Important Questions for CBSE Class 11 Chemistry Chapter 7 Equilibrium

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Equilibrium is the chapter that requires Chemical as well as Ionic Equilibrium. Different ways of attaining a reaction at equilibrium are one of the major topics of this chapter. The other vital questions covered in Chapter 7 Class 11 Chemistry important questions will be based on key topics covered in Chapter 7 as below:

• Equilibrium In Physical Processes
• Equilibrium In Chemical Processes – Dynamic Equilibrium
• Law Of Chemical Equilibrium And Equilibrium Constant
• Homogeneous Equilibria
• Heterogeneous Equilibria
• Applications Of Equilibrium Constants
• Relationship Between Equilibrium Constant K, Reaction
• Quotient Q And Gibbs Energy G
• Factors Affecting Equilibria
• Ionic Equilibrium In Solution
• Acids, Bases, And Salts
• Ionization Of Acids And Bases
• Buffer Solutions
• Solubility Equilibria Of Sparingly Soluble Salts

The set of various questions has been included in the important questions Class 11 Chemistry Chapter 7 after deep study and detailed research of all the topics in the chapter. As a result, students get a clear-cut understanding of all the topics and can work on their weak areas.

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Class 11 Chemistry Equilibrium Important Questions with Solutions

The following vital questions and their solutions are included in the Chemistry Class 11 Chapter 7 important questions:

Question 1. We know about the relationship between Kc and Kp is Kp = Kc(RT)∆n

What will be the value of ∆n for the reaction

NH4Cl (s) → NH3 (g) + HCl (g)

(i) 1

(ii) 0.5

(iii) 1.5

(iv) 2

Answer 1: Option (iv) is the correct answer.

Question 2. In the case of the reaction H2(g) + I2(g) → 2HI (g), the standard free energy is given as 

∆G > 0.

The equilibrium constant (K ) will be ___.

(i) K = 0

(ii) K > 1

(iii) K = 1

(iv) K < 1

Answer 2: Option (iv) is the correct answer.

Question 3. Which of the following is not the general feature of equilibria involving physical processes?

(i) Equilibrium is possible only in the closed system at the given temperature.

(ii) All measurable properties of the given system remain constant.

(iii) All the physical processes stop at equilibrium.

(iv) The opposing process occurs at the same rate, and there is the dynamic, however stable condition.

Answer 3: Option (iii) is the correct answer.

Question 4. PCl5, PCl3, and Cl2 are at equilibrium at 500K in the closed container, and their concentrations are given as 0.8 × 10–3mol L–1, 1.2 × 10–3 mol L–1, and 1.2 × 10–3 mol L–1 respectively. The value of Kc for the given reaction

PCl5 (g) PCl3 (g) + Cl2 (g) 

will be,

(i) 1.8 × 103mol L–1

(ii) 1.8 × 10–3

(iii) 1.8 × 10–3 L moL–1

(iv) 0.55 × 104

Answer 4: Option (ii) is the correct answer.

Question 5. If hydrochloric acid is added to the cobalt nitrate solution at room temperature, the following reaction will occur, and the reaction mixture will become blue. On cooling the mixture, it becomes pink. Based on the given information, mark the right answers.

[Co (H2O)6]3+ (aq) + 4Cl–(aq) ⇌ [CoCl4]2– (aq) + 6H2O (l)

(pink) (blue)

(i) ∆H > 0 for the reaction

(ii) ∆H < 0 for the reaction

(iii) ∆H = 0 for the reaction

(iv) The sign of ∆H can’t be predicted based on the given information.

Answer 5; Option (i) is the correct answer.

Question 6. The pH of the neutral water at 25°C is 7.0. When the temperature increases, the ionisation of water increases, but the concentration of H+ ions as well as OH– ions are the same. What would be the pH of pure water at 60°C?

(i) Equal to 7.0

(ii) Greater than 7.0

(iii) Less than 7.0

(iv) Equal to zero

Answer 6: Option (iii) is the correct answer.

Question 7. Ka, 2Ka, and 3Ka are the respective ionization constants for the following given reactions.

H2S ⇌ H+ + HS–

HS– ⇌ H+ + S2–

H2S ⇌ 2H+ + S2–

The correct relationship in between Ka1, Ka2 and Ka3 will be

(i) Ka3 = Ka1 × Ka2

(ii) Ka3 = Ka1 + Ka2

(iii) Ka3 = Ka1 – Ka2

(iv) Ka3 = Ka1 /Ka2

Answer 7: Option (i) is the correct answer.

Question 8. The acidity of the compound BF3 could be explained based on which among the following concepts?

(i) Arrhenius’s concept

(ii) Bronsted Lowry’s concept

(iii) Lewis’s concept

(iv) Bronsted Lowry as well as Lewis’s concept.

Answer 8: Option (iii) is the correct answer.

Question 9. Which among the following will produce the buffer solution if mixed in equal volumes?

(i) 0.1 mol dm–3 NH4OH as well as 0.1 mol dm–3 HCl

(ii) 0.05 mol dm–3 NH4OH as well as 0.1 mol dm–3 HCl

(iii) 0.1 mol dm–3 NH4OH as well as 0.05 mol dm–3 HCl

(iv) 0.1 mol dm–3 CH4COONa as well as 0.1 mol dm–3 NaOH

Answer 9. Option (iii) is the correct answer.

Question 10. Which among the following solvents in silver chloride is most soluble?

(i) 0.1 mol dm–3 AgNO3 solution

(ii) 0.1 mol dm–3 HCl solution

(iii) H2O

(iv) Aqueous ammonia

Answer 10. Option (iv) is the correct answer.

Question 11. The ionization for hydrochloric in the water has given below:

HCl(aq) + H2O (l ) ⇌ H3O+(aq) + Cl–(aq)

Label the two conjugate acid-base pairs in this ionization.

Answer 11: The two required conjugate acid-base pairs in the ionization of HCl are (HCl–Cl–), where HCl is the conjugate acid as well as Cl– is the conjugate base. In the same way, the second pair is (H2O–, H3O+ ), where H2O is the conjugate base and H3O+ is the conjugate acid.

Question 12. The conjugate acid of the weak base is always stronger. What would be the decreasing order of the basic strength of the following conjugate bases?

OH–, RO–, CH3COO–, Cl–

Answer 12: The conjugate base of the strong acid is weak thud the decreasing order of the basic strength would be;

RO– > OH– > CH3COO– > Cl–

Question 13. The aqueous solution of the sugar does not conduct electricity. But, when sodium chloride is added to water, it conducts electricity. How would you explain the statement based on the ionisation, as well as how is it affected by the concentration of the sodium chloride?

Answer 13; The aqueous solution of sugar does not conduct electricity as they exist as the molecule in the water. They don’t have free ions to conduct electricity; however, in the case of NaCl, free ions of Na+ as well as Cl– are present to conduct the electricity. Conductance depends on the no. of the ions present in the given solution. More will the given no. of ions of NaCl in water more will be the conductivity.

Question 14. The reaction between ammonia and boron trifluoride is given below as follows:

NH3 + BF3 → H3N: BF3

Identify the acid and base in the given reaction. Which theory gives an explanation to it? What is the hybridisation of B as w N in the given reactants?

Answer 14: NH3 is the Lewis base, while BF3 is the Lewis acid. Lewis’s electronic theory of acids, as well as bases, explains it.

The hybridisation state of the nitrogen in NH3 is sp3 hybridised, as well as Boron in BF3, which is sp2 hybridised.

Question 15. The compound BF3 does not have a proton; however, it still acts as an acid and reacts with NH3. Why is it so? What type of bond can be formed between the two?

Answer 15: According to the Lewis concept, e- deficient species are known as lewis acid. Thus BF3 will act as the lewis acid while NH3 (N=1S2 2S2 2p3) has the lone pair; hence it will act as the lewis base, and it will donate the lone pair to the empty p-orbital of the Boron through the coordinate bond to form an adduct.

Question 16. Based on the given equation pH = – log [H+], the pH of the given 10-8 mol dm-3 solution of HCl shall be 8. But, it is observed to be less than 7.0. Justify the reason.

Answer 16: The solution is given very dilute, and we know that HCl reacts with water to give hydronium ions. The decrease in the pH could be observed as the result of the large concentration of H+. Hydronium ion concentration also needs to be considered here.

Then, the total pH would be; 

[H3O+] = 10-8 + 10-7 M = 7

Thus, the solution will be acidic.

Question 17. The ionisation constant of the weak base MOH is given by the expression;

Kb = [M+][OH–]/[MOH–]

Values of the ionisation constant for some weak bases at particular temperatures give below:

Base: Di-methylamine, Urea, Pyridine, and Ammonia

Kb: 5.4 × 10-4, 1.3 × 10-14, 1.77× 10-9, 1.77 × 10-5

Arrange the following bases in the decreasing order of the extent of their ionisation at equilibrium. Which among the above base is the strongest?

Answer 17: The decreasing order of the bases based on the ionisation constant at equilibrium would be;-

Di-methylamine > Ammonia > Pyridine > Urea

The strongest base would be Di-methyl amine because its pKb value is 3.29, and we already know that the less the pKb value, the strong is the base.

Question 18. Arrange the following compounds in increasing order for pH

KNO3 (aq), CH3COONa (aq), NH4Cl (aq), C6H5COONH4 (aq)

Answer 18: The increasing order of the pH would be;

CH3COONa< KNO3< C6H5COONH4<NH4Cl

CH3COONa is the salt of a weak acid (CH3COOH) and strong base (NaOH)

KNO3 is the salt of strong acid (HNO3)-strong base (KOH)

C6H5COONH4 is the salt of a weak acid (benzoic acid) and weak base (NH4OH)

NH4Cl is the salt of a strong acid (HCl) and weak base (NH4OH)

Question 19. Conjugate acid of the weak base is always stronger. The decreasing order of the basic strength for the following conjugate bases will be?

OH−, RO−, CH3COO−, CI−

Answer 19: Conjugate acids of the given bases are H2O, ROH, CH3COOH, and HCl.

Their acidic strength has the order.

HCl > CH3COOH > H2O >ROH.

Therefore, their conjugate bases would have strength in the order.

RO− > OH− > CH3COO− > CI−

Question 20. The value of Kc for the given reaction 2HI (g) ⇋ H2 (g) + I2 (g) is 1 × 10-4

At the given time, the composition of the reaction mixture is given as follows;

[HI] =2 × 10-5 mol, [H2] =1 × 10-5 mol as well as [I2] =1 × 10-5 mol. 

In which direction would the reaction proceed?

Answer 20: Given as; 

Kc = 1×10-4

Kc = [H2][I2]/[HI]2

Qc expresses the relative ratio of the products to reactants at the given instant.

Qc = [H2][I2]/[HI]2

= (1×10-5)(1×10-5)/(2×10-4)

Qc = 1/4 = 0.25

Where; Qc >Kc, the reaction would proceed in the reverse direction.

Question 21. When 0.561 g of KOH is dissolved in water to obtain 200 mL of solution at 298 K. Calculate the concentrations for the potassium, hydrogen and hydroxyl ions. What is the pH?

Answer 21. [KOH(aq)] = 0.561 / (1/5)g/L

= 2.805 g/L

= 2.805 x (1/56.11)

= 0.05M

KOH(aq) → 10-13 (aq) + OH–(aq)

[OH–] = 0.05M = [10-13] [H+][OH–] = Kw [H+] = Kw/[OH–] [H+] = 10-14/0.05 [H+] = 2 × 10-13 M

pH = -log[H+]

pH = -log[2 × 10-13 ]

pH = 12.70

Question 22. The pH of the given 0.08 mol dm−3 HOCI solution is 2.85. Calculate the ionisation constant.

Answer 22: pH of the given HOCl = 2.85

However, – pH = log [H+]

So, -2.85 = log [H+]

-3.15 = log [H+] [H+] = 1.413 × 10−3

For the given weak monobasic acid [H+] = (Ka X C) 1 / 2

Ka = [H+]*2 / C

Ka = (1.413 × 10−3)*2 / 0.08

Ka = 24.957 × 10−6

Ka = 2.4957 × 10−5

Question 23. The sparingly soluble salt gets precipitated only if the product of the concentration of the ions in the given solution (Qsp) becomes greater than the solubility product. When the solubility of BaSO4 in water is 8 × 10-4 mol dm-3, calculate the solubility in 0.01 mol dm-3 of H2SO4.

Answer 23: Given that the standard solubility of BaSO4 in water is 8 ×10-4 g/L

The equation for the disassociation of BaSO4 would be-

BaSO4 ⇌ Ba2+ + SO42-

(S’ is the solubility of the Ba2+ in 0.01 of HCl)

S <<< 0.01, thus it could be neglected

We already know that Ksp = S2

Ksp = (8×10-8)2

= 64×10-8

Then, Ksp= (S’) (0.01)

S’ = 64.8×10-8/0.01 = 6.4×10-5

Thus, the solubility of BaSO4 in the given 0.01 mol dm-3 of H2SO4 is 6.4×10-5

Question 24. Calculate the hydrogen ion concentration for the following biological fluids which have pH as given below:

(I) For Human saliva, 6.4 

(II) For Human stomach fluid, 1.2 

(III) For Human muscle fluid, 6.83 

(IV) For Human blood, 7.38.

Answer 24. (I) For Human saliva, 6.4

pH = 6.4

6.4 = – log [H+] [H+] = 3.98 x 10-7

(II) For Human stomach fluid, 1.2

pH =1.2

1.2 = – log [H+]

∴[H+] = 0.063

(III) For Human muscle fluid, 6.83

pH = 6.83

pH = – log [H+]

∴6.83 = – log [H+] [H+] =1.48 x 10-7 M

(IV) For Human blood, 7.38

pH = 7.38 = – log [H+]

∴ [H+] = 4.17 x 10-8 M

Question 25. On the basis of the given equation pH = – log [H+], the pH of the given 10-8 mol dm-3 solution for HCI should be 8. But, it is observed as less than 7.0. Justify the reason.

Answer 25: The water concentration could not be neglected as the solution is very dilute.

[H3O+] = 10-8 +10-7 M

[H3O+] = 10-8 (1 + 10)

[H3O+] = 11 X 10-8 M

pH = – log [H3O+]

pH = – log 11 X 10−8 M

pH = 8 – log 11

pH = 8 – 1.04

pH = 6.96

pH would be less than 7.0.

Question 26. The pH of the given 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionisation constant of the given acid and the degree of ionisation in the solution.

Answer 26. c = 0.1 M

pH = 2.34

-log [H+] = pH

-log [H+] = 2.34

[H+]=4.5× 10−3

And

[H+]=cα

4.5 × 10−3 = 0.1× α

α = 0.1/(4.5 × 10−3)

α = 0.045

Ka = cα2

Ka = 0.1 × (0.045)2

Ka = 0.0002025

Ka = 2.025 x 10−4

Question 27. The value of the given Kc for the given reaction 2HI (g) ⇋ H2 (g) + I2 (g) is 1 x 10−4. At the given time, the composition of the reaction mixture is [HI] = 2 x 10−5 mol, [I2 ] = 1 x 10−5 mol. Determine the direction of how the reaction would proceed?

Answer 27: At the given time, the reaction quotient Q for the reaction would be given by the expressions.

Q = [H2 ][I2 ] / [HI]

Q = 1 X 10−5 X 1 X 10−5 / (2 X 10−5 )2

Q = 1 / 4

Q = 0.25

Q = 2.5 X 10−1

Since the value of the given reaction quotient is greater than the value of Kc, i.e. 1×10−4, the reaction would proceed in the reverse direction.

Question 28. The pH of the given solution of the strong acid is 5.0. What would be the pH of the given solution obtained after diluting the given solution 100 times?

Answer 28: pH = 5 means [H+] = 10−5

On diluting 100 times, we get,

[H+] = 10−5 / 100 = 10−7

On calculating the pH of the given equation, we get,

pH = − log [H+], pH value comes out as 7, that is impossible.

Therefore, the total H+ ion concentration = H+ ions for the given acid H+ ion from water

[H+] = 10−7 +10−7 M

[H+] = 2 X 10−7

pH = 7 – 0.3010

pH = 6.699

Question 29. Predict whether the solutions for the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF.

Answer 29.

• KBr

KBr + H2O ↔ KOH (Strong base)+ HBr (Strong acid) Therefore, it is a neutral solution.

• NH4NO3

NH4NO3 + H2O↔ NH4OH(Weak base) + HNO2 (Strong acid)

Therefore, it is an acidic solution.

• KF

KF + H2O ↔ KOH (Strong base) + HF (weak acid)

Therefore, it is a basic solution.

• NaNO2

NaNO2 + H2O↔ NH4OH(Strong base) + HNO2(Weak acid)

Therefore, it is a basic solution.

• NaCN

NaCN + H2O ↔ HCN (Weak acid) + NaOH (Strong base)

Therefore, it is a basic solution.

• NaCl

NaCl + H2O ↔ NaOH (Strong base) + HCL (Strong acid)

Therefore, it is a neutral solution.

Question 30. Calculate the pH of the given solution formed by mixing equal volumes of the two solutions A and B of the strong acid having pH = 6 and pH = 4, respectively.

Answer 30: The pH of the given solution A = 6,

Thus, the concentration of the [H+] ion in solution A=10−6 mol / L

The pH of the given solution B = 4

Thus, the concentration of the [H+] ion in solution B=10−4mol / L.

On mixing, one litre of each given solution, the total volume = 1L + 1L = 2L.

Amount of the given H+ ions in 1L of the solution A = concentration × Volume (V) =

Amount of the given H+ ions in 1L of the solution A = 10−6 X 1 = 10−6

Amount of the given H+ ions in 1L of the solution B =10−4 X 1 = 10−4

The total amount of the given H+ ions in the solution formed by mixing solutions A and B is (10−5 mol + 10−4 mol)

This amount is present in the given 2L solution,

So, Total [H+] = 10−4 (1 + 1.01) / 2 mol / L

Total [H+] = 1.01 X 10−4 / 2 mol / L

Total [H+] = 0.5 X 10−4 mol / L

Total [H+] = 5 X 10−5 mol / L

pH = – log [H+]

pH = – log [5 X 10−5]

pH = – [ log 5 – 5 log 10]

pH = – log 5 + 5

pH = 5 – log 5

pH = 5 – 0.6990

pH = 4.3010

pH = 4.3

Therefore, the pH will be 4.3.

Question 31. The solubility product of the given compound Al(OH)3 is given as 2.7 x 10−11. Calculate the solubility in g / L and also find out the pH of the given solution. (Atomic mass of Al = 27 u).

Answer 31: Assume S be the solubility of Al(OH)3

Ksp = [Al3+] [OH−]3

Ksp = (S) (3S)3

Ksp = 27S4

S4 = Ksp / 27

S4 = 27 × 10−11 / 27 x 10

S4 = 1 x 10−12 mol / L.

S = 1 x 10−3 mol / L.

(i) Solubility for Al(OH)3: The molar mass of the given compound Al(OH)3 is 78g.

Hence, the solubility of the given compound Al(OH)3 in g / L = 1×103 × 78 g / L

Solubility of the given compound Al(OH)3 in g / L = 78 × 10−3 g / L

Solubility of the given compound Al(OH)3 in g / L = 7.8 × 10−2 g / L

(ii) pH of the given solution: S = 1 × 10−3 mol / L

[OH] = 3S = 3 × 1 × 10−3 = 3 × 10−3

pOH =3 − log 3

pH = 14 − pOH = 11 + log 3 = 11.4771.

Question 32. The first ionization constant for H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in the 0.1M solution. How would this concentration be affected when the solution is 0.1M in HCl also? If the second dissociation constant for H2S is 1.2 × 10–13, calculate the required concentration of S2– under both conditions.

Answer 32. To calculate for [HS–]

To find [HS–]:

Case 1 – HCl is absent.

Then,

Ka = ([H+][HS–])/[H2S] = 9.1×10-8 (given)

Thus, x2/(0.1-x) = 9.1×10-8 

However, 0.1-x is approximately equal to 0.1. Putting the value in the given equation:

x2/0.1 = 9.1×10-8

x2 = 9.1×10-9

x = 9.54× 10-5 M

So, the concentration of HS– is 9.54×10-5 M.

Case 2 – HCl is present

Then,

Ka = ([H+][HS–])/[H2S] = (y× (0.1+y))/(0.1-y) = 9.1×10-8 (given)

However, (0.1 + y) and (0.1 – y) can be approximated to 0.1.

9.1× 10-8 = (0.1*y)/0.1

Thus, y = [HS–] = 9.1× 10-8 M

To calculate for [S2-]:

Case 1 – HCl is absent.

The dissociation of HS– is given by the following equation below:

HS– ⇌ H+ + S2-[HS–] = 9.1×10-5 M

[H+] = 9.54×10-5 M

Ka = ([H+][S2-])/[HS–] = 1.2×10-13 (given)

Ka = (9.1× 10-5 × [S2-])/9.1×10-5

Thus, [S2-] = 1.2×10-13 M

Case 2 – HCl is present

[HS–] = 9.1×10-8 M [H+] = 0.1 M

Ka = 1.2×10-13 M

= (0.1×[S2-])/ 9.1×10-8

Hence, [S2-] = 1.092×10-19 M

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