NCERT Class 11 Chemistry Equilibrium Long Answer Questions

NCERT Class 11 Chemistry Equilibrium Long Question And Answers

Question 1. Equilibria involving physical and chemical changes are dynamic in nature—why?
Answer:

In a system, if two processes occur simultaneously at the same rate, then the system is said to be in a state of dynamic equilibrium. When a chemical reaction remains at equilibrium, the forward and the backward reactions occur simultaneously at the same rate.

Due to this, the equilibrium established in a chemical reaction is regarded as a dynamic equilibrium.

Similarly, when a physical process remains at equilibrium, the two opposite processes occur simultaneously at the same rate.

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For this reason, a physical equilibrium is also dynamic. For example, at equilibrium established during the evaporation of a liquid in a closed container, the evaporation of the liquid and the condensation of its vapor take place simultaneously at the same rate.

Question 2. The equilibrium established in the dissolution of a solid in a liquid or a gas in a liquid is dynamic. Explain
Answer:

If a solid solute is continuously dissolved in a suitable solvent, eventually a saturated solution of the solute is obtained. In this solution, an equilibrium exists between dissolved solute and undissolved solute.

At this equilibrium, the solute particles from undissolved solid solute get dissolved at the same rate as the solute particles from the solution get deposited on the surface of undissolved solid solute. Since these two processes occur at the same rate, the equilibrium established between dissolved solute and undissolved solute is dynamic.

When a gas is brought in contact with a suitable solvent, the gas keeps on dissolving in the solvent, and eventually, a saturated solution of the gas is formed. At this state, the gas over the liquid surface remains in equilibrium with the dissolved gas in the liquid.

In this state of equilibrium, the gas molecules from the gas phase enter the liquid phase at the same rate as the dissolved gas molecules from the liquid enter the gas phase. Two opposite processes therefore occur at the same rate. For this reason, the equilibrium that forms when a gas remains in equilibrium with its saturated solution is dynamic.

Question 3. Write the expressions of K_c and K_p for the following reactions:

1. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

2. 2BrF₅(g)⇌  Br₂(g) + 5F₂(g)

3. 3O₂(g) ⇌ 2O₃(g)

4. 4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)

5. P₄(g) + 3O₂(g) ⇌ P₄O₆(s)

6. CO(g) + 2H₂(g)⇌  CH₃OH(l)

Answer:

1.  \(K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} ; K_p=\frac{p_{\mathrm{SO}_3}^2}{p_{\mathrm{SO}_2}^2 \times p_{\mathrm{O}_2}}\)

2.  \(K_c=\frac{\left[\mathrm{Br}_2\right]\left[\mathrm{F}_2\right]^5}{\left[\mathrm{BrF}_5\right]^2} ; K_p=\frac{p_{\mathrm{Br}_2} \times p_{\mathrm{F}_2}^5}{p_{\mathrm{BrF}_5}^2}\)

3.  \(K_c=\frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]^3} ; K_p=\frac{p_{\mathrm{O}_3}^2}{p_{\mathrm{O}_2}^3}\)

4.  \(K_c=\frac{\left[\mathrm{Cl}_2\right]^2\left[\mathrm{H}_2 \mathrm{O}\right]^2}{[\mathrm{HCl}]^4\left[\mathrm{O}_2\right]} ; K_p=\frac{p_{\mathrm{Cl}_2}^2 \times p_{\mathrm{H}_2 \mathrm{O}}^2}{p_{\mathrm{HCl}}^4 \times p_{\mathrm{O}_2}}\)

5.  \(K_c=\frac{\left[\mathrm{P}_4 \mathrm{O}_6(s)\right]}{\left[\mathrm{P}_4\right]\left[\mathrm{O}_2\right]^3}=\frac{1}{\left[\mathrm{P}_4\right]\left[\mathrm{O}_2\right]^3}\)

∵ [P₄O₆(s)]=1]

⇒  \(K_p=\frac{1}{p_{\mathrm{P}_4} \times p_{\mathrm{O}_2}^3}\)

6.  \(K_c=\frac{\left[\mathrm{CH}_3 \mathrm{OH}(l)\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^2}=\frac{1}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^2}\)

∵ [CH₃OH(l)]=1

⇒ \(K_p=\frac{1}{p_{\mathrm{CO}} \times p_{\mathrm{H}_2}^2}\)

Question 4. Write the expression of K_c the following reactions:

1. 2Ag⁺(aq) + Cu(s) ⇌ Cu²⁺(aq) + 2Ag(s)

2. NH₄NO₃(s) ⇌ N₂O(g) + 2H₂O(g)

3. Au⁺(aq) + 2CN^–(aq) ⇌ [Au(CN)₂]^–(aq)

4. 3Cu(s) + 2NO₃^–(aq) + 8H⁺(aq) ⇌ 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l)

5. Cl₂(g) + 2Br^–(aq) ⇌ Br₂(l) + 2Cl^–(aq)

Answer:

1.  \(K_c=\frac{\left[\mathrm{Cu}^{2+}(a q)\right][\mathrm{Ag}(s)]^2}{\left[\mathrm{Ag}^{+}(a q)\right]^2[\mathrm{Cu}(s)]}=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]}{\left[\mathrm{Ag}^{+}(a q)\right]^2}\)

∵  [Ag(s)]=1 and [Cu(s)]=1

2.  \(K_c=\frac{\left[\mathrm{N}_2 \mathrm{O}(g)\right]\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2}{\left[\mathrm{NH}_4 \mathrm{NO}_3(s)\right]}\)

=\(\left[\mathrm{N}_2 \mathrm{O}(g)\right]\left[\mathrm{H}_2 \mathrm{O}(g)\right]^2\)

∵ [NH₄NO₃(s)]=1

3. \(K_c=\frac{\left[\mathrm{Au}(\mathrm{CN})_2^{-}(a q)\right]}{\left[\mathrm{Au}^{+}(a q)\right]\left[\mathrm{CN}^{-}(a q)\right]^2}\)

4.  \(K_c=\frac{\left[\mathrm{Cu}^{2+}(a q)\right]^3[\mathrm{NO}(g)]^2\left[\mathrm{H}_2 \mathrm{O}(l)\right]^4}{[\mathrm{Cu}(s)]^3\left[\mathrm{NO}_3^{-}(a q)\right]^2\left[\mathrm{H}^{+}(a q)\right]^8}\)

⇒ \(\frac{\left[\mathrm{Cu}^{2+}(a q)\right]^3[\mathrm{NO}(g)]^2}{\left[\mathrm{NO}_3^{-}(a q)\right]^2\left[\mathrm{H}^{+}(a q)\right]^8}\)

∵ Cu(s)=1 and H₂O(l)]=1

5.  \(K_c=\frac{\left[\mathrm{Br}_2(l)\right]\left[\mathrm{Cl}^{-}(a q)\right]^2}{\left[\mathrm{Cl}_2(g)\right]\left[\mathrm{Br}^{-}(a q)\right]^2}=\frac{\left[\mathrm{Cl}^{-}(a q)\right]^2}{\left[\mathrm{Cl}_2(g)\right]\left[\mathrm{Br}^{-}(a q)\right]^2}\)

 ∵ [Br₂(l)]=1

Question 5. Establish the relation between Kp and Kc for the following: aA + bB₂⇌ IL + mM [where the terms have their significance]
Answer:

⇒  \(K_c=\frac{[L]^l \times[M]^m}{[A]^a \times[B]^b}\)

Where [A],[B],[L] and [M] are Molar Concentrations Of A, B, L, and M At Equilibrium

⇒ \(K_p=\frac{p_L^l \times p_M^m}{p_A^a \times p_B^b}; \text { where } p_A, p_B, p_L \text { and } p_M\)

Partial pressures of A, B, L, and M respectively at equilibrium. Now, p_A = [A]RT, p_B = [B]RT, p_L = [L]RT and P_M = [M]RT.

Putting the values of p_A, p_B, p_L, and PM in the expression of, K_p we have—

⇒ \(K_p=\frac{[L]^l \times[M]^m}{[A]^a \times[B]^b} \times(R T)^{(l+m)-(a+b)}\)

Or \(K_p=K_c \times(R T)^{(l+m)-(a+b)}\)

Question 6. Find the relation between K_p &K_c for the given reactions:

1. NH₃(g) +HCl(g) ⇌NH₄Cl(s)

2. C(s) + CO₂(g) + 2Cl₂(g)⇌2COCl₂(g)

3. ½N₂(g) + O₂(g) ⇌NO₂(g)

4. C(s) + H₂O(g) ⇌CO(g) + H₂(g)

5. Fe(s) + H₂O(g)⇌ FeO(s) + H₂(g)

6. CH₄(g) + H₂O(l) ⇌CO(g) + 3H₂(g)

7. CO(g) + 2H₂(g)⇌ CH₃OH(Z)

Answer:

1. \(\Delta n=0-(1+1)=-2 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-2}\)

2. \(\Delta n=2-(1+2)=-1 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-1}\)

3.  \(\Delta n=1-\left(\frac{1}{2}+1\right)=\left(-\frac{1}{2}\right) ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-\frac{1}{2}}\)

4.  \(\Delta n=(1+1)-1=+1 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)\)

5.  \(\Delta n=1-1=0 ; K_p=K_c(R T)^{\Delta n}=K_c\)

6.  \(\Delta n=(1+3)-1=+3 ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^3\)

7. \(\Delta n=0-(1+2)=(-3) ; K_p=K_c(R T)^{\Delta n}=K_c(R T)^{-3}\)

Question 7. At a constant temperature, the equilibrium constant of the reaction

1. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is K, and that of the reaction

2. ½ N₂(g) + H₂(g) ⇌ NH₃(g) is R.

Explain this difference in K values even though the reactants and products in both the reactions are same.
Answer:

The value of the equilibrium constant of a reaction depends upon the mode of writing the balanced equation of the reaction.

1. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Equilibrium constant

⇒ \(K=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\)

2. \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g}) ;\)

Equilibrium constant

⇒ \(K_1=\frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{N}_2\right]^{1 / 2} \times\left[\mathrm{H}_2\right]^{3 / 2}}\)

∴ \(K_1^2=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\) = K

⇒ \(K_1=\sqrt{K}\)

In both reactions 1 and 2 reactants and products are the same, but in the equations of these two reactions, the stoichiometric coefficients of reactants and products are different, resulting in different values of the corresponding equilibrium constant.

Question 8. If the value of the equilibrium constant of the reaction 2SO₂(g) + O₂(g)→2SO₃(g) is K, then what will be the lues of equilibrium constants of the following reactions?

1. 4SO₂(g) + 2O₂(g) ⇌4SO₃(g)
2. 2SO₃(g) ⇌ 2SO₂(g) + O₂(g)
3. SO₂(g) + ½O₂(g) ⇌ SO₃(g)
4. SO₃(g) ⇌ SO₂(g) + ½O₂(g)

Answer:

⇒ \(2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) 2 \mathrm{SO}_3(g) ; K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}\)

1.  \(\text { (I) } 4 \mathrm{SO}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g})\rightleftharpoons 4 \mathrm{SO}_3(\mathrm{~g})\)

Equilibrium constant = \(\frac{\left[\mathrm{SO}_3\right]^4}{\left[\mathrm{SO}_2\right]^4\left[\mathrm{O}_2\right]^2}=K^2\)

2.  2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

Equilibrium constant =\(\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}=\frac{1}{K}\)

3.  SO(g)+½O₂(g)⇌ SO₃(g);

Equilibrium constant = \(\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}}=\sqrt{\mathrm{K}}\)

4.  SO₃(g) ⇌ SO₂(g) +½ O(g)

Equilibrium constant = [SO₂][O₂]^½/[SO₃]

= \(\frac{1}{\sqrt{K}}\)

Question 9. A few reactions and their equilibrium constants are given:

1. A ⇌ B + C; K_c = 2;
2. C ⇌ B +D; K_c  = 3
3. D ⇌ B +E; K_c  = 4.

Find the equilibrium constant ofthe reaction, A ⇌3B +E

Answer:

1. \(K_c=2=\frac{|B|[C \mid}{|A|}\)

2. \(K_c=3=\frac{[B \| D]}{[C]}\)

3. \(K_c=4=\frac{[B][E]}{[D]}\)

For the reaction A ⇌ 3B + E

Equilibrium constant K_c=  \(\frac{[B]^3[E]}{[A]}\) ……………………..(1)

Multiplying equilibrium constants of reactions (1),(2)and (30), we have 2× 3× 4

= \(\frac{[B \mid[C]}{\mid A]} \times \frac{[B][D]}{[C]} \times \frac{[B] \mid E]}{[D]}\)

Or,  24 = \(\frac{[B]^3[E]}{[A]}\)…………………….(24)

From the equations (1) and (2), we get  K_c =  24.

Therefore, the equilibrium constant for the reaction A ⇌ 3B + E is 24

Question 10. If the values of Kc for the reactions, A + 2B ⇌ C and C ⇌ 2D are 2 and 4, respectively, then what will be the value of  K_c for the reaction, 2D ⇌ A + 2B

Answer:

A + 2B ⇌ C; K_c = 2 = \(\frac{[C]}{[A][B]^2}\) ………………(1)

C ⇌ 2D ; K_c = 4 =  \(\frac{[D]^2}{[C]}\) ………………(2)

Adding equations 11 1 and [2), A + 26 s=± 2D ………………(3)

The equilibrium constant for equation (3)= Product of The equilibrium constants for reactions (1) and (2)

⇒ \(=\frac{[D]^2}{[A][B]^2}=2 \times 4=8\)

The equilibrium constant For the reaction 2D ⇒ A + 2B, the equilibrium constant = reciprocal of the equilibrium constant for reaction (3)

⇒  \(\frac{1}{8}=0.125\)

Question 11. For the reaction A + B C + D, the equilibrium constant is K. What would be the value of the reaction quotient (Q) when the reaction just starts and when it reaches equilibrium?
Answer:

For the given reaction, the equilibrium constant,

⇒ \(K=\frac{[C]_{e q} \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}} \text { where }[A]_{e q},[B]_{e q},[C]_{e q}\)

[D]_eq are the equilibrium concentrations of A, B, C, and D respectively.

For the given reaction, the reaction quotient (Q) at any moment during the reaction.

⇒ \(Q=\frac{[C] \times[D]}{[A] \times[B]} ; \text { where }[A],[B],[C] \text { and }[D]\)

Are the molar concentrations of.A, B, C, and D respectively at the moment under consideration.

1. Initially: [C] = 0, [D] = 0. Therefore, Q = 0.

2.  At The Equilibrium state;

⇒ \([A]=[A]_{e q},[B]=[B]_{e q}\)

⇒ \([C]=[C]_{e q} \text { and }[D]=[D]_{e q}\)

Hence, Q = K

Question 12. Consider the reaction, A(g) + 2B(g) 2D(g) + 3E(g) + heat, and state how the following changes at equilibrium will affect the yield of the product, D(g)— Temperature is increased, Pressure is increased at a constant temperature, Some amount of E (g) is removed from the reaction system at constant temperature and volume, and Some amount of D(g) is added to the reaction system, keeping temperature and volume constant.
Answer:

The reaction is exothermic. So, the increasing temperature at the equilibrium of the reaction will cause the equilibrium to shift to tire left. This decreases the yield of D(g). As a result, the concentration of D(g) at the new equilibrium will be lower than that at the original equilibrium.

The reaction involves an increase in the number of gas molecules [Δn = (2 + 3)-(l + 2) =+2] in the forward direction. Therefore, at a constant temperature, if the pressure is increased at the equilibrium of the reaction, then, according to Le Chatelier’s principle, the equilibrium will shift to the left. As a result, the yield of D(g) will decrease and the concentration of D(g) at the new equilibrium will be lower than that at the original equilibrium.

At constant temperature and volume, if some amount of E(g) [product] is removed from the reaction system, then, according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right. As a result, the yield of D(g) will increase, and the concentration of D(g) will increase. Thus, the concentration of D(g) at the new equilibrium will be more than that at the original equilibrium.

At constant temperature and volume, if some amount of D(g) [product] is added to the reaction system at equilibrium, then, according to Le Chatelier’s principle, the equilibrium will shift to the left. As a result, the yield of D(g) will decrease. However, the concentration of D(g) at the new equilibrium will be greater than that at the original equilibrium.

Question 13. Consider the following reactions; What will be the effect on the following if the temperature is increased?

1. The equilibrium constant
2. The position of equilibrium
3. The yield of products.

Answer:

1. The reaction is endothermic as ΔH > 0.

As the reaction is endothermic, the value of its equilibrium constant will increase with the rise in temperature

According to Le Chatelier’s principle, a rise in Na₂CO₃ is a salt of strong base NaOH and weak acid H₂CO₂.

In its aqueous solution, Na₂CO₃ dissociates completely and forms Na⁺ and CO₃ ions. In water, COl- is a stronger base than HaO, and hence it reacts with water, forming OH^– and HCO₂ ions. the temperature at the equilibrium of an endothermic.

Reaction shifts the equilibrium to the right. As a result, the yields of the products increase.

2. The reaction is exothermic as ΔH < 0.

As the reaction is exothermic, the value of its equilibrium constant will decrease with the temperature rise. According to Le Chatelier’s principle, increasing temperature at the equilibrium of an exothermic reaction shifts the equilibrium to the left. Consequently, the yields of the products decrease.

Question 14. For these reactions, predict the effect on the following if pressure is increased at equilibrium Position of equilibrium, Yield of the products.
Answer:

The reaction involves no change in volume. Therefore, pressure has no effect either on the position of the equilibrium or on the yield of the product.

The reaction involves a decrease in volume in the forward direction. So, at a constant temperature increasing pressure at the equilibrium of the reaction favors the shifting of equilibrium to the right, thereby increasing the yield of the product.

The reaction involves a decrease in volume in the reverse direction. So, at constant temperature, increasing pressure at the equilibrium of the reaction shifts the equilibrium to the left, thereby decreasing the yield ofthe product

Question 15. At constant temperature, in a closed vessel, the following equilibrium is established during the decomposition of

NH₄Cl(S); NH₄Cl(s)⇌NH₃(g) + HCl(g).

If pressure of. the equilibrium mixture =P and value of equilibrium =K , then show that

⇒ \(P=2 \sqrt{K_p} \text {. }\)

Answer 

Total number of moles at equilibrium = x + x = 2x

∴ \(p_{\mathrm{NH}_3}=\left(\frac{x}{2 x}\right) P ; p_{\mathrm{HCl}}=\left(\frac{x}{2 x}\right) P\)

∴ \(K_p=p_{\mathrm{NH}_3} \times p_{\mathrm{HCl}}=\left(\frac{x}{2 x}\right) P \times\left(\frac{x}{2 x}\right) P=\frac{p^2}{4}\)

∴ \(K_p=\frac{P^2}{4} \quad \text { or, } P^2=4 K_p\)

∴ \(P=2 \sqrt{K_p}\)

Question 16. A weak tribasic acid, H3A, in its aqueous solution, ionizes in the following three steps: 

1. H₃A(aq) + H₂O(l) ⇌ H₃O+(aq)+ H₂A^–(aq)
2. H₂A^–(aq) + H₂O(l) ⇌ H₃O+(aq) + HA^2-(aq)
3. HA^2-(aq) + H₂O(l) ⇌ H₃O+(aq) + A^3-(aq)

If the ionization constants of the steps are K₁, K₂, and K₃ respectively, then determine the overall ionization constant of HgA.

Answer:

Step – 1:  \(K_1=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{H}_2 \mathrm{~A}^{-}\right]}{\left[\mathrm{H}_3 \mathrm{~A}\right]} \text {; }\)

Step – 2: \( K_2=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{HA}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~A}^{-}\right]}\)

Step – 3: \(K_3=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{3-}\right]}{\left[\mathrm{HA}^{2-}\right]}\)

The equation for the overall ionisation of H₃A is:

⇒ \(\mathrm{H}_3 \mathrm{~A}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{A}^{3-}(a q)+3 \mathrm{H}_3 \mathrm{O}^{+}(a q)\)

If the overall ionization constant of H₃A is K, then

⇒ \(K=\frac{\left[\mathrm{A}^{3-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{~A}\right]}\)

Multiplying ionization constants of steps 1,2 and 3

We have, \(K_1 \times K_2 \times K_3=\frac{\left[\mathrm{A}^{3-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{~A}\right]}\)

Therefore, K = K₁ × K₂ × K₃

Question 17. Between 0.1(M) and 0.01(M) acetic acid solutions which one will have a higher pH, and why?
Answer:

In an aqueous solution of a weak monoprotic add

Example CH₃COOH )

⇒  \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{c \times K_a}\)

Where c = molar concentration of the solution and Ka = ionization constant of the weak acid.

Therefore, in 0.1(M) CH₃COOH solution,

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{0.1 \times K_a} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and in 0.01(M) CHgCOOH solution,

[H₃O⁺] \(\sqrt{0.01 \times K_a} \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

Hence, the concentration of H₃O⁺ ions in 0.1(M) CH₃COOH solution will be more than that in 0.01(M) CH₆COOH solution.

Therefore, the pH of 0.01 (M) CH₃COOH solution will be greater than that of O.l (M) CH₃COOH solution.

Question 18. What color change does a blue or red litmus paper exhibit when it is put separately in each of the following aqueous solutions? CH₃COONa₂CH₃COONH₄, NH₄Cl
Answer:

An aqueous solution of CH₃COONa is alkaline because in solution CH₃COO^– ions resulting from the complete dissociation of CH₃COONa undergo hydrolysis, thereby increasing the concentration of OH^–. So, if a red litmus paper is dipped into this solution, it will turn blue

In its aqueous solution, CH₃COONH₄ dissociates completely forming NH⁺(aqr) and CH₃COO^–(aq) ions. Both these ions undergo hydrolysis in water. The important fact is that in water the strength of CH₃COO^– ion as a base and the strength of NH₂ ion as an acid are the same. Consequently, the concentration of OH^– ions produced due to hydrolysis of CH₃COO^– ions:

⇒ \(\left[\mathrm{CH}_3 \mathrm{COO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q)+\mathrm{OH}^{-}(a q)\right]\)

Almost the same as the concentration of H₃O⁺ ions produced due to hydrolysis of NH^– ions:

⇒ \(\left[\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)\right]\)

As a result, an aqueous solution of CH₃COONH₄ is neutral. So, if a litmus paper is dipped into this solution, it will not exhibit any color charge.

In its aqueous solution, NH₄Cl dissociates completely forming NH₂(aq) and Cl^–(aq). NH₂ ions so formed undergo hydrolysis and thereby increase the concentration of H₃O⁺ ions in the solution. As a result, an aqueous solution of NH₄Cl is acidic. Therefore, if a blue litmus paper is dipped into this solution, its color will change to red.

Question 19. Arrange the following aqueous solutions in order of their increasing pH values: Na₂CO₃, CH₃COONH₄, and CuSO₄.
Answer:

Na₂CO₃ is a salt of strong base NaOH and weak acid H₂CO₂. In its aqueous solution, Na₂CO₃ dissociates completely and forms Na⁺ and CO₃^2- ions. In water, CO₃^2- is a stronger base than HaO, and hence it reacts with water, forming OH^– and HCO^–₃ ions.

This results in an increase in the concentration of OH ions in the solution of Na₂CO₃ and makes the solution alkaline.

In its aqueous solution, CuSO₄ dissociates completely, forming [Cu(H₂O)₆]₂⁺ and SO^– ions. SO^–ion, being a conjugate base of strong acid H₂SO₄, is a very weak base and cannot react with water.

In [Cu(H₂O)g]²⁺, the charge density of the Cu²⁺ ion is very high, and consequently, H₂O molecules bonded to the Cu²⁺ ion get polarized.

This causes the weakening of the O — H bonds in the H₂O molecule. Consequently, the O — H bond gets ionized easily, thereby forming an H+ ion. This H+ ion is then accepted by H₂O, which forms the H₃O⁺ ion.

⇒ [Cu(H₂O)6]²⁺(aq) + H₂O(l) ⇌ [Cu(H₂O)₅OH]⁺(aq) + H₃O⁺(aq)

Therefore, the hydrolysis of CuSO₄ in its solution increases the concentration of H₃O⁺ ions, thereby making the solution acidic. So, the increasing order of pH values of the given aqueous solutions: CuSO₄ < CH₃COONH₄ < Na₂CO₃

Question 20. Which of the following are buffer solutions? Give reasons:

1. 100 mL 0.1 (M)NH₃ + 50 mL 0.1(M) HCI
2. 100 mL 0.1 (M)CH₃COOH + 50 m L 0.2 (M) NaOH
3. 100 mL 0.1 (M) CH₃COOH +100 mL 0.15 (M) NaOH
4. 100 mL 0.1 (M) CH₃COONa +25mL 0.1(M) HCI
5. 50 mL 0.2 (M) NH₄Cl + 50 mL 0.1 (M) NaOH

Answer:

1. 100 mL of 0.1(M) NH₃ = O.Olmol of NH₃ and 50mL of 0.1(M) HCI = 0.005mol of HCl. In the mixed solution, 0.005 mol of HCl reacts completely with the same amount of NH₃, forming 0.005 mol of NH₄Cl, and 0.005 mol of NH₃ remains in the solution. Therefore, the mixed solution contains weak base NH₃ and its salt NH₄Cl. Hence, it is a buffer solution.

2. 100mL of 0.1(M) CH₃COOH = 0.01 mol of CH₃COOH and 50mL of 0.2 (M) NaOH = 0.01 mol of NaOH. In the mixed solution, 0.01 mol of CH₃COOH reacts completely with 0.01 mol of NaOH, forming 0.01 mol of CH₃COONa. Therefore, the mixed solution is a solution of CH₃COONa. Hence, it is not a buffer solution

3.100mL of 0.1 (M) CH₃COOH = 0.01 mol of CH₃COOH and 100mL of 0.15(M) NaOH = 0.015 mol of NaOH. In the mixed solution, 0.01 mol CH₃COOH reacts completely with 0.01 mol of NaOH, forming 0.01 mol of CH₃COONa, and 0.005 mol of NaOH remains in the solution. Therefore, the mixed solution is a solution of NaOH. Hence, it cannot be a buffer solution.

4. 100mL of 0.1(M) CH₃COONa = O.Olmol of CH₃COONa and 25mL of 0.1(M) HCI = 0.0025 mol of HCI. In the mixed solution, 0.0025 mol of HCl reacts completely with the same amount of CH₃COONa, forming CH₃COOH and NaCl.

The number of moles of CH₃COONa remaining in the solution is (0.01- 0.0025) = 7.5 × 10^-3 mol. Therefore, the mixed solution contains CH₃COOH and its salt CH₃COONa. Hence, it is a buffer solution.

5. 50mL of 0.2 (M) NH₄Cl EE 0.01 mol of NH₄Cl and 50mL of 0.1 (M) NaOH = 0.005 mol of NaOH . In the mixed solution, 0.005mol of NaOH reacts completely with the same amount of NH₄Cl, thereby forming NH₃ and NaCl. The number of moles of NH₄Cl left behind in the solution is (0.01-0.005) = 0.005 mol. Therefore, the mixed solution contains NH₃ and NH₄Cl. Hence, it is a buffer solution.

Question 21. The buffer capacity of a buffer solution consisting of a weak acid, HA, and its salt, NaA becomes maximum when its pH is 5.0. At this pH, what is the relation between the molar concentrations of HA and NaA? Abo finds the value of a of HA.
Answer:

In the case of a buffer solution consisting of a weak acid and its salt, the buffer capacity of the solution is maximum when pH = pK_a.

It is given that the buffer capacity of the given buffer is maximum when its pH = 5. Therefore, the pK_a ofthe weak acid present in the buffer is 5.

Now, the pH of a buffer solution consisting of a weak acid and its salt is given by

⇒  \(p H=p K_a+\log \frac{[\text { salt }]}{[\text { acid }]}\)

Since, \(p H=p K_a=5, \log \frac{[\text { salt }]}{[\text { acid }]}=0\) , [salt] = [acid]

∴ [NaA]=[HA]

Question 22. At a given temperature, if the solubility product and the
solubility of a sparingly soluble salt M₂X₃ are Ksp and S, respectively, then prove that S \(S=\left(\frac{K_{s p}}{108}\right)^{1 / 5}.\)
Answer:

In its saturated solution, M₂X₃ ionizes partially, forming M₃ (aq) and X₂ (aq) ions. Eventually, the following equilibrium is established.

⇒ \(\mathrm{M}_2 \mathrm{X}_3(s) \rightleftharpoons 2 \mathrm{M}^{3+}(a q)+3 \mathrm{X}^{2-}(a q)\)

If the solubility of M₂X₃ in its saturated solution is S mol.L^-1 , then in the solution [M³⁺] = 2S mol.L^-1 and [X^2-] = 3S mol.L^-1

Therefore, the solubility product of M₂X₃

⇒ \(K_{s p}=\left[\mathrm{M}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3=(2 S)^2(3 S)^3=108 S^5\)

∴ \(S=\left(\frac{K_{s p}}{108}\right)^{\frac{1}{5}}\)

Question 23. Find out the equilibrium constant for the reaction, XeO₄(g) + 2HF(g)⇌ XeO₃F₂(g) + HaO(g) Consider K₁ as the equilibrium constant for the reaction, XeF₆(g) + HaO(g) ⇌ XeOF₄(g) + 2HF(g) and K₂ as the equilibrium constant for the reaction, XeO₄(g) + XeF₆(g) ⇌ XeOF₄(g) + XeO₃F₂(g).
Answer:

According to the given condition

⇒ \(K_1=\frac{\left[\mathrm{XeOF}_4\right] \times[\mathrm{HF}]^2}{\left[\mathrm{XeF}_6\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]} \text { and } K_2=\frac{\left[\mathrm{XeOF}_4\right] \times\left[\mathrm{XeO}_3 \mathrm{~F}_2\right]}{\left[\mathrm{XeO}_4\right] \times\left[\mathrm{XeF}_6\right]}\)

∴ \(\text { For } \mathrm{XeO}_4(g)+2 \mathrm{HF}(g) \rightleftharpoons \mathrm{XeO}_3 \mathrm{~F}_2(g)+\mathrm{H}_2 \mathrm{O}(g)\)

Equilibrium constant ,\(K=\frac{\left[\mathrm{XeO}_3 \mathrm{~F}_2\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{XeO}_4\right] \times\left[\mathrm{HF}^2\right.}\)

⇒ \(\text { Now, } \frac{K_2}{K_1}=\frac{\left[\mathrm{XeO}_3 \mathrm{~F}_2\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{XeO}_4\right] \times[\mathrm{HF}]^2}=K\)

∴ K= \(\frac{K_2}{K_1}\)

Question 24.  At a particular temperature, for the reaction, aA + bB dC + cD, the equilibrium constant is K. Find out the equilibrium constants for the following reactions at the same temperature.

1. m a A +  m b B m d C+m c D

2. \(\frac{1}{m} a A+\frac{1}{m} b B\frac{1}{m} d C+\frac{1}{m} c D\)

Answer:

For the reaction, aA + bB ⇌ dC+cD the equilibrium constant

⇒ \(K=\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\)

1. Now,  For the reaction maA + mbB ⇌ mdC + mcD equilibrium constant,

⇒ \(K_1=\frac{[C]^{m d} \times[D]^{m c}}{[A]^{m a} \times[B]^{m b_3}}\)

= \(\left\{\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\right\}^m=K^m\)

2. For the reaction

⇒  \(\frac{1}{m} a A+\frac{1}{m} b B \rightleftharpoons \frac{1}{m} d C+\frac{1}{m} c D\) equilibrium constant,

⇒ \(K_2=\frac{[C]^{d / m} \times[D]^{c / m}}{[A]^{a / m} \times[B]^{b / m}}\) \(=\left\{\frac{[C]^d \times[D]^c}{[A]^a \times[B]^b}\right\}^{1 / m}\)

=\((K)^{1 / m}\)

Question 25. State the nature of aqueous solutions containing the following ions with reason:

1. NH⁺₄,
2. F^–
3. Cl^–.

Answer:

1. NH₄⁺:

NH₄⁺ ion is the conjugate acid of a weak base, NH₃. In an aqueous solution, NH₂ is stronger than H₂O [weak Bronsted acid]. Hence, in an aqueous solution, NH₂ ion reacts with water to produce H₃0+ ions, leaving the die solution acidic.

NH⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)

2. F^–:

F^– ion is the conjugate base of a weak acid, HF. F^– is stronger titan H₂O [weak Bronsted base] in aqueous solution. For this reason, in an aqueous solution, the F^– ion reacts with water to give OH^– ions, and as a result, the solution becomes basic.

⇒  \(\mathrm{F}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HF}(a q)+\mathrm{OH}^{-}(a q)\)

3. Cl^–:

Cl^– is the die conjugate base of strong acid, HCl. Hence, it is very weak and cannot react with water. Consequently, the die aqueous solution of Cl^– is neutral.

Question 26. The solubility of zinc phosphate in water is S mol. L^-1 . Derive the mathematical expression of the solubility product of the compound.
Answer:

The following equilibrium is established by zinc phosphate, [Zn3(P04)2] in its saturated aqueous solution.

⇒ \(\mathrm{Zn}_3\left(\mathrm{PO}_4\right)_2(s) \rightleftharpoons 3 \mathrm{Zn}^{2+}(a q)+2 \mathrm{PO}_4^{3-}(a q)\)

∴ \(K_{s p}=\left[\mathrm{Zn}^{2+}\right]^3 \times\left[\mathrm{PO}_4^{3-}\right]^2\)

As given, the solubility of Zn₃(PO₄)₂ is S mol. L^-1 .

Hence, \(\left[\mathrm{Zn}^{2+}\right]=3 S \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }\left[\mathrm{PO}_4^{3-}\right]=2 S \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ K_sp = (3S)³ × (2S)²= 108 S⁵; this is the mathematical expression of the solubility product of zinc phosphate.

Question 27. When H₂S gas Is passed through an acidified solution of Cu²⁺ and Zn²⁺, only CuS is precipitated—why?
Answer:

H₂S is a very weak acid. In its aqueous solution, only a small fraction of it ionizes to produce H₃O⁺ and S^2- ions

⇒\(\left[\mathrm{H}_2 \mathrm{~S}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q)\right]\)

In an acidified solution of H₂S, due to the common ion (H₃O⁺) effect, the ionization of H₂S is further reduced, and as a result, the concentration of S^2- ions becomes extremely low.

When H₂S gas is passed through an acidified solution of Cu²⁺ and Zn²⁺ ions, the concentration of S^-2 ions becomes so low that only the product of the concentrations of Cu²⁺ and S^2- ions exceeds the solubility product of CuS. However, the product of the concentrations of Zn²⁺ and S^2- ions lies well below the value of the solubility product of ZnS. This is why only CuS is precipitated in preference to ZnS.

Question 28. Why is the aqueous solution of Cu(NO₃)₂ acidic?
Answer:

Being a strong electrolyte, Cu(NO₃)₂ dissociates almost completely in its aqueous solution to form [Cu(H₂O)₆]²⁺ and NO₂ ions. In the solution, H₂O also ionizes slightly to form H₃O⁺ and OH^– ions

Cu(NO₃)₂(aq) + 6H₂O(l) → [Cu(H₂O)g]₂+(aq) + 2NO(aq)

2H₂O(l) ⇌  + H₃O⁺(aq) + OH

NO^–₃ is the conjugate base of a strong acid, HNO₃. Hence it is a very weak base. So, NO₂ ions cannot react with water. On the other hand, since the charge density of Cu²⁺ ion in the cationf[Cu(H₂O)₆]⁺² is very high, the H₂O molecules attached to it get polarised, and their O—H bonds become weaker.; These O —H bonds dissociate to produce H+ ions, which are accepted by H₂O to give H₃O⁺ ions.

Cu(H₂O)₆]2+(aq) + H₂O(l) ⇌  [Cu(H₂O)₅OH]⁺(aq) + H₃O⁺(aq)

As a result, the concentration of H₃O⁺ ions in the aqueous solution of Cu(NO₃)₂ becomes higher than that of OH- ions (from H₂O), and consequently, the solution becomes acidic.

Question 29. Despite being a neutral salt, the aqueous solution of Na₂CO₃ is alkaline—why?
Answer:

1. Na₂CO₃, being a strong electrolyte, dissociates almost completely in aqueous solution, forming Na⁺ and CO₃ ions:

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3(a q) \rightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_3^{2-}(a q)\)

2. Water, being a very poor electrolyte, ionizes slightly to form H3O+ and OH- ions:

⇒ \(2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

3. Hydrated Na+ ion (Na⁺(aq) ] is a very weak acid and cannot react with water.

4. CO₃^2- the conjugate base of weak acid HCO₃^–, is a stronger base than H₂O (which is a Bronsted base). Therefore, it reacts with water in aqueous solution, forming the following equilibrium.

⇒ \(\mathrm{CO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(a q) \rightleftharpoons \mathrm{HCO}_3^{-}(a q)+\mathrm{OH}^{-}(a q)\)

HCO₃^– so formed undergoes ionization

⇒ \(\mathrm{HCO}_3^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CO}_3^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)]\)

But, due to the common ion (CO₂^-3) effect, this ionization occurs to a very small extent.

As a result, [OH^–] in solution is much higher than [H⁺]. So, aqueous solution of Na₂CO₃ is alkaline.