PSEB 11th Class Chemistry Solutions Chapter 7 Equilibrium

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 7 Equilibrium Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 7 Equilibrium

PSEB 11th Class Chemistry Guide Equilibrium InText Questions and Answers

Question 1.
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?
Answer:
(a) If the volume of the container is suddenly increa50sed, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Question 2.
What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂] = 0.60M,[O₂] = 0.82M and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Answer:
The given reaction is
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Equilibrium constant
K_c = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=\frac{(1.90 \mathrm{M})^{2}}{(0.60 \mathrm{M})^{2}(0.82 \mathrm{M})}\)
= 12.238 M^-1

Question 3.
At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms I₂(g) ⇌ 2I(g)
Calculate K_p for the equilibrium.
Answer:
Given, I₂(g) ⇌ 2I(g)
I atoms in iodine vapours = 40% by volume
So, iodine vapours of I₂ molecules = 60% by volume

Question 4.
Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO (g) + Cl₂(g)
(ii) 2CU(NO₃)₂(S) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(oq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(ag)
(iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
(v) I₂(s) + 5F₂ ⇌ 2IF₅
Answer:

Question 5.
Find out the value of K_c for each of the following equilibria from the value of Kp.
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
The relation between K_p and K_c is given as
K_p = K_c(RT)^Δn
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K.
Δn = 3 – 2 = 1
R = 0.0831 bar L mol^-1K^-1
T = 500 K
K_p =1.8 x 10^-2
K_p = K_c( RT)^Δn
1.8 x 10^-2 = K_c(0.0831 x 500)¹
K_c = \(\frac{1.8 \times 10^{-2}}{0.0831 \times 500}\) = 4.33 x 10^-4

(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Δn = 2 -1 = 1
R = 0.0831 bar L mol^-1K^-1
T = 1073 K
K_p =167
Now, K_p = K_c(RT)^Δn
⇒ 167 = K_c(0.0831 x 1073)¹
⇒ K_c = \(\frac{167}{0.0831 \times 1073}\) = 1.87

Question 6.
For the following equilibriuih, K_c = 6.3 x 10¹⁴ at 1000 K
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?
Answer:
It is given that K_c for the forward reaction is 6.3 x 10¹⁴ at 1000 K.
Then, K_c for the reverse reaction will be,
NO₂(g) + O₃(g) ⇌ NO(g) + O₃(g)
K_c = \(\frac{1}{K_{c}}=\frac{1}{6.3 \times 10^{14}}\) = 1.59 x 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Answer:
For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Question 8.
Reaction between N₂ and O₂ takes place as follows:
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37 , determine the composition of equilibrium mixture.
Answer:
Let the concentration of N₂O at equilibrium be x.
The given reaction is :

The value of equilibrium constant i.e., K_c = 2.0 x 10^-37 is very small which means negligible amounts of N₂ and O₂ react.

Question 9.
Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO andBr₂.
Answer:
The balanced chemical equation is

Given, 2x = 0.0518
x = 0.0259 mol
Moles of NO at equilibrium = 0.087 – 2x
= 0.087-0.0518
= 0.0352 mol
Moles of Br2 at equilibrium = 0.0437 – x
= 0.0437 – 0.0259
= 0.0178 mol

Question 10.
At 450 K, K_p = 2.0 x 10¹⁰/bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature?
Answer:
The given reaction is
2SO₂(g) + O₂(g) ⇌ 2SO₃Cg)
Δn = 2 – 3 = -1
T = 450 K
R = 0.0831 bar L K^-1 mol^-1
K_p = 2.0 x 10¹⁰ bar^-1
We know that,
K_p = K_c(RT)^Δn
=> 2.0 x 10¹⁰ bar^-1 = k_c(0.0831 L bar K^-1 mol^-1 x 450 K)^-1
\(K_{c}=\frac{2.0 \times 10^{10} \mathrm{bar}^{-1}}{\left(0.0831 \mathrm{~L} \mathrm{barK} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)^{-1}}\)
K_c = (2.0 x 10¹⁰ bar^-1) (0.0831 L bar K^-1mol^-1450 K)
= 74.79 x 10¹⁰ L mol^-1 = 7.48 x 10¹¹ L mol^-1

Question 11.
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(g) ⇌ H₂(g) + I₂(g)
Answer:
The given reaction is

∵ Decrease is pressure of HI = 0.2 – 0.04 = 0.16 atm;
So equilibrium pressure of H₂ is \(\frac{0.16}{2}\) = 0.08 atm and for I₂ is \(\frac{0.16}{2}\) = 0.08 atm
as two moles of HI on dissociation gives 1 mol of H₂ and 1 mol of I₂.
Therefore,
K_p = \(\frac{p_{\mathrm{H}_{2}} \times p_{\mathrm{I}_{2}}}{\left(p_{\mathrm{HI}}\right)^{2}}=\frac{0.08 \times 0.08}{(0.04)^{2}}=\frac{0.0064}{0.0016}\) = 4.0
Hence, the value of K_p is 4.0.

Question 12.
A mixture of 1.57 mol of N₂,1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1. 7 x 10².
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer:
The given reaction is :
N₂(g) + 3H₂(g) 2NH₃(g)
Given, [N₂] = \(\frac{1.57}{20}\) = 0.0785 M
[H₂] = \(\frac{1.92}{20}\) = 0.096 M
[NH₃] = \(\frac{8.13}{20}\) = 0.4065 M
Now, reaction quotient Q_c. is :
Q_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.4065)^{2}}{(0.0785)(0.096)^{3}}\) = 2.4 x 10³M^-2
Since Q_c ≠ K_c the reaction mixture is not in equilibrium.
Again Q_c > K_c. Hence, the reaction will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}{\left[\mathrm{NO}^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}\right.}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
The balanced chemical equation corresponding to the given expression can be written as :
4NO(g) + 6H₂O(l) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is

H₂O reacted = 40% of 1 mol of H₂O = 0.4 mol
x = 0.4 mol 1 – x = 1 – 0.4 = 0.6 mol
Therefore, the equilibrium constant for the reaction,
K_c = \(\) = 0.444

Question 15.
At 700 K, equilibrium constant for the reaction
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L^-1 of Hl(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Answer:
The given reaction is
H₂(g) + I₂(g) ⇌ 2HI(g); K_c = 54.8
Or the reaction
2HI(g) ⇌ H₂(g) + I₂(g); K_c‘ = \(\)
Given, [HI] = 0.5 mol L^-1
According to equation
[H₂] = [I₂] = x mol L^-1
Therefore,
\(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=K_{c}^{\prime}\)
⇒ \(\frac{x \times x}{(0.5)^{2}}=\frac{1}{54.8}\)
⇒ x² = \(\frac{0.25}{54.8}\)
⇒ x = 0.06754
x = 0.068 mol L^-1
Hence, at equilibrium, [H₂] = [I₂] = x = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of IC1 was 0.78 M?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:

Question 17.
K_p = 0.04 atm at9 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
Answer:

Question 18.
Ethyl acetate is formed by the reaction between ethanol acid and acetic acid and the equilibrium is represented as :
CH₃COOH(l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅(Z) + H₂O(l)
(i) 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)
(ii) 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.
(iii) 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?
Answer:

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If value of K is 8.3 x 10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
The given reaction is

It is given that the value of equilibrium constant, K = 8.3 x 10^-3.
K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
[Given, [PCl₅]_equili = 0.5 x 10^-1 mol L^-1]
\(\frac{x \times x}{0.5 \times 10^{-1}}\) = 8.3 x 10^-3
⇒ x² = 4.15 x10^-4
⇒ x = 2.04 x 10^-2 = 0.0204 mol L^-1 = 0.02 mol L^-1
Therefore, at equilibrium,
[Pcl₃] = [Cl₂] = 0.02 mol L^-1

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (H) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂ (g); K_p = 0.265 atm at 1050 K. What are the equilibrium partial pressures of CO and ₂ at 1050 K if the initial partial pressures are P_Co = 1.4 atm and pCO₂ = 0.80 atm?
Answer:
(i) The given reaction is

Since Q_p > K_p, the reaction will proceed in the backward direction.
Therefore, we can say that the pressure of CO will increase while the pressure of CO₂ will decrease.
Now,let the increase in pressure of CO = decrease in pressure of CO₂ be p.
Hence pCO₂ = 0.80 – p and P_CO = 1.4 + p
and K_p = \(\frac{p_{\mathrm{CO}_{2}}}{p_{\mathrm{CO}}}\)
0.265 = \(\frac{0.80-p}{1.4+p}\)
0.371 + 0.265p = 0.80 — p= 1.265p= 0.429
p = 0.339atm
Hence, at equilibrium
P_CO2 = 0.80 – 0.339 = 0.461 atm
And, equilibrium partial pressure of
P_CO = 1.4 + 0.339 = 1.739 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.06 1.
At a particular time, the analysis shows that composition of the
reaction mixture is 3.0 mol L^-1 N₂,2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃ Is the reaction at equilibrium? if not in which direction
does the reaction tend to proceed to reach equilibrium?
Answer:
The given reaction is
N₂(g) + 3H₂(g) ⇌ 2NH₃(g);K_c = 0.061 at 500K
Given, [N₂] = 3.0mol L^-1, [H₂] = 2.0 mol L^-1, [NH₃] = 0.5 mol L^-1
So, Q = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\) = 0.0104
It is given that K_c = 0.06 1
Since Q_c ≠ K_c, the reaction is not at equilibrium.
Since Q_c < K_c, the reaction will proceed in the forward direction to reach equilibrium.

Question 22.
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

x = 11.312(3.30 x 10^-3 – x)
x = 0.03732 – 11.312x
x + 11.312x = 0.03732
x = \(\frac{0.03732}{12.312}\)= 3.0321 x 10^-3 mol L^-1
[BrCl]_equili = (3.30 x 10^-3 – 3.032 x 10^-3) mol L^-1
= 2.68 x 10^-4 mol L^-1

Question 23.
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature.
Answer:
Let the total mass of the gaseous mixture be 100g.
Mass of CO = 90.55 g
and, mass of CO₂ = (100 – 90.55) = 9.45 g
Now, number of moles of CO,
n_CO = \(\frac{90.55}{28}\) = 3.234 mol
(Molar mass of CO = 28 g mol^-1 )
Now, number of moles of CO₂,
n_CO = \(\)
(Molar mass of CO₂ = 44 g mol^-1 )

For the given reaction, Δn = 2 -1 = 1
We know that,
K_p = K_c(RT)^Δn
⇒ 14.19 = K_c(0.0831 x 1127)¹
⇒ K_c = 0.154 (approximately)

Question 24.
Calculate (a) \(\Delta \boldsymbol{G}^{\ominus}\) and (b) the equilibrium constant for the formation of N02 from NO and Oa at 298K
NO(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)
where \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (N02)= 52.0 kJ/mol; \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (NO) = 87.0kJ/mol;
\(\Delta_{f} \boldsymbol{G}^{\ominus}\) (O2) = 0 kJ/mol
Answer:
(a) The given reaction is
N0(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)

= (52.0 – 87.0 + \(\frac{1}{2}\) x 0 )kJ mol^-1 = -35.0 kJ mol^-1

(b) \(\Delta_{r} G^{\ominus}\) = – 2.303 RT logK_c
-35.0 = – 2.303 x 0.0831 x 298 log K_c
∴ log K_c= \(\frac{35}{5.7058}\)= 6.134
∴ K_c = antilog 6.134 = 1.361 x 10⁶.

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl₅(g) ⇌ PCl₃(g) +Cl₂(g)
(b) CaO (s) + CO₂ (g) ⇌ CaCO₃ (s)
(c) 3Fe(s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₅(g)
Answer:
(a)The number of moles of reaction products will increase. According to Le-Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(b) The number of moles of reaction products will decrease.
(c) The number of moles of reaction products remains the same

Question 26.
Which of the following reactions will get affected hy increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Answer:
In all the above reactions, the reaction no. (ii) proceeds with the same no. of moles on both sides
i.e., n_p = n_r = 3 .
∴ This reaction will not be affected by the increase in pressure i. e., the direction of equilibrium will not be affected by the increase in pressure. All other reactions will be affected by the increase in pressure.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
n_p > n_r , n_p = 2; nr = 1
∴ Equilibrium will shift to the left increasing pressure.
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
Here, n_r – 1; n_p = 2, therefore n_p > n_r
∴ Equilibrium will go to left on increase of pressure.
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
Here, n_r = 3; n_p = 1 therefore n_p < n_r
∴ Equilibrium will shift to the right on increasing pressure.
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Here n_r = 0; n_p = 1, therefore n_p > n_r
∴ Equilibrium will shift backwards (left) on increasing the pressure.
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Here n_r = 9; n_p = 10, therefore n_p > n_r
∴ Equilibrium will shift backwards on increasing the pressure.

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Answer:
Given reaction is H₂(g) + Br₂(g)⇌ 2HBr(g); K_p = 1.6 x 10⁵ at 1024 K
Therefore, for the reaction 2HBr(g) ⇌ H₂(g)+Br₂(g), the equilibrium constant will be,

p = 2.5 x 1^-2(5.0 x 10^-3)p
p+(5.0 x 10^-3)p = 2.5 x 10^-2
(1005 x 10^-3)p = 2.5 x 10^-2
p = 2.49 x 10^-2 bar = 2.5 x 10^-2 bar
rherefore, at equilibrium,
[H₂] = [Br₂] = 2.49 x 10^-2 bar
[HBr] =10 — 2 x (2.49 x 10^-2) bar
= 9.95 bar = 10 bar

Question 28.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Write an 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?
Answer:
(a) The given reaction is
CH₄(g) + H₄O(g) ⇌ CO(g) + 3H₂(g)
\(K_{p}=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_{2}}^{3}}{p_{\mathrm{CH}_{4}} \times p_{\mathrm{H}_{2} \mathrm{O}}}\)
(b) (1) According to LeChatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le-Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium 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.

Question 29.
Describe the effect of:
(a) addition of H₂
(b) addition of CH₃OH
(e) removal of CO (d) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO (g) ⇌ CH₃OH(g)
Answer:
2H₂(g) + CO(g) ⇌ CH₃OH(g)
According to Le Chatelier’s principle,
(a) Addition of H₂ (increase in concentration of reactants) shifts the equilibrium in forward direction (more product is formed).
(b) Addition of CH₃OH (increase in concentration of product) shifts the equilibrium in backward direction.
(c) Removal of CO also shifts the equilibrium in backward direction.
(d) Removal of CH₃OH shifts the equilibrium in forward direction.

Question 30.
At 473K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 x 10^-3. If decomposition is depicted as,
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = 1240 kJ mol^-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 PCl₅ is added
(ii) pressure is increased?
(iii) the temperature is increased?
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 x 10^-3
(a) K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
(b) Value of K_c for the reverse reaction at the same temperature is
K’_c = \(\frac{1}{K_{c}}=\frac{1}{8.3 \times 10^{-3}}\) = 1.2048 x 10² = 120.48
(c) (i) Addition pf PCl₅ have no effect on K_c because K_c is constant at constant temperature.
(ii) K_c does not change with pressure.
(iii) The given reaction is endothermic, hence on increasing the temperature, K_c will increase.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO(g) +H₂O (g) ⇌ CO₂(g) + H₂(g)
If a reaction vessel at 400° C is charged with an equimolar mixture of CO and steam such that p_CO = P_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_P = 10.1 at 400°C
Answer:
The given reaction is
CO(g) + H20(g) ⇌ C02(g) + H2(g)

p = 12.71 – 3.17p
4.17 p = 12.71
p = \(\frac{12.71}{4.17}\) = 3.04 bar
Hence PH₂ = 3.04 bar

Question 32.
Predict which of the following reaction will have appreciable concentration of reactants and products:
(a) Cl₂(g) ⇌ 2Cl(g);K_c = 5 x 10^-39
(b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 x 10⁸
(c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g); K_c = 1.8
Answer:
Following conclusions can be drawn from the values of K_c:
(a) Since the value of K_c is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K_c is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K_c is 1.8, this means that both the products and reactants have appreciable concentration.

Question 33.
The value of K_c for the reaction
3O₂(g) ⇌ 2O₃(g)
is 2.0 x 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 x 10^-2, what is the concentration of O₃?
Answer:
The given reaction is
3O₂(g) ⇌ 2O₃(g)
Then K
It is given that Kc = 2.0 x 10^-50 and [02(g)] = 1.6 x 10^-2
Then, we have,
\(2.0 \times 10^{-50}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[1.6 \times 10^{-2}\right]^{3}}\)
⇒ [O₃]² = 2.0 x 10^-50 x (1.6 x 10^-2)³
⇒ [O₃]² = 8.192 x 10^-56
⇒ [O₃] = 2.86 x 10^-28 M
Hence, the concentration of O₃ is 2.86 x 10^-28 M.

Question 34.
The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H20 and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:
The given equation is
CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)
Therefore,
\(\frac{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2}\right]^{3}}=K_{c}\)
Given, K_c = 3.90, [CO] = 0.30 mol, [H₂] = 0.10 mol and [H₂O] \(\frac{\left[\mathrm{CH}_{4}\right] \times 0.02}{0.3 \times(0.1)^{3}}\) = 3.90
[CH₄] = \(\frac{3.90 \times 0.3 \times(0.1)^{3}}{0.02}=\frac{0.00117}{0.02}\)
= 0.0585 M= 5.85 x 10^-2M
Hence, the concentration of CH4 at equilibrium is 5.85 x 10^-2 M.

Question 35.
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species :
\(\mathrm{HNO}_{2}, \mathrm{CN}^{-}, \mathrm{HClO}_{4}, \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_{3}^{2-} \text { and } \mathrm{S}^{2-}\)
Answe:
A conjugate acid-base pair is a pair that differs only by one proton.
The conjugate acid-base for the given species is mentioned in the table below:

Question 36.
Which of the followings are Lewis acids
\(\mathbf{H}_{2} \mathbf{O}, \mathbf{B F}_{3}, \mathrm{H}^{+} \text {and } \mathrm{NH}_{4}^{+}\)
Answer:
Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺ and \(\mathrm{NH}_{4}^{+}\) are Lewis acids.

Question 37.
What will be the conjugate bases for the Bronsted acids : HF, H₂SO₄ and \(\mathrm{HCO}_{3}^{-}\)?
Answer:
The table below lists the conjugate bases for the given Bronsted acids :

Question 38.
Write the conjugate acids for the following Bronsted bases: \(\mathbf{N H}_{2}^{-}\), NH₃ andHCOC^–.
Answer:
The table below lists the conjugate acids for the given Bronsted bases : Bronsted base Conjugate acid

Question 39.
The species : H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:
The table below lists the conjugate acids and conjugate bases for the given species :

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^– (b)F^– (c)H⁺ (d) BCl₃
OH^– and F^– are electron rich species and can donate electron pair. Hence, these act as Lewis base.

H⁺ and BCl₃ are electron deficient species and can accept electron pair. Hence, these act as Lewis acid.

Question 41.
The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3 M. What is its pH?
Answer:
Given,
[H⁺] = 3.8 x 10^-3 M
∴ pH value of soft drink = – log[H⁺] = – log(3.8 x 10^-3)
= – log3.8 – log10^-3 = – log3.8 + 3 log10
= – log3.8 + 3
= -0.58 + 3
= 2.42

Question 42.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
Given, pH = 3.76
We know that,
pH = – log[H⁺]
⇒ log[H⁺] = -pH
⇒ [H⁺] = antilog (-pH)
= antilog (-3.76) -1 +1 = antilog \(\overline{4} .24\) = 1.74 x 10-4 M Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 x 10^-4 M.

Question 43.
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1.8 x 10^-4 and 4 8 x 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
If K_a is the ionization constant of a weak acid and Kb is the ionization constant of its conjugate base then K_a.K_b = K_w
or K_b = \(\frac{K_{w}}{K_{a}}\)
Given, K_a of HF = 6.8 x 10^-4
Hence, K_b of its conjugate base F^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\)= 1.5 x 10^-11
(K_w = ionic product of water =1 x 10^-14 at 298 K)
Given, K_a of HCOOH = 1.8 x 10^-4
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}\) = 5.6 x 10^-11
Hence, K_b of its coagulate base CN^–
Given, K_a of HCN = 4.8 x 10^-9
Hence, K_b of its coagulate base HCOO^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-9}}\) = 2.08 x 10^-6

Question 44.
The ionization constant of phenol is 1.0 x 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?
Answer:
Ionization of phenol :
C₆H₅OH + H₂O ⇌ C₆H₅O^– + H₃O⁺

Question 45.
The first ionization constant of H2S is 9.1 x 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 HCI also? (ifi) If the second dissociation constant of H₂S is 1.2 x 10^-13, calculate the concentration of S^2- under both conditions.
Answer:

Hence, concentration of [HS^–] is decreased in the presence of 0.1 M
HCI due to common-ion effect.
(iii) For second dissociation constant,
HS + H₂O ⇌ H₃O⁺ + S^2- (In absence of HCl)
[HS^–] = 9.54 x 10^-5 M
\(K_{a_{2}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}\)

Question 46.
The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:
CH₃COOH CH₃COO^– + H⁺
K_a for CH₃COOH = 1.74 x 10^-5
[CH₃COOH] = c = 0.05 M
CH₃COOH CH₃COO^– + H⁺ [where α = degree of dissociation and c = molar concentration]

[CH₃COO^–] = 0.933 x 10^-3 = 9.33 x 10^-4 M
pH = – log[H⁺] = – log (9.33 x 10^-4)
= – (-4) – log9.3 = 4 – 0.9 = 3.03

Question 47.
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.
Answer:
Let the organic acid be HA.
⇒ HA ⇌ H⁺ + A^–
Concentration of HA = 0.01 M
pH = 4.15
-log[H⁺] = pH= 4.15
log[H⁺] = – 4.15
log[H⁺] = 5.85
[H⁺] = antilog \(\overline{5} .85\)
= 7.080 x 10^-5
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Now, [H⁺] = [A^–] = 7.08 x 10^-5 M
Then K_a = \(
K_a = 5.01 x 10^-7
PK_a = – logK_a = – log(5.01 x 10^-7)
pK_a = 7 – 0.699 = 6.301

Question 48.
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
Answer:
(a) HCl (aq) ⇌ H⁺ (aq) + Cl^–(aq)
[HCl]= 0.003 M
As HC1 is completely dissociated into H⁺ ions
∴ [H⁺] = [HCl] = 0.003 M
pH = – log[H⁺] = – log [3 x 10^-3]
= 3 + (-0.4771) = 2.523
(b) NaOH(aq) ⇌ Na⁺(aq) + OH^– (aq)
[NaOH] = 0.005 = 5 x 10^-3 M
[OH^–] = [NaOH] = 5 x 10^-3 M
∴ [latex]\left[\mathrm{H}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{5.0 \times 10^{-3}}\)
[H⁺]= 2.0 x 10^-12
∴ pH = – log(2 x 10^-12) = – (-12) – log2
= 12 – 0.30 = 11.70
[log2 = 0.30]

(c)

[HBr] = 0.002 M
[H⁺]= [HBr] = 0.002 M= 2.0 x 10^-3 M
pH = – log[H⁺] = – log[2 x 10^-3]
=- (-3) – log2
= 3 – log2
= 3 – 0.3 = 2.70

(d)

[OH^–] = 0.002 M
[H⁺] = \(\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.002}\) = 2 x 10^-12
pH = – log[H⁺] = -(-12) – log 5 = 12 – 0.70 = 11.30

Question 49.
Calculate the pH of the following solutions:
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 niL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution.
Answer:

(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution HC1 is completely dissociated to give H+ ions
[HCl] = ?
M₁V₁ = M₂V₂
1 mL of 13.6 M HCl = 1000 mL of M₂
M₂ = \(\frac{1 \times 13.6}{1000}\) = 0.0136 M
[HC1] = [H+] = 0.0136 M pH = – log[H+] = – log(1.36 x 10-2)
= – (-2) – log 1.36 = 2 – 0.13 = 1.87

Question 50.
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.
Answer:
α (Degree of ionization) = 0.132
c (molar cone.) = 0.1 M

∴ H⁺ = c x α = 0.1 x 0.132 = 0.0132
pH = – logH⁺ = – log(1.32 x 10^-2)
= – (-2) – log 1.32 = 2 – 0.12 = 1.88
pK_a = -logK_a
Now, K_a = cα²
K_a = 0.1 x (0.132)² = 1.74 x 10^-3
∴ pK_a = – log (1.74 x 10^-3) = – (-3) – log1.74 = 3 – 0.24 = 2.76

Question 51.
The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95.
Calculate its ionization constant and pK_b.
Answer:
Molar cone, of codeine, c = 0.005 = 5 x 10^-3
pH = 9.95
pOH = 14 – 9.95 = 4.05 (∵ pH + pOH = 14)
pOH = – log [OH^–]
log[OH^–] = -4.05= \(\overline{5} .95\)
[OH^–] = antilog \(\overline{5} .95\)
= 8.91 x 10-5
k_b = \(\left(\frac{8.91 \times 10^{-5}}{5 \times 10^{-3}}\right)^{2}\) = 1.588 x 10^-6
pK_b = – logK_b = – log(1.588 x 10^-6)
= 6 + (-0.2009) = 5.7991 = 5.80

Question 52.
What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from table 7.7 (427 x 10^-10). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
Given, K_b = 4.27 x 10^-10, c = 0.001 M
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\)

[0H] = 6.534 x 10^-7
pOH = — log(6.534 x 10^-7)
= 7+ (-0.8152)= 6.18
pH + pOH =14
pH 14—6.18=7.82

Thus, the ionization constant of the conjugate acid of aniline is 2.34 x 10^-5.

Question 53.
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?
Answer:
PK_a = – log K_a,
4.74 =-logK_a
log K_a = -4.74 = \(\overline{5} .26\)
Ka = antilog \(\overline{5} .26\) = 1.82 x 10^-5

[CH₃COOH is a weak acid and HC1 is a strong acid, so we can assume
that (cα + 0.01) 0.01]

In the presence of strong acid, dissociation of weak acid i.e., CH₃COOH decreases due to common ion effect.

Q.54. The Ionization constant of dimethylanilne is 54 x 10.
Calculate Its degree of ionization in its 0.02 M solution. What
percentage of dimethylamine is ionized if the solution is also
0.1MInNaOH?
Ans. Given, Kb = 5.4 x 10
c=0.02M

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

Question 55.
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.
Answer:
(a) pH of Human muscle fluid = 6.83
pH = – log[H+] log[H⁺] = -6.83 = \(\overline{7} .17\)
[H⁺] = antilog \(\overline{7} .17\)
[H⁺] = 1.48 x 10^-7 M

(b) pH of Human stomach fluid =1.2
log[H⁺] = -1.2 = \(\overline{2} .80\)
[H⁺] = antilog \(\overline{2} .80\)
.-. [H⁺] = 6.3 x 10^-2 M

(c) pH of Human blood = 7.38
log[H⁺] = – 7.38 = \(\overline{8} .62\)
.-. [H⁺] = antilog \(\overline{8} .62\) = 4.17 x 10^-8 M

(d) pH of Human saliva = 6.4
log[H⁺] =-6.4 = \(\overline{7} .60\)
[H⁺] = antilog \(\overline{7} .60\) = 3.98 x 10^-7 M

Question 56.
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.
Answer:
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = – log[H⁺]
(i) pH of milk = 6.8
Since, pH = -log[H⁺]
6.8 = -log[H+] log[H⁺] = -6.8 = \(\overline{7} .20\)
[H+] = antilog (\(\overline{7} .20\)) = 1.5 x 10~7 M

(ii) pH of black coffee = 5.0
Since, pH = – log[H⁺]
5.0 = – log[H+] log[H⁺] = – 5.0
[H⁺] = antilog (-5.00) = 10-5 M

(iii) pH of tomato juice = 4.2
Since, pH = – log[H⁺]
4.2 = – log[H⁺]
log[H⁺] = – 4.2 = \(\overline{5} .80\)
[H⁺] = antilog (\(\overline{5} .80\)) = 6.31 x 10^-5M

(iv) pH of lemon juice = 2.2
Since, pH = – log[H⁺]
2.2 = – log[H+]
log[H⁺] = -2.2 = \(\overline{3} .8\)
[H⁺] – antilog (\(\overline{3} .8\)) = 6.31 x 10^-3 M

(v) pH of egg white = 7.8
Since, pH = -log[H⁺]
7.8 = – log[H⁺]
log[H⁺]= -7.8 = \(\overline{8} .20\)
[H⁺] = antilog (\(\overline{8} .20\)) = 1.58 x 10^-8 M

Question 57.
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?
Answer:
Molar cone, of KOH = \(\frac{0.561 \times 1000}{56.1 \times 200}\) = 0.05M
56.1 x 200
KOH being a strong electrolyte, is completely ionized in aqueous solution.
KOH(aq) ⇌ K⁺(aq) + OH^–(aq)
[OH^–] = 0.05 M = [K⁺]
[H⁺][OH^–] = k_w
[H⁺] = \(\) = 2 x 10^-13
pH = – log[H⁺] = – log[2 x 10^-13]
= – (-13) – log2 = 13 – 0.03
∴ pH = 12.70

Question 58.
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.
Answer:
Solubility of Sr(OH)₂ = 19.23 g/L
Then, concentration of Sr(OH)₂ = \(\frac{19.23}{121.63 \times 1}\) M = 0.1581 M
Sr(OH)₂(aq) Sr²⁺(aq) + 2(OH^–)(aq)
∴ [Sr²⁺] = 0.1581 M
[OH^–] – 2 x 0.1581 M = 0.3162 M
Now
K_w = [OH^–] [H⁺]
\(\frac{1 \times 10^{-14}}{0.3162}\) = [H⁺]
[H⁺] = 3.16 x 10^-14
pH = – log[H⁺]
pH = 14 – 0.4997 = 13.5003 ≈ 13.5

Question 60.
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.
Answer:
Given, pH = 2.34
Molar cone, (c) = 0.1 M
HCNO H⁺ + CNO^–
pH = – log[H⁺]
2.34 = -log[H⁺]
log[H⁺] =-2.34 = \(\overline{3} .66\)
.-. [H⁺] = antilog \(\overline{3} .66\) = 4.57 x 10^-3 M
[H⁺] = \(\sqrt{K_{a}^{c}}\)
4.57 x 10^-3 = \(\sqrt{K_{a}^{c}}\)
Ionization constant,
K_a = 2.088 x 10^-4
Degree of ionization α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{2.088 \times 10^{-14}}{0.1}}\)
α = 0.0457

Question 61.
The ionization constant of nitrous acid is 45 x 104. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Hydrolysis constant K_h = \(\frac{K_{w}}{K_{a}}\)
where K_w = Ionic product of water, K_a = Ionisation constant of the acid

pOH = -log(9.42 x 10^-7)= 7-0.97= 6.03
∴ pH = 14 – pOH = 14 – 6.03 = 7.97

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Given, pH = 3.44
We know that,
PH = – log[H⁺]
.-. [H⁺]= 3.63 x 10^-4
Then K_h = \(\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}\) (Concentration = 0.02M)
=> K_h = 6.6 x 10^-6
Now, K_h = \(\frac{K_{w}}{K_{a}}\)
K_a = \(\frac{K_{w}}{K_{h}}=\frac{1 \times 10^{-14}}{6.6 \times 10^{6}}\)
= 1.51 x 10^-9

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Answer:

Question 64.
The ionization constant of chloroacetic acid is 1.35 x 10^-3 . What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?
Answer:
Given that Ka = 1.35 x 10^-3.
=> Ka = cα²
α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}}\)
(∵ Concentration of acid = 0.1 M)
= \(\sqrt{1.35 \times 10^{-2}}\) =0.116
.-. [H⁺]= cα = 0.1 x 0.116 = 0.0116
=> pH = – log[H⁺] = – log[0.0116] = 1.94
To find pH of 0.1 M sodium salt, we use the formula
pH = – \(\frac{1}{2}\)[log C_w + log K_a – log c]
= –\(\frac{1}{2}\)[log1 x 10^-14 + log(1.35 x 10^-3) – log(0.1)]
= –\(\frac{1}{2}\)[-14 + (-3 + 0.1303) – (-1)]
= – \(\frac{1}{2}\) [-15.8697] = 7.93485 ≈ 7.94

Question 65.
Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?
Answer:
Ionic product,
K_w = [H₃O⁺] [OH^–]
= 2.7 x 10^-14 at 310 K
H₂O + H₂O *=* [H₃0⁺][OH^–]
[H₃0⁺]= [OH^–]
Therefore, [H₃0⁺] = \(\sqrt{2.7 \times 10^{-14}}\)
⇒ = 1.64 x 10^-7 M
⇒ [H₃0⁺] = 1.64 x 10^-7
⇒ pH = – log[H₃0⁺] = – log[1.64 x 10^-7 = 6.78
Hence, the pH of neutral water is 6.78.

Question 66.
Calculate the pH of the resultant mixtures :
(a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HC1
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH
Answer:

Question 67.
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants.
[K_sp(Ag₂CrO₄) = 1.1 x 10^-12, K_sp(BaCrO₄) = 1.2 x 10^-10,
K_sp[Fe(OH)₃] = 1.0 x 10^-38, K_sp(PbCl₂) = 1.6 x 10^-5
K_sp(Hg₂I₂) = 4.5 x 10^-29]
Determine also the molarities of individual ions.
Answer:
(i) Silver chromate : Ag₂CrO₄⇌ 2Ag⁺ + CrO₄^2-, K_sp = 1.1 x 10^-12
Then, K_sp = [Ag⁺]²[CrO₄^2-]
Let the solubility of Ag₂CrO₄ be s.
⇒ [Ag⁺] = 2s and [CrO₄^2-] = s
Then, K_sp = (2s)² s – 4s³
⇒ 1.1 x 10^-12 = 4s³
0.275 x 10^-12 = s³
s = 0.65 x 10^-4 M
Molarity of Ag⁺ = 2s = 2 x 0.65 x 10^-4
= 1.30 x 10^-4 M
Molarity of CrO₄^2- = s = 0.65 x 10^-4 M

(ii) Barium chromate : BaCrO₄ ⇌ Ba²⁺ + \(\mathrm{CrO}_{4}^{2-}\); K_sp = 1.2 x 10^-10
Then, K_sp = [Ba²⁺] [latex]\mathrm{CrO}_{4}^{2-}[/latex]
Let s be the solubility of BaCrO₄.
⇒ [Ba²⁺] = s and [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s
⇒ K_sp = s²
⇒ 1.2 x 10^-10 = s²
⇒ s = 1.09 x 10^-5 M
Molarity of [Ba²⁺] = Molarity of [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s = 1.09 x 10^-5 M

(iii) Ferric hydroxide: Fe(OH)₃ ⇌ Fe²⁺ + 3OH^–; K_sp = 1.0.x 10^-38

K_sp = [Fe²⁺][OH^–]³
Let s be the solubility of Fe(OH)₃
⇒ [Fe³⁺] = s and [OH^–] = 3s
⇒ K_sp = s. (3s)³ = s x 27x³
K_sp = 27x⁴
1.0 x 10^-38 = 27x⁴
0.037 x 10^-38 = s⁴
0.00037 x 10^-36 = s⁴
s = 1.39 x 10^-10 M
Molarity of [Fe³⁺] = s = 1.39 x 10^-10 M
Molarity of [OH^–] = 3s = 4.17 x 10^-10 M

(iv) Lead chloride : PbCl₂ ⇌ Pb²⁺ + 2Cl^–; K_sp = 1.6 x 10^-5
K_sp = [Pb²⁺][Cl^–]²
Let s be the solubility of PbCl₂.
⇒ [Pb²⁺] = s and [Cl^–] = 2s
Thus, K_sp = s. (2s)² = 4s³
⇒ 1.6 x 10^-5 = 4s³
⇒ 0.4 x 10^-5 = s³
4 x 10^-6 = s³
⇒ s = 1.59 x 10^-2 M
Molarity of [Pb²⁺] = s = 1.59 x 10^-2 M
Molarity of [Cl^–] = 2s = 3.18 x 10^-2 M

(v) Mercurous iodide : Hg₂I₂ ⇌ \(\mathrm{Hg}_{2}^{2+}\) + = 4.5 x 10^-29
K_sp = [\(\mathrm{Hg}_{2}^{2+}\) ] []I^–]
Let s be the solubility of [Hg²I²]
⇒ [Hg₂] = s and [I^–] = 2s
K_sp = (s).(2s)² = 4s³
⇒ 4s³ = 4.5 x 10^-29
⇒ s³ = 1.125 x 10^-29
s = 2.24 x 10^-10 M
Molarity of [ \(\mathrm{Hg}_{2}^{2+}\) ] = s = 2.24 x 10^-10 M
Molarity of [I^–] = 2s = 4.48 x 10^-10 M.

Question 68.
The solubility product constant of Ag₂CrO₄ and AgBr are
1.1 x 10^-12 and 5.0 x 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
Let s be the solubility of Ag₂CrO₄
Thus, Ag₂CrO₄ ⇌ 2Ag²⁺ + \(\mathrm{CrO}_{4}^{-}\); K_sp = 1.1 x 10^-12
K_sp = [Ag²⁺]². [latex]\mathrm{CrO}_{4}^{-}[/latex]
=> [Ag²⁺] = (2s)² and [latex]\mathrm{CrO}_{4}^{-}[/latex] = s
K_sp = (2s)². s= 4s³
1.1 x 10^-12 = 4s³
s = 6.5 x 10^-5 M

Let s be the solubility of AgBr.
AgBr(s) ⇌ Ag⁺ + Br^–; K_sp = 5.0 x 10^-13
K_sp = s² = 5.0 x 10^-13
s = \(\sqrt{5.0 \times 10^{-13}}\)
∴ s = 7.07 x 10^-7 M

Therefore, the ratio of the molarities of their saturated solution is
\(\frac{s\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)}{s(\mathrm{AgBr})}=\frac{6.5 \times 10^{-5} \mathrm{M}}{7.07 \times 10^{-7} \mathrm{M}}\) = 9.19

Question 69.
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 x 10^-8).
Ans. 2NaIO₃ + Cu(ClO₃)3 → 2NaClO₃ + Cu(IO₃)₂
Molar cone, of both solutions before mixing = 0.002 M
Molar cone, of both solution after mixing
\(\left[\mathrm{IO}_{3}^{-}\right]=\left[\mathrm{Cu}^{2+}\right]=\frac{0.002}{2}\)= 0.001 M
Cu(IO₃)₂ ⇌ Cu²⁺ + \(2 \mathrm{IO}_{3}^{-}\)
[Cu²⁺] = 0.001 M
\(\) = 0.001 M
Ionic product = [Cu²⁺]\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\)
– 1 x 10^-3 x [1 x 10^-3]² = 1 x 10^-9
K_sp = 7.4 x 10^-8
Cu(IO₃)₂ is precipitated if [Cu²⁺] .\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\) > K_sp
Since, the ionic product is less than the solubility product. Hence there will be no precipitation.

Question 70.
The ionization constant of benzoic acid is 6.46 x 10^-5 and Ksp for silver benzoate is 2.5 x 10 . How many times is silver benzoate
more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Answer:

where s is the solubility of C₆H₅COOAg
pH = 3.19
pH = -log[H⁺]
log[H⁺] = – pH = -3.19 = \(\overline{4} .81\)
[H⁺] = antilog \(\overline{4} .81\) = 6.46 x 10^-4

C₆H₅COOAg is 3.2 times more soluble in buffer than in pure water.