Chapter 9 of 9

Updated 6 August 2026

NCERT Solutions

Equilibrium — NCERT Solutions

CBSE · Class 11 · Chemistry

NCERT Solutions for Equilibrium, CBSE Class 11 Chemistry: 73 textbook questions solved step by step. Covers Exercises.

131 questions82 flashcards14 formulas & key relations5 concepts

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Exercises

6.1A 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?Show solution

Given: A liquid–vapour equilibrium in a sealed container at fixed temperature; volume is suddenly increased.

(a) Initial effect on vapour pressure:
When the volume is suddenly increased, the same number of vapour molecules now occupy a larger volume. Therefore, the concentration (and hence the partial pressure) of the vapour decreases initially.

(b) Initial change in rates:

Rate of evaporation: Evaporation depends on the nature of the liquid and temperature, not on the volume of the container. Hence, the rate of evaporation remains unchanged initially.
Rate of condensation: Condensation depends on the concentration (number density) of vapour molecules. Since the vapour pressure has decreased, the rate of condensation decreases initially.

Because rate of evaporation > rate of condensation, more liquid evaporates to restore equilibrium.

(c) Final state after equilibrium is restored:
More liquid evaporates until the rate of evaporation once again equals the rate of condensation. At the new equilibrium, the vapour pressure equals the original vapour pressure (since temperature is unchanged and vapour pressure depends only on temperature). Thus, the final vapour pressure is the same as the original vapour pressure.

6.2What is

K_c

for the following equilibrium when the equilibrium concentration of each substance is:

[\mathrm{SO}_2] = 0.60\,\mathrm{M}

[\mathrm{O}_2] = 0.82\,\mathrm{M}

and

[\mathrm{SO}_3] = 1.90\,\mathrm{M}

2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g})

Show solution

Given:
$[\mathrm{SO}_2] = 0.60\,\mathrm{M},\quad [\mathrm{O}_2] = 0.82\,\mathrm{M},\quad [\mathrm{SO}_3] = 1.90\,\mathrm{M}$

Expression for $K_c$ :
$K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}$

Calculation:
$K_c = \frac{(1.90)^2}{(0.60)^2 \times (0.82)}$
$= \frac{3.61}{0.36 \times 0.82} = \frac{3.61}{0.2952}$
$\boxed{K_c \approx 12.23}$

6.3At a certain temperature and total pressure of

10^5\,\mathrm{Pa}

, iodine vapour contains

40\%

by volume of I atoms.

\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{I}(\mathrm{g})

Calculate

K_p

for the equilibrium.Show solution

Given: Total pressure $P = 10^5\,\mathrm{Pa}$ ; iodine vapour contains 40% by volume of I atoms.

Concept: % by volume = % by moles (for ideal gases).

So mole fraction of I atoms, $\chi_{\mathrm{I}} = 0.40$
Mole fraction of $\mathrm{I}_2$ , $\chi_{\mathrm{I}_2} = 1 - 0.40 = 0.60$

Partial pressures:
$p_{\mathrm{I}} = 0.40 \times 10^5 = 4.0 \times 10^4\,\mathrm{Pa}$
$p_{\mathrm{I}_2} = 0.60 \times 10^5 = 6.0 \times 10^4\,\mathrm{Pa}$

$K_p$ expression:
$K_p = \frac{(p_{\mathrm{I}})^2}{p_{\mathrm{I}_2}} = \frac{(4.0 \times 10^4)^2}{6.0 \times 10^4}$
$= \frac{16 \times 10^8}{6.0 \times 10^4} = \frac{16}{6} \times 10^4$
$\boxed{K_p = 2.67 \times 10^4\,\mathrm{Pa}}$

6.4Write the expression for the equilibrium constant,

K_c

for each of the following reactions:
(i)

2\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})

(ii)

2\mathrm{Cu(NO_3)_2(s)} \rightleftharpoons 2\mathrm{CuO(s)} + 4\mathrm{NO_2(g)} + \mathrm{O_2(g)}

(iii)

\mathrm{CH_3COOC_2H_5(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{CH_3COOH(aq)} + \mathrm{C_2H_5OH(aq)}

(iv)

\mathrm{Fe^{3+}(aq)} + 3\mathrm{OH^-(aq)} \rightleftharpoons \mathrm{Fe(OH)_3(s)}

(v)

\mathrm{I_2(s)} + 5\mathrm{F_2} \rightleftharpoons 2\mathrm{IF_5}

Show solution

Concept: Pure solids and pure liquids are excluded from the equilibrium constant expression. Only gaseous and aqueous species are included.

(i) $2\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$
$K_c = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2}$

(ii) $2\mathrm{Cu(NO_3)_2(s)} \rightleftharpoons 2\mathrm{CuO(s)} + 4\mathrm{NO_2(g)} + \mathrm{O_2(g)}$

Solids are excluded:
$K_c = [\mathrm{NO_2}]^4[\mathrm{O_2}]$

(iii) $\mathrm{CH_3COOC_2H_5(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{CH_3COOH(aq)} + \mathrm{C_2H_5OH(aq)}$

Pure liquid water is excluded:
$K_c = \frac{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}{[\mathrm{CH_3COOC_2H_5}]}$

(iv) $\mathrm{Fe^{3+}(aq)} + 3\mathrm{OH^-(aq)} \rightleftharpoons \mathrm{Fe(OH)_3(s)}$

Solid $\mathrm{Fe(OH)_3}$ is excluded:
$K_c = \frac{1}{[\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3}$

(v) $\mathrm{I_2(s)} + 5\mathrm{F_2(g)} \rightleftharpoons 2\mathrm{IF_5(g)}$

Solid $\mathrm{I_2}$ is excluded:
$K_c = \frac{[\mathrm{IF_5}]^2}{[\mathrm{F_2}]^5}$

6.5Find out the value of

K_c

for each of the following equilibria from the value of

K_p

:
(i)

2\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})

;

K_p = 1.8 \times 10^{-2}

500\,\mathrm{K}

(ii)

\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}

;

K_p = 167

1073\,\mathrm{K}

Show solution

Formula: $K_p = K_c(RT)^{\Delta n}$ , so $K_c = \dfrac{K_p}{(RT)^{\Delta n}}$

where $\Delta n$ = (moles of gaseous products) $-$ (moles of gaseous reactants), $R = 0.0831\,\mathrm{bar\,L\,mol^{-1}\,K^{-1}}$ .

(i) $2\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$

$\Delta n = (2+1) - 2 = 1$

$K_c = \frac{K_p}{(RT)^1} = \frac{1.8 \times 10^{-2}}{0.0831 \times 500} = \frac{1.8 \times 10^{-2}}{41.55}$
$\boxed{K_c = 4.33 \times 10^{-4}}$

(ii) $\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$

Only $\mathrm{CO_2}$ is gaseous; $\Delta n = 1 - 0 = 1$

$K_c = \frac{K_p}{(RT)^1} = \frac{167}{0.0831 \times 1073} = \frac{167}{89.17}$
$\boxed{K_c = 1.87}$

6.6For the following equilibrium,

K_c = 6.3 \times 10^{14}

1000\,\mathrm{K}

\mathrm{NO}(\mathrm{g}) + \mathrm{O_3}(\mathrm{g}) \rightleftharpoons \mathrm{NO_2}(\mathrm{g}) + \mathrm{O_2}(\mathrm{g})

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is

K_c

for the reverse reaction?Show solution

Given: $K_c(\text{forward}) = 6.3 \times 10^{14}$ at $1000\,\mathrm{K}$

Concept: For the reverse reaction, the equilibrium constant is the reciprocal of the forward equilibrium constant.

$K_c(\text{reverse}) = \frac{1}{K_c(\text{forward})} = \frac{1}{6.3 \times 10^{14}}$

$\boxed{K_c(\text{reverse}) = 1.59 \times 10^{-15}}$

6.7Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?Show solution

Explanation:

The equilibrium constant expression is written in terms of activities of the species involved. For an ideal solution or gas, the activity is proportional to molar concentration or partial pressure. However, for a pure solid or pure liquid, the activity is defined as unity (1) because their molar concentration (density/molar mass) remains essentially constant throughout the reaction and does not change with the extent of reaction.

For example, the molar concentration of water (pure liquid) is:
$[\mathrm{H_2O}] = \frac{1000\,\mathrm{g/L}}{18\,\mathrm{g/mol}} \approx 55.5\,\mathrm{mol/L} = \text{constant}$

Since their concentrations are constant, they are incorporated into the equilibrium constant itself and do not appear explicitly in the $K_c$ expression. Therefore, pure liquids and solids are ignored (their activity = 1) while writing the equilibrium constant expression.

6.8Reaction between

\mathrm{N_2}

and

\mathrm{O_2}

takes place as follows:

2\mathrm{N_2}(\mathrm{g}) + \mathrm{O_2}(\mathrm{g}) \rightleftharpoons 2\mathrm{N_2O}(\mathrm{g})

If a mixture of

0.482\,\mathrm{mol}\,\mathrm{N_2}

and

0.933\,\mathrm{mol}

\mathrm{O_2}

is placed in a

10\,\mathrm{L}

reaction vessel and allowed to form

\mathrm{N_2O}

at a temperature for which

K_c = 2.0 \times 10^{-37}

, determine the composition of equilibrium mixture.Show solution

Given:

Initial moles: $n(\mathrm{N_2}) = 0.482\,\mathrm{mol}$ , $n(\mathrm{O_2}) = 0.933\,\mathrm{mol}$ , Volume $= 10\,\mathrm{L}$
$K_c = 2.0 \times 10^{-37}$ (extremely small)

Initial concentrations:
$[\mathrm{N_2}]_0 = \frac{0.482}{10} = 0.0482\,\mathrm{M}$
$[\mathrm{O_2}]_0 = \frac{0.933}{10} = 0.0933\,\mathrm{M}$

ICE Table: Let $x$ mol/L of $\mathrm{N_2O}$ be formed at equilibrium.

$2\mathrm{N_2} + \mathrm{O_2} \rightleftharpoons 2\mathrm{N_2O}$

	$\mathrm{N_2}$	$\mathrm{O_2}$	$\mathrm{N_2O}$
I	0.0482	0.0933	0
C	$-x$	$-x/2$	$+x$
E	$0.0482-x$	$0.0933-x/2$	$x$

$K_c$ expression:
$K_c = \frac{[\mathrm{N_2O}]^2}{[\mathrm{N_2}]^2[\mathrm{O_2}]} = 2.0 \times 10^{-37}$

Since $K_c$ is extremely small, $x$ is negligibly small compared to initial concentrations. So:
$K_c \approx \frac{x^2}{(0.0482)^2(0.0933)}$
$x^2 = 2.0 \times 10^{-37} \times (0.0482)^2 \times 0.0933$
$x^2 = 2.0 \times 10^{-37} \times 2.323 \times 10^{-3} \times 0.0933$
$x^2 = 2.0 \times 10^{-37} \times 2.167 \times 10^{-4}$
$x^2 = 4.334 \times 10^{-41}$
$x = 6.58 \times 10^{-21}\,\mathrm{mol/L}$

Equilibrium composition (essentially unchanged):
$[\mathrm{N_2}] \approx 0.0482\,\mathrm{M} \Rightarrow n(\mathrm{N_2}) = 0.482\,\mathrm{mol}$
$[\mathrm{O_2}] \approx 0.0933\,\mathrm{M} \Rightarrow n(\mathrm{O_2}) = 0.933\,\mathrm{mol}$
$[\mathrm{N_2O}] = 6.58 \times 10^{-21}\,\mathrm{mol/L} \Rightarrow n(\mathrm{N_2O}) = 6.58 \times 10^{-20}\,\mathrm{mol}$

The equilibrium mixture essentially contains the same amounts of $\mathrm{N_2}$ and $\mathrm{O_2}$ as initially, with a negligible amount of $\mathrm{N_2O}$ .

6.9Nitric oxide reacts with

\mathrm{Br_2}

and gives nitrosyl bromide as per reaction given below:

2\mathrm{NO}(\mathrm{g}) + \mathrm{Br_2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NOBr}(\mathrm{g})

When

0.087\,\mathrm{mol}

of NO and

0.0437\,\mathrm{mol}

\mathrm{Br_2}

are mixed in a closed container at constant temperature,

0.0518\,\mathrm{mol}

of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and

\mathrm{Br_2}

.Show solution

Given:

Initial: $n(\mathrm{NO}) = 0.087\,\mathrm{mol}$ , $n(\mathrm{Br_2}) = 0.0437\,\mathrm{mol}$ , $n(\mathrm{NOBr}) = 0$
At equilibrium: $n(\mathrm{NOBr}) = 0.0518\,\mathrm{mol}$

From stoichiometry:

For every 2 mol of NOBr formed, 2 mol of NO and 1 mol of $\mathrm{Br_2}$ are consumed.

Moles of NO consumed $= 0.0518\,\mathrm{mol}$ (same as NOBr formed, 2:2 ratio)

Moles of $\mathrm{Br_2}$ consumed $= \dfrac{0.0518}{2} = 0.0259\,\mathrm{mol}$

Equilibrium amounts:
$n(\mathrm{NO})_{\mathrm{eq}} = 0.087 - 0.0518 = \boxed{0.0352\,\mathrm{mol}}$
$n(\mathrm{Br_2})_{\mathrm{eq}} = 0.0437 - 0.0259 = \boxed{0.0178\,\mathrm{mol}}$

6.10At

450\,\mathrm{K}

K_p = 2.0 \times 10^{10}\,\mathrm{bar}^{-1}

for the given reaction at equilibrium.

2\mathrm{SO_2}(\mathrm{g}) + \mathrm{O_2}(\mathrm{g}) \rightleftharpoons 2\mathrm{SO_3}(\mathrm{g})

What is

K_c

at this temperature?Show solution

Given: $K_p = 2.0 \times 10^{10}\,\mathrm{bar}^{-1}$ , $T = 450\,\mathrm{K}$ , $R = 0.0831\,\mathrm{bar\,L\,mol^{-1}\,K^{-1}}$

$\Delta n$ for the reaction:
$\Delta n = 2 - (2+1) = -1$

Relation between $K_p$ and $K_c$ :
$K_p = K_c(RT)^{\Delta n}$
$K_c = K_p \times (RT)^{-\Delta n} = K_p \times (RT)^{1}$

$K_c = 2.0 \times 10^{10} \times (0.0831 \times 450)$
$= 2.0 \times 10^{10} \times 37.395$
$\boxed{K_c = 7.48 \times 10^{11}\,\mathrm{mol^{-1}\,L}}$

6.11A 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?

2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H_2}(\mathrm{g}) + \mathrm{I_2}(\mathrm{g})

Show solution

Given: Initial pressure of HI $= 0.2\,\mathrm{atm}$ ; equilibrium pressure of HI $= 0.04\,\mathrm{atm}$

Pressure decrease of HI $= 0.2 - 0.04 = 0.16\,\mathrm{atm}$

From stoichiometry: 2 mol HI decomposes to give 1 mol $\mathrm{H_2}$ and 1 mol $\mathrm{I_2}$ .

$p_{\mathrm{H_2}} = p_{\mathrm{I_2}} = \frac{0.16}{2} = 0.08\,\mathrm{atm}$

$K_p$ expression:
$K_p = \frac{p_{\mathrm{H_2}} \cdot p_{\mathrm{I_2}}}{(p_{\mathrm{HI}})^2} = \frac{0.08 \times 0.08}{(0.04)^2} = \frac{6.4 \times 10^{-3}}{1.6 \times 10^{-3}}$

$\boxed{K_p = 4.0}$

Note: Since $\Delta n = 0$ for this reaction, $K_c = K_p = 4.0$ .

6.12A mixture of

1.57\,\mathrm{mol}

\mathrm{N_2}

1.92\,\mathrm{mol}

\mathrm{H_2}

and

8.13\,\mathrm{mol}

\mathrm{NH_3}

is introduced into a

20\,\mathrm{L}

reaction vessel at

500\,\mathrm{K}

. At this temperature, the equilibrium constant,

K_c

for the reaction

\mathrm{N_2}(\mathrm{g}) + 3\mathrm{H_2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NH_3}(\mathrm{g})

1.7 \times 10^2

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

Given: Volume $= 20\,\mathrm{L}$ , $K_c = 1.7 \times 10^2$

Concentrations:
$[\mathrm{N_2}] = \frac{1.57}{20} = 0.0785\,\mathrm{M}$
$[\mathrm{H_2}] = \frac{1.92}{20} = 0.096\,\mathrm{M}$
$[\mathrm{NH_3}] = \frac{8.13}{20} = 0.4065\,\mathrm{M}$

Reaction quotient $Q_c$ :
$Q_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \frac{(0.4065)^2}{(0.0785)(0.096)^3}$
$= \frac{0.1652}{0.0785 \times 8.847 \times 10^{-4}} = \frac{0.1652}{6.945 \times 10^{-5}}$
$Q_c = 2.38 \times 10^3$

Comparison:
$Q_c = 2.38 \times 10^3 > K_c = 1.7 \times 10^2$

Since $Q_c > K_c$ , the reaction is not at equilibrium. The reaction will proceed in the reverse direction (i.e., decomposition of $\mathrm{NH_3}$ ) to reach equilibrium.

6.13The equilibrium constant expression for a gas reaction is,

K_c = \frac{[\mathrm{NH_3}]^4[\mathrm{O_2}]^5}{[\mathrm{NO}]^4[\mathrm{H_2O}]^5}

Write the balanced chemical equation corresponding to this expression.Show solution

Given: $K_c = \dfrac{[\mathrm{NH_3}]^4[\mathrm{O_2}]^5}{[\mathrm{NO}]^4[\mathrm{H_2O}]^5}$

Interpretation: Products are $\mathrm{NH_3}$ and $\mathrm{O_2}$ ; reactants are $\mathrm{NO}$ and $\mathrm{H_2O}$ .

The balanced chemical equation is:
$\boxed{4\mathrm{NO}(\mathrm{g}) + 6\mathrm{H_2O}(\mathrm{g}) \rightleftharpoons 4\mathrm{NH_3}(\mathrm{g}) + 5\mathrm{O_2}(\mathrm{g})}$

Verification: This is the reverse of the catalytic oxidation of ammonia. The $K_c$ expression matches the given expression.

6.14One mole of

\mathrm{H_2O}

and one mole of CO are taken in

10\,\mathrm{L}

vessel and heated to

725\,\mathrm{K}

. At equilibrium

40\%

of water (by mass) reacts with CO according to the equation,

\mathrm{H_2O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H_2}(\mathrm{g}) + \mathrm{CO_2}(\mathrm{g})

Calculate the equilibrium constant for the reaction.Show solution

Given: Initial moles: $n(\mathrm{H_2O}) = 1\,\mathrm{mol}$ , $n(\mathrm{CO}) = 1\,\mathrm{mol}$ ; Volume $= 10\,\mathrm{L}$ ; 40% of water reacts.

Moles reacted: $0.40 \times 1 = 0.4\,\mathrm{mol}$ of $\mathrm{H_2O}$

ICE Table (in moles):

	$\mathrm{H_2O}$	$\mathrm{CO}$	$\mathrm{H_2}$	$\mathrm{CO_2}$
I	1	1	0	0
C	$-0.4$	$-0.4$	$+0.4$	$+0.4$
E	0.6	0.6	0.4	0.4

Equilibrium concentrations (in $10\,\mathrm{L}$ vessel):
$[\mathrm{H_2O}] = \frac{0.6}{10} = 0.06\,\mathrm{M},\quad [\mathrm{CO}] = 0.06\,\mathrm{M}$
$[\mathrm{H_2}] = \frac{0.4}{10} = 0.04\,\mathrm{M},\quad [\mathrm{CO_2}] = 0.04\,\mathrm{M}$

$K_c$ :
$K_c = \frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = \frac{1.6 \times 10^{-3}}{3.6 \times 10^{-3}}$
$\boxed{K_c = 0.444}$

6.15At

700\,\mathrm{K}

, equilibrium constant for the reaction:

\mathrm{H_2}(\mathrm{g}) + \mathrm{I_2}(\mathrm{g}) \rightleftharpoons 2\mathrm{HI}(\mathrm{g})

is 54.8. If

0.5\,\mathrm{mol\,L^{-1}}

of HI(g) is present at equilibrium at

700\,\mathrm{K}

, what are the concentration of

\mathrm{H_2}(\mathrm{g})

and

\mathrm{I_2}(\mathrm{g})

assuming that we initially started with HI(g) and allowed it to reach equilibrium at

700\,\mathrm{K}

?Show solution

Given: $K_c = 54.8$ for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ ; $[\mathrm{HI}]_{\mathrm{eq}} = 0.5\,\mathrm{mol\,L^{-1}}$

Since we started with HI only, the reverse reaction occurs:
$2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H_2}(\mathrm{g}) + \mathrm{I_2}(\mathrm{g})$

For this reverse reaction:
$K_c' = \frac{1}{K_c} = \frac{1}{54.8} = 1.825 \times 10^{-2}$

Let $[\mathrm{H_2}] = [\mathrm{I_2}] = x$ at equilibrium (by symmetry, since we started with only HI).

$K_c' = \frac{[\mathrm{H_2}][\mathrm{I_2}]}{[\mathrm{HI}]^2} = \frac{x^2}{(0.5)^2} = 1.825 \times 10^{-2}$
$x^2 = 1.825 \times 10^{-2} \times 0.25 = 4.56 \times 10^{-3}$
$x = 0.0676\,\mathrm{mol\,L^{-1}}$

$\boxed{[\mathrm{H_2}] = [\mathrm{I_2}] = 0.068\,\mathrm{mol\,L^{-1}}}$

6.16What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2\mathrm{ICl}(\mathrm{g}) \rightleftharpoons \mathrm{I_2}(\mathrm{g}) + \mathrm{Cl_2}(\mathrm{g});\quad K_c = 0.14

Show solution

Given: $[\mathrm{ICl}]_0 = 0.78\,\mathrm{M}$ , $K_c = 0.14$

ICE Table:

	$2\mathrm{ICl}$	$\mathrm{I_2}$	$\mathrm{Cl_2}$
I	0.78	0	0
C	$-2x$	$+x$	$+x$
E	$0.78-2x$	$x$	$x$

$K_c$ expression:
$K_c = \frac{[\mathrm{I_2}][\mathrm{Cl_2}]}{[\mathrm{ICl}]^2} = \frac{x \cdot x}{(0.78-2x)^2} = 0.14$

$\frac{x^2}{(0.78-2x)^2} = 0.14$

Taking square root of both sides:
$\frac{x}{0.78-2x} = \sqrt{0.14} = 0.3742$

$x = 0.3742(0.78 - 2x) = 0.2919 - 0.7484x$
$x + 0.7484x = 0.2919$
$1.7484x = 0.2919$
$x = 0.167\,\mathrm{M}$

Equilibrium concentrations:
$[\mathrm{ICl}] = 0.78 - 2(0.167) = 0.78 - 0.334 = \boxed{0.446\,\mathrm{M}}$
$[\mathrm{I_2}] = [\mathrm{Cl_2}] = \boxed{0.167\,\mathrm{M}}$

6.17

K_p = 0.04\,\mathrm{atm}

899\,\mathrm{K}

for the equilibrium shown below. What is the equilibrium concentration of

\mathrm{C_2H_6}

when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

\mathrm{C_2H_6}(\mathrm{g}) \rightleftharpoons \mathrm{C_2H_4}(\mathrm{g}) + \mathrm{H_2}(\mathrm{g})

Show solution

Given: $K_p = 0.04\,\mathrm{atm}$ , initial pressure of $\mathrm{C_2H_6} = 4.0\,\mathrm{atm}$ , $T = 899\,\mathrm{K}$

ICE Table (in terms of pressure):

	$\mathrm{C_2H_6}$	$\mathrm{C_2H_4}$	$\mathrm{H_2}$
I	4.0	0	0
C	$-p$	$+p$	$+p$
E	$4.0-p$	$p$	$p$

$K_p = \frac{p_{\mathrm{C_2H_4}} \cdot p_{\mathrm{H_2}}}{p_{\mathrm{C_2H_6}}} = \frac{p^2}{4.0-p} = 0.04$

$p^2 = 0.04(4.0-p) = 0.16 - 0.04p$
$p^2 + 0.04p - 0.16 = 0$

Using quadratic formula:
$p = \frac{-0.04 + \sqrt{(0.04)^2 + 4 \times 0.16}}{2} = \frac{-0.04 + \sqrt{0.0016 + 0.64}}{2}$
$= \frac{-0.04 + \sqrt{0.6416}}{2} = \frac{-0.04 + 0.8010}{2} = \frac{0.761}{2} = 0.3805\,\mathrm{atm}$

Equilibrium pressure of $\mathrm{C_2H_6}$ :
$p_{\mathrm{C_2H_6}} = 4.0 - 0.3805 = 3.619\,\mathrm{atm}$

Equilibrium concentration (using $pV = nRT$ , $c = p/RT$ ):
$[\mathrm{C_2H_6}] = \frac{p}{RT} = \frac{3.619}{0.0821 \times 899} = \frac{3.619}{73.81}$
$\boxed{[\mathrm{C_2H_6}] = 0.049\,\mathrm{mol\,L^{-1}}}$

6.18Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

\mathrm{CH_3COOH(l)} + \mathrm{C_2H_5OH(l)} \rightleftharpoons \mathrm{CH_3COOC_2H_5(l)} + \mathrm{H_2O(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\,\mathrm{K}

, if one starts with

1.00\,\mathrm{mol}

of acetic acid and

0.18\,\mathrm{mol}

of ethanol, there is

0.171\,\mathrm{mol}

of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with

0.5\,\mathrm{mol}

of ethanol and

1.0\,\mathrm{mol}

of acetic acid and maintaining it at

293\,\mathrm{K}

0.214\,\mathrm{mol}

of ethyl acetate is found after sometime. Has equilibrium been reached?Show solution

(i) Reaction quotient $Q_c$ :
$Q_c = \frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}$

(ii) Calculation of $K_c$ :

Let volume of the system $= V\,\mathrm{L}$ .

Initial moles: $n(\mathrm{CH_3COOH}) = 1.00$ , $n(\mathrm{C_2H_5OH}) = 0.18$ , $n(\mathrm{ester}) = 0$ , $n(\mathrm{H_2O}) = 0$

At equilibrium: $n(\mathrm{ester}) = 0.171\,\mathrm{mol}$

Moles consumed: $n(\mathrm{CH_3COOH}) = 0.171$ , $n(\mathrm{C_2H_5OH}) = 0.171$

Equilibrium moles:

$n(\mathrm{CH_3COOH}) = 1.00 - 0.171 = 0.829\,\mathrm{mol}$
$n(\mathrm{C_2H_5OH}) = 0.18 - 0.171 = 0.009\,\mathrm{mol}$
$n(\mathrm{ester}) = 0.171\,\mathrm{mol}$
$n(\mathrm{H_2O}) = 0.171\,\mathrm{mol}$

Since all species are in the same volume $V$ , it cancels:
$K_c = \frac{(0.171/V)(0.171/V)}{(0.829/V)(0.009/V)} = \frac{0.171 \times 0.171}{0.829 \times 0.009}$
$= \frac{0.02924}{0.007461} = \boxed{3.92}$

(iii) Checking if equilibrium is reached:

Initial: $n(\mathrm{CH_3COOH}) = 1.0$ , $n(\mathrm{C_2H_5OH}) = 0.5$

After some time: $n(\mathrm{ester}) = 0.214\,\mathrm{mol}$

Moles at this point:

$n(\mathrm{CH_3COOH}) = 1.0 - 0.214 = 0.786$
$n(\mathrm{C_2H_5OH}) = 0.5 - 0.214 = 0.286$
$n(\mathrm{ester}) = 0.214$
$n(\mathrm{H_2O}) = 0.214$

$Q_c = \frac{0.214 \times 0.214}{0.786 \times 0.286} = \frac{0.04580}{0.2248} = 0.2036$

Since $Q_c = 0.204 \neq K_c = 3.92$ , equilibrium has not been reached. Since $Q_c < K_c$ , the reaction will proceed in the forward direction.

6.19A sample of pure

\mathrm{PCl_5}

was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of

\mathrm{PCl_5}

was found to be

0.5 \times 10^{-1}\,\mathrm{mol\,L^{-1}}

. If value of

K_c

8.3 \times 10^{-2}

, what are the concentrations of

\mathrm{PCl_3}

and

\mathrm{Cl_2}

at equilibrium?

\mathrm{PCl_5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl_3}(\mathrm{g}) + \mathrm{Cl_2}(\mathrm{g})

Show solution

Given: $[\mathrm{PCl_5}]_{\mathrm{eq}} = 0.05\,\mathrm{mol\,L^{-1}}$ , $K_c = 8.3 \times 10^{-2}$

$K_c$ expression:
$K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}$

Since the vessel was initially evacuated and only $\mathrm{PCl_5}$ was introduced, by stoichiometry:
$[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = x$

$8.3 \times 10^{-2} = \frac{x \cdot x}{0.05} = \frac{x^2}{0.05}$
$x^2 = 8.3 \times 10^{-2} \times 0.05 = 4.15 \times 10^{-3}$
$x = \sqrt{4.15 \times 10^{-3}} = 0.0644\,\mathrm{mol\,L^{-1}}$

$\boxed{[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = 6.44 \times 10^{-2}\,\mathrm{mol\,L^{-1}}}$

6.20One 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

\mathrm{CO_2}

\mathrm{FeO(s)} + \mathrm{CO(g)} \rightleftharpoons \mathrm{Fe(s)} + \mathrm{CO_2(g)};\quad K_p = 0.265\,\mathrm{atm}\text{ at }1050\,\mathrm{K}

What are the equilibrium partial pressures of CO and

\mathrm{CO_2}

1050\,\mathrm{K}

if the initial partial pressures are:

p_{\mathrm{CO}} = 1.4\,\mathrm{atm}

and

p_{\mathrm{CO_2}} = 0.80\,\mathrm{atm}

?Show solution

Given: $K_p = 0.265$ , $p_{\mathrm{CO}}^0 = 1.4\,\mathrm{atm}$ , $p_{\mathrm{CO_2}}^0 = 0.80\,\mathrm{atm}$

$K_p$ expression (solids excluded):
$K_p = \frac{p_{\mathrm{CO_2}}}{p_{\mathrm{CO}}}$

Check $Q_p$ :
$Q_p = \frac{0.80}{1.4} = 0.571 > K_p = 0.265$

Since $Q_p > K_p$ , the reaction proceeds in the reverse direction.

Let pressure of $\mathrm{CO_2}$ decrease by $x$ :

	$\mathrm{CO}$	$\mathrm{CO_2}$
I	1.4	0.80
C	$+x$	$-x$
E	$1.4+x$	$0.80-x$

$K_p = \frac{0.80-x}{1.4+x} = 0.265$
$0.80 - x = 0.265(1.4 + x) = 0.371 + 0.265x$
$0.80 - 0.371 = x + 0.265x = 1.265x$
$0.429 = 1.265x$
$x = 0.339\,\mathrm{atm}$

Equilibrium partial pressures:
$p_{\mathrm{CO}} = 1.4 + 0.339 = \boxed{1.739\,\mathrm{atm}}$
$p_{\mathrm{CO_2}} = 0.80 - 0.339 = \boxed{0.461\,\mathrm{atm}}$

6.21Equilibrium constant,

K_c

for the reaction

\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}\text{ at }500\,\mathrm{K}\text{ is }0.061

At a particular time, the analysis shows that composition of the reaction mixture is

3.0\,\mathrm{mol\,L^{-1}}\,\mathrm{N_2}

2.0\,\mathrm{mol\,L^{-1}}\,\mathrm{H_2}

and

0.5\,\mathrm{mol\,L^{-1}}\,\mathrm{NH_3}

. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?Show solution

Given: $K_c = 0.061$ ; $[\mathrm{N_2}] = 3.0\,\mathrm{M}$ , $[\mathrm{H_2}] = 2.0\,\mathrm{M}$ , $[\mathrm{NH_3}] = 0.5\,\mathrm{M}$

Reaction quotient $Q_c$ :
$Q_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \frac{(0.5)^2}{(3.0)(2.0)^3} = \frac{0.25}{3.0 \times 8} = \frac{0.25}{24}$
$Q_c = 1.04 \times 10^{-2}$

Comparison:
$Q_c = 1.04 \times 10^{-2} < K_c = 0.061$

Since $Q_c < K_c$ , the reaction is not at equilibrium. The reaction will proceed in the forward direction (towards formation of $\mathrm{NH_3}$ ) to reach equilibrium.

6.22Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2\mathrm{BrCl(g)} \rightleftharpoons \mathrm{Br_2(g)} + \mathrm{Cl_2(g)}

for which

K_c = 32

500\,\mathrm{K}

. If initially pure BrCl is present at a concentration of

3.3 \times 10^{-3}\,\mathrm{mol\,L^{-1}}

, what is its molar concentration in the mixture at equilibrium?Show solution

Given: $[\mathrm{BrCl}]_0 = 3.3 \times 10^{-3}\,\mathrm{M}$ , $K_c = 32$

ICE Table:

	$2\mathrm{BrCl}$	$\mathrm{Br_2}$	$\mathrm{Cl_2}$
I	$3.3 \times 10^{-3}$	0	0
C	$-2x$	$+x$	$+x$
E	$3.3\times10^{-3}-2x$	$x$	$x$

$K_c = \frac{[\mathrm{Br_2}][\mathrm{Cl_2}]}{[\mathrm{BrCl}]^2} = \frac{x^2}{(3.3\times10^{-3}-2x)^2} = 32$

Taking square root:
$\frac{x}{3.3\times10^{-3}-2x} = \sqrt{32} = 5.657$
$x = 5.657(3.3\times10^{-3} - 2x) = 1.867\times10^{-2} - 11.314x$
$x(1 + 11.314) = 1.867\times10^{-2}$
$12.314x = 1.867\times10^{-2}$
$x = 1.516\times10^{-3}\,\mathrm{M}$

Equilibrium concentration of BrCl:
$[\mathrm{BrCl}] = 3.3\times10^{-3} - 2(1.516\times10^{-3}) = 3.3\times10^{-3} - 3.032\times10^{-3}$
$\boxed{[\mathrm{BrCl}] = 2.68\times10^{-4}\,\mathrm{mol\,L^{-1}}}$

6.23At

1127\,\mathrm{K}

and 1 atm pressure, a gaseous mixture of CO and

\mathrm{CO_2}

in equilibrium with solid carbon has

90.55\%

CO by mass.

\mathrm{C(s)} + \mathrm{CO_2(g)} \rightleftharpoons 2\mathrm{CO(g)}

Calculate

K_c

for this reaction at the above temperature.Show solution

Given: Total pressure $= 1\,\mathrm{atm}$ ; 90.55% CO by mass; $T = 1127\,\mathrm{K}$

Step 1: Find mole fractions.

Assume 100 g of gas mixture:

Mass of CO $= 90.55\,\mathrm{g}$ , moles of CO $= \dfrac{90.55}{28} = 3.234\,\mathrm{mol}$
Mass of $\mathrm{CO_2} = 9.45\,\mathrm{g}$ , moles of $\mathrm{CO_2} = \dfrac{9.45}{44} = 0.2148\,\mathrm{mol}$

Total moles $= 3.234 + 0.2148 = 3.449\,\mathrm{mol}$

$\chi_{\mathrm{CO}} = \frac{3.234}{3.449} = 0.9377,\quad \chi_{\mathrm{CO_2}} = \frac{0.2148}{3.449} = 0.0623$

Step 2: Partial pressures.
$p_{\mathrm{CO}} = 0.9377 \times 1 = 0.9377\,\mathrm{atm}$
$p_{\mathrm{CO_2}} = 0.0623\,\mathrm{atm}$

Step 3: $K_p$ .
$K_p = \frac{(p_{\mathrm{CO}})^2}{p_{\mathrm{CO_2}}} = \frac{(0.9377)^2}{0.0623} = \frac{0.8793}{0.0623} = 14.11\,\mathrm{atm}$

Step 4: Convert to $K_c$ .
$\Delta n = 2 - 1 = 1$ (only gaseous species counted)

$K_p = K_c(RT)^{\Delta n}$
$K_c = \frac{K_p}{RT} = \frac{14.11}{0.0821 \times 1127} = \frac{14.11}{92.53}$
$\boxed{K_c = 0.153\,\mathrm{mol\,L^{-1}}}$

6.24Calculate a)

\Delta G^0

and b) the equilibrium constant for the formation of

\mathrm{NO_2}

from NO and

\mathrm{O_2}

at 298 K:

\mathrm{NO(g)} + \frac{1}{2}\mathrm{O_2(g)} \rightleftharpoons \mathrm{NO_2(g)}

where

\Delta_f G^0(\mathrm{NO_2}) = 52.0\,\mathrm{kJ/mol}

\Delta_f G^0(\mathrm{NO}) = 87.0\,\mathrm{kJ/mol}

\Delta_f G^0(\mathrm{O_2}) = 0\,\mathrm{kJ/mol}

Show solution

(a) Calculation of $\Delta G^0$ :

$\Delta G^0 = \Delta_f G^0(\mathrm{NO_2}) - \left[\Delta_f G^0(\mathrm{NO}) + \frac{1}{2}\Delta_f G^0(\mathrm{O_2})\right]$
$= 52.0 - \left[87.0 + \frac{1}{2}(0)\right] = 52.0 - 87.0$
$\boxed{\Delta G^0 = -35.0\,\mathrm{kJ\,mol^{-1}}}$

(b) Calculation of equilibrium constant $K$ :

Using the relation: $\Delta G^0 = -RT\ln K$

$\ln K = \frac{-\Delta G^0}{RT} = \frac{-(-35000\,\mathrm{J\,mol^{-1}})}{8.314 \times 298}$
$= \frac{35000}{2477.6} = 14.13$

$K = e^{14.13} = 1.36 \times 10^6$

$\boxed{K = 1.36 \times 10^6}$

6.25Does 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)

\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}

(b)

\mathrm{CaO(s)} + \mathrm{CO_2(g)} \rightleftharpoons \mathrm{CaCO_3(s)}

(c)

3\mathrm{Fe(s)} + 4\mathrm{H_2O(g)} \rightleftharpoons \mathrm{Fe_3O_4(s)} + 4\mathrm{H_2(g)}

Show solution

Concept (Le Chatelier's Principle): When pressure is decreased (volume increased), the equilibrium shifts in the direction that increases the number of moles of gas.

(a) $\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$

Gaseous moles: Reactant side = 1, Product side = 2.
Decrease in pressure → equilibrium shifts to the right (forward direction).
Number of moles of products increases.

(b) $\mathrm{CaO(s)} + \mathrm{CO_2(g)} \rightleftharpoons \mathrm{CaCO_3(s)}$

Gaseous moles: Reactant side = 1 (only $\mathrm{CO_2}$ ), Product side = 0.
Decrease in pressure → equilibrium shifts to the left (reverse direction) to produce more gas.
Number of moles of products (CaCO₃) decreases.

(c) $3\mathrm{Fe(s)} + 4\mathrm{H_2O(g)} \rightleftharpoons \mathrm{Fe_3O_4(s)} + 4\mathrm{H_2(g)}$

Gaseous moles: Reactant side = 4 ( $\mathrm{H_2O}$ ), Product side = 4 ( $\mathrm{H_2}$ ). $\Delta n_{\mathrm{gas}} = 0$ .
Decrease in pressure has no effect on the equilibrium position.
Number of moles of products remains the same.

6.26Which 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.
(i)

\mathrm{COCl_2(g)} \rightleftharpoons \mathrm{CO(g)} + \mathrm{Cl_2(g)}

(ii)

\mathrm{CH_4(g)} + 2\mathrm{S_2(g)} \rightleftharpoons \mathrm{CS_2(g)} + 2\mathrm{H_2S(g)}

(iii)

\mathrm{CO_2(g)} + \mathrm{C(s)} \rightleftharpoons 2\mathrm{CO(g)}

(iv)

2\mathrm{H_2(g)} + \mathrm{CO(g)} \rightleftharpoons \mathrm{CH_3OH(g)}

(v)

\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}

(vi)

4\mathrm{NH_3(g)} + 5\mathrm{O_2(g)} \rightleftharpoons 4\mathrm{NO(g)} + 6\mathrm{H_2O(g)}

Show solution

Concept: Increasing pressure shifts equilibrium towards the side with fewer moles of gas. Reactions with $\Delta n_{\mathrm{gas}} = 0$ are unaffected.

(i) $\mathrm{COCl_2(g)} \rightleftharpoons \mathrm{CO(g)} + \mathrm{Cl_2(g)}$
$\Delta n_{\mathrm{gas}} = 2 - 1 = +1$ . Affected. Increasing pressure → backward direction.

(ii) $\mathrm{CH_4(g)} + 2\mathrm{S_2(g)} \rightleftharpoons \mathrm{CS_2(g)} + 2\mathrm{H_2S(g)}$
$\Delta n_{\mathrm{gas}} = 3 - 3 = 0$ . Not affected by pressure change.

(iii) $\mathrm{CO_2(g)} + \mathrm{C(s)} \rightleftharpoons 2\mathrm{CO(g)}$
$\Delta n_{\mathrm{gas}} = 2 - 1 = +1$ . Affected. Increasing pressure → backward direction.

(iv) $2\mathrm{H_2(g)} + \mathrm{CO(g)} \rightleftharpoons \mathrm{CH_3OH(g)}$
$\Delta n_{\mathrm{gas}} = 1 - 3 = -2$ . Affected. Increasing pressure → forward direction.

(v) $\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$
$\Delta n_{\mathrm{gas}} = 1 - 0 = +1$ . Affected. Increasing pressure → backward direction.

(vi) $4\mathrm{NH_3(g)} + 5\mathrm{O_2(g)} \rightleftharpoons 4\mathrm{NO(g)} + 6\mathrm{H_2O(g)}$
$\Delta n_{\mathrm{gas}} = 10 - 9 = +1$ . Affected. Increasing pressure → backward direction.

6.27The equilibrium constant for the following reaction is

1.6 \times 10^5

1024\,\mathrm{K}

\mathrm{H_2(g)} + \mathrm{Br_2(g)} \rightleftharpoons 2\mathrm{HBr(g)}

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

1024\,\mathrm{K}

.Show solution

Given: $K_p = 1.6 \times 10^5$ for $\mathrm{H_2 + Br_2 \rightleftharpoons 2HBr}$ ; initial $p_{\mathrm{HBr}} = 10.0\,\mathrm{bar}$

Since we start with HBr only, the reverse reaction occurs:
$2\mathrm{HBr(g)} \rightleftharpoons \mathrm{H_2(g)} + \mathrm{Br_2(g)}$

$K_p' = \frac{1}{K_p} = \frac{1}{1.6 \times 10^5} = 6.25 \times 10^{-6}$

ICE Table (pressures in bar):

	$2\mathrm{HBr}$	$\mathrm{H_2}$	$\mathrm{Br_2}$
I	10.0	0	0
C	$-2p$	$+p$	$+p$
E	$10.0-2p$	$p$	$p$

$K_p' = \frac{p_{\mathrm{H_2}} \cdot p_{\mathrm{Br_2}}}{(p_{\mathrm{HBr}})^2} = \frac{p^2}{(10.0-2p)^2} = 6.25 \times 10^{-6}$

Taking square root:
$\frac{p}{10.0-2p} = \sqrt{6.25 \times 10^{-6}} = 2.5 \times 10^{-3}$
$p = 2.5 \times 10^{-3}(10.0 - 2p) = 0.025 - 5 \times 10^{-3}p$
$p(1 + 5 \times 10^{-3}) = 0.025$
$p = \frac{0.025}{1.005} = 0.02488\,\mathrm{bar}$

Equilibrium pressures:
$p_{\mathrm{HBr}} = 10.0 - 2(0.02488) = 10.0 - 0.04976 \approx \boxed{9.95\,\mathrm{bar}}$
$p_{\mathrm{H_2}} = p_{\mathrm{Br_2}} \approx \boxed{2.49 \times 10^{-2}\,\mathrm{bar}}$

6.28Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

\mathrm{CH_4(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO(g)} + 3\mathrm{H_2(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?Show solution

(a) Expression for $K_p$ :
$K_p = \frac{p_{\mathrm{CO}} \cdot (p_{\mathrm{H_2}})^3}{p_{\mathrm{CH_4}} \cdot p_{\mathrm{H_2O}}}$

(b) Effects:

(i) Increasing the pressure:
$\Delta n_{\mathrm{gas}} = (1+3) - (1+1) = +2$ . Increasing pressure shifts equilibrium to the left (backward direction), decreasing the yield of $\mathrm{H_2}$ and CO. However, $K_p$ remains unchanged (it depends only on temperature).

(ii) Increasing the temperature:
The reaction is endothermic ( $\Delta H > 0$ ). By Le Chatelier's principle, increasing temperature shifts equilibrium to the right (forward direction), increasing the yield of $\mathrm{H_2}$ and CO. The value of $K_p$ increases with increase in temperature for an endothermic reaction.

(iii) Using a catalyst:
A catalyst increases the rate of both forward and reverse reactions equally. It helps the system reach equilibrium faster but does not change the equilibrium composition or the value of $K_p$ .

6.29Describe the effect of: a) addition of

\mathrm{H_2}

, b) addition of

\mathrm{CH_3OH}

, c) removal of CO, d) removal of

\mathrm{CH_3OH}

on the equilibrium of the reaction:

2\mathrm{H_2(g)} + \mathrm{CO(g)} \rightleftharpoons \mathrm{CH_3OH(g)}

Show solution

Concept (Le Chatelier's Principle): Adding a reactant or removing a product shifts equilibrium to the right (forward); removing a reactant or adding a product shifts equilibrium to the left (backward).

(a) Addition of $\mathrm{H_2}$ :
$\mathrm{H_2}$ is a reactant. Adding it increases its concentration, so the equilibrium shifts to the right (forward direction) to consume the added $\mathrm{H_2}$ . More $\mathrm{CH_3OH}$ is produced.

(b) Addition of $\mathrm{CH_3OH}$ :
$\mathrm{CH_3OH}$ is a product. Adding it increases its concentration, so the equilibrium shifts to the left (backward direction) to consume the added $\mathrm{CH_3OH}$ . More $\mathrm{H_2}$ and CO are produced.

(c) Removal of CO:
CO is a reactant. Removing it decreases its concentration, so the equilibrium shifts to the left (backward direction) to replenish CO. The amount of $\mathrm{CH_3OH}$ decreases.

(d) Removal of $\mathrm{CH_3OH}$ :
$\mathrm{CH_3OH}$ is a product. Removing it decreases its concentration, so the equilibrium shifts to the right (forward direction) to produce more $\mathrm{CH_3OH}$ .

6.30At

473\,\mathrm{K}

, equilibrium constant

K_c

for decomposition of phosphorus pentachloride,

\mathrm{PCl_5}

8.3 \times 10^{-3}

. If decomposition is depicted as,

\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}\quad \Delta_f H^0 = 124.0\,\mathrm{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

\mathrm{PCl_5}

is added (ii) pressure is increased (iii) the temperature is increased?Show solution

(a) Expression for $K_c$ :
$K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}$

(b) $K_c$ for the reverse reaction:
For the reverse reaction $\mathrm{PCl_3(g)} + \mathrm{Cl_2(g)} \rightleftharpoons \mathrm{PCl_5(g)}$ :
$K_c' = \frac{1}{K_c} = \frac{1}{8.3 \times 10^{-3}} = \boxed{1.20 \times 10^2}$

(c) Effect on $K_c$ :

(i) More $\mathrm{PCl_5}$ is added:
$K_c$ depends only on temperature. Adding more $\mathrm{PCl_5}$ shifts the equilibrium to the right but $K_c$ remains unchanged.

(ii) Pressure is increased:
Increasing pressure shifts the equilibrium to the left (fewer moles of gas), but $K_c$ remains unchanged as it depends only on temperature.

(iii) Temperature is increased:
The reaction is endothermic ( $\Delta H^0 = +124.0\,\mathrm{kJ\,mol^{-1}}$ ). Increasing temperature favours the forward (endothermic) reaction. Therefore, $K_c$ increases with increase in temperature.

6.31In the water gas shift reaction:

\mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)}

If a reaction vessel at

400^\circ\mathrm{C}

is charged with an equimolar mixture of CO and steam such that

p_{\mathrm{CO}} = p_{\mathrm{H_2O}} = 4.0\,\mathrm{bar}

, what will be the partial pressure of

\mathrm{H_2}

at equilibrium?

K_p = 10.1

400^\circ\mathrm{C}

Show solution

Given: $p_{\mathrm{CO}}^0 = p_{\mathrm{H_2O}}^0 = 4.0\,\mathrm{bar}$ ; $K_p = 10.1$

ICE Table (pressures in bar):

	CO	$\mathrm{H_2O}$	$\mathrm{CO_2}$	$\mathrm{H_2}$
I	4.0	4.0	0	0
C	$-x$	$-x$	$+x$	$+x$
E	$4.0-x$	$4.0-x$	$x$	$x$

$K_p = \frac{p_{\mathrm{CO_2}} \cdot p_{\mathrm{H_2}}}{p_{\mathrm{CO}} \cdot p_{\mathrm{H_2O}}} = \frac{x \cdot x}{(4.0-x)(4.0-x)} = \frac{x^2}{(4.0-x)^2} = 10.1$

Taking square root:
$\frac{x}{4.0-x} = \sqrt{10.1} = 3.178$
$x = 3.178(4.0-x) = 12.71 - 3.178x$
$x(1+3.178) = 12.71$
$4.178x = 12.71$
$x = 3.042\,\mathrm{bar}$

$\boxed{p_{\mathrm{H_2}} = 3.04\,\mathrm{bar}}$

6.32Predict which of the following reaction will have appreciable concentration of reactants and products:
a)

\mathrm{Cl_2(g)} \rightleftharpoons 2\mathrm{Cl(g)}

;

K_c = 5 \times 10^{-39}

\mathrm{Cl_2(g)} + 2\mathrm{NO(g)} \rightleftharpoons 2\mathrm{NOCl(g)}

;

K_c = 3.7 \times 10^8

\mathrm{Cl_2(g)} + 2\mathrm{NO_2(g)} \rightleftharpoons 2\mathrm{NO_2Cl(g)}

;

K_c = 1.8

Show solution

Concept:

If $K_c \gg 1$ (very large): reaction goes nearly to completion; mostly products present.
If $K_c \ll 1$ (very small): reaction barely proceeds; mostly reactants present.
If $K_c \approx 1$ : appreciable concentrations of both reactants and products are present.

(a) $K_c = 5 \times 10^{-39}$ — extremely small. Reaction barely proceeds. Mostly reactants present; negligible products.

(b) $K_c = 3.7 \times 10^8$ — very large. Reaction goes nearly to completion. Mostly products present; negligible reactants.

(c) $K_c = 1.8$ — close to 1. Appreciable concentrations of both reactants and products are present at equilibrium.

Answer: Reaction (c) will have appreciable concentrations of both reactants and products.

6.33The value of

K_c

for the reaction

3\mathrm{O_2(g)} \rightleftharpoons 2\mathrm{O_3(g)}

2.0 \times 10^{-50}

25^\circ\mathrm{C}

. If the equilibrium concentration of

\mathrm{O_2}

in air at

25^\circ\mathrm{C}

1.6 \times 10^{-2}\,\mathrm{M}

, what is the concentration of

\mathrm{O_3}

?Show solution

Given: $K_c = 2.0 \times 10^{-50}$ ; $[\mathrm{O_2}] = 1.6 \times 10^{-2}\,\mathrm{M}$

$K_c$ expression:
$K_c = \frac{[\mathrm{O_3}]^2}{[\mathrm{O_2}]^3}$

$[\mathrm{O_3}]^2 = K_c \times [\mathrm{O_2}]^3 = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^3$
$= 2.0 \times 10^{-50} \times 4.096 \times 10^{-6}$
$= 8.192 \times 10^{-56}$

$[\mathrm{O_3}] = \sqrt{8.192 \times 10^{-56}} = 2.86 \times 10^{-28}\,\mathrm{M}$

$\boxed{[\mathrm{O_3}] = 2.86 \times 10^{-28}\,\mathrm{mol\,L^{-1}}}$

6.34The reaction,

\mathrm{CO(g)} + 3\mathrm{H_2(g)} \rightleftharpoons \mathrm{CH_4(g)} + \mathrm{H_2O(g)}

is at equilibrium at

1300\,\mathrm{K}

in a 1 L flask. It also contains

0.30\,\mathrm{mol}

of CO,

0.10\,\mathrm{mol}

\mathrm{H_2}

and

0.02\,\mathrm{mol}

\mathrm{H_2O}

and an unknown amount of

\mathrm{CH_4}

in the flask. Determine the concentration of

\mathrm{CH_4}

in the mixture. The equilibrium constant,

K_c

for the reaction at the given temperature is 3.90.Show solution

Given: Volume $= 1\,\mathrm{L}$ , so concentrations equal moles.
$[\mathrm{CO}] = 0.30\,\mathrm{M},\quad [\mathrm{H_2}] = 0.10\,\mathrm{M},\quad [\mathrm{H_2O}] = 0.02\,\mathrm{M},\quad [\mathrm{CH_4}] = ?$
$K_c = 3.90$

$K_c$ expression:
$K_c = \frac{[\mathrm{CH_4}][\mathrm{H_2O}]}{[\mathrm{CO}][\mathrm{H_2}]^3}$

$3.90 = \frac{[\mathrm{CH_4}] \times 0.02}{0.30 \times (0.10)^3} = \frac{[\mathrm{CH_4}] \times 0.02}{0.30 \times 0.001}$
$= \frac{[\mathrm{CH_4}] \times 0.02}{3.0 \times 10^{-4}}$

$[\mathrm{CH_4}] = \frac{3.90 \times 3.0 \times 10^{-4}}{0.02} = \frac{1.17 \times 10^{-3}}{0.02}$

$\boxed{[\mathrm{CH_4}] = 0.0585\,\mathrm{mol\,L^{-1}}}$

6.35What 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-}}

, and

\mathrm{S^{2-}}

Show solution

Conjugate Acid-Base Pair:
A conjugate acid-base pair consists of two species that differ by a single proton ( $\mathrm{H^+}$ ). When a Brønsted-Lowry acid donates a proton, the species formed is its conjugate base. When a Brønsted-Lowry base accepts a proton, the species formed is its conjugate acid.

Conjugate bases (for acids — remove one $\mathrm{H^+}$ ):

$\mathrm{HNO_2}$ : conjugate base = $\mathrm{NO_2^-}$
$\mathrm{HClO_4}$ : conjugate base = $\mathrm{ClO_4^-}$

Conjugate acids (for bases — add one $\mathrm{H^+}$ ):

$\mathrm{CN^-}$ : conjugate acid = $\mathrm{HCN}$
$\mathrm{F^-}$ : conjugate acid = $\mathrm{HF}$
$\mathrm{OH^-}$ : conjugate acid = $\mathrm{H_2O}$
$\mathrm{CO_3^{2-}}$ : conjugate acid = $\mathrm{HCO_3^-}$
$\mathrm{S^{2-}}$ : conjugate acid = $\mathrm{HS^-}$

6.36Which of the followings are Lewis acids?

\mathrm{H_2O}

\mathrm{BF_3}

\mathrm{H^+}

, and

\mathrm{NH_4^+}

Show solution

Lewis acid: A species that can accept a pair of electrons.

$\mathrm{H_2O}$ : Has lone pairs; acts as a Lewis base (electron pair donor). Not a Lewis acid.
$\mathrm{BF_3}$ : Boron has an incomplete octet (only 6 electrons); it can accept an electron pair. Lewis acid. ✓
$\mathrm{H^+}$ : A bare proton with no electrons; it readily accepts an electron pair. Lewis acid. ✓
$\mathrm{NH_4^+}$ : Nitrogen has a complete octet and no vacant orbital to accept electrons. Not a Lewis acid.

Answer: $\mathrm{BF_3}$ and $\mathrm{H^+}$ are Lewis acids.

6.37What will be the conjugate bases for the Brönsted acids: HF,

\mathrm{H_2SO_4}

and

\mathrm{HCO_3^-}

?Show solution

Concept: Conjugate base = acid $-$ one proton ( $\mathrm{H^+}$ ).

Brønsted Acid	Conjugate Base
$\mathrm{HF}$	$\mathrm{F^-}$
$\mathrm{H_2SO_4}$	$\mathrm{HSO_4^-}$
$\mathrm{HCO_3^-}$	$\mathrm{CO_3^{2-}}$

6.38Write the conjugate acids for the following Brönsted bases:

\mathrm{NH_2^-}

\mathrm{NH_3}

and

\mathrm{HCOO^-}

Concept: Conjugate acid = base $+$ one proton ( $\mathrm{H^+}$ ).

Brønsted Base	Conjugate Acid
$\mathrm{NH_2^-}$	$\mathrm{NH_3}$
$\mathrm{NH_3}$	$\mathrm{NH_4^+}$
$\mathrm{HCOO^-}$	$\mathrm{HCOOH}$

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6.39The species:

\mathrm{H_2O}

\mathrm{HCO_3^-}

\mathrm{HSO_4^-}

and

\mathrm{NH_3}

can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Concept: Amphoteric species can donate or accept a proton.

Conjugate base = species $-$ $\mathrm{H^+}$
Conjugate acid = species $+$ $\mathrm{H^+}$

Species	Conjugate Base (as acid)	Conjugate Acid (as base)
$\mathrm{H_2O}$	$\mathrm{OH^-}$	$\mathrm{H_3O^+}$
$\mathrm{HCO_3^-}$	$\mathrm{CO_3^{2-}}$	$\mathrm{H_2CO_3}$
$\mathrm{HSO_4^-}$	$\mathrm{SO_4^{2-}}$	$\mathrm{H_2SO_4}$
$\mathrm{NH_3}$	$\mathrm{NH_2^-}$	$\mathrm{NH_4^+}$

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6.40Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a)

\mathrm{OH^-}

(b)

\mathrm{F^-}

(c)

\mathrm{H^+}

(d)

\mathrm{BCl_3}

Lewis acid: electron pair acceptor. Lewis base: electron pair donor.

(a) $\mathrm{OH^-}$ : Has lone pairs of electrons on oxygen. It donates an electron pair. → Lewis base.
$\mathrm{OH^-} + \mathrm{H^+} \rightarrow \mathrm{H_2O}$

(b) $\mathrm{F^-}$ : Has lone pairs of electrons on fluorine. It donates an electron pair. → Lewis base.
$\mathrm{F^-} + \mathrm{BF_3} \rightarrow \mathrm{BF_4^-}$

(c) $\mathrm{H^+}$ : A bare proton with no electrons; it accepts an electron pair. → Lewis acid.
$\mathrm{H^+} + \mathrm{OH^-} \rightarrow \mathrm{H_2O}$

(d) $\mathrm{BCl_3}$ : Boron has an incomplete octet (6 electrons, vacant $p$ orbital); it accepts an electron pair. → Lewis acid.
$\mathrm{BCl_3} + \mathrm{NH_3} \rightarrow \mathrm{BCl_3 \cdot NH_3}$

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6.41The concentration of hydrogen ion in a sample of soft drink is

3.8 \times 10^{-3}\,\mathrm{M}

. What is its pH?

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6.42The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

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6.43The ionization constants of HF, HCOOH and HCN at 298 K are

6.8 \times 10^{-4}

1.8 \times 10^{-4}

and

4.8 \times 10^{-9}

respectively. Calculate the ionization constants of the corresponding conjugate bases.

Concept: For a conjugate acid-base pair: $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at 298 K.

So $K_b = \dfrac{K_w}{K_a}$

(i) Conjugate base of HF is $\mathrm{F^-}$ :
$K_b(\mathrm{F^-}) = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = \boxed{1.47 \times 10^{-11}}$

(ii) Conjugate base of HCOOH is $\mathrm{HCOO^-}$ :
$K_b(\mathrm{HCOO^-}) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = \boxed{5.56 \times 10^{-11}}$

(iii) Conjugate base of HCN is $\mathrm{CN^-}$ :
$K_b(\mathrm{CN^-}) = \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-9}} = \boxed{2.08 \times 10^{-6}}$

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6.44The ionization constant of phenol is

1.0 \times 10^{-10}

. What is the concentration of phenolate ion in

0.05\,\mathrm{M}

solution of phenol? What will be its degree of ionization if the solution is also

0.01\,\mathrm{M}

in sodium phenolate?

Given: $K_a = 1.0 \times 10^{-10}$ , $C = 0.05\,\mathrm{M}$

Part 1: Concentration of phenolate ion in 0.05 M phenol:

Let $x = [\mathrm{C_6H_5O^-}]$ at equilibrium.
$K_a = \frac{x^2}{C - x} \approx \frac{x^2}{C}$ (since $K_a$ is very small)
$x^2 = K_a \times C = 1.0 \times 10^{-10} \times 0.05 = 5.0 \times 10^{-12}$
$x = \sqrt{5.0 \times 10^{-12}} = 2.24 \times 10^{-6}\,\mathrm{M}$
$\boxed{[\mathrm{C_6H_5O^-}] = 2.24 \times 10^{-6}\,\mathrm{M}}$

Part 2: Degree of ionization in presence of 0.01 M sodium phenolate (common ion effect):

Sodium phenolate provides $[\mathrm{C_6H_5O^-}] = 0.01\,\mathrm{M}$ initially.

Let $\alpha$ = degree of ionization. Then $[\mathrm{C_6H_5O^-}] \approx 0.01 + 0.05\alpha \approx 0.01$ (since $\alpha$ is very small).

$K_a = \frac{(0.01)(0.05\alpha)}{0.05(1-\alpha)} \approx \frac{0.01 \times 0.05\alpha}{0.05} = 0.01\alpha$
$\alpha = \frac{K_a}{0.01} = \frac{1.0 \times 10^{-10}}{0.01} = 1.0 \times 10^{-8}$
$\boxed{\alpha = 1.0 \times 10^{-8}}$

The degree of ionization is drastically reduced due to the common ion effect.

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6.45The first ionization constant of

\mathrm{H_2S}

9.1 \times 10^{-8}

. Calculate the concentration of

\mathrm{HS^-}

ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of

\mathrm{H_2S}

1.2 \times 10^{-13}

, calculate the concentration of

\mathrm{S^{2-}}

under both conditions.

Given: $K_{a1} = 9.1 \times 10^{-8}$ , $K_{a2} = 1.2 \times 10^{-13}$ , $C = 0.1\,\mathrm{M}$

Condition 1: Pure 0.1 M $\mathrm{H_2S}$ solution

$[\mathrm{HS^-}]$ :
$K_{a1} = \frac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]} \approx \frac{x^2}{0.1}$
$x^2 = 9.1 \times 10^{-8} \times 0.1 = 9.1 \times 10^{-9}$
$x = [\mathrm{HS^-}] = 9.54 \times 10^{-5}\,\mathrm{M} \approx \boxed{9.54 \times 10^{-5}\,\mathrm{M}}$

Also $[\mathrm{H^+}] = 9.54 \times 10^{-5}\,\mathrm{M}$

$[\mathrm{S^{2-}}]$ in pure $\mathrm{H_2S}$ :
$K_{a2} = \frac{[\mathrm{H^+}][\mathrm{S^{2-}}]}{[\mathrm{HS^-}]}$
$[\mathrm{S^{2-}}] = \frac{K_{a2} \times [\mathrm{HS^-}]}{[\mathrm{H^+}]} = \frac{1.2 \times 10^{-13} \times 9.54 \times 10^{-5}}{9.54 \times 10^{-5}} = K_{a2}$
$\boxed{[\mathrm{S^{2-}}] = 1.2 \times 10^{-13}\,\mathrm{M}}$

Condition 2: 0.1 M $\mathrm{H_2S}$ + 0.1 M HCl

HCl provides $[\mathrm{H^+}] = 0.1\,\mathrm{M}$ (common ion effect suppresses ionization of $\mathrm{H_2S}$ ).

$[\mathrm{HS^-}]$ :
$K_{a1} = \frac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]} = \frac{0.1 \times [\mathrm{HS^-}]}{0.1}$
$[\mathrm{HS^-}] = K_{a1} = \boxed{9.1 \times 10^{-8}\,\mathrm{M}}$

The concentration of $\mathrm{HS^-}$ is drastically reduced.

$[\mathrm{S^{2-}}]$ in presence of HCl:
$[\mathrm{S^{2-}}] = \frac{K_{a2} \times [\mathrm{HS^-}]}{[\mathrm{H^+}]} = \frac{1.2 \times 10^{-13} \times 9.1 \times 10^{-8}}{0.1}$
$= \frac{1.092 \times 10^{-20}}{0.1} = \boxed{1.09 \times 10^{-19}\,\mathrm{M}}$

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6.46The ionization constant of acetic acid is

1.74 \times 10^{-5}

. Calculate the degree of dissociation of acetic acid in its

0.05\,\mathrm{M}

solution. Calculate the concentration of acetate ion in the solution and its pH.

Given: $K_a = 1.74 \times 10^{-5}$ , $C = 0.05\,\mathrm{M}$

Degree of dissociation $\alpha$ :
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}} = \sqrt{3.48 \times 10^{-4}} = 1.865 \times 10^{-2}$
$\boxed{\alpha = 1.865 \times 10^{-2} \approx 1.87\%}$

Concentration of acetate ion $[\mathrm{CH_3COO^-}]$ :
$[\mathrm{CH_3COO^-}] = C\alpha = 0.05 \times 1.865 \times 10^{-2} = 9.325 \times 10^{-4}\,\mathrm{M}$
$\boxed{[\mathrm{CH_3COO^-}] = 9.33 \times 10^{-4}\,\mathrm{M}}$

pH:
$[\mathrm{H^+}] = C\alpha = 9.33 \times 10^{-4}\,\mathrm{M}$
$\mathrm{pH} = -\log(9.33 \times 10^{-4}) = -(\log 9.33 - 4) = -(0.9699 - 4) = 3.03$
$\boxed{\mathrm{pH} = 3.03}$

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6.47It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its

\mathrm{pK_a}

Given: $C = 0.01\,\mathrm{M}$ , $\mathrm{pH} = 4.15$

Concentration of anion $[\mathrm{A^-}]$ :
$[\mathrm{H^+}] = [\mathrm{A^-}] = 10^{-4.15} = 7.08 \times 10^{-5}\,\mathrm{M}$
$\boxed{[\mathrm{A^-}] = 7.08 \times 10^{-5}\,\mathrm{M}}$

Ionization constant $K_a$ :
$[\mathrm{HA}] = C - [\mathrm{A^-}] = 0.01 - 7.08 \times 10^{-5} \approx 9.929 \times 10^{-3}\,\mathrm{M}$
$K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{(7.08 \times 10^{-5})^2}{9.929 \times 10^{-3}}$
$= \frac{5.013 \times 10^{-9}}{9.929 \times 10^{-3}} = 5.05 \times 10^{-7}$
$\boxed{K_a = 5.05 \times 10^{-7}}$

$\mathrm{pK_a}$ :
$\mathrm{pK_a} = -\log(5.05 \times 10^{-7}) = 7 - \log 5.05 = 7 - 0.703 = \boxed{6.30}$

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6.48Assuming 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

Concept: For strong acids: $[\mathrm{H^+}] = C$ ; for strong bases: $[\mathrm{OH^-}] = C$ , then $\mathrm{pOH} = -\log[\mathrm{OH^-}]$ , $\mathrm{pH} = 14 - \mathrm{pOH}$ .

(a) 0.003 M HCl:
$[\mathrm{H^+}] = 0.003 = 3 \times 10^{-3}\,\mathrm{M}$
$\mathrm{pH} = -\log(3 \times 10^{-3}) = 3 - \log 3 = 3 - 0.477 = \boxed{2.52}$

(b) 0.005 M NaOH:
$[\mathrm{OH^-}] = 0.005 = 5 \times 10^{-3}\,\mathrm{M}$
$\mathrm{pOH} = -\log(5 \times 10^{-3}) = 3 - \log 5 = 3 - 0.699 = 2.301$
$\mathrm{pH} = 14 - 2.301 = \boxed{11.70}$

(c) 0.002 M HBr:
$[\mathrm{H^+}] = 2 \times 10^{-3}\,\mathrm{M}$
$\mathrm{pH} = -\log(2 \times 10^{-3}) = 3 - \log 2 = 3 - 0.301 = \boxed{2.70}$

(d) 0.002 M KOH:
$[\mathrm{OH^-}] = 2 \times 10^{-3}\,\mathrm{M}$
$\mathrm{pOH} = 3 - 0.301 = 2.699$
$\mathrm{pH} = 14 - 2.699 = \boxed{11.30}$

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6.49Calculate the pH of the following solutions:
a)

2\,\mathrm{g}

of TlOH dissolved in water to give 2 litre of solution.
b)

0.3\,\mathrm{g}

\mathrm{Ca(OH)_2}

dissolved in water to give

500\,\mathrm{mL}

of solution.
c)

0.3\,\mathrm{g}

of NaOH dissolved in water to give

200\,\mathrm{mL}

of solution.
d)

1\,\mathrm{mL}

13.6\,\mathrm{M}

HCl is diluted with water to give 1 litre of solution.

(a) TlOH (Molar mass = 221 g/mol):
$[\mathrm{TlOH}] = \frac{2/221}{2} = \frac{0.00905}{2} = 4.52 \times 10^{-3}\,\mathrm{M}$
TlOH is a strong base: $[\mathrm{OH^-}] = 4.52 \times 10^{-3}\,\mathrm{M}$
$\mathrm{pOH} = -\log(4.52 \times 10^{-3}) = 2.345$
$\boxed{\mathrm{pH} = 14 - 2.345 = 11.65}$

(b) $\mathrm{Ca(OH)_2}$ (Molar mass = 74 g/mol):
$[\mathrm{Ca(OH)_2}] = \frac{0.3/74}{0.5} = \frac{4.054 \times 10^{-3}}{0.5} = 8.108 \times 10^{-3}\,\mathrm{M}$
$\mathrm{Ca(OH)_2}$ gives 2 $\mathrm{OH^-}$ per formula unit:
$[\mathrm{OH^-}] = 2 \times 8.108 \times 10^{-3} = 1.622 \times 10^{-2}\,\mathrm{M}$
$\mathrm{pOH} = -\log(1.622 \times 10^{-2}) = 1.790$
$\boxed{\mathrm{pH} = 14 - 1.790 = 12.21}$

(c) NaOH (Molar mass = 40 g/mol):
$[\mathrm{NaOH}] = \frac{0.3/40}{0.2} = \frac{7.5 \times 10^{-3}}{0.2} = 0.0375\,\mathrm{M}$
$[\mathrm{OH^-}] = 0.0375\,\mathrm{M}$
$\mathrm{pOH} = -\log(0.0375) = 1.426$
$\boxed{\mathrm{pH} = 14 - 1.426 = 12.57}$

(d) HCl diluted:
$[\mathrm{HCl}] = \frac{13.6 \times 1}{1000} = 0.0136\,\mathrm{M}$
$[\mathrm{H^+}] = 0.0136\,\mathrm{M}$
$\mathrm{pH} = -\log(0.0136) = -\log(1.36 \times 10^{-2}) = 2 - \log 1.36 = 2 - 0.134$
$\boxed{\mathrm{pH} = 1.87}$

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6.50The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the

\mathrm{pK_a}

of bromoacetic acid.

Given: $C = 0.1\,\mathrm{M}$ , $\alpha = 0.132$

$[\mathrm{H^+}]$ :
$[\mathrm{H^+}] = C\alpha = 0.1 \times 0.132 = 0.0132\,\mathrm{M}$

pH:
$\mathrm{pH} = -\log(0.0132) = -\log(1.32 \times 10^{-2}) = 2 - \log 1.32 = 2 - 0.121 = \boxed{1.88}$

$K_a$ :
$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.1 \times (0.132)^2}{1 - 0.132} = \frac{0.1 \times 0.01742}{0.868} = \frac{1.742 \times 10^{-3}}{0.868} = 2.007 \times 10^{-3}$

$\mathrm{pK_a}$ :
$\mathrm{pK_a} = -\log(2.007 \times 10^{-3}) = 3 - \log 2.007 = 3 - 0.3025 = \boxed{2.70}$

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6.51The pH of 0.005 M codeine (

\mathrm{C_{18}H_{21}NO_2}

) solution is 9.95. Calculate its ionization constant and

\mathrm{pK_b}

Given: $C = 0.005\,\mathrm{M}$ , $\mathrm{pH} = 9.95$

$[\mathrm{OH^-}]$ :
$\mathrm{pOH} = 14 - 9.95 = 4.05$
$[\mathrm{OH^-}] = 10^{-4.05} = 8.91 \times 10^{-5}\,\mathrm{M}$

$K_b$ :
Codeine is a weak base (B). $[\mathrm{BH^+}] = [\mathrm{OH^-}] = 8.91 \times 10^{-5}\,\mathrm{M}$
$[\mathrm{B}] = C - [\mathrm{OH^-}] = 0.005 - 8.91 \times 10^{-5} \approx 4.911 \times 10^{-3}\,\mathrm{M}$

$K_b = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]} = \frac{(8.91 \times 10^{-5})^2}{4.911 \times 10^{-3}} = \frac{7.939 \times 10^{-9}}{4.911 \times 10^{-3}}$
$\boxed{K_b = 1.62 \times 10^{-6}}$

$\mathrm{pK_b}$ :
$\mathrm{pK_b} = -\log(1.62 \times 10^{-6}) = 6 - \log 1.62 = 6 - 0.210 = \boxed{5.79}$

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6.52What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Given: $C = 0.001\,\mathrm{M}$ ; $K_b(\text{aniline}) = 4.27 \times 10^{-10}$ (from NCERT Table 6.7)

$[\mathrm{OH^-}]$ :
$[\mathrm{OH^-}] = \sqrt{K_b \times C} = \sqrt{4.27 \times 10^{-10} \times 10^{-3}} = \sqrt{4.27 \times 10^{-13}}$
$= 6.53 \times 10^{-7}\,\mathrm{M}$

pOH and pH:
$\mathrm{pOH} = -\log(6.53 \times 10^{-7}) = 7 - \log 6.53 = 7 - 0.815 = 6.185$
$\boxed{\mathrm{pH} = 14 - 6.185 = 7.82}$

Degree of ionization $\alpha$ :
$\alpha = \frac{[\mathrm{OH^-}]}{C} = \frac{6.53 \times 10^{-7}}{10^{-3}} = \boxed{6.53 \times 10^{-4}}$

$K_a$ of conjugate acid (anilinium ion $\mathrm{C_6H_5NH_3^+}$ ):
$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{4.27 \times 10^{-10}} = \boxed{2.34 \times 10^{-5}}$

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6.53Calculate the degree of ionization of 0.05 M acetic acid if its

\mathrm{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?

Given: $C = 0.05\,\mathrm{M}$ , $\mathrm{pK_a} = 4.74$
$K_a = 10^{-4.74} = 1.82 \times 10^{-5}$

Degree of ionization in pure acetic acid:
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.82 \times 10^{-5}}{0.05}} = \sqrt{3.64 \times 10^{-4}} = 1.908 \times 10^{-2}$
$\boxed{\alpha = 1.91 \times 10^{-2} \approx 1.91\%}$

With common ion (HCl):

In presence of HCl, $[\mathrm{H^+}] \approx [\mathrm{HCl}]$ (since HCl is strong acid and suppresses acetic acid ionization).

Let $\alpha'$ = degree of ionization of acetic acid in presence of HCl.
$K_a = \frac{([\mathrm{HCl}] + C\alpha') \cdot C\alpha'}{C(1-\alpha')} \approx \frac{[\mathrm{HCl}] \cdot C\alpha'}{C} = [\mathrm{HCl}] \cdot \alpha'$
$\alpha' = \frac{K_a}{[\mathrm{HCl}]}$

(a) 0.01 M HCl:
$\alpha' = \frac{1.82 \times 10^{-5}}{0.01} = \boxed{1.82 \times 10^{-3}}$

(b) 0.1 M HCl:
$\alpha' = \frac{1.82 \times 10^{-5}}{0.1} = \boxed{1.82 \times 10^{-4}}$

The degree of ionization decreases significantly due to the common ion effect.

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6.54The ionization constant of dimethylamine is

5.4 \times 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?

Given: $K_b = 5.4 \times 10^{-4}$ , $C = 0.02\,\mathrm{M}$

Degree of ionization in pure 0.02 M solution:
$\alpha = \sqrt{\frac{K_b}{C}} = \sqrt{\frac{5.4 \times 10^{-4}}{0.02}} = \sqrt{0.027} = 0.164$
$\boxed{\alpha = 0.164 \approx 16.4\%}$

(Note: Since $\alpha$ is not negligible, we should use the exact formula:
$K_b = \frac{C\alpha^2}{1-\alpha} \Rightarrow 5.4 \times 10^{-4} = \frac{0.02\alpha^2}{1-\alpha}$
$0.02\alpha^2 + 5.4 \times 10^{-4}\alpha - 5.4 \times 10^{-4} = 0$
Solving: $\alpha = 0.157$ , i.e., $\approx 15.7\%$ )

In presence of 0.1 M NaOH (common ion $\mathrm{OH^-}$ ):

$[\mathrm{OH^-}] \approx 0.1\,\mathrm{M}$ (from NaOH, dominant)

$K_b = \frac{[\mathrm{(CH_3)_2NH_2^+}][\mathrm{OH^-}]}{[\mathrm{(CH_3)_2NH}]} = \frac{C\alpha' \times 0.1}{C(1-\alpha')} \approx \frac{0.1 \times C\alpha'}{C} = 0.1\alpha'$
$\alpha' = \frac{K_b}{0.1} = \frac{5.4 \times 10^{-4}}{0.1} = 5.4 \times 10^{-3}$
$\boxed{\% \text{ ionization} = 0.54\%}$

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6.55Calculate 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.

Formula: $[\mathrm{H^+}] = 10^{-\mathrm{pH}}$

(a) pH = 6.83:
$[\mathrm{H^+}] = 10^{-6.83} = 1.48 \times 10^{-7}\,\mathrm{M}$

(b) pH = 1.2:
$[\mathrm{H^+}] = 10^{-1.2} = 6.31 \times 10^{-2}\,\mathrm{M}$

(c) pH = 7.38:
$[\mathrm{H^+}] = 10^{-7.38} = 4.17 \times 10^{-8}\,\mathrm{M}$

(d) pH = 6.4:
$[\mathrm{H^+}] = 10^{-6.4} = 3.98 \times 10^{-7}\,\mathrm{M}$

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6.56The 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.

Formula: $[\mathrm{H^+}] = 10^{-\mathrm{pH}}$

Fluid	pH	$[\mathrm{H^+}]\,\mathrm{(M)}$
Milk	6.8	$10^{-6.8} = 1.58 \times 10^{-7}$
Black coffee	5.0	$10^{-5.0} = 1.0 \times 10^{-5}$
Tomato juice	4.2	$10^{-4.2} = 6.31 \times 10^{-5}$
Lemon juice	2.2	$10^{-2.2} = 6.31 \times 10^{-3}$
Egg white	7.8	$10^{-7.8} = 1.58 \times 10^{-8}$

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6.57If

0.561\,\mathrm{g}

of KOH is dissolved in water to give

200\,\mathrm{mL}

of solution at

298\,\mathrm{K}

. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Given: Mass of KOH $= 0.561\,\mathrm{g}$ ; Volume $= 200\,\mathrm{mL} = 0.2\,\mathrm{L}$ ; Molar mass of KOH $= 56.1\,\mathrm{g/mol}$

Moles of KOH:
$n = \frac{0.561}{56.1} = 0.01\,\mathrm{mol}$

Concentration of KOH:
$[\mathrm{KOH}] = \frac{0.01}{0.2} = 0.05\,\mathrm{M}$

KOH is a strong base, completely dissociating:
$[\mathrm{K^+}] = 0.05\,\mathrm{M}$
$[\mathrm{OH^-}] = 0.05\,\mathrm{M}$

$[\mathrm{H^+}]$ :
$[\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]} = \frac{1.0 \times 10^{-14}}{0.05} = 2.0 \times 10^{-13}\,\mathrm{M}$

pH:
$\mathrm{pOH} = -\log(0.05) = 1.301$
$\boxed{\mathrm{pH} = 14 - 1.301 = 12.70}$

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6.58The solubility of

\mathrm{Sr(OH)_2}

298\,\mathrm{K}

19.23\,\mathrm{g/L}

of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Given: Solubility $= 19.23\,\mathrm{g/L}$ ; Molar mass of $\mathrm{Sr(OH)_2} = 87.6 + 34 = 121.6\,\mathrm{g/mol}$

Molar solubility:
$s = \frac{19.23}{121.6} = 0.1581\,\mathrm{mol/L}$

Dissociation: $\mathrm{Sr(OH)_2} \rightarrow \mathrm{Sr^{2+}} + 2\mathrm{OH^-}$

$[\mathrm{Sr^{2+}}] = s = \boxed{0.158\,\mathrm{M}}$
$[\mathrm{OH^-}] = 2s = 2 \times 0.1581 = \boxed{0.316\,\mathrm{M}}$

pH:
$\mathrm{pOH} = -\log(0.316) = 0.500$
$\boxed{\mathrm{pH} = 14 - 0.500 = 13.50}$

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6.59The ionization constant of propanoic acid is

1.32 \times 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?

Given: $K_a = 1.32 \times 10^{-5}$ , $C = 0.05\,\mathrm{M}$

Degree of ionization in pure 0.05 M solution:
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} = 1.625 \times 10^{-2}$
$\boxed{\alpha = 1.63 \times 10^{-2} \approx 1.63\%}$

pH:
$[\mathrm{H^+}] = C\alpha = 0.05 \times 1.625 \times 10^{-2} = 8.125 \times 10^{-4}\,\mathrm{M}$
$\mathrm{pH} = -\log(8.125 \times 10^{-4}) = 4 - \log 8.125 = 4 - 0.910 = \boxed{3.09}$

Degree of ionization in presence of 0.01 M HCl:

With common ion $[\mathrm{H^+}] \approx 0.01\,\mathrm{M}$ :
$\alpha' = \frac{K_a}{[\mathrm{H^+}]} = \frac{1.32 \times 10^{-5}}{0.01} = \boxed{1.32 \times 10^{-3}}$

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6.60The 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.

Given: $C = 0.1\,\mathrm{M}$ , $\mathrm{pH} = 2.34$

$[\mathrm{H^+}]$ :
$[\mathrm{H^+}] = 10^{-2.34} = 4.57 \times 10^{-3}\,\mathrm{M}$

Degree of ionization:
$\alpha = \frac{[\mathrm{H^+}]}{C} = \frac{4.57 \times 10^{-3}}{0.1} = \boxed{4.57 \times 10^{-2} \approx 4.57\%}$

$K_a$ :
$[\mathrm{HCNO}] = C - [\mathrm{H^+}] = 0.1 - 4.57 \times 10^{-3} = 0.09543\,\mathrm{M}$
$K_a = \frac{[\mathrm{H^+}][\mathrm{CNO^-}]}{[\mathrm{HCNO}]} = \frac{(4.57 \times 10^{-3})^2}{0.09543}$
$= \frac{2.088 \times 10^{-5}}{0.09543} = \boxed{2.19 \times 10^{-4}}$

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6.61The ionization constant of nitrous acid is

4.5 \times 10^{-4}

. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Given: $K_a(\mathrm{HNO_2}) = 4.5 \times 10^{-4}$ , $C = 0.04\,\mathrm{M}$ sodium nitrite ( $\mathrm{NaNO_2}$ )

$\mathrm{NaNO_2}$ is a salt of strong base (NaOH) and weak acid ( $\mathrm{HNO_2}$ ). The $\mathrm{NO_2^-}$ ion undergoes hydrolysis:
$\mathrm{NO_2^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{HNO_2} + \mathrm{OH^-}$

$K_b$ for $\mathrm{NO_2^-}$ :
$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11}$

$[\mathrm{OH^-}]$ :
$[\mathrm{OH^-}] = \sqrt{K_b \times C} = \sqrt{2.22 \times 10^{-11} \times 0.04} = \sqrt{8.89 \times 10^{-13}}$
$= 9.43 \times 10^{-7}\,\mathrm{M}$

pH:
$\mathrm{pOH} = -\log(9.43 \times 10^{-7}) = 6.025$
$\boxed{\mathrm{pH} = 14 - 6.025 = 7.97}$

Degree of hydrolysis $h$ :
$h = \frac{[\mathrm{OH^-}]}{C} = \frac{9.43 \times 10^{-7}}{0.04} = \boxed{2.36 \times 10^{-5}}$

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6.62A 0.02 M solution of pyridinium hydrochloride has

\mathrm{pH} = 3.44

. Calculate the ionization constant of pyridine.

Given: $C = 0.02\,\mathrm{M}$ , $\mathrm{pH} = 3.44$

Pyridinium hydrochloride is a salt of weak base (pyridine) and strong acid (HCl). The pyridinium ion $\mathrm{C_5H_5NH^+}$ hydrolyses:
$\mathrm{C_5H_5NH^+} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_5H_5N} + \mathrm{H_3O^+}$

$[\mathrm{H^+}]$ :
$[\mathrm{H^+}] = 10^{-3.44} = 3.63 \times 10^{-4}\,\mathrm{M}$

$K_a$ of pyridinium ion:
$K_a = \frac{[\mathrm{H^+}]^2}{C - [\mathrm{H^+}]} = \frac{(3.63 \times 10^{-4})^2}{0.02 - 3.63 \times 10^{-4}} = \frac{1.318 \times 10^{-7}}{0.01964} = 6.71 \times 10^{-6}$

$K_b$ of pyridine:
$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.71 \times 10^{-6}} = \boxed{1.49 \times 10^{-9}}$

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6.63Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN,

\mathrm{NH_4NO_3}

\mathrm{NaNO_2}

and KF

Concept:

Salt of strong acid + strong base → neutral
Salt of strong acid + weak base → acidic
Salt of weak acid + strong base → basic

Salt	Acid	Base	Nature of Solution
NaCl	HCl (strong)	NaOH (strong)	Neutral
KBr	HBr (strong)	KOH (strong)	Neutral
NaCN	HCN (weak)	NaOH (strong)	Basic
$\mathrm{NH_4NO_3}$	$\mathrm{HNO_3}$ (strong)	$\mathrm{NH_4OH}$ (weak)	Acidic
$\mathrm{NaNO_2}$	$\mathrm{HNO_2}$ (weak)	NaOH (strong)	Basic
KF	HF (weak)	KOH (strong)	Basic

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6.64The ionization constant of chloroacetic acid is

1.35 \times 10^{-3}

. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Given: $K_a = 1.35 \times 10^{-3}$ , $C = 0.1\,\mathrm{M}$

pH of 0.1 M chloroacetic acid:

Let $x = [\mathrm{H^+}]$ :
$K_a = \frac{x^2}{C - x}$
$1.35 \times 10^{-3} = \frac{x^2}{0.1 - x}$
$x^2 + 1.35 \times 10^{-3}x - 1.35 \times 10^{-4} = 0$
$x = \frac{-1.35 \times 10^{-3} + \sqrt{(1.35 \times 10^{-3})^2 + 4 \times 1.35 \times 10^{-4}}}{2}$
$= \frac{-1.35 \times 10^{-3} + \sqrt{1.823 \times 10^{-6} + 5.4 \times 10^{-4}}}{2}$
$= \frac{-1.35 \times 10^{-3} + \sqrt{5.418 \times 10^{-4}}}{2} = \frac{-1.35 \times 10^{-3} + 2.328 \times 10^{-2}}{2}$
$= \frac{2.193 \times 10^{-2}}{2} = 1.097 \times 10^{-2}\,\mathrm{M}$
$\mathrm{pH} = -\log(1.097 \times 10^{-2}) = 2 - \log 1.097 = 2 - 0.040 = \boxed{1.96}$

pH of 0.1 M sodium chloroacetate (salt of weak acid + strong base):

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.35 \times 10^{-3}} = 7.41 \times 10^{-12}$

$[\mathrm{OH^-}] = \sqrt{K_b \times C} = \sqrt{7.41 \times 10^{-12} \times 0.1} = \sqrt{7.41 \times 10^{-13}} = 8.61 \times 10^{-7}\,\mathrm{M}$

$\mathrm{pOH} = -\log(8.61 \times 10^{-7}) = 6.065$
$\boxed{\mathrm{pH} = 14 - 6.065 = 7.94}$

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6.65Ionic product of water at

310\,\mathrm{K}

2.7 \times 10^{-14}

. What is the pH of neutral water at this temperature?

Given: $K_w = 2.7 \times 10^{-14}$ at $310\,\mathrm{K}$

For neutral water: $[\mathrm{H^+}] = [\mathrm{OH^-}]$

$[\mathrm{H^+}]^2 = K_w = 2.7 \times 10^{-14}$
$[\mathrm{H^+}] = \sqrt{2.7 \times 10^{-14}} = 1.643 \times 10^{-7}\,\mathrm{M}$

$\mathrm{pH} = -\log(1.643 \times 10^{-7}) = 7 - \log 1.643 = 7 - 0.2156$
$\boxed{\mathrm{pH} = 6.78}$

Note: At $310\,\mathrm{K}$ (body temperature), neutral pH is 6.78, which is slightly less than 7.

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6.66Calculate the pH of the resultant mixtures:
a)

10\,\mathrm{mL}

0.2\,\mathrm{M}\,\mathrm{Ca(OH)_2} + 25\,\mathrm{mL}

0.1\,\mathrm{M}

HCl
b)

10\,\mathrm{mL}

0.01\,\mathrm{M}\,\mathrm{H_2SO_4} + 10\,\mathrm{mL}

0.01\,\mathrm{M}\,\mathrm{Ca(OH)_2}

10\,\mathrm{mL}

0.1\,\mathrm{M}\,\mathrm{H_2SO_4} + 10\,\mathrm{mL}

0.1\,\mathrm{M}

KOH

(a) $10\,\mathrm{mL}$ of $0.2\,\mathrm{M}\,\mathrm{Ca(OH)_2} + 25\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ HCl:

Millimoles of $\mathrm{OH^-}$ from $\mathrm{Ca(OH)_2}$ : $10 \times 0.2 \times 2 = 4\,\mathrm{mmol}$
Millimoles of $\mathrm{H^+}$ from HCl: $25 \times 0.1 = 2.5\,\mathrm{mmol}$

Excess $\mathrm{OH^-} = 4 - 2.5 = 1.5\,\mathrm{mmol}$
Total volume $= 35\,\mathrm{mL}$
$[\mathrm{OH^-}] = \frac{1.5}{35} = 0.04286\,\mathrm{M}$
$\mathrm{pOH} = -\log(0.04286) = 1.368$
$\boxed{\mathrm{pH} = 14 - 1.368 = 12.63}$

(b) $10\,\mathrm{mL}$ of $0.01\,\mathrm{M}\,\mathrm{H_2SO_4} + 10\,\mathrm{mL}$ of $0.01\,\mathrm{M}\,\mathrm{Ca(OH)_2}$ :

Millimoles of $\mathrm{H^+}$ from $\mathrm{H_2SO_4}$ : $10 \times 0.01 \times 2 = 0.2\,\mathrm{mmol}$
Millimoles of $\mathrm{OH^-}$ from $\mathrm{Ca(OH)_2}$ : $10 \times 0.01 \times 2 = 0.2\,\mathrm{mmol}$

Exact neutralization. The solution is neutral.
$\boxed{\mathrm{pH} = 7}$

(c) $10\,\mathrm{mL}$ of $0.1\,\mathrm{M}\,\mathrm{H_2SO_4} + 10\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ KOH:

Millimoles of $\mathrm{H^+}$ from $\mathrm{H_2SO_4}$ : $10 \times 0.1 \times 2 = 2\,\mathrm{mmol}$
Millimoles of $\mathrm{OH^-}$ from KOH: $10 \times 0.1 = 1\,\mathrm{mmol}$

Excess $\mathrm{H^+} = 2 - 1 = 1\,\mathrm{mmol}$
Total volume $= 20\,\mathrm{mL}$
$[\mathrm{H^+}] = \frac{1}{20} = 0.05\,\mathrm{M}$
$\mathrm{pH} = -\log(0.05) = 1.301$
$\boxed{\mathrm{pH} = 1.30}$

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6.67Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.

Using $K_{sp}$ values from NCERT Table 6.9:

$K_{sp}(\mathrm{Ag_2CrO_4}) = 1.1 \times 10^{-12}$
$K_{sp}(\mathrm{BaCrO_4}) = 1.2 \times 10^{-10}$
$K_{sp}(\mathrm{Fe(OH)_3}) = 1.0 \times 10^{-38}$
$K_{sp}(\mathrm{PbCl_2}) = 1.6 \times 10^{-5}$
$K_{sp}(\mathrm{Hg_2I_2}) = 4.5 \times 10^{-29}$

(i) Silver chromate $\mathrm{Ag_2CrO_4}$ :
$\mathrm{Ag_2CrO_4} \rightleftharpoons 2\mathrm{Ag^+} + \mathrm{CrO_4^{2-}}$
$K_{sp} = (2s)^2(s) = 4s^3$
$s = \left(\frac{1.1 \times 10^{-12}}{4}\right)^{1/3} = (2.75 \times 10^{-13})^{1/3} = 6.52 \times 10^{-5}\,\mathrm{M}$
$[\mathrm{Ag^+}] = 2s = 1.30 \times 10^{-4}\,\mathrm{M};\quad [\mathrm{CrO_4^{2-}}] = 6.52 \times 10^{-5}\,\mathrm{M}$

(ii) Barium chromate $\mathrm{BaCrO_4}$ :
$\mathrm{BaCrO_4} \rightleftharpoons \mathrm{Ba^{2+}} + \mathrm{CrO_4^{2-}}$
$K_{sp} = s^2$
$s = \sqrt{1.2 \times 10^{-10}} = 1.095 \times 10^{-5}\,\mathrm{M}$
$[\mathrm{Ba^{2+}}] = [\mathrm{CrO_4^{2-}}] = 1.10 \times 10^{-5}\,\mathrm{M}$

(iii) Ferric hydroxide $\mathrm{Fe(OH)_3}$ :
$\mathrm{Fe(OH)_3} \rightleftharpoons \mathrm{Fe^{3+}} + 3\mathrm{OH^-}$
$K_{sp} = s(3s)^3 = 27s^4$
$s = \left(\frac{1.0 \times 10^{-38}}{27}\right)^{1/4} = (3.70 \times 10^{-40})^{1/4} = 1.39 \times 10^{-10}\,\mathrm{M}$
$[\mathrm{Fe^{3+}}] = 1.39 \times 10^{-10}\,\mathrm{M};\quad [\mathrm{OH^-}] = 3s = 4.17 \times 10^{-10}\,\mathrm{M}$

(iv) Lead chloride $\mathrm{PbCl_2}$ :
$\mathrm{PbCl_2} \rightleftharpoons \mathrm{Pb^{2+}} + 2\mathrm{Cl^-}$
$K_{sp} = s(2s)^2 = 4s^3$
$s = \left(\frac{1.6 \times 10^{-5}}{4}\right)^{1/3} = (4.0 \times 10^{-6})^{1/3} = 1.587 \times 10^{-2}\,\mathrm{M}$
$[\mathrm{Pb^{2+}}] = 1.59 \times 10^{-2}\,\mathrm{M};\quad [\mathrm{Cl^-}] = 2s = 3.17 \times 10^{-2}\,\mathrm{M}$

(v) Mercurous iodide $\mathrm{Hg_2I_2}$ :
$\mathrm{Hg_2I_2} \rightleftharpoons \mathrm{Hg_2^{2+}} + 2\mathrm{I^-}$
$K_{sp} = s(2s)^2 = 4s^3$
$s = \left(\frac{4.5 \times 10^{-29}}{4}\right)^{1/3} = (1.125 \times 10^{-29})^{1/3} = 2.24 \times 10^{-10}\,\mathrm{M}$
$[\mathrm{Hg_2^{2+}}] = 2.24 \times 10^{-10}\,\mathrm{M};\quad [\mathrm{I^-}] = 2s = 4.48 \times 10^{-10}\,\mathrm{M}$

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6.68The solubility product constants of

\mathrm{Ag_2CrO_4}

and AgBr are

1.1 \times 10^{-12}

and

5.0 \times 10^{-13}

respectively. Calculate the ratio of the molarities of their saturated solutions.

For $\mathrm{Ag_2CrO_4}$ : $\mathrm{Ag_2CrO_4} \rightleftharpoons 2\mathrm{Ag^+} + \mathrm{CrO_4^{2-}}$
$K_{sp} = 4s_1^3 = 1.1 \times 10^{-12}$
$s_1 = \left(\frac{1.1 \times 10^{-12}}{4}\right)^{1/3} = (2.75 \times 10^{-13})^{1/3} = 6.52 \times 10^{-5}\,\mathrm{M}$

For AgBr: $\mathrm{AgBr} \rightleftharpoons \mathrm{Ag^+} + \mathrm{Br^-}$
$K_{sp} = s_2^2 = 5.0 \times 10^{-13}$
$s_2 = \sqrt{5.0 \times 10^{-13}} = 7.07 \times 10^{-7}\,\mathrm{M}$

Ratio:
$\frac{s_1}{s_2} = \frac{6.52 \times 10^{-5}}{7.07 \times 10^{-7}} = \boxed{92.2}$

The molar solubility of $\mathrm{Ag_2CrO_4}$ is about 92 times that of AgBr.

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6.69Equal 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 \times 10^{-8}

Given: $[\mathrm{NaIO_3}] = 0.002\,\mathrm{M}$ , $[\mathrm{Cu(ClO_3)_2}] = 0.002\,\mathrm{M}$ ; equal volumes mixed.

When equal volumes are mixed, concentrations are halved:
$[\mathrm{IO_3^-}] = \frac{0.002}{2} = 0.001\,\mathrm{M} = 1.0 \times 10^{-3}\,\mathrm{M}$
$[\mathrm{Cu^{2+}}] = \frac{0.002}{2} = 0.001\,\mathrm{M} = 1.0 \times 10^{-3}\,\mathrm{M}$

Ionic product $Q_{sp}$ for $\mathrm{Cu(IO_3)_2}$ :
$\mathrm{Cu(IO_3)_2} \rightleftharpoons \mathrm{Cu^{2+}} + 2\mathrm{IO_3^-}$
$Q_{sp} = [\mathrm{Cu^{2+}}][\mathrm{IO_3^-}]^2 = (1.0 \times 10^{-3})(1.0 \times 10^{-3})^2 = 1.0 \times 10^{-9}$

Comparison:
$Q_{sp} = 1.0 \times 10^{-9} < K_{sp} = 7.4 \times 10^{-8}$

Since $Q_{sp} < K_{sp}$ , no precipitation of copper iodate will occur.

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6.70The ionization constant of benzoic acid is

6.46 \times 10^{-5}

and

K_{sp}

for silver benzoate is

2.5 \times 10^{-13}

. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Given: $K_a(\mathrm{C_6H_5COOH}) = 6.46 \times 10^{-5}$ ; $K_{sp}(\mathrm{AgC_6H_5COO}) = 2.5 \times 10^{-13}$ ; buffer pH $= 3.19$

Solubility in pure water:
$\mathrm{AgC_6H_5COO} \rightleftharpoons \mathrm{Ag^+} + \mathrm{C_6H_5COO^-}$
$K_{sp} = s_0^2 \Rightarrow s_0 = \sqrt{2.5 \times 10^{-13}} = 5.0 \times 10^{-7}\,\mathrm{M}$

Solubility in buffer (pH = 3.19):
$[\mathrm{H^+}] = 10^{-3.19} = 6.46 \times 10^{-4}\,\mathrm{M}$

In the buffer, benzoate ion is partially converted to benzoic acid:
$[\mathrm{C_6H_5COO^-}] = \frac{K_a}{[\mathrm{H^+}] + K_a} \times [\text{total benzoate}]$

Let $s$ = solubility in buffer. Total benzoate $= s$ .
$[\mathrm{C_6H_5COO^-}] = \frac{K_a \cdot s}{[\mathrm{H^+}] + K_a}$

$K_{sp} = [\mathrm{Ag^+}][\mathrm{C_6H_5COO^-}] = s \times \frac{K_a \cdot s}{[\mathrm{H^+}] + K_a}$
$2.5 \times 10^{-13} = \frac{s^2 \times 6.46 \times 10^{-5}}{6.46 \times 10^{-4} + 6.46 \times 10^{-5}} = \frac{s^2 \times 6.46 \times 10^{-5}}{7.106 \times 10^{-4}}$
$s^2 = \frac{2.5 \times 10^{-13} \times 7.106 \times 10^{-4}}{6.46 \times 10^{-5}} = \frac{1.777 \times 10^{-16}}{6.46 \times 10^{-5}} = 2.75 \times 10^{-12}$
$s = 1.658 \times 10^{-6}\,\mathrm{M}$

Ratio:
$\frac{s}{s_0} = \frac{1.658 \times 10^{-6}}{5.0 \times 10^{-7}} \approx \boxed{3.32}$

Silver benzoate is approximately 3.32 times more soluble in the buffer of pH 3.19 than in pure water.

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6.71What 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 \times 10^{-18}

Given: $K_{sp}(\mathrm{FeS}) = 6.3 \times 10^{-18}$ ; equimolar solutions mixed in equal volumes.

Let concentration of each solution $= C\,\mathrm{M}$ .

When equal volumes are mixed:
$[\mathrm{Fe^{2+}}] = \frac{C}{2},\quad [\mathrm{S^{2-}}] = \frac{C}{2}$

For no precipitation: $Q_{sp} \leq K_{sp}$
$[\mathrm{Fe^{2+}}][\mathrm{S^{2-}}] \leq K_{sp}$
$\frac{C}{2} \times \frac{C}{2} \leq 6.3 \times 10^{-18}$
$\frac{C^2}{4} \leq 6.3 \times 10^{-18}$
$C^2 \leq 4 \times 6.3 \times 10^{-18} = 2.52 \times 10^{-17}$
$C \leq \sqrt{2.52 \times 10^{-17}} = 1.587 \times 10^{-8.5} \approx 1.587 \times 10^{-8.5}$

Let me recalculate:
$C \leq \sqrt{2.52 \times 10^{-17}} = \sqrt{25.2 \times 10^{-18}} = 5.02 \times 10^{-9}\,\mathrm{M}$

$\boxed{C_{\max} = 5.02 \times 10^{-9}\,\mathrm{M}}$

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6.72What is the minimum volume of water required to dissolve

1\,\mathrm{g}

of calcium sulphate at

298\,\mathrm{K}

? (For calcium sulphate,

K_{sp}

9.1 \times 10^{-6}

Given: $K_{sp}(\mathrm{CaSO_4}) = 9.1 \times 10^{-6}$ ; mass $= 1\,\mathrm{g}$ ; Molar mass of $\mathrm{CaSO_4} = 136\,\mathrm{g/mol}$

Molar solubility:
$\mathrm{CaSO_4} \rightleftharpoons \mathrm{Ca^{2+}} + \mathrm{SO_4^{2-}}$
$K_{sp} = s^2 \Rightarrow s = \sqrt{9.1 \times 10^{-6}} = 3.017 \times 10^{-3}\,\mathrm{mol/L}$

Moles of $\mathrm{CaSO_4}$ in 1 g:
$n = \frac{1}{136} = 7.35 \times 10^{-3}\,\mathrm{mol}$

Minimum volume of water:
$V = \frac{n}{s} = \frac{7.35 \times 10^{-3}}{3.017 \times 10^{-3}} = 2.44\,\mathrm{L}$

$\boxed{V_{\min} = 2.44\,\mathrm{L}}$

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6.73The concentration of sulphide ion in

0.1\,\mathrm{M}

HCl solution saturated with hydrogen sulphide is

1.0 \times 10^{-19}\,\mathrm{M}

. If

10\,\mathrm{mL}

of this is added to

5\,\mathrm{mL}

0.04\,\mathrm{M}

solution of the following:

\mathrm{FeSO_4}

\mathrm{MnCl_2}

\mathrm{ZnCl_2}

and

\mathrm{CdCl_2}

in which of these solutions precipitation will take place?

Given: $[\mathrm{S^{2-}}]$ in HCl solution $= 1.0 \times 10^{-19}\,\mathrm{M}$ ; metal ion concentration $= 0.04\,\mathrm{M}$

When $10\,\mathrm{mL}$ of $\mathrm{S^{2-}}$ solution is mixed with $5\,\mathrm{mL}$ of metal ion solution:

$[\mathrm{S^{2-}}] = \frac{1.0 \times 10^{-19} \times 10}{15} = 6.67 \times 10^{-20}\,\mathrm{M}$
$[\mathrm{M^{2+}}] = \frac{0.04 \times 5}{15} = 1.33 \times 10^{-2}\,\mathrm{M}$

Ionic product $Q_{sp}$ for each sulphide:
$Q_{sp} = [\mathrm{M^{2+}}][\mathrm{S^{2-}}] = 1.33 \times 10^{-2} \times 6.67 \times 10^{-20} = 8.87 \times 10^{-22}$

$K_{sp}$ values (from NCERT Table 6.9):

$K_{sp}(\mathrm{FeS}) = 6.3 \times 10^{-18}$
$K_{sp}(\mathrm{MnS}) = 2.5 \times 10^{-13}$
$K_{sp}(\mathrm{ZnS}) = 1.6 \times 10^{-24}$
$K_{sp}(\mathrm{CdS}) = 8.0 \times 10^{-27}$

Comparison:

Salt	$K_{sp}$	$Q_{sp}$	Precipitation?
FeS	$6.3 \times 10^{-18}$	$8.87 \times 10^{-22}$	No ( $Q < K_{sp}$ )
MnS	$2.5 \times 10^{-13}$	$8.87 \times 10^{-22}$	No ( $Q < K_{sp}$ )
ZnS	$1.6 \times 10^{-24}$	$8.87 \times 10^{-22}$	Yes ( $Q > K_{sp}$ )
CdS	$8.0 \times 10^{-27}$	$8.87 \times 10^{-22}$	Yes ( $Q > K_{sp}$ )

Precipitation will take place in $\mathrm{ZnCl_2}$ and $\mathrm{CdCl_2}$ solutions.

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Key topics in Equilibrium include Nature of Equilibrium, Law of Chemical Equilibrium and Equilibrium Constant, Relation Between Kp and Kc, Extent and Direction of Reaction. Study these first, then practise questions on each for Class 11 exams.

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