 10^{–29 }= 4 s^{3}

⇒ s = 2.24 x 10^{–10} M

Molarity of [Hg_{2}^{2+}] = 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 Ag _{2}CrO_{4} 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 Ag_{2}CrO_{4} = s

\begin{array}{l}A{g}_{2}Cr{O}_{4}\underset{}{\rightleftharpoons}2A{g}^{+}+Cr{O}_{4}^{2-}\\ {K}_{sp}={\left[A{g}^{+}\right]}^{2}\left[Cr{O}_{4}^{2-}\right]\end{array}

Let the solubility of Ag_{2}CrO_{4} be s.

For [Ag^{+}] = 2s

[CrO_{4}^{2–}] = s

Then,

K_{sp}= (2s^{2}). s = 4 s^{3}

⇒ 1.1 x 10^{–12 }= 4 s^{3}

⇒ s = 0.65 x 10^{–4} M = 6.5 x 10^{–5} M

Let s’ be the solubility of AgBr.

AgBr\left(s\right)\underset{}{\rightleftharpoons}A{g}^{+}+B{r}^{-}

Thus, [Ag^{+}] = [Br^{–}] = s

⇒K_{sp} = (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

\frac{s}{s\u2018}=\frac{6.5\times {10}^{-5}M}{7.07\times {10}^{-7}M}=91.9

**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 K _{sp}= 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:

\begin{array}{l}\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{NaI{O}_{3}}\stackrel{}{\to}N{a}^{+}+\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{I{O}_{3}^{-}}\\ \underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{Cu{\left(Cl{O}_{3}\right)}_{2}}\stackrel{}{\to}C{u}^{2+}+\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{2C{l}_{3}^{-}}\\ \text{The solubility equilibrium for copper iodate can be written as}:\\ Cu{\left(I{O}_{3}\right)}_{2}\left(aq\right)\underset{}{\rightleftharpoons}C{u}^{2+}\left(aq\right)+2I{O}_{3}^{-}\left(aq\right)\end{array}

Ionic product of copper iodate:

= [Cu^{2+}] [IO_{3}^{–}]^{2}

= (0.001) (0.001)^{2}

= 1 x 10^{–9}

The K_{sp }value for copper iodate, K_{sp} = 7.4 × 10^{–8}

Since the ionic product (1 × 10^{–9}) is less than K_{sp }(7.4 × 10^{–8}), precipitation will not occur.

**Q.52** **The ionization constant of benzoic acid is 6.46 × 10 ^{–5} and K_{sp} 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

\begin{array}{l}{C}_{6}{H}_{5}COOH\left(aq\right)\underset{}{\rightleftharpoons}{C}_{6}{H}_{5}CO{O}^{-}\left(aq\right)+{H}^{+}\left(aq\right)\\ {K}_{a}=\frac{\left[{C}_{6}{H}_{5}CO{O}^{-}\left(aq\right)\right]\left[{H}^{+}\left(aq\right)\right]}{\left[{C}_{6}{H}_{5}COOH\left(aq\right)\right]}\\ \Rightarrow \frac{\left[{C}_{6}{H}_{5}COOH\left(aq\right)\right]}{\left[{C}_{6}{H}_{5}CO{O}^{-}\left(aq\right)\right]}=\frac{\left[{H}^{+}\left(aq\right)\right]}{{K}_{a}}=\frac{6.46\times {10}^{-4}}{6.46\times {10}^{-5}}=10\end{array}

Let the solubility of C_{6}H_{5}COOAg be x mol/L

Then,

[Ag^{+}] = x

⇒ [C_{6}H_{5}COOH] + [C_{6}H_{5}COO^{–}] = x

⇒ 10 [C_{6}H_{5}COO^{–}] + [C_{6}H_{5}COO^{–}] = x

⇒11 [C_{6}H_{5}COO^{–}] = x

⇒ [C_{6}H_{5}COO^{–}] = x/11

\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[{C}_{6}{H}_{5}CO{O}^{-}\right]\\ \Rightarrow 2.5\times {10}^{-13}=x.\left(x/11\right)\end{array}

⇒ 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 C_{6}H_{5}COOAg be y mol/L in pure water.

[Ag^{+}] =y M

and [CH_{3}COO^{–}]= y M

\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[{C}_{6}{H}_{5}CO{O}^{-}\right]\\ \Rightarrow y=\sqrt{Ksp}=\sqrt{\left(2.5\times {10}^{-13}\right)}=5\times {10}^{-7}mol/L\\ \frac{x}{y}=\frac{1.66\times {10}^{-6}}{5\times {10}^{-7}}=3.32\end{array}

Hence, C_{6}H_{5}COOAg 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, K _{sp}= 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.

\begin{array}{l}\therefore \left[FeS{O}_{4}\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M\\ \therefore \left[F{e}^{2+}\right]=\therefore \left[FeS{O}_{4}\right]=\frac{x}{2}M\\ \therefore \left[{S}^{2-}\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M\\ FeS\left(s\right)\underset{}{\rightleftharpoons}F{e}^{2+}\left(aq\right)+{S}^{2-}\left(aq\right)\\ {K}_{sp}=\left[F{e}^{2+}\right]\left[{S}^{2-}\right]\\ \Rightarrow 6.3\times {10}^{-18}=\frac{x}{2}\times \frac{x}{2}\\ \Rightarrow 6.3\times {10}^{-18}=\frac{{x}^{2}}{4}\\ \Rightarrow x=5.02\times {10}^{-9}\end{array}

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, K _{sp} is 9.1 × 10^{–6}).**

**Ans.**

\begin{array}{l}CaS{O}_{4}\left(s\right)\underset{}{\rightleftharpoons}C{a}^{2+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)\\ {K}_{sp}=\left[C{a}^{2+}\right]\left[S{O}_{4}^{2-}\right]\end{array}

Let the solubility of CaSO_{4} be s.

Then, K_{sp} = s^{2}

⇒ 9.1 x 10^{–6} = s^{2}

⇒ s = 3.02 x 10^{–3} mol/L

Molecular mass of CaSO_{4 }= 136 g/mol

Solubility of CaSO_{4} in gram/L = 3.02 x 10^{–3} x 136 g/L

= 0.411 g/L

Thus, to dissolve 0.411 g of CaSO_{4}, we need water = 1 L

And to dissolve 1 g of CaSO_{4}, 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: FeSO_{4}, MnCl_{2}, ZnCl_{2 }and CdCl_{2}. in which of these solutions precipitation will take place?**

**Given, K _{sp} 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 K_{sp} value.

10mL of solution containing S^{2-} ion is mixed with 5 mL of metal salt solution.

Before mixing,

\begin{array}{l}\left[{S}^{2-}\right]=1.0\times {10}^{-19}M\text{}\left(\text{Volume}=\text{1}0\text{mL}\right)\\ \left[{M}^{2+}\right]=0.04M\text{}\left(\text{Volume}=\text{5 mL}\right)\end{array}

After mixing,

[S^{2–}] = 1.0 x 10^{–19} x (10/15) = 6.67 x 10^{–20}

[M^{2+}]= [Fe^{3+}] = [Mn^{2+}] = [Zn^{2+}] = [Cd^{2+}]

= 0.04 x (5/15) = 1.33 x 10^{–2} M

∴ Ionic product of each will be=

[M^{2+}][S^{2–}] = (1.33 x 10^{–2}) (6.67 x 10^{–20})

= 8.87 x 10^{–22}

This ionic product exceeds the K_{sp} of ZnS and CdS. Therefore, precipitation will occur in ZnCl_{2} and CdCl_{2} 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 K _{c }for the following equilibrium when the equilibrium concentration of each substance is: [SO_{2}] = 0.60 M, [O_{2}] = 0.82 M and [SO_{3}] =1.90 M?**

2S{O}_{2}\left(g\right)+{O}_{2}\underset{}{\rightleftharpoons}2S{O}_{3}\left(g\right)

**Ans.**

The expression for the equilibrium constant (K_{c}) for the give reaction is:

\begin{array}{l}{K}_{c}=\frac{{\left[S{O}_{3}\right]}^{2}}{{\left[S{O}_{2}\right]}^{2}\left[{O}_{2}\right]}\\ {K}_{c}=\frac{{\left(1.90\right)}^{2}{M}^{2}}{{\left(0.60\right)}^{2}\left(0.821\right){M}^{3}}=12.239{\text{M}}^{-1}\left(approx\right)\end{array}

Hence, K_{c} for the equilibrium K_{c} is 12.239 M^{-1}.

**Q.58** **At a certain temperature and total pressure of 10 ^{5} Pa, iodine vapour contains 40% by volume of I atoms**

{I}_{2}\left(g\right)\underset{}{\rightleftharpoons}2I\left(g\right)

Calculate K_{p} for the equilibrium.

**Ans.**

Partial pressure of I atoms,

\begin{array}{l}{p}_{I}=\frac{40}{100}\times {p}_{total}\\ =\frac{40}{100}\times {10}^{5}=4\times {10}^{4}Pa\\ {\text{Partial pressure of I}}_{\text{2}}\text{molecules}\\ {p}_{{I}_{2}}=\frac{60}{100}\times {p}_{total}\\ =\frac{60}{100}\times {10}^{5}=6\times {10}^{4}Pa\\ \text{Let us write}{\text{the expression of K}}_{\text{p}}\\ {K}_{p}=\frac{{\left(pI\right)}^{2}}{p{I}_{2}}\\ \text{Place the value of partial pressure of iodine atom and iodine molecule in this equation}.\text{We get}\\ {K}_{p}=\frac{{\left(4\times {10}^{4}\right)}^{2}P{a}^{2}}{6\times {10}^{6}\text{}Pa}\\ =2.67\times {10}^{4}\text{Pa}\end{array}

**Q.59** **Write the expression for the equilibrium constant, K _{c} for each of the following reactions:**

\begin{array}{l}(i)\text{}2NOCl\left(g\right)\stackrel{}{\rightleftharpoons}2NO\left(g\right)+C{l}_{2}\left(g\right)\\ (ii)\text{}2Cu{\left(N{O}_{3}\right)}_{2}\left(s\right)\stackrel{}{\rightleftharpoons}2CUO\left(s\right)+4N{O}_{2}\left(g\right)+{O}_{2}\left(g\right)\\ (iii)\text{}C{H}_{3}COO{C}_{2}{H}_{5}\left(aq\right)+{H}_{2}O\left(I\right)\stackrel{}{\rightleftharpoons}C{H}_{3}COOH\left(aq\right)+{C}_{2}{H}_{5}OH\left(aq\right)\\ (iv)\text{}F{e}^{3+}\left(aq\right)+3O{H}^{-}\left(aq\right)\stackrel{}{\rightleftharpoons}Fe{\left(OH\right)}_{3}\left(s\right)\\ (v)\text{}{I}_{2}\left(s\right)+5{F}_{2}\stackrel{}{\rightleftharpoons}2I{F}_{5}\end{array}

**Ans.**

\begin{array}{l}(i)\text{}{K}_{c}=\frac{{\left[N{O}_{\left(g\right)}\right]}^{2}{\left[C{l}_{2\left(g\right)}\right]}^{2}}{{\left[NOC{l}_{\left(g\right)}\right]}^{2}}\\ (ii)\text{}{K}_{c}={\left[N{O}_{2\left(g\right)}\right]}^{4}\left[{O}_{2\left(g\right)}\right]\\ \left(\text{The concentration of a solid is considered as constant}.\right)\\ (iii)\text{}{K}_{c}=\frac{\left[C{H}_{3}COOH\left(aq\right)\right]\left[{C}_{2}{H}_{5}OH\left(aq\right)\right]}{\left[C{H}_{3}COO{C}_{2}{H}_{5}\left(aq\right)\right]}\\ \left(\text{When water is used as solvent},\text{its concentration remains constant}.\right)\\ (iv)\text{}{K}_{c}=\frac{1}{\left[F{e}^{3+}{}_{\left(aq\right)}\right]{\left[O{H}^{-}{}_{\left(aq\right)}\right]}^{3}}\\ \left(\text{The concentration of a solid is considered as constant}.\right)\\ (v)\text{}{K}_{c}={\frac{\left[I{F}_{5}\right]}{{\left[{F}_{2}\right]}^{5}}}^{2}\\ \left(\text{The concentration of a solid is considered as constant}.\right)\end{array}

**Q.60** **Find out the value of K _{c} for each of the following equilibrium from the value of K_{p}:**

\begin{array}{l}(i)\text{}2NOCl\left(g\right)\underset{}{\rightleftharpoons}2NO\left(g\right)+C{l}_{2}\left(g\right);\text{}{K}_{p}=1.8\times {10}^{-2}at\text{500 K}\\ (ii)\text{}CaC{O}_{3}\left(s\right)\underset{}{\rightleftharpoons}CaO\left(s\right)+C{O}_{2}\left(g\right);\text{}{K}_{p}=167\text{}at\text{1073 K}\end{array}

**Ans.**

The relationship between K_{p} and K_{c} is written as:

{K}_{p}={K}_{c}{\left(RT\right)}^{\Delta n}

(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

K_{p}= 1.8 × 10^{–2}

Let us place these values in the following equation,

\begin{array}{l}{K}_{p}={K}_{c}{\left(RT\right)}^{\Delta n}\\ \Rightarrow 1.8\times {10}^{-2}={K}_{c}{\left(0.0831\times 500\right)}^{1}\\ \Rightarrow {K}_{c}=\frac{1.8\times {10}^{-2}}{0.0831\times 500}\end{array}

= 4.33 x 10^{-4} (approx)

ii) Δn = 2 – 1 = 1

R = 0.0831 bar L mol^{–1 }K^{–1}

T = 1073 K

K_{p}= 167

Now,

{K}_{p}={K}_{c}{\left(RT\right)}^{\Delta n} \begin{array}{l}\Rightarrow 167={K}_{c}{\left(0.0831\times 1073\right)}^{1}\\ \Rightarrow {K}_{c}=\frac{167}{0.0831\times 1073}\\ =1.87\left(approx\right)\end{array}

**Q.61** **For the following equilibrium K _{c}= 6.3 x 10^{14} at 1000K,**

NO\left(g\right)+{O}_{3}\left(g\right)\underset{}{\rightleftharpoons}N{O}_{2}\left(g\right)+{O}_{2}\left(g\right)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_{c}, for the reverse reaction?

**Ans.**

For the forward reaction, K_{c} = 6.3 x 10^{14}

K_{c} for the backward (reverse) reaction, K_{c}’ = 1/K_{c}

=\frac{1}{6.3\times {10}^{14}}

= 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-

\begin{array}{c}Pure\text{substance}=\frac{Number\text{of moles}}{\text{Volume}}\\ =\frac{\text{Mass/Molecular mass}}{Volume}\\ =\frac{Mass}{Volume\times Molecular\text{mass}}\\ =\frac{Density}{Molecular\text{mass}}\end{array}

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 N _{2} and O_{2} takes place as follows:**

2{N}_{2}\left(g\right)+{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}2{N}_{2}O\left(g\right)

If a mixture of 0.482 mol of N_{2 }and 0.933 mol of O_{2 }is placed in a 10 L reaction vessel and allowed to form N_{2}O at a temperature for which

K_{c}= 2.0 × 10^{–37}, determine the composition of equilibrium mixture.

**Ans.**

Let us assume that the concentration Of N_{2}O at equilibrium be x.

\begin{array}{l}{\text{Let us assume that the concentration Of N}}_{\text{2}}\text{O at equilibrium be x}.\\ \underset{\begin{array}{l}\\ \text{Initial conc}\text{. 0}\text{.482mol}\\ \\ At\text{equilibrium}\left(0.482-x\right)mol\end{array}}{2{N}_{2}\left(g\right)}+\underset{\begin{array}{l}\\ 0.933\text{mol}\\ \\ \left(0.933-x\right)mol\end{array}}{{O}_{2}\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ x\text{}mol\end{array}}{2{N}_{2}O\left(g\right)}\\ \text{The reaction is carried out in 1}0\text{L vessel},\text{at equilibrium}\\ \left[{N}_{2}\right]=\frac{0.482-x}{10}\\ \left[{O}_{2}\right]=\frac{0.933-x}{10}\\ \left[{N}_{2}O\right]=x/10\\ \text{The value of equilibrium constant is very small}\left(\text{as given}\right).\\ \text{Therefore},{\text{the amount of N}}_{\text{2}}{\text{and O}}_{\text{2}}\text{reacted is also very small}.\\ {\text{Thus x can be neglected from the molar concentration terms of N}}_{\text{2}}{\text{and O}}_{\text{2}}.\\ \left[{N}_{2}\right]=\frac{0.482}{10}=0.0482{\text{mol L}}^{-1}\\ \left[{O}_{2}\right]=\frac{0.933}{10}=0.0933{\text{mol L}}^{-1}\\ {K}_{c}=\frac{{\left[{N}_{2}{O}_{\left(g\right)}\right]}^{2}}{{\left[{N}_{2\left(g\right)}\right]}^{2}\left[{O}_{2\left(g\right)}\right]}\\ \Rightarrow 2.0\times {10}^{-37}=\frac{{\left(\frac{x}{10}\right)}^{2}}{{\left(0.0482\right)}^{2}\left(0.0933\right)}\\ \Rightarrow \frac{{x}^{2}}{100}=2.0\times {10}^{-37}\times {\left(0.0482\right)}^{2}\times \left(0.0933\right)\\ \Rightarrow {x}^{2}=43.35\times {10}^{-40}\\ \Rightarrow x=6.6\times {10}^{-20}\\ \left[{N}_{2}O\right]=\frac{x}{10}=\frac{6.6\times {10}^{-20}}{10}=6.6\times {10}^{-21}\end{array}

**Q.64** **Nitric oxide reacts with Br _{2 }and gives nitrosyl bromide as per reaction given below:**

2NO\left(g\right)+B{r}_{2}\left(g\right)\underset{}{\rightleftharpoons}2NOBr\left(g\right)

When 0.087 mol of NO and 0.0437 mol of Br_{2 }are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br_{2}.

**Ans.**

The given reaction is:

\underset{\begin{array}{l}\\ 2\text{mol}\end{array}}{2NO\left(g\right)}+\underset{\begin{array}{l}\\ 1\text{mol}\end{array}}{B{r}_{2}\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 2\text{mol}\end{array}}{2NOBr\left(g\right)}

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

[Br_{2}] = 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:

[Br_{2}] = 0.0437 – 0.0259 = 0.0178 mol

**Q.65** **At 450 K, K _{p}= 2.0 × 10^{10}/bar for the given reaction at equilibrium.**

2S{O}_{2}\left(g\right)+{O}_{2}\left(g\right)\underset{}{\rightleftharpoons}2S{o}_{3}\left(g\right)

What is K_{c} at this temperature?

**Ans.**

For the given reaction,

Δn = 2 – 3 = -1

T= 450 K

K_{p} = 2.0 x 10^{10} bar^{-1}

R= 0.0831 L bar K^{–1} mol^{–1}

From the relation,

\begin{array}{l}{K}_{p}={K}_{c}{\left(RT\right)}^{\Delta n}\\ \Rightarrow 2.0\times {10}^{10}ba{r}^{-1}={K}_{c}{\left(0.0831{\text{L bar K}}^{-1}mo{l}^{-1}\times 450K\right)}^{-1}\\ {K}_{c}=\frac{\left(2.0\times {10}^{10}ba{r}^{-1}\right)}{{\left(0.0831{\text{L bar K}}^{-1}mo{l}^{-1}\times 450\text{}K\right)}^{-1}}\\ =7.48\times {10}^{11}{\text{L mol}}^{-1}\\ =7.48\times {10}^{11}{\text{M}}^{-1}\end{array}

**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 K _{p} for the given equilibrium?**

2HI\left(g\right)\underset{}{\rightleftharpoons}{H}_{2}\left(g\right)+{I}_{2}\left(g\right)

**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.

\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.2 atm}\\ \\ At\text{equilibrium 0}\text{.04 atm}\end{array}}{2HI\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ 0.16/2\text{}atm\\ =0.08\text{}atm\end{array}}{{H}_{2}\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ 0.16/2\text{}atm\\ =0.08\text{}atm\end{array}}{{I}_{2}\left(g\right)}\\ {K}_{p}=\frac{{p}_{{H}_{2}}\times {p}_{{I}_{2}}}{{p}_{HI}^{2}}\\ =\frac{\left(0.08\text{}atm\times 0.08\text{atm}\right)}{{\left(0.04\text{atm}\right)}^{2}}\\ =\frac{.0064}{.0016}\\ =4.0\end{array}

**Q.67** **A mixture of 1.57 mol of N _{2}, 1.92 mol of H_{2} and 8.13 mol of NH_{3} is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_{c} for the reaction is 1.7 x 10^{2}.**

{N}_{2}\left(g\right)+3{H}_{2}\left(g\right)\underset{}{\rightleftharpoons}2N{H}_{3}\left(g\right)

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

**Ans.**

For the given reaction

{N}_{2}\left(g\right)+3{H}_{2}\left(g\right)\underset{}{\rightleftharpoons}2N{H}_{3}\left(g\right)

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

[N_{2}]= (1.57/20) mol L^{-1}

[H_{2}]= (1.92/20) mol L^{-1}

[NH_{3}]= (8.13/20) mol L^{-1}

Now, let us calculate reaction quotient Q_{c}.

\begin{array}{l}{Q}_{c}=\frac{{\left[N{H}_{3}\right]}^{2}}{\left[{N}_{2}\right]{\left[{H}_{2}\right]}^{3}}\\ =\frac{{\left[\frac{8.13}{20}mol{\text{L}}^{-1}\right]}^{2}}{\left[\frac{1.57}{20}mol{\text{L}}^{-1}\right]{\left[\frac{1.92}{20}mol{\text{L}}^{-1}\right]}^{3}}\\ =2.38\times {10}^{3}\end{array}

As Q_{c} ≠ K_{c}, the reaction mixture is not at equilibrium.

And Q_{c} >K_{c}, the net reaction will proceed in the backward (reverse) direction.

**Q.68**

\begin{array}{l}\text{The equilibrium constant expression for a gas reaction is},\\ {K}_{c}=\frac{{\left[NH{}_{3}\right]}^{4}{\left[{O}_{2}\right]}^{5}}{{\left[NO\right]}^{4}{\left[{H}_{2}O\right]}^{6}}\\ \text{Write the balanced chemical equation corresponding to this expression}.\end{array}

**Ans.**

\begin{array}{l}\text{The balanced chemical equation can be written as}:\\ 4NO\left(g\right)+6{H}_{2}O\left(g\right)\underset{}{\rightleftharpoons}4N{H}_{3}\left(g\right)+5{O}_{2}\left(g\right)\end{array}

**Q.69** **One mole of H _{2}O 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\left(g\right)+CO\left(g\right)\underset{}{\rightleftharpoons}{H}_{2}\left(g\right)+C{O}_{2}\left(g\right)

Calculate the equilibrium constant for the reaction.

**Ans.**

\underset{\begin{array}{l}\\ Initial\text{conc}\text{.}\frac{1}{10}M\\ \\ At\text{equilibrium}\frac{1-0.4}{10}M\\ \\ =0.06\text{M}\end{array}}{{H}_{2}O\left(g\right)}+\underset{\begin{array}{l}\\ \frac{1}{10}M\\ \\ \frac{1-0.4}{10}M\\ \\ =0.06\text{M}\end{array}}{CO\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ \frac{0.4}{10}M\\ \\ =0.04\text{M}\end{array}}{{H}_{2}\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ \frac{0.4}{10}M\\ \\ =0.04\text{M}\end{array}}{C{O}_{2}\left(g\right)}

At equilibrium, the given concentrations are:

[H_{2}O]= (1 – 0.40)/10 mol L^{-1 }= 0.06 mol L^{-1}

[CO]= 0.06 mol L^{-1}

[H_{2}]=0.4/10 mol L^{-1 }= 0.04 mol L^{-1}

[CO_{2}]= 0.04 mol L^{-1}

\begin{array}{l}{K}_{c}=\frac{\left[{H}_{2}\right]\left[C{O}_{2}\right]}{\left[CO\right]\left[{H}_{2}O\right]}=\frac{0.04\times 0.04}{0.06\times 0.06}\\ =0.444\end{array}

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

{H}_{2}\left(g\right)+{I}_{2}\left(g\right)\underset{}{\rightleftharpoons}2HI\left(g\right)

**is 54.8. If 0.5 mol L ^{–1}of HI (g) is present at equilibrium at 700 K, what are the concentration of H_{2} (g) and I_{2} (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 K_{c }= 54.8

Thus, the equilibrium constant of the backward reaction K’_{c} = 1/54.8

The backward reaction can be written as:

2HI\left(g\right)\underset{}{\rightleftharpoons}{H}_{2}\left(g\right)+{I}_{2}\left(g\right)

[HI]= 0.5 mol L^{-1}

Let the concentrations of hydrogen and iodine at equilibrium be x mol L^{-1 }

[H_{2}] = [I_{2}] = x mol L^{-1 }

\begin{array}{l}\frac{\left[{H}_{2}\right]\left[{I}_{2}\right]}{{\left[HI\right]}^{2}}=K{\u2018}_{c}\\ \Rightarrow \frac{x\cdot x}{{\left(0.5\right)}^{2}}=\frac{1}{54.8}\\ \Rightarrow {x}^{2}=\frac{0.25}{54.8}\\ Or,\text{}x=0.068{\text{mol L}}^{-1}\end{array}

Hence, at equilibrium, [H_{2}] = [I_{2}] = 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\left(g\right)\underset{}{\rightleftharpoons}{I}_{2}\left(g\right)+C{l}_{2}\left(g\right);\text{}{K}_{c}=0.14

**Ans.**

For the given reaction,

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.78 M}\\ \\ At\text{equilibrium}\left(\text{0}\text{.78-2x}\right)\text{M}\end{array}}{2ICl\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ \text{x M}\end{array}}{{I}_{2}\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ \text{x M}\end{array}}{C{l}_{2}\left(g\right)}

Now we can write from the expression,

**Q.72** **K _{p} = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C_{2}H_{6 }when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?**

{C}_{2}{H}_{6}\left(g\right)\underset{}{\rightleftharpoons}{C}_{2}{H}_{4}\left(g\right)+{H}_{2}\left(g\right)

**Ans.**

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

\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 4}\text{.0 atm}\\ \\ \text{At equilibrium}\left(4.0-p\right)atm\end{array}}{{C}_{2}{H}_{6}\left(g\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ p\text{atm}\end{array}}{{C}_{2}{H}_{4}\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ p\text{atm}\end{array}}{{H}_{2}\left(g\right)}\\ \frac{{p}_{{C}_{2}{H}_{4}}\times {p}_{{H}_{2}}}{{p}_{{C}_{2}{H}_{6}}}={K}_{p}\\ \Rightarrow \frac{p\times p}{\left(40-p\right)}=0.04\\ \Rightarrow {p}^{2}=0.16-0.04p\\ \Rightarrow {p}^{2}+0.04p-0.16=0\\ p=\frac{-0.04\pm \sqrt{{\left(0.04\right)}^{2}-4\times 1\times \left(0.16\right)}}{2\times 1}\\ =\frac{-0.04\pm 0.80}{2}\\ =0.76/2\\ p=0.38\\ \text{Hence},\text{at equilibrium},\\ \left[{C}_{2}{H}_{6}\right]=4-p=4-0.38=3.62\text{atm}\end{array}

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

C{H}_{3}COOH\left(l\right)+{C}_{2}{H}_{5}OH\left(l\right)\underset{}{\rightleftharpoons}C{H}_{3}COO{C}_{2}{H}_{5}\left(l\right)+{H}_{2}O\left(l\right)

- Write the concentration ratio (reaction quotient), Q
_{c}, for this reaction (note: water is not in excess and is not a solvent in this reaction) - 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.
- 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,

\begin{array}{l}{Q}_{c}=\frac{\left[C{H}_{3}COO{C}_{2}{H}_{5}\right]\left[{H}_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[{C}_{2}{H}_{5}OH\right]}\\ \left(\text{ii}\right)\text{Let us assume that the volume of the reaction mixture is V}.\\ \text{Also},\text{we will consider that water is a solvent and is present in excess}.\\ \text{The given chemical reaction is}:\\ \underset{\begin{array}{l}\\ Initial\text{conc}\text{. 1/V M}\\ \text{At equilibrium}\left(1-0.171\right)/V\\ =0.829/V\end{array}}{C{H}_{3}COOH\left(l\right)}+\underset{\begin{array}{l}\text{0}\text{.18/V M}\\ \\ \left(\text{0}\text{.18-0}\text{.171}\right)/V\\ =0.009/V\end{array}}{{C}_{2}{H}_{5}OH\left(l\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ 0.171/V\end{array}}{C{H}_{3}COO{C}_{2}{H}_{5}\left(l\right)}+\underset{\begin{array}{l}\\ 0\\ \\ 0.171/V\end{array}}{{H}_{2}O\left(l\right)}\\ \text{Therefore},\text{equilibrium constant for the given reaction is}:\\ {K}_{c}=\frac{\left[C{H}_{3}COO{C}_{2}{H}_{5}\right]\left[{H}_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[{C}_{2}{H}_{5}OH\right]}\\ =\frac{\frac{0.171}{V}\times \frac{0.171}{V}}{\frac{0.829}{V}\times \frac{0.829}{V}}\\ =3.92\left(approx\right)\\ (iii)\text{}\text{Let us assume again that the volume of the reaction mixture as V}.\\ \underset{\begin{array}{l}\\ Initial\text{conc}\text{. 1}\text{.0/V M}\\ \text{At equilibrium}\left(1-2.14\right)/V\\ =0.786/V\text{M}\end{array}}{C{H}_{3}COOH\left(l\right)}+\underset{\begin{array}{l}\text{0}\text{.5/V M}\\ \\ \left(\text{0}\text{.5-0}\text{.214}\right)/V\\ =0.286/V\end{array}}{{C}_{2}{H}_{5}OH\left(l\right)}\underset{}{\rightleftharpoons}\underset{\begin{array}{l}\\ 0\\ \\ 0.214/V\text{M}\end{array}}{C{H}_{3}COO{C}_{2}{H}_{5}\left(l\right)}+\underset{\begin{array}{l}\\ 0\\ \\ 0.214/V\text{M}\end{array}}{{H}_{2}O\left(l\right)}\\ \text{The expression for the quotient is},\\ {Q}_{c}=\frac{\left[C{H}_{3}COO{C}_{2}{H}_{5}\right]\left[{H}_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[{C}_{2}{H}_{5}OH\right]}\\ =\frac{\frac{0.214}{V}\times \frac{0.214}{V}}{\frac{0.786}{V}\times \frac{0.786}{V}}\\ =0.2037\\ =0.204\left(approx\right)\\ {\text{As Q}}_{\text{c}}<{\text{K}}_{\text{c}},\text{equilibrium has not been reached}.\end{array}

## 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.