NCERT Solutions Class 11 Chemistry Chapter 7

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

PC{l}_5(g)\underset{}{\rightleftharpoons }PC{l}_3(g)+C{l}_2(g)

Ans.

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

\underset{At\text{equilibrium 0}{\text{.5×10}}^{-1}mol{L}^{-1}}{PC{l}_5(g)}\underset{}{\rightleftharpoons }\underset{x{\text{molL}}^{-1}}{PC{l}_3(g)}+\underset{x{\text{molL}}^{-1}}{C{l}_2(g)}

The value of equilibrium constant K_c = 8.3 x 10^-3

Now we can write the expression for equilibrium as:

\begin{array}{l}\Rightarrow \frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=K_c\\ \Rightarrow \frac{x\cdot +x}{\left(0.5\times {10}^{-1}\right)}=8.3\times {10}^{-3}\end{array}

⇒ x²= 4.15 x 10^-4

⇒ x = 2.04 x 10^-2

= 0.0204

= 0.02 (approx)

At equilibrium, we can write as-

[PCl₃] = [Cl₂] = 0.02 mol L^-1

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

FeO\left(s\right)+CO\left(g\right)\underset{}{\rightleftharpoons }Fe\left(s\right)+C{O}_2\left(g\right),K_p=0.265at1050K

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

Ans.

Pco=1.4atm and Pco₂ = 0.80 atm ?

\begin{array}{l}\underset{Initially}{FeO\left(s\right)}+\underset{1.4\text{atm}}{CO\left(g\right)}\underset{}{\rightleftharpoons }Fe\left(s\right)+\underset{0.80\text{atm}}{C{O}_2\left(g\right)}\\ Q_p=\frac{p_{C{O}_2}}{p_{CO}}=\frac{0.80}{1.4}\end{array}

= 0.571

Given K_p = 0.265 at 1050 K

Q_p > K_p

Thus, the reaction will proceed in the backward direction.

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

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

\begin{array}{l}{K}_p=\frac{p_{C{O}_2}}{p_{CO}}\\ \Rightarrow 0.265=\frac{0.80-p}{1.4+p}\\ \Rightarrow 0.371+0.265p=0.80-p\\ \Rightarrow 1.265p=0.429\end{array}

⇒ p=0.339 atm

At equilibrium, partial pressure of CO₂,

p_{C{O}_2}=0.80-0.339=0.461\text{atm}

And at equilibrium, partial pressure of CO,

p_CO= 1.4 + 0.339 = 1.739 atm

Q.3 Equilibrium constant, K_c for the reaction

N_2\left(g\right)+3H_2\left(g\right)\underset{}{\rightleftharpoons }2N{H}_3\left(g\right)\text{at}500K\text{is 0}\text{.061}

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

Ans.

\begin{array}{l}\underset{At\text{a particular time: 3}{\text{.0 molL}}^{-1}}{N_2\left(g\right)}+\underset{\text{2}{\text{.0 molL}}^{-1}}{3H_2\left(g\right)}\underset{}{\rightleftharpoons }\underset{\text{0}{\text{.5 molL}}^{-1}}{2N{H}_3\left(g\right)}\\ Now,we\text{know that,}\\ Q_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}\\ =\frac{{\left(0.5\right)}^2}{\left(3.0\right){\left(2.0\right)}^3}=0.0104\end{array}

As given, K_c = 0.061

Since Q_c ≠ K_c the reaction is not at equilibrium.

and Q_c < K_c , reaction will proceed in the forward direction to reach equilibrium.

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

2BrCl\left(g\right)\underset{}{\rightleftharpoons }B{r}_2\left(g\right)+C{l}_2\left(g\right)

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

Ans.

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

\begin{array}{l}\begin{array}{ccccc}2BrCl\left(g\right)& \underset{}{\rightleftharpoons }& B{r}_2\left(g\right)& +& C{l}_2\left(g\right)\\ Initial\text{Conc}\text{. 3}{\text{.3×10}}^{-3}& & 0& & 0\\ At\text{equilibrium}\left(3.3\times {10}^{-3}-2x\right)& & x& & x\end{array}\\ Thus,\\ \Rightarrow \frac{\left[B{r}_2\right]\left[C{l}_2\right]}{\left[BrC{l}^2\right]}=K_c\\ \Rightarrow \frac{x\cdot x}{{\left(3.3\times {10}^{-3}-2x\right)}^2}=32\\ \Rightarrow \frac{x}{\left(3.3\times {10}^{-3}-2x\right)}=5.66\\ \Rightarrow x=18.678\times {10}^{-3}-11.32x\\ \Rightarrow 12.32x=18.678\times {10}^{-3}\\ \Rightarrow x=1.5\times {10}^{-3}\\ Therefore,\text{at equilibrium,}\\ \left[BrCl\right]=3.3\times {10}^{-3}-\left(2\times 1.5\times {10}^{-3}\right)\\ =3.3\times {10}^{-3}-3.0\times {10}^{-3}\\ =0.3\times {10}^{-3}\\ =3.0\times {10}^{-4}mol{L}^{-1}\end{array}

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

C\left(s\right)+C{O}_2\left(g\right)\underset{}{\rightleftharpoons }2CO\left(g\right)

Calculate K_c for this reaction at the above temperature.

Ans.

Let us assume that the total mass of the gaseous mixture = 100 g
Mass of CO = 90.55 g and
Mass of CO₂ = (100 – 90.55) g = 9.45 g
Number of moles of CO (n_CO) = 90.55/28 = 3.234 mol
[Molecular mass of CO = (12+16) = 28]
Number of moles of CO₂ (n_CO2) = 9.45/44 = 0.215 mol
[Molecular mass of CO₂ = (12+32) = 44]
Partial pressure of CO,

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

NO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\underset{}{\rightleftharpoons }N{O}_2\left(g\right)

Ans.

Where Δ_fG° (NO₂) = 52.0 kJ/mol

Δ_fG° (NO) = 87.0 kJ/mol

Δ_fG° (O₂) = 0 kJ/mol

Ans. (a) For the above reaction,

ΔG°= ΔG° (products) – ΔG° (reactants)

= 52.0 – (87.0 + 0)

= -35.0 KJ mol^-1

(b) -ΔG° = RT ln K_c

or – ΔG° = 2.303 RT log K_c

or log K_c = -ΔG°/2.303 RT

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

= 1.36 x 10⁶

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

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

\begin{array}{l}a)PC{l}_5\left(g\right)\underset{}{\rightleftharpoons }PC{l}_3\left(g\right)+C{l}_2\left(g\right)\\ b)CaO\left(s\right)+C{O}_2\left(g\right)\underset{}{\rightleftharpoons }CaC{O}_3\left(s\right)\\ c)3Fe\left(s\right)+4H_{2}O\left(I\right)\underset{}{\rightleftharpoons }F{e}_3{O}_4\left(s\right)+4H_2\left(g\right)\end{array}

Ans.

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

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

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

\begin{array}{l}(i)COC{l}_2\left(g\right)\underset{}{\rightleftharpoons }CO\left(g\right)+C{l}_2\left(g\right)\\ (ii)C{H}_4\left(g\right)+2S_2\left(g\right)\underset{}{\rightleftharpoons }C{S}_2\left(g\right)+2H_{2}S\left(g\right)\\ (iii)C{O}_2\left(g\right)+C\left(s\right)\underset{}{\rightleftharpoons }2CO\left(g\right)\\ (iv){\text{2H}}_2\left(g\right)+CO\left(g\right)\underset{}{\rightleftharpoons }C{H}_{3}OH\left(g\right)\\ (v)CaC{O}_3\left(s\right)\underset{}{\rightleftharpoons }CaO\left(s\right)+C{O}_2\left(g\right)\\ (vi)4N{H}_3\left(g\right)+5O_2\left(g\right)\underset{}{\rightleftharpoons }4NO\left(g\right)+6H_{2}O\left(g\right)\end{array}

Ans.

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

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

H_2\left(g\right)+B{r}_2\left(g\right)\underset{}{\rightleftharpoons }2HBr\left(g\right)

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

Ans.

Equilibrium constant for the forward reaction,

K_p = 1.6 x 10⁵

Let us calculate equilibrium constant for the backward reaction.

K_p^’ = 1/K_p

\begin{array}{l}=\frac{1}{1.6\times {10}^5}=6.25\times {10}^{-6}\\ \text{The backward reaction can be written as}\\ 2HBr\left(g\right)\underset{}{\rightleftharpoons }{H}_2\left(g\right)+B{r}_2\left(g\right)\end{array}

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

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 10}\\ \\ \text{At Equilibrium}\left(10-2p\right)\end{array}}{2HBr\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ p\end{array}}{H_2\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ p\end{array}}{B{r}_2\left(g\right)}

Thus,

\begin{array}{l}\frac{p_{H_2}\times p_{B{r}_2}}{p^{2}HBr}=K{‘}_p\\ \Rightarrow \frac{p\times p}{{\left(10-2p\right)}^2}=6.25\times {10}^{-6}\\ \Rightarrow \frac{p\times p}{\left(10-2p\right)}=2.5\times {10}^{-3}\\ \Rightarrow p=2.5\times {10}^{-2}-\left(5.0\times {10}^{-3}\right)p\\ \Rightarrow p+\left(5.0\times {10}^{-3}\right)p=2.5\times {10}^{-2}\\ \Rightarrow p=2.49\times {10}^{-2}bar\end{array}

Therefore, at equilibrium,

[H₂] = [Br₂] = 2.49 x 10^-2 bar

[HBr] = 10 – 2 x (2.49 x 10^-2) bar

= 10 bar (approx)

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

C{H}_4\left(g\right)+H_{2}O\left(g\right)\underset{}{\rightleftharpoons }CO\left(g\right)+3H_2\left(g\right)

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

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

(i) Increasing the pressure

(ii) Increasing the temperature

(iii) Using a catalyst

Ans.

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

K_p=\frac{p_{CO}\times p_{H_2}^3}{p_{C{H}_4}\times p_{H_{2}O}}

(b)

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

Q.11 Describe the effect of:

a) Addition of H₂

b) Addition of CH₃OH

c) Removal of CO

d) Removal of CH₃OH

On the equilibrium of the reaction:

2H_2\left(g\right)+CO\left(g\right)\underset{}{\rightleftharpoons }C{H}_{3}OH\left(g\right)

Ans.

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

b) On addition of CH₃OH, the rate of backward reaction increases than the rate of forward reaction. Hence, the equilibrium will shift in the backward direction. . To maintain, the equilibrium constant some amount of CH₃OH breaks into CO and H_2.

c) On removing CO, the rate of forward reaction decreases than the rate of backward reaction. Hence, the equilibrium will shift in the backward direction. To maintain the equilibrium constant, some amount of CH₃OH breaks into CO and H_2.

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

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

PC{l}_5\left(g\right)\underset{}{\rightleftharpoons }P{C}_3\left(g\right)+C{l}_2\left(g\right),{\Delta }_{f}Hº=124.0{\text{kJmol}}^{-1}

a) Write an expression for K_c for the reaction.

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

c) What would be the effect on K_c

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

Ans.

a) The expression for K_c for the given reaction

K_c=\frac{\left[PC{l}_3\left(g\right)\right]\left[C{l}_2\left(g\right)\right]}{\left[PC{l}_5\left(g\right)\right]}

b) The value of K_c for the backward reaction:

K’_c= 1/ K_c

=1 / (8.3 x 10^-3)

= 1.2048 x 10²

= 120.48

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

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

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

As K_c= k_f/k_b

Thus, the value of K_c also increases.

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

CO\left(g\right)+H_{2}O\left(g\right)\underset{}{\rightleftharpoons }C{O}_2\left(g\right)+H_2\left(g\right)

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

p_{CO}=p_{H_{2}O}=4.0ba\mathrm{r,}

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

Ans.

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

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 4}\text{.0 bar}\\ \\ \text{At Equilibrium}\left(4.0-p\right)\end{array}}{Co\left(g\right)}+\underset{\begin{array}{l}\\ \text{4}\text{.0 bar}\\ \\ \left(4.0-p\right)\end{array}}{H_{2}O\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{p}\end{array}}{C{O}_2\left(g\right)}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{p}\end{array}}{H_2\left(g\right)}

And K_p = 10.1

Thus,

\begin{array}{l}\frac{p_{C{O}_2}\times p_{H_2}}{p_{CO}\times p_{H_{2}O}}=K_p\\ \Rightarrow \frac{p\times p}{{\left(4.0-p\right)}^2}=10.1\\ \Rightarrow \frac{p\times p}{\left(4.0-p\right)}=3.178\\ \Rightarrow p=12.712-3.178\text{p}\\ \text{4}\text{.178 p}=12.712\\ \Rightarrow p=3.04\end{array}

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

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

\begin{array}{l}a)C{l}_2\left(g\right)\underset{}{\rightleftharpoons }2Cl\left(g\right),K_c=5\times {10}^{-39}\\ b)C{l}_2\left(g\right)+2NO\left(g\right)\underset{}{\rightleftharpoons }NOCl\left(g\right),K_c=3.7\times {10}^8\\ c)C{l}_2\left(g\right)+2N{O}_2\left(g\right)\underset{}{\rightleftharpoons }2N{O}_{2}Cl\left(g\right),K_c=1.8\end{array}

Ans.

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

Q.15 The value of K_c for the reaction

3O_2\left(g\right)\underset{}{\rightleftharpoons }2O_3\left(g\right)

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

Ans.

The given reaction is:

\begin{array}{l}3O_2\left(g\right)\underset{}{\rightleftharpoons }2O_3\left(g\right)\\ K_c=\frac{{\left[O_3\left(g\right)\right]}^2}{{\left[O_2\left(g\right)\right]}^3}\end{array}

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

\begin{array}{l}\Rightarrow 2.0\times {10}^{-50}=\frac{{\left[O_3\left(g\right)\right]}^2}{{\left(1.6\times {10}^{-2}\right)}^3}\\ \Rightarrow {\left[O_3\left(g\right)\right]}^2=\left(2.0\times {10}^{-50}\right){\left(1.6\times {10}^{-2}\right)}^3\\ \Rightarrow {\left[O_3\left(g\right)\right]}^2=8.192\times {10}^{-56}\\ \Rightarrow \left[O_3\left(g\right)\right]=2.86\times {10}^{28}\text{M}\end{array}

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

Q.16 The reaction,

CO\left(g\right)+3H_2\left(g\right)\underset{}{\rightleftharpoons }C{H}_4\left(g\right)+H_{2}O\left(g\right)

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

Ans.

Equilibrium

Conc. 0.30 0.10 x 0.02

CO\left(g\right)+3H_2\left(g\right)\underset{}{\rightleftharpoons }C{H}_4\left(g\right)+H_{2}O\left(g\right)

The equilibrium constant for the reaction,

\begin{array}{l}{K}_c=\frac{\left[C{H}_4\left(g\right)\right]\left[H_{2}O\left(g\right)\right]}{\left[CH\left(g\right)\right]{\left[H_{2}O\left(g\right)\right]}^3}\\ \Rightarrow \frac{x\times 0.02}{0.3\times {\left(0.1\right)}^3}=3.90\\ \Rightarrow x=\frac{3.90\times 0.3\times {\left(0.1\right)}^3}{0.02}\end{array}

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

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

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

HNO₂, CN^–, HClO₄^–, F^–, OH^–, CO₃^2- and S^2-

Ans.

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

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

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

Ans.

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

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

Ans.

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

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

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

Ans.

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

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

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

Ans.

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

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

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

1. OH^– (b) F^– (c) H⁺ and (d) BCl₃

Ans.

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

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

Ans.

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

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

Ans.

Given pH= 3.76

From the relation,

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

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

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

Ans.

From the relation,

K_w=K_a\times K_b

Here,

K_w= ionic product of water

K_b = ionization constant of conjugate base

K_a = ionization constant of conjugate acid

K_b=\frac{K_w}{K_a}

K_a of HF = 6.8 × 10^–4

K_w of water = 1 x 10^-14

Hence, K_b of its conjugate base F^–

\begin{array}{l}=\frac{K_w}{K_a}\\ =\frac{1\times {10}^{-14}}{6.8\times {10}^{-4}}=1.5\times {10}^{-11}\end{array}

Again, K_a of HCOOH = 1.8 × 10^–4

So, K_b of its conjugate base HCOO^–

=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-4}}=5.6\times {10}^{-11}

Given, K_a of HCN = 4.8 × 10^–9

=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{4.8\times {10}^{-9}}=2.08\times {10}^{-6}

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

Ans.

Ionization of phenol involves the following reaction:

\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.05}\\ \\ \text{At equilibrium}\left(0.05-x\right)\end{array}}{C_6{H}_{5}OH}+H_{2}O\underset{}{\rightleftharpoons }{\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{x}\end{array}}{C_6{H}_{5}O}}^{-}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{x}\end{array}}{H_3{O}^{+}}\\ K_a=\frac{\left[C_6{H}_5{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_6{H}_{5}OH\right]}\\ \Rightarrow K_a\frac{x\cdot x}{\left(0.05-x\right)}\\ \text{The value of the ionization constant is very less},\text{x will be very small}.\text{We can ignore x in the denominator}.\\ x=\sqrt{\left(1\times {10}^{-10}\times 0.05\right)}\\ x=\sqrt{\left(5\times {10}^{-12}\right)}\\ =2.2\times {10}^{-6}M=\left[H_3{O}^{+}\right]\\ Since\left[H_3{O}^{+}\right]=\left[C_6{H}_5{O}^{-}\right]\\ \left[C_6{H}_5{O}^{-}\right]=2.2\times {10}^{-6}M\end{array}

Let the degree of ionization of phenol be α.

\begin{array}{l}\underset{\begin{array}{l}\\ Conc.\left(0.05-0.05\alpha \right)\end{array}}{C_6{H}_{5}OH}+H_{2}O\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{C_6{H}_5{O}^{-}}+\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{H_3{O}^{+}}\\ \underset{\begin{array}{l}\\ Conc.\end{array}}{C_6{H}_{5}ONa}\underset{}{\rightleftharpoons }{C}_6{H}_5{O}^{-}+\underset{\begin{array}{l}\\ 0.01\end{array}}{N{a}^{+}}\end{array}

[C₆H₅OH]= 0.05 – 0.05α; 0.05 M

[C₆H₅O^–] = 0.01 + 0.05α; 0.01 M

[H₃O⁺] = 0.05α

\begin{array}{l}{K}_a=\frac{\left[C_6{H}_5{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_6{H}_{5}OH\right]}\\ K_a=\frac{\left(0.01\right)\left(0.05\alpha \right)}{\left(0.05\right)}\\ 1\times {10}^{-10}=0.01\alpha \\ \alpha =1\times {10}^{-8}\end{array}

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

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

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

(iii)Also if the second dissociation constant of H₂S is 1.2 × 10^–13, calculate the concentration of S^2– under both conditions.

Ans.

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

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

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

0.1 – x = 0.1

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

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

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

Thus, at equilibrium,

[H₂S] = 0.1 – y» 0.1,

[H⁺]= 0.1 + y » 0.1

[HS^–]= y M

K_a=\frac{0.1\times y}{0.1}

= 9.1 x 10^-8

Thus, y = 9.1 x 10^-8 M

(iii)The ionization of H₂S is as follows –

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. (M) 9}{\text{.4×10}}^{-5}\\ \\ final\text{conc}\text{. (M) 9}{\text{.4×10}}^{-5}-X\end{array}}{H_{2}S}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ \text{9}{\text{.4×10}}^{-5}\\ \\ \text{X}\end{array}}{2H^{+}}+\underset{\begin{array}{l}\\ \text{0}\\ \\ \text{X}\end{array}}{S^{2-}}

Therefore,

\begin{array}{l}{K}_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}\\ K_{a_2}=\frac{\left(9.54\times {10}^{-5}\right)\left(X\right)}{9.54\times {10}^{-5}}\\ X=1.2\times {10}^{-13}\end{array}

Hence,

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

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

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

And

HCl\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ \text{0}\text{.1M}\end{array}}{H^{+}}+\underset{\begin{array}{l}\\ 0.1M\end{array}}{C{l}^{-}} \begin{array}{l}{K}_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}\\ 1.2\times {10}^{-13}=\frac{\left(0.1\right)\left(Y\right)}{9.1\times {10}^{-8}}\\ 10.92\times {10}^{-21}=0.1Y\\ \frac{10.92\times {10}^{-21}}{0.1}=Y\\ Y=1.092\times {10}^{-19}M\\ \left[S^{2-}\right]=1.092\times {10}^{-19}M\end{array}

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

Ans.

The given reaction is

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

K_a= ionization constant

C = concentration

C = 0.05 M K_a= 1.74 × 10^–5

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

= 1.86 x 10^-2

[CH₃COO^–]= = 0.05 x 1.86 x 10^-2

= 0.093 x 10^-2 M

[CH₃COO^–] = [H⁺] = 0.093 x 10^-2 M

pH = – log [H⁺]= – log (0.093 x 10^-2)

or, pH = 3.03

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

Ans.

Let the organic acid be HA.

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

Concentration of HA = 0.01 M

pH = 4.15

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

Let us calculate ionization constant by the expression mentioned here.

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

Then,

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

⇒K_a= 5.01 x 10^-7

⇒pK_a = – log K_a

= – log (5.01 x 10^-7)

= 6.3001

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

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

Ans.

(a) 0.003 M HCl:

Considering the equilibrium: <

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

Thus, [H⁺] = [HCl] = 3 x 10^–3

pH= – log [H⁺]= –log (3 x 10^–3)= 2.52

(b) 0.005 M NaOH:

Considering the equilibrium:

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

Thus, [OH^–] = 5 x 10^–3 M

From the relation,

[H⁺] [OH^–] = 1 x 10^-14

Or, [H⁺] = (1 x 10^–14) / (5 x 10^–3)

= 2 x 10^–12

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

(c) 0.002 M HBr:

Considering the equilibrium:

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

[H⁺] = 2 x10^–3 M

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

(d) 0.002 M KOH:

Considering the equilibrium:

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

Thus, [OH^–] = 2 x 10^–3 M

From the relation,

[H⁺] [OH^–] = 1 x 10^-14

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

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

Q.31 Calculate the pH of the following solutions:

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

Ans.

(a)

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

[OH^–]= [TlOH] = 4.52 x 10^–3 M

[H⁺] [OH^–] = 1 x 10^-14

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

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

(b)

\begin{array}{l}Molar\text{conc}\text{. of Ca}{\left(OH\right)}_2\text{=}\frac{0.3g}{74gmo{l}^{-1}}\times \frac{1}{0.5L}\\ =8.11\times {10}^{-3}\\ Ca{\left(OH\right)}_2\left(aq\right)\underset{}{\rightleftharpoons }C{a}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\end{array}

[OH^–] = 2 [Ca(OH)₂]

= 2 x (8.11 x 10^–3) M

= 16.22 x 10^–3 M

pOH = – log (16.22 x 10^–3)

= 3 – 1.2101

= 1.79

pH = 14 – 1.79

(c)

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

[OH^–]= [NaOH] = 3.75 x 10^–2 M

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

pH = 14 – 1.43 = 12.57

(d) From the relation,

M₁V₁ = M₂V₂

or 13.6 M x 1 mL = M₂ x 1000 mL

or M₂ = 1.36 x 10^–2 M

[H⁺] = [OH^–] = 1.36 x 10^–2 M

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

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

Ans.

Degree of ionization, a = 0.132

Concentration, c = 0.1 M

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

= 0.0132

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

= 0.1 x (0.132)²

K_a = 0.0017

and pK_a = – log (0.0017)

= 2.77

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

Ans.

Considering the reaction

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

pH = 9.95, pOH = 14 – pH

= 14 – 9.95

= 4.05

or – log [OH^–] = 4.05

or log[OH^–] = – 4.05

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

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

pK_b = – log (1.588 x 10^–6) = 6 – 0.1987 = 5.8

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

Ans.

K_b= 4.27 × 10^–10 and c = 0.001 M

K_b= cα²

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

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

= 0.065 x 10^–5

pOH = – log [OH]

= – log (0.065 x 10^–5)

= – log (65 x 10^–8)

= 6.187

We know,

pH + pOH = 14

or pH = 14 – 6.187 = 7.813

Again,

K_a x K_b = K_w

or K_a x (4.27 × 10^–10) = 10^–14

K_a=\frac{{10}^{-14}}{4.27\times {10}^{-10}}=2.34\times {10}^{-5}

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

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

Ans.

Given values are as follows-

C = 0.05 M = 5 x 10^-2

pK_a = 4.74

pK_a = – log K_a = 4.74

or K_a = antilog (– 4.74) = 1.82 x 10^–5

From the relation,

K_a = cα²

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

= 1.908 x 10^–2

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

1. When 0.01 M HCl is taken:

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

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

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

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

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

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

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

Ans.

c = 0.02 M and K_b = 5.4 × 10^–4

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

=0.1643

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

\begin{array}{l}NaOH\left(aq\right)\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0.1\text{M}\end{array}}{N{a}^{+}\left(aq\right)}+\underset{\begin{array}{l}\\ 0.1\text{M}\end{array}}{O{H}^{-}\left(aq\right)}\\ \underset{\left(0.02-x\right)~0.02}{\underset{}{{\left(C{H}_3\right)}_{2}NH}}+H_{2}O\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ x\end{array}}{{\left(C{H}_3\right)}_{2}N{H}_2^{+}}+\underset{\begin{array}{l}\\ \left(x+0.1\right)~0.1\end{array}}{O{H}^{-}}\\ \left[{\left(C{H}_3\right)}_{2}N{H}_2^{+}\right]=x\\ \left[{\text{OH}}^{–}\right]=\text{x}+0.\text{1}~0.\text{1}\\ {\text{K}}_b=\frac{\left[{\left(C{H}_3\right)}_{2}N{H}_2^{+}\right]\left[O{H}^{-}\right]}{{\left(C{H}_3\right)}_{2}NH}\\ \Rightarrow 5.4\times {10}^{-4}=\frac{x\times 0.1}{0.02}\end{array}

⇒x = 0.0054

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

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

(a) Human muscle-fluid, 6.83

(b) Human stomach-fluid, 1.2

(c) Human blood, 7.38

(d) Human saliva, 6.4.

Ans.

(a) For Human muscle fluid,

pH = 6.83

or pH = – log [H⁺]

or 6.83 = – log [H⁺]

or [H⁺] = antilog (-6.83)

or [H⁺] = 1.48 × 10^–7 M

(b) For Human stomach fluid,

pH = 1.2 = – log [H⁺]

or [H⁺] = antilog (-1.2)

or [H⁺] = 0.063 M

(c) For Human blood,

pH = 7.38 = – log [H⁺]

or [H⁺] = antilog (-7.38)

or [H⁺] = 4.17 × 10^–8 M

(d) For Human saliva,

pH = 6.4 = – log [H⁺]

or [H⁺] = antilog (-6.4)

or [H⁺] = 3.98 × 10^–7 M

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

Ans.

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

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

1. pH of milk = 6.8

Using the above relation,

pH = – log [H⁺]

⇒ log [H⁺] = –6.8

⇒ [H⁺] = antilog (–6.8)

= 1.5 x 10^–7 M

2. pH of black coffee = 5.0

Since, pH = – log [H⁺]

⇒ 5.0 = – log [H⁺]

⇒ log [H⁺] = –5.0

⇒ [H⁺] = antilog(–5.0) = 10^–5 M

3. pH of tomato juice = 4.2

pH = – log [H⁺]

⇒ 4.2 = – log [H⁺]

⇒ log [H⁺] = –4.2

⇒ [H⁺] = antilog(–4.2)

=6.31 x10^–5 M

4. pH of lemon juice = 2.2

pH = – log [H⁺]

⇒ 2.2 = – log [H⁺]

⇒ log [H⁺] = –2.2

⇒[H⁺] = antilog(–2.2)

=6.31 x10^–3 M

5. pH of egg white = 7.

pH = – log [H⁺]

⇒7.8 = – log [H⁺]

⇒log [H⁺] = –7.8

⇒ [H⁺] = antilog(–7.8)

=1.58 x 10^–8 M

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

Ans.

Let us calculate the concentration of KOH in the solution:

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

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

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

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

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

Ans.

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

Concentration of Sr(OH)₂

\begin{array}{l}=\frac{19.23g{L}^{-1}}{121.6gmo{l}^{-1}}\\ =0.1581\text{M}\\ \text{Assuming complete dissociation of Sr}{\left(\text{OH}\right)}_{\text{2}}\\ Sr{\left(OH\right)}_2\underset{}{\rightleftharpoons }S{r}^{2+}+2O{H}^{-}\end{array}

[Sr²⁺] = 0.1581 M

[OH^–] = 2 x 0.1581 M = 0.3162 M

pOH = – log (0.3162) = 0.5

pH = 14 – 0.5 = 13.5

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

Ans.

Let the degree of ionization of propanoic acid be α.

Assume propionic acid as HA, we have:

\begin{array}{l}\underset{\begin{array}{l}\\ 0.05-0.05\alpha \end{array}}{HA+H_{2}O}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{H_3{O}^{+}}+\underset{\begin{array}{l}\\ 0.05\alpha \end{array}}{A^{-}}\\ \text{Let us assumeas 0}\text{.05}-0.05\alpha \text{as}0.0\text{5}.\\ K_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}\\ =\frac{\left(0.05\alpha \right)\left(0.05\alpha \right)}{0.05}=0.05{\alpha }^2\\ \alpha =\sqrt{\frac{K_a}{0.05}}\\ =\sqrt{\frac{1.32\times {10}^{-5}}{.05}}\\ =1.63\times {10}^{-2}\end{array}

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

pH = – log [H⁺]

= – log [H₃O⁺]

= – log (8.15 x 10^–4)

= 3.09

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

[H₃O⁺] = 0.01

[A^–] = 0.05 α’

[HA]= 0.05

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

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

Ans.

Concentration c = 0.1M

pH = 2.34

– log [H⁺] = 2.34

or [H⁺] = antilog (-2.34)

or [H⁺] = 4.5 x 10^–3

Also [H⁺] = c . α

⇒ 4.5 x 10^–3 = 0.1 x α

⇒ α = 45 x 10^–3= 0.045

K_a = c . α²

= 0.1 (45 x 10^–3)²

= 2.02 x 10^–4

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

Ans.

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

\underset{\begin{array}{l}\\ 0.04-x\end{array}}{N{O}_2^{-}+H_{2}O}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ x\end{array}}{HN{O}_2}+\underset{\begin{array}{l}\\ x\end{array}}{O{H}^{-}} \begin{array}{l}{K}_h=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[N{O}_2^{-}\right]}\\ \Rightarrow \frac{K_w}{K_a}=\frac{{10}^{-14}}{4.5\times {10}^{-4}}=0.22\times {10}^{-10}\end{array}

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

[NO₂^–]=0.04 – x ~ 0.04

[HNO₂] = x

[OH^–] = x

\begin{array}{l}{K}_h=\frac{\left[HN{O}_2\right]\left[O{H}^{-}\right]}{\left[N{O}_2^{-}\right]}\\ K_h=\frac{x^2}{0.04}=0.22\times {10}^{-10}\\ \Rightarrow x^2=0.0088\times {10}^{-10}\\ \Rightarrow x=0.093\times {10}^{-5}\end{array}

[OH^–] = 0.093 x 10^–5 M

[H⁺]= K_w /(0.093 x 10^–5)

= 10^–14 / (0.093 x 10^–5)

= 10.75 x 10^–9 M

pH = – log (10.75 x 10^–9) = 7.96

Therefore, degree of hydrolysis h

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

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

Ans.

pH = – log [H⁺]

⇒3.44 = – log [H⁺]

⇒[H⁺] = – antilog pH

⇒[H⁺] = – antilog (3.44)

⇒[H⁺] = 3.63 x 10^–4

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

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

Ans.

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

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

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

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

Ans.

It is given that K_a for ClCH₂COOH is 1.35 × 10^–3

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

= 0.116

[H⁺] = c . a

= 0.1 x 0.116

= 0.0116

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

= – log (.0116)

=1.94

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

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

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

Ans.

Ionic product of water

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

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

⇒2.7 × 10^–14 = [H⁺]²

⇒ [H⁺] = 1.64 x 10^–7

pH = – log [H⁺]

= – log (1.64 x 10^–7)

= 6.78

Hence, the pH of neutral water is 6.78.

Q.48 Calculate the pH of the resultant mixtures:

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

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

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

Ans.

(a)

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

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

Ca{\left(OH\right)}_{2}2HCl\stackrel{}{\to }CaC{l}_2+2H_{2}O

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

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

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

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

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

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

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

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

pH = 14 – 1.37 = 12.63

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

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

Ca{\left(OH\right)}_2+H_{2}S{O}_4\stackrel{}{\to }CaS{O}_4+2H_{2}O

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

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

10 mL 0.1 M KOH = 1 millimole of KOH

2KOH+H_{2}S{O}_4\stackrel{}{\to }{K}_{2}S{O}_4+2H_{2}O

2 millimole of KOH reacts with = 1 millimole of H₂SO₄

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

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

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

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

[H⁺] = 2 x (2.5 x 10^–2) = 5 x 10^–2

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

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

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

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

Determine also the molarities of individual ions.

Ans.

1. Silver chromate:

\begin{array}{l}A{g}_{2}Cr{O}_4\underset{}{\rightleftharpoons }2A{g}^{+}+Cr{O}_4^{2-}\\ K_{sp}={\left[A{g}^{+}\right]}^2\left[Cr{O}_4^{2-}\right]\end{array}

Let the solubility of Ag₂CrO₄ be s.

For [Ag⁺] = 2s

[CrO₄^2–] = s

Then,

K_sp = (2s) ².s = 4 s³

⇒ 1.1 x 10^–12 = 4 s³

⇒ s = 0.65 x 10^–4 M

Molarity of Ag⁺ = 2 s

= 2 x (0.65 x 10^–4)

= 1.3 x 10^–4 M

Molarity of CrO₄^2– = s = 0.65 x 10^–4 M

2. Barium chromate:

\begin{array}{l}BaCr{O}_4\underset{}{\rightleftharpoons }B{a}^{2+}+Cr{O}_4^{2-}\\ K_{sp}=\left[B{a}^{+}\right]\left[Cr{O}_4^{2-}\right]\end{array} K_{sp}=\left[B{a}^{+}\right]\left[Cr{O}_4^{2-}\right]

The solubility of BaCrO₄ is s.

Thus, [Ba²⁺] = s

and [CrO₄^2–] = s

K_sp = s . s = s²

⇒ s² = 1.2 x 10^–10

⇒ s = 1.09 x 10^–5 M

Molarity of Ba²⁺ = Molarity of CrO₄^2– = 1.09 x 10^–5 M

3. Ferric hydroxide:

\begin{array}{l}Fe{\left(OH\right)}_3\underset{}{\rightleftharpoons }F{e}^{3+}+3O{H}^{-}\\ K_{sp}=\left[F{e}^{3+}\right]{\left[O{H}^{-}\right]}^3\end{array}

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

Thus, [Fe³⁺] = s

and [OH^–] = 3s

∴ K_sp = s. (3s)³ = 27 s⁴

⇒ K_sp = 27 s⁴

⇒ 1.0 x 10^–38 = 27 s⁴

⇒ 0.037 x 10^–38 = s⁴

⇒ s = 1.39 x 10^–10 M

Molarity of Fe³⁺ = s = 1.39 x 10^–10 M

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

= 4.17 x 10^–10 M

4. Lead Chloride:

\begin{array}{l}PbC{l}_2\underset{}{\rightleftharpoons }P{b}^{2+}+2C{l}^{-}\\ K_{sp}=\left[P{b}^{2+}\right]{\left[C{l}^{-}\right]}^2\end{array}

Let the solubility product of PbCl₂ be s.

Thus, [Pb²⁺] = s

and [Cl^–] = 2s

∴ K_sp = s. (2s)²

⇒ K_sp = 4 s³

⇒ 1.6 x 10^–5 = 4 s³

⇒ s³ = 4 x 10^–6

⇒ s = 1.58 x 10^–2 M

Molarity of Pb²⁺ = s = 1.58 x 10^–2 M

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

= 3.16 x 10^–2 M

5. Mercurous iodide:

\begin{array}{l}H{g}_2{I}_2\underset{}{\rightleftharpoons }H{g}_2^{2+}+2I^{-}\\ K_{sp}=\left[H{g}_2^{2+}\right]{\left[I^{-}\right]}^2\end{array}

Let the solubility product of Hg₂I₂ be s.

[Hg₂²⁺] = s

and [I^–] = 2s

∴ K_sp= s . (2s)²

⇒ K_sp= 4 s³

⇒4.5 x 10^–29 = 4 s³

⇒ s = 2.24 x 10^–10 M

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

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

= 4.48 x 10^–10 M

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

Ans.

Let the solubility of Ag₂CrO₄ = s

\begin{array}{l}A{g}_{2}Cr{O}_4\underset{}{\rightleftharpoons }2A{g}^{+}+Cr{O}_4^{2-}\\ K_{sp}={\left[A{g}^{+}\right]}^2\left[Cr{O}_4^{2-}\right]\end{array}

Let the solubility of Ag₂CrO₄ be s.

For [Ag⁺] = 2s

[CrO₄^2–] = s

Then,

K_sp= (2s²). s = 4 s³

⇒ 1.1 x 10^–12 = 4 s³

⇒ s = 0.65 x 10^–4 M = 6.5 x 10^–5 M

Let s’ be the solubility of AgBr.

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

Thus, [Ag⁺] = [Br^–] = s

⇒K_sp = (s’)² = 5.0 x 10^–13

⇒ s’ = 7.07 x 10^–7 M

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

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

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

Ans.

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

Considering the ionization of both compounds:

\begin{array}{l}\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{NaI{O}_3}\stackrel{}{\to }N{a}^{+}+\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{I{O}_3^{-}}\\ \underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{Cu{\left(Cl{O}_3\right)}_2}\stackrel{}{\to }C{u}^{2+}+\underset{\begin{array}{l}\\ 0.001\text{M}\end{array}}{2C{l}_3^{-}}\\ \text{The solubility equilibrium for copper iodate can be written as}:\\ Cu{\left(I{O}_3\right)}_2\left(aq\right)\underset{}{\rightleftharpoons }C{u}^{2+}\left(aq\right)+2I{O}_3^{-}\left(aq\right)\end{array}

Ionic product of copper iodate:

= [Cu²⁺] [IO₃^–]²

= (0.001) (0.001)²

= 1 x 10^–9

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

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

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

Ans.

Given,

pH = – log [H⁺] = 3.19

⇒[H⁺] = antilog (–3.19)

= 6.457 x 10^–4 M

\begin{array}{l}{C}_6{H}_{5}COOH\left(aq\right)\underset{}{\rightleftharpoons }{C}_6{H}_{5}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)\\ K_a=\frac{\left[C_6{H}_{5}CO{O}^{-}\left(aq\right)\right]\left[H^{+}\left(aq\right)\right]}{\left[C_6{H}_{5}COOH\left(aq\right)\right]}\\ \Rightarrow \frac{\left[C_6{H}_{5}COOH\left(aq\right)\right]}{\left[C_6{H}_{5}CO{O}^{-}\left(aq\right)\right]}=\frac{\left[H^{+}\left(aq\right)\right]}{K_a}=\frac{6.46\times {10}^{-4}}{6.46\times {10}^{-5}}=10\end{array}

Let the solubility of C₆H₅COOAg be x mol/L

Then,

[Ag⁺] = x

⇒ [C₆H₅COOH] + [C₆H₅COO^–] = x

⇒ 10 [C₆H₅COO^–] + [C₆H₅COO^–] = x

⇒11 [C₆H₅COO^–] = x

⇒ [C₆H₅COO^–] = x/11

\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[C_6{H}_{5}CO{O}^{-}\right]\\ \Rightarrow 2.5\times {10}^{-13}=x.\left(x/11\right)\end{array}

⇒ x = 1.66 x 10^–6 mol/L

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

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

[Ag⁺] =y M

and [CH₃COO^–]= y M

\begin{array}{l}{K}_{sp}=\left[A{g}^{+}\right]\left[C_6{H}_{5}CO{O}^{-}\right]\\ \Rightarrow y=\sqrt{Ksp}=\sqrt{\left(2.5\times {10}^{-13}\right)}=5\times {10}^{-7}mol/L\\ \frac{x}{y}=\frac{1.66\times {10}^{-6}}{5\times {10}^{-7}}=3.32\end{array}

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

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

Ans.

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

\begin{array}{l}\therefore \left[FeS{O}_4\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M\\ \therefore \left[F{e}^{2+}\right]=\therefore \left[FeS{O}_4\right]=\frac{x}{2}M\\ \therefore \left[S^{2-}\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M\\ FeS\left(s\right)\underset{}{\rightleftharpoons }F{e}^{2+}\left(aq\right)+S^{2-}\left(aq\right)\\ K_{sp}=\left[F{e}^{2+}\right]\left[S^{2-}\right]\\ \Rightarrow 6.3\times {10}^{-18}=\frac{x}{2}\times \frac{x}{2}\\ \Rightarrow 6.3\times {10}^{-18}=\frac{x^2}{4}\\ \Rightarrow x=5.02\times {10}^{-9}\end{array}

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

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

Ans.

\begin{array}{l}CaS{O}_4\left(s\right)\underset{}{\rightleftharpoons }C{a}^{2+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)\\ K_{sp}=\left[C{a}^{2+}\right]\left[S{O}_4^{2-}\right]\end{array}

Let the solubility of CaSO₄ be s.

Then, K_sp = s²

⇒ 9.1 x 10^–6 = s²

⇒ s = 3.02 x 10^–3 mol/L

Molecular mass of CaSO₄ = 136 g/mol

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

= 0.411 g/L

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

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

= 2.43 L

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

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

Ans.

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

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

Before mixing,

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

After mixing,

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

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

= 0.04 x (5/15) = 1.33 x 10^–2 M

∴ Ionic product of each will be=

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

= 8.87 x 10^–22

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

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

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

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

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

Ans.

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

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

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

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

2S{O}_2\left(g\right)+O_2\underset{}{\rightleftharpoons }2S{O}_3\left(g\right)

Ans.

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

\begin{array}{l}{K}_c=\frac{{\left[S{O}_3\right]}^2}{{\left[S{O}_2\right]}^2\left[O_2\right]}\\ K_c=\frac{{\left(1.90\right)}^2{M}^2}{{\left(0.60\right)}^2\left(0.821\right)M^3}=12.239{\text{M}}^{-1}\left(approx\right)\end{array}

Hence, K_c for the equilibrium K_c is 12.239 M^-1.

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

I_2\left(g\right)\underset{}{\rightleftharpoons }2I\left(g\right)

Calculate K_p for the equilibrium.

Ans.

Partial pressure of I atoms,

\begin{array}{l}{p}_I=\frac{40}{100}\times p_{total}\\ =\frac{40}{100}\times {10}^5=4\times {10}^{4}Pa\\ {\text{Partial pressure of I}}_{\text{2}}\text{molecules}\\ p_{I_2}=\frac{60}{100}\times p_{total}\\ =\frac{60}{100}\times {10}^5=6\times {10}^{4}Pa\\ \text{Let us write}{\text{the expression of K}}_{\text{p}}\\ K_p=\frac{{\left(pI\right)}^2}{p{I}_2}\\ \text{Place the value of partial pressure of iodine atom and iodine molecule in this equation}.\text{We get}\\ K_p=\frac{{\left(4\times {10}^4\right)}^{2}P{a}^2}{6\times {10}^{6}Pa}\\ =2.67\times {10}^4\text{Pa}\end{array}

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

\begin{array}{l}(i)2NOCl\left(g\right)\stackrel{}{\rightleftharpoons }2NO\left(g\right)+C{l}_2\left(g\right)\\ (ii)2Cu{\left(N{O}_3\right)}_2\left(s\right)\stackrel{}{\rightleftharpoons }2CUO\left(s\right)+4N{O}_2\left(g\right)+O_2\left(g\right)\\ (iii)C{H}_{3}COO{C}_2{H}_5\left(aq\right)+H_{2}O\left(I\right)\stackrel{}{\rightleftharpoons }C{H}_{3}COOH\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\\ (iv)F{e}^{3+}\left(aq\right)+3O{H}^{-}\left(aq\right)\stackrel{}{\rightleftharpoons }Fe{\left(OH\right)}_3\left(s\right)\\ (v)I_2\left(s\right)+5F_2\stackrel{}{\rightleftharpoons }2I{F}_5\end{array}

Ans.

\begin{array}{l}(i)K_c=\frac{{\left[N{O}_{\left(g\right)}\right]}^2{\left[C{l}_{2\left(g\right)}\right]}^2}{{\left[NOC{l}_{\left(g\right)}\right]}^2}\\ (ii)K_c={\left[N{O}_{2\left(g\right)}\right]}^4\left[O_{2\left(g\right)}\right]\\ \left(\text{The concentration of a solid is considered as constant}.\right)\\ (iii)K_c=\frac{\left[C{H}_{3}COOH\left(aq\right)\right]\left[C_2{H}_{5}OH\left(aq\right)\right]}{\left[C{H}_{3}COO{C}_2{H}_5\left(aq\right)\right]}\\ \left(\text{When water is used as solvent},\text{its concentration remains constant}.\right)\\ (iv)K_c=\frac{1}{\left[F{e}^{3+}{}_{\left(aq\right)}\right]{\left[O{H}^{-}{}_{\left(aq\right)}\right]}^3}\\ \left(\text{The concentration of a solid is considered as constant}.\right)\\ (v)K_c={\frac{\left[I{F}_5\right]}{{\left[F_2\right]}^5}}^2\\ \left(\text{The concentration of a solid is considered as constant}.\right)\end{array}

Q.60 Find out the value of K_c for each of the following equilibrium from the value of K_p:

\begin{array}{l}(i)2NOCl\left(g\right)\underset{}{\rightleftharpoons }2NO\left(g\right)+C{l}_2\left(g\right);K_p=1.8\times {10}^{-2}at\text{500 K}\\ (ii)CaC{O}_3\left(s\right)\underset{}{\rightleftharpoons }CaO\left(s\right)+C{O}_2\left(g\right);K_p=167at\text{1073 K}\end{array}

Ans.

The relationship between K_p and K_c is written as:

K_p=K_c{\left(RT\right)}^{\Delta n}

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

Δn= 3 – 2 = 1

R= 0.0831 bar L mol^–1 K^–1

T= 500 K

K_p= 1.8 × 10^–2

Let us place these values in the following equation,

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

= 4.33 x 10^-4 (approx)

ii) Δn = 2 – 1 = 1

R = 0.0831 bar L mol^–1 K^–1

T = 1073 K

K_p= 167

Now,

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

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

NO\left(g\right)+O_3\left(g\right)\underset{}{\rightleftharpoons }N{O}_2\left(g\right)+O_2\left(g\right)

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

Ans.

For the forward reaction, K_c = 6.3 x 10¹⁴

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

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

= 1.59 x 10^-15

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

Ans.

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

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

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

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

2N_2\left(g\right)+O_2\left(g\right)\underset{}{\rightleftharpoons }2N_{2}O\left(g\right)

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 × 10^–37, determine the composition of equilibrium mixture.

Ans.

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

\begin{array}{l}{\text{Let us assume that the concentration Of N}}_{\text{2}}\text{O at equilibrium be x}.\\ \underset{\begin{array}{l}\\ \text{Initial conc}\text{. 0}\text{.482mol}\\ \\ At\text{equilibrium}\left(0.482-x\right)mol\end{array}}{2N_2\left(g\right)}+\underset{\begin{array}{l}\\ 0.933\text{mol}\\ \\ \left(0.933-x\right)mol\end{array}}{O_2\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ xmol\end{array}}{2N_{2}O\left(g\right)}\\ \text{The reaction is carried out in 1}0\text{L vessel},\text{at equilibrium}\\ \left[N_2\right]=\frac{0.482-x}{10}\\ \left[O_2\right]=\frac{0.933-x}{10}\\ \left[N_{2}O\right]=x/10\\ \text{The value of equilibrium constant is very small}\left(\text{as given}\right).\\ \text{Therefore},{\text{the amount of N}}_{\text{2}}{\text{and O}}_{\text{2}}\text{reacted is also very small}.\\ {\text{Thus x can be neglected from the molar concentration terms of N}}_{\text{2}}{\text{and O}}_{\text{2}}.\\ \left[N_2\right]=\frac{0.482}{10}=0.0482{\text{mol L}}^{-1}\\ \left[O_2\right]=\frac{0.933}{10}=0.0933{\text{mol L}}^{-1}\\ K_c=\frac{{\left[N_2{O}_{\left(g\right)}\right]}^2}{{\left[N_{2\left(g\right)}\right]}^2\left[O_{2\left(g\right)}\right]}\\ \Rightarrow 2.0\times {10}^{-37}=\frac{{\left(\frac{x}{10}\right)}^2}{{\left(0.0482\right)}^2\left(0.0933\right)}\\ \Rightarrow \frac{x^2}{100}=2.0\times {10}^{-37}\times {\left(0.0482\right)}^2\times \left(0.0933\right)\\ \Rightarrow x^2=43.35\times {10}^{-40}\\ \Rightarrow x=6.6\times {10}^{-20}\\ \left[N_{2}O\right]=\frac{x}{10}=\frac{6.6\times {10}^{-20}}{10}=6.6\times {10}^{-21}\end{array}

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

2NO\left(g\right)+B{r}_2\left(g\right)\underset{}{\rightleftharpoons }2NOBr\left(g\right)

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

Ans.

The given reaction is:

\underset{\begin{array}{l}\\ 2\text{mol}\end{array}}{2NO\left(g\right)}+\underset{\begin{array}{l}\\ 1\text{mol}\end{array}}{B{r}_2\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 2\text{mol}\end{array}}{2NOBr\left(g\right)}

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

Thus, 0.0518 mole of NOBr are formed from

= (0.0518/2) mole of Br or NO

= 0.0259 mole of Br or 0.0259 mole of NO.

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

[NO] = 0.087 mol

[Br₂] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518 = 0.0352 mol

And the amount of Br present at equilibrium is:

[Br₂] = 0.0437 – 0.0259 = 0.0178 mol

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

2S{O}_2\left(g\right)+O_2\left(g\right)\underset{}{\rightleftharpoons }2S{o}_3\left(g\right)

What is K_c at this temperature?

Ans.

For the given reaction,

Δn = 2 – 3 = -1

T= 450 K

K_p = 2.0 x 10¹⁰ bar^-1

R= 0.0831 L bar K^–1 mol^–1

From the relation,

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

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

2HI\left(g\right)\underset{}{\rightleftharpoons }{H}_2\left(g\right)+I_2\left(g\right)

Ans.

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

\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.2 atm}\\ \\ At\text{equilibrium 0}\text{.04 atm}\end{array}}{2HI\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ 0.16/2atm\\ =0.08atm\end{array}}{H_2\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ 0.16/2atm\\ =0.08atm\end{array}}{I_2\left(g\right)}\\ K_p=\frac{p_{H_2}\times p_{I_2}}{p_{HI}^2}\\ =\frac{\left(0.08atm\times 0.08\text{atm}\right)}{{\left(0.04\text{atm}\right)}^2}\\ =\frac{.0064}{.0016}\\ =4.0\end{array}

Q.67 A mixture of 1.57 mol of 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 is 1.7 x 10².

N_2\left(g\right)+3H_2\left(g\right)\underset{}{\rightleftharpoons }2N{H}_3\left(g\right)

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

Ans.

For the given reaction

N_2\left(g\right)+3H_2\left(g\right)\underset{}{\rightleftharpoons }2N{H}_3\left(g\right)

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

[N₂]= (1.57/20) mol L^-1

[H₂]= (1.92/20) mol L^-1

[NH₃]= (8.13/20) mol L^-1

Now, let us calculate reaction quotient Q_c.

\begin{array}{l}{Q}_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}\\ =\frac{{\left[\frac{8.13}{20}mol{\text{L}}^{-1}\right]}^2}{\left[\frac{1.57}{20}mol{\text{L}}^{-1}\right]{\left[\frac{1.92}{20}mol{\text{L}}^{-1}\right]}^3}\\ =2.38\times {10}^3\end{array}

As Q_c ≠ K_c, the reaction mixture is not at equilibrium.

And Q_c >K_c, the net reaction will proceed in the backward (reverse) direction.

Q.68

\begin{array}{l}\text{The equilibrium constant expression for a gas reaction is},\\ K_c=\frac{{\left[NH{}_3\right]}^4{\left[O_2\right]}^5}{{\left[NO\right]}^4{\left[H_{2}O\right]}^6}\\ \text{Write the balanced chemical equation corresponding to this expression}.\end{array}

Ans.

\begin{array}{l}\text{The balanced chemical equation can be written as}:\\ 4NO\left(g\right)+6H_{2}O\left(g\right)\underset{}{\rightleftharpoons }4N{H}_3\left(g\right)+5O_2\left(g\right)\end{array}

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

H_{2}O\left(g\right)+CO\left(g\right)\underset{}{\rightleftharpoons }{H}_2\left(g\right)+C{O}_2\left(g\right)

Calculate the equilibrium constant for the reaction.

Ans.

\underset{\begin{array}{l}\\ Initial\text{conc}\text{.}\frac{1}{10}M\\ \\ At\text{equilibrium}\frac{1-0.4}{10}M\\ \\ =0.06\text{M}\end{array}}{H_{2}O\left(g\right)}+\underset{\begin{array}{l}\\ \frac{1}{10}M\\ \\ \frac{1-0.4}{10}M\\ \\ =0.06\text{M}\end{array}}{CO\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ \frac{0.4}{10}M\\ \\ =0.04\text{M}\end{array}}{H_2\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ \frac{0.4}{10}M\\ \\ =0.04\text{M}\end{array}}{C{O}_2\left(g\right)}

At equilibrium, the given concentrations are:

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

[CO]= 0.06 mol L^-1

[H₂]=0.4/10 mol L^-1 = 0.04 mol L^-1

[CO₂]= 0.04 mol L^-1

\begin{array}{l}{K}_c=\frac{\left[H_2\right]\left[C{O}_2\right]}{\left[CO\right]\left[H_{2}O\right]}=\frac{0.04\times 0.04}{0.06\times 0.06}\\ =0.444\end{array}

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

H_2\left(g\right)+I_2\left(g\right)\underset{}{\rightleftharpoons }2HI\left(g\right)

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

Ans.

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

2HI\left(g\right)\underset{}{\rightleftharpoons }{H}_2\left(g\right)+I_2\left(g\right)

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

\begin{array}{l}\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}=K{‘}_c\\ \Rightarrow \frac{x\cdot x}{{\left(0.5\right)}^2}=\frac{1}{54.8}\\ \Rightarrow x^2=\frac{0.25}{54.8}\\ Or,x=0.068{\text{mol L}}^{-1}\end{array}

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

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

2ICl\left(g\right)\underset{}{\rightleftharpoons }{I}_2\left(g\right)+C{l}_2\left(g\right);K_c=0.14

Ans.

For the given reaction,

\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 0}\text{.78 M}\\ \\ At\text{equilibrium}\left(\text{0}\text{.78-2x}\right)\text{M}\end{array}}{2ICl\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ \text{x M}\end{array}}{I_2\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ \text{x M}\end{array}}{C{l}_2\left(g\right)}

Now we can write from the expression,

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

C_2{H}_6\left(g\right)\underset{}{\rightleftharpoons }{C}_2{H}_4\left(g\right)+H_2\left(g\right)

Ans.

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

\begin{array}{l}\underset{\begin{array}{l}\\ Initial\text{conc}\text{. 4}\text{.0 atm}\\ \\ \text{At equilibrium}\left(4.0-p\right)atm\end{array}}{C_2{H}_6\left(g\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ p\text{atm}\end{array}}{C_2{H}_4\left(g\right)}+\underset{\begin{array}{l}\\ 0\\ \\ p\text{atm}\end{array}}{H_2\left(g\right)}\\ \frac{p_{C_2{H}_4}\times p_{H_2}}{p_{C_2{H}_6}}=K_p\\ \Rightarrow \frac{p\times p}{\left(40-p\right)}=0.04\\ \Rightarrow p^2=0.16-0.04p\\ \Rightarrow p^2+0.04p-0.16=0\\ p=\frac{-0.04\pm \sqrt{{\left(0.04\right)}^2-4\times 1\times \left(0.16\right)}}{2\times 1}\\ =\frac{-0.04\pm 0.80}{2}\\ =0.76/2\\ p=0.38\\ \text{Hence},\text{at equilibrium},\\ \left[C_2{H}_6\right]=4-p=4-0.38=3.62\text{atm}\end{array}

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

C{H}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\underset{}{\rightleftharpoons }C{H}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)

1. 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)
2. At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
3. Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Ans.

Reaction quotient,

\begin{array}{l}{Q}_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}\\ \left(\text{ii}\right)\text{Let us assume that the volume of the reaction mixture is V}.\\ \text{Also},\text{we will consider that water is a solvent and is present in excess}.\\ \text{The given chemical reaction is}:\\ \underset{\begin{array}{l}\\ Initial\text{conc}\text{. 1/V M}\\ \text{At equilibrium}\left(1-0.171\right)/V\\ =0.829/V\end{array}}{C{H}_{3}COOH\left(l\right)}+\underset{\begin{array}{l}\text{0}\text{.18/V M}\\ \\ \left(\text{0}\text{.18-0}\text{.171}\right)/V\\ =0.009/V\end{array}}{C_2{H}_{5}OH\left(l\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ 0.171/V\end{array}}{C{H}_{3}COO{C}_2{H}_5\left(l\right)}+\underset{\begin{array}{l}\\ 0\\ \\ 0.171/V\end{array}}{H_{2}O\left(l\right)}\\ \text{Therefore},\text{equilibrium constant for the given reaction is}:\\ K_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}\\ =\frac{\frac{0.171}{V}\times \frac{0.171}{V}}{\frac{0.829}{V}\times \frac{0.829}{V}}\\ =3.92\left(approx\right)\\ (iii)\text{Let us assume again that the volume of the reaction mixture as V}.\\ \underset{\begin{array}{l}\\ Initial\text{conc}\text{. 1}\text{.0/V M}\\ \text{At equilibrium}\left(1-2.14\right)/V\\ =0.786/V\text{M}\end{array}}{C{H}_{3}COOH\left(l\right)}+\underset{\begin{array}{l}\text{0}\text{.5/V M}\\ \\ \left(\text{0}\text{.5-0}\text{.214}\right)/V\\ =0.286/V\end{array}}{C_2{H}_{5}OH\left(l\right)}\underset{}{\rightleftharpoons }\underset{\begin{array}{l}\\ 0\\ \\ 0.214/V\text{M}\end{array}}{C{H}_{3}COO{C}_2{H}_5\left(l\right)}+\underset{\begin{array}{l}\\ 0\\ \\ 0.214/V\text{M}\end{array}}{H_{2}O\left(l\right)}\\ \text{The expression for the quotient is},\\ Q_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}\\ =\frac{\frac{0.214}{V}\times \frac{0.214}{V}}{\frac{0.786}{V}\times \frac{0.786}{V}}\\ =0.2037\\ =0.204\left(approx\right)\\ {\text{As Q}}_{\text{c}}<{\text{K}}_{\text{c}},\text{equilibrium has not been reached}.\end{array}