NCERT Solutions for Class 11 Chemistry Chapter 7 - Equilibrium

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

1. What is the initial effect of the change on vapour pressure?

Ans: The vapour pressure would first fall if the container's volume was rapidly increased. This is due to the fact that the amount of vapour remains constant while the volume increases rapidly. As a result, the same amount of vapour is spread over a wider surface area.

2. How do rates of evaporation and condensation change initially? 

Ans: The rate of evaporation is also constant because the temperature is constant. The density of the vapour phase reduces as the container volume increases. As a result, the rate of vapour particle collisions drops as well. As a result, the rate of condensation initially slows.

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

Ans: The rate of evaporation equals the rate of condensation when equilibrium is eventually restored. Only the volume changes in this situation, but the temperature remains fixed. Temperature, not volume, determines vapour pressure. As a result, the final vapour pressure will be the same as the system's initial vapour pressure.

2. What is ${{\text{K}}_{\text{c}}}$  for the following equilibrium when the equilibrium concentration of each substance is: $[{\text{S}}{{\text{O}}_{\text{2}}}] = 0.60{\text{M}}$ , $[{{\text{O}}_{\text{2}}}] = 0.82{\text{M}}$ and $[{\text{S}}{{\text{O}}_3}] = 1.90{\text{M}}$ ? $2{\text{S}}{{\text{O}}_{\text{2}}}{\text{(g}}) + {{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$ 

Ans: The equilibrium constant for the given reaction will be:

${{{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {{\text{S}}{{\text{O}}_{\text{3}}}} \right]}^2}}}{{{{\left[ {{\text{S}}{{\text{O}}_{\text{2}}}} \right]}^2}\left[ {{{\text{O}}_2}} \right]}}} $

${ = \dfrac{{{{(1.90)}^2}{{\text{M}}^2}}}{{{{(0.60)}^2}(0.821){{\text{M}}^3}}}{\text{}}}$

${{\text{K}}_{\text{c}}} = 12.239{{\text{M}}^{ - 1}}$ (approximately)

Therefore, the equilibrium constant for the given reaction is $12.239{{\text{M}}^{ - 1}}$.

3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of iodine atoms. ${{\text{I}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{I (g)}}$ . Calculate ${{\text{K}}_{\text{P}}}$ for the equilibrium.

Ans: Partial pressure for ${\text{I}}$  atoms:

${{{\text{p}}_{\text{1}}} = \dfrac{{40}}{{100}} \times {{\text{p}}_{{\text{total}}}}}$ 

= $\dfrac{40}{100} \times 10^5$ 

= $4 \times 10^4 Pa$ 

Partial pressure for ${{\text{I}}_{\text{2}}}$ molecules:

${{\text{p}}_{{{\text{l}}_{\text{2}}}}} = \dfrac{{60}}{{100}} \times {{\text{p}}_{{\text{total}}}}$

 $ = \dfrac{{60}}{{100}} \times {10^5}$

 $ = 6 \times {10^4}{\text{Pa}}$ 

Now for, the given reaction:

${{\text{K}}_{\text{p}}} = \dfrac{{{{({\text{pI}})}^2}}}{{{{\text{p}}_{{{\text{1}}_{\text{2}}}}}}}$

 $ = \dfrac{{\left( {4 \times {{10}^4}} \right){\text{P}}{{\text{a}}^2}}}{{6 \times {{10}^4}{\text{Pa}}}}$

$ = 2.67 \times {10^4}{\text{Pa}}$ 

4. Write the expression for the equilibrium constant, ${{\text{K}}_{\text{c}}}$  for each of the following reactions:

1. $2{\text{NOCl (g}}) \rightleftharpoons 2{\text{NO (g}}) + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$ 

Ans: The expression for the equilibrium constant is:

${{\text{K}}_{\text{c}}}=\dfrac{{{{[{\text{NO}}]}^2}{\text{[C}}{{\text{l}}_2}]}}{{{{[{\text{NOCl}}]}^2}}}$ 

2. $2{\text{Cu(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{(s}}) \rightleftharpoons 2{\text{CuO (s)}} + 4{\text{N}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g)}}$ 

Ans: The expression is:

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{\text{CuO}}]}^2}{{[{\text{N}}{{\text{O}}_2}]}^4}[{{\text{O}}_{\text{2}}}]}}{{{{[{\text{Cu(N}}{{\text{O}}_3}{)_2}]}^2}}}$ 

${{\text{K}}_{\text{c}}} = {[{\text{N}}{{\text{O}}_2}]^4}[{{\text{O}}_{\text{2}}}]$ 

3. ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{(aq}}) + {{\text{H}}_{\text{2}}}{\text{O (l)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH (aq)}} + {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH (aq)}}$ 

Ans: The expression will be:

${{\text{K}}_{\text{c}}} = \dfrac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{COOH] [}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH]}}}}{{[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{] [}}{{\text{H}}_{\text{2}}}{\text{O]}}}}$ 

${K_c} = \dfrac{{[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH] [}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH]}}}}{{[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}]}}$ 

4. ${\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{(aq}}) + 3{\text{O}}{{\text{H}}^ - }({\text{aq)}} \rightleftharpoons {\text{Fe(OH}}{{\text{)}}_{\text{3}}}{\text{(s)}}$ 

Ans: The expression will be:

${{\text{K}}_{\text{c}}} = \dfrac{{[{\text{Fe(OH}}{{\text{)}}_3}]}}{{[{\text{F}}{{\text{e}}^{3 + }}]{{[{\text{O}}{{\text{H}}^ - }]}^3}}}$ 

${{\text{K}}_{\text{c}}} = \dfrac{1}{{[{\text{F}}{{\text{e}}^{3 + }}]{{[{\text{O}}{{\text{H}}^ - }]}^3}}}$ 

5. ${{\text{I}}_{\text{2}}}{\text{(s)}} + 5{{\text{F}}_2} \rightleftharpoons 2{\text{I}}{{\text{F}}_5}$ 

Ans: The expression will be:

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{\text{I}}{{\text{F}}_5}]}^2}}}{{[{{\text{I}}_2}]{{[{{\text{F}}_2}]}^2}}}$ 

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{\text{I}}{{\text{F}}_5}]}^2}}}{{{{[{{\text{F}}_2}]}^2}}}$ 

5. Find out the value of ${{\text{K}}_{\text{c}}}$  for each of the following equilibria from the value of ${{\text{K}}_{\text{p}}}$ :

1. $2{\text{NOCl (g}}) \rightleftharpoons 2{\text{NO (g}}) + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}})$ ; ${K_p} = 1.8 \times {10^{ - 2}}$ at 500K

Ans: The relationship between ${{\text{K}}_{\text{p}}}$ and ${{\text{K}}_{\text{c}}}$ is given as:

${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{({\text{RT}})^{{{\Delta n}}}}$ 

$\Delta {\text{n}} = 3 - 2 = 1$ 

${\text{R}} = 0.0831{\text{ barLmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ 

T = 500K

${K_p} = 1.8 \times {10^{ - 2}}$

Now, ${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{({\text{RT}})^{{{\Delta n}}}}$

$1.8 \times {10^{ - 2}} = {{\text{K}}_{\text{c}}}{(0.0831 \times 500)^1}$ 

${{\text{K}}_{\text{c}}} = \dfrac{{1.8 \times {{10}^{ - 2}}}}{{0.0831 \times 500}}$

${{\text{K}}_{\text{c}}} = 4.33 \times {10^{ - 4}}$ (Approximately)

2. ${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s}}) \rightleftharpoons {\text{CaO (s)}} + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}}$ ; ${K_p} = 167$ at 1073K

Ans: Here,

$\Delta {\text{n}} = 2 - 1 = 1$ 

${\text{R}} = 0.0831{\text{ barLmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$

T = 1073K

${K_p} = 167$

${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{({\text{RT}})^{{{\Delta n}}}}$

$167 = {{\text{K}}_{\text{c}}}{(0.0831 \times 1073)^1}$ 

${{\text{K}}_{\text{c}}} = \dfrac{{167}}{{0.0831 \times 1073}}$ 

${{\text{K}}_{\text{c}}} = 1.87$ (Approximately)

6. For the following equilibrium, ${{\text{K}}_{\text{c}}} = 6.3 \times {10^{14}}$ at 1000K ${\text{NO (g)}} + {{\text{O}}_{\text{3}}}{\text{(g}}) \rightleftharpoons {\text{N}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g)}}$ Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is ${{\text{K}}_{\text{c}}}$  for the reverse reaction?

Ans: According to the question:

The ${{\text{K}}_{\text{c}}}$ for the forward reaction is $6.3 \times {10^{14}}$

Then, ${{\text{K}}_{\text{c}}}$for the reverse reaction will be:

${{\text{K}}^{\text{'}}}_{\text{c}} = \dfrac{1}{{{{\text{K}}_{\text{c}}}}}$ 

${{\text{K}}^{\text{'}}}_{\text{c}} = \dfrac{1}{{6.3 \times {{10}^{14}}}}$

${{\text{K}}^{\text{'}}}_{\text{c}} = 1.59 \times {10^{ - 15}}$

The equilibrium constant for the reverse reaction will be: $1.59 \times {10^{ - 15}}$.

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

Ans: For a pure substance which is both a solid and a liquid:

${\text{[Pure substance}}] = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume}}}}$ 

${\text{[Pure substance}}] = \dfrac{{{\text{Mass / Molecular weight}}}}{{{\text{Volume}}}}$

${\text{[Pure substance}}] = \dfrac{{{\text{Mass}}}}{{{\text{Volume }} \times {\text{ Molecular mass}}}}$

${\text{[Pure substance}}] = \dfrac{{{\text{Density}}}}{{{\text{Molecular mass}}}}$

The molecular mass and density of a pure substance (at a given temperature) are now always fixed and accounted for in the equilibrium constant. As a result, the equilibrium constant statement does not include the values of pure substances.

8. Reaction between nitrogen and oxygen takes place as follows: $2{{\text{N}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{{\text{N}}_{\text{2}}}{\text{O (g)}}$ If a mixture of 0.482 mol of ${{\text{N}}_{\text{2}}}$  and 0.933 mol of ${{\text{O}}_{\text{2}}}$  is placed in a 10 L reaction vessel and allowed to form ${{\text{N}}_{\text{2}}}{\text{O}}$  at a temperature for which ${{\text{K}}_{\text{c}}} = 2.0 \times {10^{ - 37}}$ , determine the composition of equilibrium mixture.

Ans: Let the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ at    equilibrium be x.

The initial concentration of ${{\text{N}}_{\text{2}}}$ is 0.482mol.

The initial concentration of ${{\text{O}}_{\text{2}}}$ is 0.933mol.

At equilibrium the concentration of ${{\text{N}}_{\text{2}}}$ is (0.482-x)mol, the concentration of ${{\text{O}}_{\text{2}}}$is (0.933-x)mol and the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ is x mol.

The reaction occurs in a 10L reaction vessel:

$[{{\text{N}}_{\text{2}}}] = \dfrac{{0.482 - x}}{{10}}$ , $[{O_2}] = \dfrac{{0.933 - x/2}}{{10}}$ , $[{{\text{N}}_{\text{2}}}{\text{O}}] = \dfrac{x}{{10}}$ 

The value of equilibrium constant i.e. ${{\text{K}}_{\text{c}}} = 2.0 \times {10^{ - 37}}$  is very small. Therefore, the amount of nitrogen and oxygen reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of nitrogen and oxygen. Then,

$[{{\text{N}}_{\text{2}}}] = \dfrac{{0.482}}{{10}} = 0.048{\text{2 mol}}{{\text{L}}^{ - 1}}$ and $[{{\text{O}}_{\text{2}}}] = \dfrac{{0.933}}{{10}} = 0.093{\text{3 mol}}{{\text{L}}^{{\text{ - 1}}}}$

Now, according to the equation:

${{\text{K}}_{\text{c}}} = \dfrac{{{{[{{\text{N}}_{\text{2}}}{\text{O}}]}^2}}}{{{{[{{\text{N}}_2}]}^2}[{{\text{O}}_{\text{2}}}]}}$ 

$2.0 \times {10^{ - 37}} = \dfrac{{{{(\dfrac{x}{2})}^2}}}{{{{(0.0482)}^2}(0.0933)}}$ 

$\dfrac{{{x^2}}}{{100}} = 2.0 \times {10^{ - 37}} \times {(0.0482)^2}(0.0933)$ 

${x^2} = 43.35 \times {10^{ - 40}}$ 

$x = 6.6 \times {10^{ - 20}}$ 

$[{{\text{N}}_{\text{2}}}{\text{O}}] = \dfrac{x}{{10}} = \dfrac{{6.6 \times {{10}^{ - 20}}}}{{10}}$ 

$[{{\text{N}}_{\text{2}}}{\text{O}}] = 6.6 \times {10^{ - 21}}$ 

9. Nitric oxide reacts with ${\text{B}}{{\text{r}}_{\text{2}}}$  and gives nitrosyl bromide as per reaction given below: $2{\text{NO (g}}) + {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{NOBr (g)}}$ When 0.087 mol of ${\text{NO}}$  and 0.0437 mol of ${\text{B}}{{\text{r}}_{\text{2}}}$  are mixed in a closed container at constant temperature, 0.0518 mol of ${\text{NOBr}}$  is obtained at equilibrium. Calculate equilibrium amount of ${\text{NO}}$  and ${\text{B}}{{\text{r}}_{\text{2}}}$ .

Ans: The given reaction is:

$2{\text{NO (g}}) + {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{NOBr (g)}}$

Now, 2 mol of ${\text{NOBr}}$  are formed from 2 mol of ${\text{NO}}$ . Therefore, 0.0518 mol of ${\text{NOBr}}$  are formed from 0.0518 mol of ${\text{NO}}$. Again, 2 mol of ${\text{NOBr}}$  are formed from 1 mol of ${\text{Br}}$ .

Therefore, 0.0518 mol of ${\text{NOBr}}$ are formed from $\dfrac{{0.0581}}{2}$ mol of ${\text{Br}}$, or 0.0259 mol of ${\text{NO}}$.

The amount of ${\text{NO}}$ and ${\text{Br}}$ present initially is as follows: 

$[{\text{NO}}] = 0.087{\text{mol}}$ and $[{\text{B}}{{\text{r}}_2}] = 0.0437{\text{mol}}$ 

Therefore, the amount of ${\text{NO}}$ present at equilibrium is:

$[{\text{NO}}] = 0.087 - 0.0518 = 0.0352{\text{mol}}$ 

And, the amount of ${\text{Br}}$  present at equilibrium is:

$[{\text{B}}{{\text{r}}_2}] = 0.0437 - 0.0259 = 0.0178{\text{mol}}$ 

10.  At 450K, ${{\text{K}}_{\text{p}}} = 2.0 \times {10^{10}}{\text{ba}}{{\text{r}}^{{\text{ - 1}}}}$ for the given reaction reaction at equilibrium. $2{\text{S}}{{\text{O}}_{\text{2}}}{\text{(g}}) + {{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{S}}{{\text{O}}_{\text{3}}}{\text{(g)}}$ , What is ${{\text{K}}_{\text{c}}}$ at this temperature?

Ans: For the given reaction:

${{\Delta}n = 2 - 3 =  - 1} $ 

${T = 450{\text{K}}} $ 

${R = 0.0831{\text{bar L bar }}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}} $ 

${{K_p} = 2.0 \times {{10}^{10}}{\text{ba}}{{\text{r}}^{ - 1}}}$ 

We know that,

${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{\text{(RT}})\Delta {\text{n}}$ 

$2.0 \times {10^{10}} = {{\text{K}}_{\text{c}}}{(0.0831 \times 450)^{ - 1}}{\text{}}$

${{\text{K}}_{\text{c}}} = \dfrac{{2.0 \times {{10}^{10}}}}{{(0.0831 \times 450)}}$

${{\text{K}}_{\text{c}}} = 74.79 \times {10^{14}}{\text{Lmol}}$

11. A sample of ${\text{HI (g)}}$  is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of ${\text{HI (g)}}$  is 0.04 atm. What is ${{\text{K}}_{\text{p}}}$  for the given equilibrium? $2{\text{HI (g}}) \rightleftharpoons {{\text{H}}_{\text{2}}}{\text{(g)}} + {{\text{I}}_{\text{2}}}{\text{(g)}}$ 

Ans: The initial concentration of ${\text{HI}}$  is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of ${\text{HI}}$  is 0.2 – 0.04 = 0.16. The given reaction is: 

At equilibrium the pressure of ${\text{HI}}$ is 0.04 atm, the pressure for ${{\text{H}}_{\text{2}}} = \dfrac{{0.16}}{2} = 0.08{\text{atm}}$ and the pressure for ${{\text{O}}_{\text{2}}} = \dfrac{{0.16}}{2} = 0.08{\text{atm}}$.

Therefore, the value of ${{\text{K}}_{\text{p}}}$ at equilibrium is 4.0.

12.  A mixture of 1.57 mol of ${{\text{N}}_{\text{2}}}$ , 1.92 mol of ${{\text{H}}_{\text{2}}}$  and 8.13 mol of ${\text{N}}{{\text{H}}_{\text{3}}}$  is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, ${{\text{K}}_{\text{c}}}$  for the reaction is $1.7 \times {10^2}$ . The reaction is: ${{\text{N}}_{\text{2}}}{\text{(g)}} + 3{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$ . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans: For the given reaction: ${{\text{N}}_{\text{2}}}{\text{(g)}} + 3{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$

The given concentration of various species is

${\left[ {{{\text{N}}_2}} \right] = \dfrac{{1.57}}{{20}}{\text{mol}}{{\text{L}}^{ - 1}}\quad \left[ {{{\text{H}}_2}} \right] = \dfrac{{1.92}}{{20}}{\text{mol}}{{\text{L}}^{ - 1}}}$ 

${\left[ {{\text{N}}{{\text{H}}_3}} \right] = \dfrac{{8.13}}{{20}}{\text{mol}}{{\text{L}}^{ - 1}}}$ 

Now, reaction quotient ${{\text{Q}}_{\text{c}}}$  is:

${{Q_C}} { = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}} $ 

${{Q_C}} { = \dfrac{{{{\left( {\dfrac{{(8.13)}}{{20}}} \right)}^2}}}{{\left( {\dfrac{{1.57}}{{20}}} \right){{\left( {\dfrac{{1.92}}{{20}}} \right)}^3}}}} $ 

${{Q_C}} { = 2.4 \times {{10}^3}} $

Since, 𝑄𝑐 ≠ 𝐾𝑐, the reaction mixture is not at equilibrium. 

Again, 𝑄𝑐 >𝐾𝑐. Hence, the reaction will proceed in the reverse direction

13. The equilibrium constant expression for a gas reaction is: 

${{\text{K}}_{\text{c}}} = \dfrac{{{{{\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}}^{\text{4}}}{\text{ [}}{{\text{O}}_{\text{2}}}{{\text{]}}^{\text{5}}}}}{{{{{\text{[NO]}}}^{\text{4}}}{\text{ [}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{]}}^{\text{6}}}}}$ 

Write the balanced chemical equation corresponding to this expression.

Ans: The balanced chemical equation corresponding to the given expression can be written as:

$4{\text{NO (g}}) + 6{{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons 4{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g}}) + 5{{\text{O}}_{\text{2}}}{\text{(g)}}$ 

14. One mole of ${{\text{H}}_{\text{2}}}{\text{O}}$  and one mole of ${\text{CO}}$  are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with ${\text{CO}}$  according to the equation: ${{\text{H}}_{\text{2}}}{\text{(g}}) + {\text{CO(g}}) \rightleftharpoons {{\text{H}}_{\text{2}}}{\text{(g}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}}$ Calculate the equilibrium constant for the reaction.

Ans: The given reaction is:

${{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(g)}}}}{\text{ + C}}{{\text{O}}_{{\text{(g)}}}}}{ \rightleftharpoons {{\text{H}}_{{\text{2(g)}}}}}{{\text{ + C}}{{\text{O}}_{{\text{2(g)}}}}} $ 

${\dfrac{{\text{1}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{\text{1}}}{{{\text{10}}}}{\text{M}}} {\text{0}} {\text{0}} $ 

${\dfrac{{{\text{1 - 0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{{\text{1 - 0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{{\text{0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}}{\dfrac{{{\text{0}}{\text{.4}}}}{{{\text{10}}}}{\text{M}}} $ 

${{\text{ = 0}}{\text{.06M}}}{{\text{ = 0}}{\text{.06M}}}{{\text{ = 0}}{\text{.04M}}}{{\text{ = 0}}{\text{.04M}}} $

Therefore, the equilibrium constant for the reaction:

${{\text{K}}_{\text{c}}} = \dfrac{{[{{\text{H}}_{\text{2}}}][{\text{C}}{{\text{O}}_{\text{2}}}]}}{{[{{\text{H}}_{\text{2}}}{\text{O}}][{\text{CO}}]}}$

${{\text{K}}_{\text{c}}} = \dfrac{{0.04 \times 0.04}}{{0.06 \times 0.06}}$

${{\text{K}}_{\text{c}}} = 0.444$

15.  At 700 K, equilibrium constant for the reaction ${{\text{H}}_{\text{2}}}{\text{(g}}) + {{\text{I}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{HI (g)}}$ is 54.8. If $0.5{\text{mol}}{{\text{L}}^{ - 1}}$ of ${\text{HI (g)}}$  is present at equilibrium at 700 K, what are the concentration of ${{\text{H}}_{\text{2}}}{\text{(g)}}$  and ${{\text{I}}_{\text{2}}}{\text{(g)}}$  assuming that we initially started with ${\text{HI (g)}}$  and allowed it to reach equilibrium at 700 K?

Ans: It is given that equilibrium constant ${{\text{K}}_{\text{c}}}$  for the reaction

${{\text{H}}_{\text{2}}}{\text{(g}}) + {{\text{I}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{HI (g)}}$ is 54.8

Therefore, at equilibrium, the equilibrium constant ${{\text{K}}^{\text{'}}}_{\text{c}}$  for the reaction

$2{\text{HI (g}}) \rightleftharpoons {{\text{H}}_{\text{2}}}{\text{(g)}} + {{\text{I}}_{\text{2}}}{\text{(g}})$ 

$[{\text{HI}}] = 0.5{\text{mol}}{{\text{L}}^{ - 1}}$ will be: $\dfrac{1}{{54.8}}$ 

Let the concentrations of hydrogen and iodine at equilibrium be $x\;{\text{mol}}{{\text{L}}^{ - 1}}$ 

$[{{\text{H}}_{\text{2}}}] = [{{\text{I}}_{\text{2}}}] = x\;{\text{mol}}{{\text{L}}^{ - 1}}$ 

Therefore, $\dfrac{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{] [}}{{\text{I}}_{\text{2}}}{\text{]}}}}{{{{{\text{[HI]}}}^{\text{2}}}}} = {{\text{K}}^{\text{'}}}_{\text{c}}$ 

${\text{}}\dfrac{{x \times x}}{{{{(0.5)}^2}}} = \dfrac{1}{{54.8}}$ 

$ \Rightarrow {x^2} = \dfrac{{0.25}}{{54.8}} \Rightarrow x = 0.06754$

$x = 0.06{\text{8 mol}}{{\text{L}}^{{\text{  -  1}}}}$ (approximately)

Hence, at equilibrium, $[{{\text{H}}_{\text{2}}}] = [{{\text{I}}_2}] = 0.06{\text{8 mol}}{{\text{L}}^{{\text{  -  1}}}}$

16.  What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ${\text{ICI}}$  was 0.78 M?

$2{\text{ICI (g}}) \rightleftharpoons {{\text{I}}_{\text{2}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$ ; ${{\text{K}}_{\text{c}}} = 0.14$ 

Ans: For the given reaction:

$2{\text{ICI (g}}) \rightleftharpoons {{\text{I}}_{\text{2}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$

The initial concentration of ${\text{ICI}}$ is 0.78M and for ${{\text{I}}_{\text{2}}}$ and ${\text{C}}{{\text{l}}_{\text{2}}}$ is zero (0).

At equilibrium, the concentration for ${\text{ICI}}$ is $(0.78 - 2x){\text{M}}$ and for ${{\text{I}}_{\text{2}}}$ and ${\text{C}}{{\text{l}}_{\text{2}}}$is $x\;{\text{M}}$.

Now, we can write:

$\dfrac{{[{{\text{I}}_{\text{2}}}][{\text{C}}{{\text{l}}_{\text{2}}}]}}{{{{[{\text{ICI}}]}^2}}} = {{\text{K}}_{\text{c}}}$

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

$\dfrac{x}{{0.78 - 2x}} = 0.374$

$x = 0.292 - 0.748x$

$1.748x = 0.292$

$x = 0.167$

Therefore, at equilibrium

$[{{\text{H}}_2}]$ = $[{{\text{I}}_2}]$ = $0.167{\text{}}$

${\text{M}}[{\text{HI}}] = (0.78 - 2 \times 0.167)$

${\text{M}}[{\text{HI}}] = 0.446{\text{M}}$

17. ${{\text{K}}_{\text{p}}} = 0.04{\text{atm}}$  at 899 K for the equilibrium shown below. What is the equilibrium concentration of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$  when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{(g}}) \rightleftharpoons {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{(g)}}$ 

Ans: Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction

${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{(g}}) \rightleftharpoons {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{(g)}}$

The initial concentration of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ is 4.0 atm and that of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ and ${{\text{H}}_{\text{2}}}$ is zero (0).

At equilibrium, the concentration of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ is (4.0 – p) and that of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ and ${{\text{H}}_{\text{2}}}$is p. 

Then we can write,

$ \Rightarrow {K_p}\dfrac{{p \times p}}{{40 - p}} = 0.04$

${p^2} = 0.16 - 0.04p{\text{}}$

${p^2} + 0.04p - 0.16 = 0$

$p = \dfrac{{ - 0.04 \pm \sqrt {{{(0.04)}^2} - 4 \times 1 \times ( - 0.16)} }}{{2 \times 1}}$

$p = \dfrac{{ - 0.04 \pm 0.08}}{2}$

If we take positive value:

$ p=\dfrac{0.76}{2}$

$p=0.38 $

So, at equilibrium:

$[{C_2}{H_6}] - 4 - p = 4 - 0.38 = 3.62{\text{ atm}}$ 

18.  A sample of pure ${\text{PC}}{{\text{l}}_{\text{5}}}$  was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of ${\text{PC}}{{\text{l}}_{\text{5}}}$  was found to be $0.5 \times {10^{ - 1}}\;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$  . If value of ${{\text{K}}_{\text{c}}}$  is $8.3 \times {10^{ - 3}}$  , what are the concentrations of ${\text{PC}}{{\text{l}}_{\text{3}}}$  and ${\text{C}}{{\text{l}}_{\text{2}}}$  at equilibrium? ${\text{PC}}{{\text{l}}_{\text{5}}}{\text{(g}}) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$ 

Ans: Let the concentrations of both ${\text{PC}}{{\text{l}}_{\text{3}}}$  and ${\text{C}}{{\text{l}}_{\text{2}}}$  at equilibrium be $x{\kern 1pt} \;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ . The given reaction is:

${\text{PC}}{{\text{l}}_{\text{5}}}{\text{(g}}) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$

At equilibrium the concentration of ${\text{PC}}{{\text{l}}_{\text{5}}}$is $0.5 \times {10^{ - 1}}\;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$

It is given that the value of equilibrium constant ${{\text{K}}_{\text{c}}}$  is $8.3 \times {10^{ - 3}}$

The expression for equilibrium will be:

$ \Rightarrow \dfrac{[\text{PC}{{\text{l}}_{\text{3}}}][\text{C}{{\text{l}}_{\text{2}}}]}{[\text{PC}{{\text{l}}_{\text{5}}}]}={{\text{K}}_{\text{c}}}$

$ \Rightarrow \dfrac{x\times x}{0.5\times {{10}^{-1}}}=8.3\times {{10}^{-3}} $

$ \Rightarrow {{x}^{2}}=4.15\times {{10}^{-4}} $

$ \Rightarrow x=2.04\times {{10}^{-2}} $

$ \Rightarrow x=0.0204 $

Therefore at equilibrium:

$[{\text{PC}}{{\text{l}}_3}] = [{\text{C}}{{\text{l}}_2}] = 0.02{\kern 1pt} {\text{mol}}{{\text{L}}^{ - 1}}$ 

19. 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 ${\text{C}}{{\text{O}}_{\text{2}}}$ .${\text{FeO (s}}) + {\text{CO (g}}) \rightleftharpoons {\text{Fe (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}})$ ; ${{\text{K}}_{\text{p}}} = 0.0265$ at 1050 K.

What are the equilibrium partial pressures of ${\text{CO}}$  and ${\text{C}}{{\text{O}}_{\text{2}}}$  at 1050 K if the initial partial pressures are: ${{\text{p}}_{{\text{CO}}}} = 1.4\;{\text{atm}}$  and ${{\text{p}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} = 0.80\;{\text{atm}}$ ?

Ans: For the given reaction:

${\text{FeO (s}}) + {\text{CO (g}}) \rightleftharpoons {\text{Fe (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}})$

The initial concentration of ${\text{FeO}}$ is 1.4 atm and the initial concentration of ${\text{C}}{{\text{O}}_{\text{2}}}$ is 0.80 atm. 

It is given in the that, ${{\text{K}}_{\text{p}}} = 0.265$.

 Since ${{\text{Q}}_{\text{p}}} > {{\text{K}}_{\text{p}}}$ , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of ${\text{CO}}$  will increase while the pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$  will decrease. Now, let the increase in pressure of ${\text{CO}}$  be equal to the  decrease in pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$  be ‘p’. Then, we can write:

${{\text{Q}}_{\text{p}}} = \dfrac{{{{\text{p}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}}}{{{{\text{p}}_{{\text{CO}}}}}} = \dfrac{{0.80}}{{1.4}} = 0.571{\text{}}$

Therefore, equilibrium partial of ${\text{C}}{{\text{O}}_{\text{2}}}$ will be, ${{\text{p}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} = 0.080 - 0.339 = 0.46{\text{1 atm}}$ 

And, equilibrium partial pressure of ${\text{CO}}$ will be, ${{\text{p}}_{{\text{CO}}}} = 1.4 - 0.339 = 1.73{\text{9 atm}}$ 

20. Equilibrium constant, ${{\text{K}}_{\text{c}}}$  for the reaction ${{\text{N}}_{\text{2}}}{\text{(g}}) + 3{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g}})$ at 500 K os 0.061. At a particular time, the analysis shows that composition of the reaction mixture is $3.{\text{0 mol}}{{\text{L}}^{ - 1}}$  ${{\text{N}}_{\text{2}}}$ , $2.{\text{0 mol}}{{\text{L}}^{ - 1}}$  ${{\text{H}}_{\text{2}}}$  and $0.{\text{5 mol}}{{\text{L}}^{ - 1}}$  ${\text{N}}{{\text{H}}_{\text{3}}}$ . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Ans: The given reaction is:

${{\text{N}}_{\text{2}}}{\text{(g}}) + 3{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g}})$

At a particular time, the concentration of ${{\text{N}}_{\text{2}}}$ is $3.{\text{0 mol}}{{\text{L}}^{ - 1}}$ , for ${{\text{H}}_{\text{2}}}$ is $2.{\text{0 mol}}{{\text{L}}^{ - 1}}$ and for ${\text{N}}{{\text{H}}_{\text{3}}}$ is $0.{\text{5 mol}}{{\text{L}}^{ - 1}}$.

Now, we know that:

${{\text{Q}}_{\text{c}}} = \dfrac{{{{[{\text{N}}{{\text{H}}_{\text{3}}}]}^{\text{2}}}}}{{[{{\text{N}}_{\text{2}}}]{{[{{\text{H}}_{\text{2}}}]}^3}}} = \dfrac{{{{(0.5)}^2}}}{{(3.0){{(2.0)}^3}}} = 0.0104{\text{}}$

It is given that ${{\text{K}}_{\text{c}}} = 0.061$ 

Since ${{\text{Q}}_{\text{c}}} \ne {{\text{K}}_{\text{c}}}$  , the reaction is not at equilibrium. 

Since ${{\text{Q}}_{\text{c}}} < {{\text{K}}_{\text{c}}}$  , the reaction will proceed in the forward direction to reach equilibrium.

21. Bromine monochloride, ${\text{BrCl}}$  decomposes into bromine and chlorine and reaches the equilibrium: $2{\text{BrCl (g}}) \rightleftharpoons {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$ for which ${{\text{K}}_{\text{c}}} = 32$ at 500 K. If initially pure ${\text{BrCl}}$  is present at a concentration of $3.3 \times {10^{ - 3}}{\text{ mol}}{{\text{L}}^{ - 1}}$  , what is its molar concentration in the mixture at equilibrium?

Ans: Let the amount of bromine and chlorine formed at equilibrium be ‘x’. The given reaction is:

$2{\text{BrCl (g}}) \rightleftharpoons {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g)}} + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$

At equilibrium the concentration of ${\text{BrCl}}$ is $(3.3 \times {10^{ - 3}} - 2x)$ 

Now, we can write.

$ \Rightarrow \dfrac{{[{\text{B}}{{\text{r}}_{\text{2}}}][{\text{C}}{{\text{l}}_2}]}}{{[{\text{BrCl}}]}} = {{\text{K}}_{\text{c}}}$

$ \Rightarrow \dfrac{{x \times x}}{{{{(3.3 \times {{10}^{ - 3}} - 2x)}^2}}} = 32$

$ \Rightarrow \dfrac{x}{{3.3 \times {{10}^{ - 3}} - 2x}} = 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, at equilibrium

$[{\text{BrCl}}] = 3.3 \times {10^{ - 3}} - (2 \times 1.5 \times {10^{ - 3}}){\text{}}$

$[{\text{BrCl}}] = 3.3 \times {10^{ - 3}} - 3.0 \times {10^{ - 3}}$

$[{\text{BrCl}}] = 3.0 \times {10^{ - 4}}{\text{ mol}}{{\text{L}}^{ - 1}}$

22. At 1127 K and 1 atm pressure, a gaseous mixture of ${\text{CO}}$  and ${\text{C}}{{\text{O}}_{\text{2}}}$  in equilibrium with solid carbon has 90.55% ${\text{CO}}$ by mass: ${\text{C (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{CO (g)}}$. Calculate ${{\text{K}}_{\text{c}}}$  for this reaction at the above temperature.

Ans: Let, the total mass of the gas mixture = 100 gm

Mass of ${\text{CO}}$ = 90.55 gm

And, mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ = (100 – 90.55) = 9.45 gm

Now, the number of moles of ${\text{CO}}$ , ${n_{CO}} = \dfrac{{90.55}}{{28}} = 3.23{\text{4 mol}}$ 

Number of moles of ${\text{C}}{{\text{O}}_{\text{2}}}$, ${{\text{n}}_{{\text{C}}{{\text{O}}_{\text{2}}}}} = \dfrac{{9.45}}{{44}} = 0.21{\text{5 mol}}$ 

Partial pressure of ${\text{CO}}$ , ${p_{CO}} = \dfrac{{{n_{CO}}}}{{{n_{CO}} + {n_{C{O_2}}}}} \times {p_{{\text{total}}}}$ 

${p_{CO}} = \dfrac{{3.234}}{{3.234 + 0.215}} \times 1{\text{}}$

${p_{CO}} = 0.938\;{\text{atm}}$

Partial pressure of ${\text{C}}{{\text{O}}_{\text{2}}}$ ,

${p_{C{O_2}}} = \dfrac{{{n_{C{O_2}}}}}{{{n_{CO}} + {n_{C{O_2}}}}} \times {p_{{\text{total}}}}{\text{}}$

${p_{C{O_2}}} = \dfrac{{0.215}}{{3.234 + 0.215}} \times 1$

${p_{C{O_2}}} = 0.06{\text{2 atm}}$

Therefore, 

For the given reaction:

$\Delta n = 2 - 1 = 1$ 

We know that,

${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}{{\text{(RT}})^{\Delta {\text{n}}}}14.19{\text{}}$ 

$ = {{\text{K}}_{\text{c}}}{(0.082 \times 1127)^1}$

${{\text{K}}_{\text{c}}} = \dfrac{{14.19}}{{0.082 \times 1127}}$

${{\text{K}}_{\text{c}}} = 0.154$

23. Calculate:

1. $\Delta {{\text{G}}^\circ }$ and

2. The equilibrium constant for the formation of ${\text{N}}{{\text{O}}_{\text{2}}}$  from ${\text{NO}}$  and ${{\text{O}}_{\text{2}}}$  at 298 K ${\text{NO (g}}) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{N}}{{\text{O}}_{\text{2}}}{\text{(g}})$ 

Where:

${\Delta _{\text{f}}}{{\text{G}}^\circ }({\text{N}}{{\text{O}}_{\text{2}}}) = 52\;{\text{kJ/mol}}$

${\Delta _{\text{f}}}{{\text{G}}^\circ }({\text{NO)}} = 87.0\;{\text{kJ/mol}}$

${\Delta _{\text{f}}}{{\text{G}}^\circ }({{\text{O}}_{\text{2}}}) = 0\;{\text{kJ/mol}}$

Ans: For the given reaction: 

${\text{NO (g}}) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{N}}{{\text{O}}_{\text{2}}}{\text{(g}})$

1. The $\Delta {{\text{G}}^\circ }$ for the given reaction will be:

$\Delta {{\text{G}}^\circ } = \Delta {{\text{G}}^\circ }({\text{Products}}) - \Delta {{\text{G}}^\circ }({\text{Reactants}}){\text{}}$ 

$\Delta {{\text{G}}^\circ } = 52.0 - [87.0 + 0]$

$\Delta {{\text{G}}^\circ } = {\text{}} - 35.{\text{0 kJmo}}{{\text{l}}^{ - 1}}$

2. We know that:

$\Delta {{\text{G}}^\circ } = {\text{RT}}\log {{\text{K}}_{\text{c}}}{\text{}}$

$\Delta {{\text{G}}^\circ } = 2.303{\text{RT}}\log {{\text{K}}_{\text{c}}}$

$\log {{\text{K}}_{\text{c}}} = \dfrac{{ - 35.0 \times {{10}^{ - 3}}}}{{ - 2.303 \times 8.314 \times 298}}$

$\log {{\text{K}}_{\text{c}}} = 6.134$

${{\text{K}}_{\text{c}}} = {\text{antilog }}(6.134)$

${{\text{K}}_{\text{c}}} = 1.36 \times {10^6}$

Hence, the equilibrium constant for the given reaction ${{\text{K}}_{\text{c}}}$ is $1.36 \times {10^6}$ .

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

1. ${\text{PC}}{{\text{l}}_{\text{5}}}{\text{(g}}) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}{\text{(g}}) + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}})$ 

Ans: The amount of moles produced by the reaction will rise. According to Le Chatelier's principle, as pressure is reduced, the equilibrium swings in the direction of a greater number of moles of gas. The number of moles of gaseous products in the reaction is greater than the number of moles of gaseous reactants. As a result, the reaction will be propelled ahead. As a result, there will be more moles of reaction products.

2. ${\text{CaO (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s}})$ 

Ans: When the given equation is subjected to a decrease in pressure the number of moles of reaction products will decrease.

3. $3{\text{Fe (s}}) + 4{{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons {\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}{\text{(s}}) + 4{{\text{H}}_{\text{2}}}{\text{(g}})$ 

Ans: When the given equation is subjected to a decrease in pressure the number of moles of reaction products will remain the same.

25. 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.

1. ${\text{COC}}{{\text{l}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{CO (g}}) + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}$ 

2. ${\text{C}}{{\text{H}}_{\text{4}}}{\text{(g}}) + 2{{\text{S}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{C}}{{\text{S}}_{\text{2}}}{\text{(g)}} + 2{{\text{H}}_{\text{2}}}{\text{S (g)}}$ 

3. ${\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{C (s}}) + 2{\text{CO (g)}}$ 

4. ${\text{2}}{{\text{H}}_{\text{2}}}{\text{(g}}) + {\text{CO (g}}) \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH (g}})$ 

5. ${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s}}) \rightleftharpoons {\text{CaO (s}}) + {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g}})$ 

6. $4{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g}}) + 5{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 4{\text{NO (g}}) + 6{{\text{H}}_{\text{2}}}{\text{O (g)}}$ 

Ans: By increasing the pressure, the reactions in (i), (iii), (iv), (v), and (vi) will be changed. Because the number of moles of gaseous reactants is greater than the number of moles of gaseous products, the reaction in (iv) will proceed in the forward direction. Because the number of moles of gaseous reactants is smaller than that of gaseous products, the reactions in (i), (iii), (v), and (vi) will shift backward.

26. The equilibrium constant for the following reaction is $1.6 \times {10^5}$  at 1024 K. ${{\text{H}}_{\text{2}}}{\text{(g)}} + {\text{B}}{{\text{r}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{HBr (g}})$ , Find the equilibrium pressure of all gases if 10.0 bar of ${\text{HBr}}$  is introduced into a sealed container at 1024 K.

Ans: Given,

$K_{p}$ for the reaction i.e.,

$H_{2(g)}+B r_{2(g)} \leftrightarrow 2 H B r_{(g)}$ is $1.6 \times 10^{5}$

Therefore, for the reaction $2 H B r_{(g)} \leftrightarrow H_{2(g)}+B r_{2(g)}$ the equilibrium constant will be, $K_{p}^{\prime}=\dfrac{1}{K_{p}}$

=$\dfrac{1}{1.6 \times 10^{5}}$

=$6.25 \times 10^{-6}$

Now, let $\mathrm{p}$ be the pressure of both $\mathrm{H}_{2}$ and $\mathrm{Br}_{2}$ at equilibrium.

$\dfrac{p_{H B r} \times p_{2}}{p^{2} H B r}=K_{p}^{\prime}$

$\dfrac{p \times p}{(10-2 p)}=6.25 \times 10^{-6}$

$\dfrac{p}{10-2 p}=2.5 \times 10^{-3}$

$p=2.5 \times 10^{-2}-\left(5.0 \times 10^{-}\right) p$

$p+\left(5.0 \times 10^{-3}\right) p=2.5 \times 10^{-2}$

$\left(1005 \times 10^{-3}\right) p=2.5 \times 10^{-2}$

$p=2.49 \times 10^{-2}$ bar $=2.5 \times 10^{-2}$ (approximately) 

Therefore, at equilibrium,

${\left[\mathrm{H}_{2}\right]=\left[\mathrm{Br}_{2}\right]=2.49 \times 10^{-2} \text { bar }} $

${[\mathrm{HBr}]=10-2 \times\left(2.49 \times 10^{-1}\right) \text { bar }} $

=$9.95 \mathrm{bar}=10 \text { bar (approximately) }$

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

${\text{C}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons {\text{CO (g}}) + 3{{\text{H}}_{\text{2}}}{\text{(g}})$ 

1. Write an expression for ${{\text{K}}_{\text{p}}}$  for the above reaction.

Ans: For the given equation:

${K_p} = \dfrac{{{p_{CO}} \times {p^3}_{{H_2}}}}{{{p_{C{H_4}}} \times {p_{{H_2}O}}}}$ 

2. How will the values of ${{\text{K}}_{\text{p}}}$ and composition of equilibrium mixture be affected by 

1. Increasing the pressure 

Ans: The equilibrium will shift in the backward direction, according to Le Chatelier's principle.

As the reaction shifts in the backward direction as the pressure is increased, the value of ${{\text{K}}_{\text{p}}}$ decreases.

2. Increasing the temperature 

Ans: Because the reaction is endothermic, the equilibrium will shift forward, according to Le Chatelier's principle.

The value of ${{\text{K}}_{\text{p}}}$ will grow as the temperature rises because the reaction will shift forward.

3. Using a catalyst?

Ans: The existence of a catalyst has no effect on the reaction's equilibrium. A catalyst does nothing but speed up a reaction. As a result, equilibrium will be soon achieved.

Because the catalyst only changes the rate of reaction, the value of ${{\text{K}}_{\text{p}}}$ is unaffected.

28. Describe the effect of the following on the equilibrium of the reaction: $2{{\text{H}}_{\text{2}}}{\text{(g}}) + {\text{CO (g}}) \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH (g}})$ 

1. Addition of ${{\text{H}}_{\text{2}}}$ 

Ans: According to Le Chatelier's principle, when ${{\text{H}}_{\text{2}}}$  is added to a reaction, the equilibrium shifts in the forward direction.

2. Addition of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$  

Ans: The equilibrium will shift backwards with the addition of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ .

3. Removal of ${\text{CO}}$  

Ans: When ${\text{CO}}$  is removed from the equation, the equilibrium shifts backward.

4. Removal of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ 

Ans: When ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$  is removed from the equation, the equilibrium shifts forward.

29. At 473 K, equilibrium constant ${{\text{K}}_{\text{r}}}$  for decomposition of phosphorus pentachloride, ${\text{PC}}{{\text{l}}_{\text{5}}}$  is $8.3 \times {10^{ - 3}}$  . If decomposition is depicted as: 

${\text{PC}}{{\text{l}}_{\text{5}}}{\text{(g}}) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}{\text{(g}}) + {\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}})$ , ${\Delta _r}{H^ \circ } = 124.0\;{\text{kJmo}}{{\text{l}}^{ - 1}}$ 

1. Write an expression for ${{\text{K}}_{\text{c}}}$  for the reaction. 

Ans: The expression for ${{\text{K}}_{\text{c}}}$ will be:

${{\text{K}}_{\text{c}}} = \dfrac{{[{\text{PC}}{{\text{l}}_{\text{3}}}{\text{] [C}}{{\text{l}}_{\text{2}}}]}}{{[{\text{PC}}{{\text{l}}_{\text{5}}}{\text{]}}}}$ 

2. What is the value of ${{\text{K}}_{\text{c}}}$  for the reverse reaction at the same temperature? 

Ans: Value of ${{\text{K}}_{\text{c}}}$  for the reverse reaction at the same temperature is:

$K_{c^{\text{prime}}}=\dfrac{1}{K_{c}}$

$=\dfrac{1}{8.3 \times 10^{-3}}=1.2048 \times 10^{2}$

$=120-48$

3. What would be the effect on ${{\text{K}}_{\text{c}}}$  if 

1. More ${\text{PC}}{{\text{l}}_{\text{5}}}$  is added?

Ans:  Because the temperature is constant in this scenario, ${{\text{K}}_{\text{c}}}$ will remain constant.

2. Pressure is increased? 

Ans: At constant temperature, ${{\text{K}}_{\text{c}}}$ remains constant. As a result, ${{\text{K}}_{\text{c}}}$ would not alter in this scenario.

3. The temperature has increased?

Ans: In an endothermic process, the value of ${{\text{K}}_{\text{c}}}$ rises as the temperature rises. Because the reaction is an endothermic reaction, the value of ${{\text{K}}_{\text{c}}}$ will rise as the temperature rises.

30. 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 ${\text{CO}}$  and ${{\text{H}}_{\text{2}}}$ . In second stage, ${\text{CO}}$  formed in first stage is reacted with more steam in water gas shift reaction:  ${\text{CO (g}}) + {{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{H}}_{\text{2}}}{\text{(g)}}$ If a reaction vessel at 400°C is charged with an equimolar mixture of ${\text{CO}}$  and steam such that ${p_{{\text{CO}}}} = {p_{{{\text{H}}_{\text{2}}}{\text{O}}}} = 4.{\text{0 bar}}$ , what will be the partial pressure of ${{\text{H}}_{\text{2}}}$  at equilibrium? ${{\text{K}}_{\text{p}}} = 10.1$  at 400°C.

Ans: Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is: 

${\text{CO (g}}) + {{\text{H}}_{\text{2}}}{\text{O (g}}) \rightleftharpoons {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{H}}_{\text{2}}}{\text{(g)}}$

At equilibrium the concentration of ${\text{CO}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ will be $(4.0 - p)$.

It is given that, ${{\text{K}}_{\text{p}}} = 10.1$ 

Now,

${\dfrac{{{p_{c{o_2}}} \times {p_2}}}{{{p_{co}} \times {p_{{H_2}o}}}} = {K_p}} $

${ \Rightarrow \dfrac{{p \times p}}{{(4.0 - p)(4.0 - p)}}} $

${ \Rightarrow \dfrac{p}{{40 - p}} = 3.178} $

${ \Rightarrow p = 12.712 - 3.178p} $

${ \Rightarrow 4.178p = 12.712} $

${ \Rightarrow p = 3.04}$

Hence, at equilibrium, the partial pressure of ${{\text{H}}_{\text{2}}}$  will be 3.04 bar.

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

1. ${\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{Cl (g)}}$ ; ${K_c} = 5 \times {10^{ - 39}}$ 

2. ${\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}}) + 2{\text{NO (g}}) \rightleftharpoons 2{\text{NOCl (g}})$ ; ${K_c} = 3.7 \times {10^8}$ 

3. ${\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}}) + 2{\text{N}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{O}}_{\text{2}}}{\text{Cl (g}})$ ; ${K_c} = 1.8$ 

Ans: If the value of ${{\text{K}}_{\text{c}}}$  lies between ${10^{ - 3}}$  and ${10^3}$  , a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

32. The value of Kc for the reaction  ${\text{3}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{{\text{O}}_{\text{3}}}{\text{(g}})$ is $2.0 \times {10^{ - 50}}$ at 25°C. If the equilibrium concentration of ${{\text{O}}_{\text{2}}}$  in air at 25°C is $1.6 \times {10^{ - 2}}$  , what is the concentration of ${{\text{O}}_{\text{3}}}$ ?

Ans: The given reaction is:

${\text{3}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{{\text{O}}_{\text{3}}}{\text{(g}})$

${K_C} = \dfrac{{{{\left[ {{O_{3(g)}}} \right]}^2}}}{{{{\left[ {{O_{2(g)}}} \right]}^3}}}$

 ${K_C} = 2.0 \times {10^{ - 50}}$ and  $\left[ {{O_{2(g)}}} \right] = 1.6 \times {10^{ - 2}}$

${2.0 \times {{10}^{ - 50}} = \dfrac{{{{\left[ {{O_{3(g)}}} \right]}^2}}}{{{{\left[ {1.6 \times {{10}^{ - 2}}} \right]}^3}}}} $ 

${ \Rightarrow {{\left[ {{O_{3(g)}}} \right]}^2} = 2.0 \times {{10}^{ - 50}} \times {{\left( {1.6 \times {{10}^{ - 2}}} \right)}^3}} $ 

${ \Rightarrow {{\left[ {{O_{3(g)}}} \right]}^2} = 8.192 \times {{10}^{ - 50}}} $ 

${ \Rightarrow \left[ {{O_{3(g)}}} \right] = 2.86 \times {{10}^{ - 28}}M}$ 

 Hence, the concentration of ${{\text{O}}_{\text{3}}}$  is $2.86 \times {10^{ - 28}}{\text{M}}$ .

33. The reaction ${\text{CO (g}}) + {\text{3}}{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{C}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{O(g}})$ is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of ${\text{CO}}$ , 0.10 mol of ${{\text{H}}_{\text{2}}}$  and 0.02 mol of ${{\text{H}}_{\text{2}}}{\text{O}}$  and an unknown amount of ${\text{C}}{{\text{H}}_{\text{4}}}$  in the flask. Determine the concentration of ${\text{C}}{{\text{H}}_{\text{4}}}$  in the mixture. The equilibrium constant, ${{\text{K}}_{\text{c}}}$  for the reaction at the given temperature is 3.90.

Ans: Let the concentration of methane at equilibrium be x.

${\text{CO (g}}) + {\text{3}}{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{C}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{O(g}})$

At equilibrium, the concentration of ${\text{CO}}$  is 0.3 M , ${{\text{H}}_{\text{2}}}$ is 0.1 M and ${{\text{H}}_{\text{2}}}{\text{O}}$  is 0.02 M 

Therefore,

${\dfrac{{\left[ {C{H_{4(g)}}} \right]\left[ {{H_2}{O_{(g)}}} \right]}}{{\left[ {C{O_{(g)}}} \right]{{\left[ {{H_{2(g)}}} \right]}^3}}} = {K_c}} $ 

${ \Rightarrow \dfrac{{x \times 0.02}}{{0.3 \times {{(0.1)}^3}}} = 3.90} $

${ \Rightarrow x = \dfrac{{3.90 \times 0.3 \times {{(0.1)}^3}}}{{0.02}}} $ 

 = $\dfrac{0.00117}{0.02} $ 

$ = 0.0585M$ 

${ = 5.85 \times {{10}^{ - 2}}M}$ 

 Hence, the concentration of ${\text{C}}{{\text{H}}_{\text{4}}}$  at equilibrium is $5.85 \times {10^{ - 2}}{\text{M}}$. 

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

${\text{HN}}{{\text{O}}_2},{\text{C}}{{\text{N}}^ - },{\text{HCl}}{{\text{O}}_4},{{\text{F}}^ - },{\text{O}}{{\text{H}}^ - },{\text{CO}}_3^{2 - },{{\text{S}}^ - }$ 

Ans: A conjugate acid-base pair varies from one other by only one proton. In the table below, the conjugate acid-base for each species is listed. 

35. Which of the followings are Lewis acids? ${{\text{H}}_{\text{2}}}{\text{O, B}}{{\text{F}}_{\text{3}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, N}}{{\text{H}}_4}^ + $. 

Ans: Lewis acids are those acids which can accept a pair of electrons. For example: ${\text{B}}{{\text{F}}_{\text{3}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ are Lewis acids.

36. What will be the conjugate bases for the Brönsted acids: ${\text{HF, }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{, HC}}{{\text{O}}_{\text{3}}}$ 

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

37. Write the conjugate acids for the following Brönsted bases: ${\text{NH}}_2^ - ,\;{\text{N}}{{\text{H}}_3},\;{\text{HCO}}{{\text{O}}^ - }$ 

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

38. The species: ${{\text{H}}_{\text{2}}}{\text{O}},{\text{HCO}}_3^ - ,{\text{HSO}}_4^ - ,{\text{N}}{{\text{H}}_3}$ can act both as Brönsted 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.

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

1. ${\text{O}}{{\text{H}}^ - }$ 

Ans:  ${\text{O}}{{\text{H}}^ - }$is a Lewis base since it can donate its lone pair of electrons

2. ${{\text{F}}^ - }$ 

Ans: ${{\text{F}}^ - }$ is a Lewis base since it can donate a pair of electrons.

3. ${{\text{H}}^{\text{ + }}}$ 

Ans: ${{\text{H}}^{\text{ + }}}$ is a Lewis acid since it can accept a pair of electrons.

4. ${\text{BC}}{{\text{l}}_{\text{3}}}$ 

Ans: ${\text{BC}}{{\text{l}}_{\text{3}}}$  is a Lewis acid since it can accept a pair of electrons.

40. The concentration of hydrogen ion in a sample of soft drink is $3.8 \times {10^{ - 3}}{\text{ M}}$ . what is its pH?

Ans: Given, 

$[{{\text{H}}^{\text{ + }}}] = 3.8 \times {10^{ - 3}}{\text{ M}}$ 

Therefore, the pH value of the soft drink:

${\text{pH}} = {\text{}} - \log [{H^ + }]{\text{}}$

${\text{pH}} = {\text{}} - \log (3.8 \times {10^{ - 3}})$

${\text{pH}} = {\text{}} - \log 3.8 - \log {10^{ - 3}}$

${\text{pH}} = {\text{}} - \log 3.8 + 3$

${\text{pH}} = {\text{}} - 0.58 + 3$

${\text{pH}} = 2.42$

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

Ans: Given the, ${\text{pH}} = 3.76$ 

It is known as:

${\text{pH}} = {\text{}} - \log [{{\text{H}}^ + }]{\text{}}$

$\log [{{\text{H}}^ + }] = {\text{}} - {\text{pH}}$

$[{{\text{H}}^ + }] = {\text{antilog}}( - {\text{pH}})$

$[{{\text{H}}^ + }] = {\text{antilog}}( - 3.76)$

$[{{\text{H}}^ + }] = 1.74 \times {10^{ - 4}}{\text{M}}$

Hence, the concentration of hydrogen ion in the given sample of vinegar is $1.74 \times {10^{ - 4}}{\text{ M}}$. 

42. The ionization constant of ${\text{HF, HCOOH, HCN}}$ 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 base.

Ans: It is known as:

${K_b} = \dfrac{{{K_w}}}{{{K_a}}}$ 

Given, ${{\text{K}}_{\text{a}}}$ of ${\text{HF}}$ is $6.8 \times {10^{ - 4}}$

Hence, ${{\text{K}}_{\text{b}}}$ of its conjugate base ${{\text{F}}^ - }$:

${{\text{K}}_{\text{b}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}{\text{}}$ 

${{\text{K}}_{\text{b}}} = \dfrac{{{{10}^{ - 14}}}}{{6.8 \times {{10}^{ - 4}}}}$

${{\text{K}}_{\text{b}}} = 1.5 \times {10^{ - 11}}$

Given ${{\text{K}}_{\text{a}}}$ of ${\text{HCOOH}}$ is $1.8$$\times$$10^{-4}$ 

Hence, ${{\text{K}}_{\text{b}}}$ of its

conjugate base ${\text{HCO}}{{\text{O}}^ - }$:

${{\text{K}}_{\text{b}}}$

=$\dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}{{\text{K}}_{\text{b}}}{\text{}}$

$ = \dfrac{{{{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 4}}}}$

${{\text{K}}_{\text{b}}} = 5.6 \times {10^{ - 11}}$

Given ${{\text{K}}_{\text{a}}}$ of ${\text{HCN}}$ is $4.8$$\times $$10^{-9}$ 

Hence, ${{\text{K}}_{\text{b}}}$ of its

conjugate base ${\text{C}}{{\text{N}}^ - }$:

${{\text{K}}_{\text{b}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}{\text{}}$

${{\text{K}}_{\text{b}}} = \dfrac{{{{10}^{ - 14}}}}{{4.8 \times {{10}^{ - 9}}}}$

${{\text{K}}_{\text{b}}} = 2.08 \times {10^{ - 6}}$

43. The ionization constant of phenol is $1.0 \times {10^{ - 10}}$ . What is the concentration of phenolate ion in 0.05 M Ans:of phenol? What will be its degree of ionization if the Ans:is also 0.01M in sodium phenolate?

Ans: According to the reaction of ionization of phenol:

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - } + {{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ 

The initial concentration of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ is 0.05 M. 

At equilibrium, the concentration of $C_{6}H_{5}O^{ - }$ and $H_{3}O^{+ }$ will be x, then the concentration of $C_{6}H_{5}OH$ will be $(0.05 - x)$.

$K_{a} = \dfrac{{[C_{6}H_5O^{- }][H_{3}O^{+ }]}}{{[C_{6}H_{5}OH]}} $ 

${{K_a} = \dfrac{{x \times x}}{{0.05 - x}}} $

 As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.

${\therefore x = \sqrt {1 \times {{10}^{ - 10}} \times 0.05} } $ 

${\quad  = \sqrt {5 \times {{10}^{ - 12}}} } $ 

${\quad  = 2.2 \times {{10}^{ - 6}}{\text{M}} = \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]} $ 

${{\text{Since}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = \left[ {{{\text{C}}_6}{{\text{H}}_5}{{\text{O}}^ - }} \right]} $ 

${\left[ {{{\text{C}}_6}{{\text{H}}_5}{{\text{O}}^ - }} \right] = 2.2 \times {{10}^{ - 6}}{\text{M}}}$ 

Now, let α be the degree of ionization of phenol in the presence of 0.01 M ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - } + {{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ 

The concentrations are as follows:

$[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}] = 0.05 - 0.05\alpha {\text{}}$

$0.05{\text{ M}}[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - }] = 0.01 + 0.05\alpha {\text{}}$

$0.01{\text{ M}}[{{\text{H}}_3}{{\text{O}}^ + }] = 0.05\alpha$

$K_{a} = \dfrac{{[C_{6}H_5O^{- }][H_{3}O^{+ }]}}{{[C_{6}H_{5}OH]}} $ 

${{\text{K}}_{\text{a}}} = \dfrac{{(0.01)(0.05\alpha )}}{{0.05}}$

$1.0 \times {10^{ - 10}} = 0.01\alpha $

$\alpha {\text{}} = 1 \times {10^{ - 8}}$

44. The first ionization constant of ${{\text{H}}_{\text{2}}}{\text{S}}$  is $9.1 \times {10^{ - 8}}$  . Calculate the concentration of ${\text{H}}{{\text{S}}^ - }$  ion in its 0.1 M solution. How will this concentration be affected if the Ans:is 0.1 M in ${\text{HCl}}$  also? If the second dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}$  is $1.2 \times {10^{ - 13}}$ , calculate the concentration of ${{\text{S}}^{2 - }}$ under both conditions.

Ans: To calculate the concentration of ${\text{H}}{{\text{S}}^ - }$ ion:

Case I (in the absence of ${\text{HCl}}$ ):

Let the concentration of ${\text{H}}{{\text{S}}^ - }$be x M:

${{\text{H}}_{\text{2}}}{\text{S}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{H}}{{\text{S}}^ - }$ 

The final concentration of ${{\text{H}}_{\text{2}}}{\text{S}}$ will be $(0.1 - x)$ 

Then,

${{K_{{a_i}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]}}{{\left[ {{H_2}S} \right]}}} $ 

${9.1 \times {{10}^{ - 8}} = \dfrac{{(x)(x)}}{{0.1 - x}}} $ 

${\left( {9.1 \times {{10}^{ - 8}}} \right)(0.1 - x) = {x^2}} $

Taking,  $0.1 - xM;0.1{\text{M}}$ we have $\left( {9.1 \times {{10}^{ - 8}}} \right)(0.1) = {x^2}$ 

${9.1 \times {{10}^{ - 9}} = {x^2}} $ 

$x{ = \sqrt {9.1 \times {{10}^{ - 9}}} } $ 

${ = 9.54 \times {{10}^{ - 5}}{\text{M}}} $ 

$\Rightarrow {\left[ {H{S^ - }} \right] = 9.54 \times {{10}^{ - 5}}{\text{M}}} $

Case II (in the presence of ${\text{HCl}}$):

In the presence of $0.1{\text{ M}}$ of ${\text{HCl}}$, let ${\text{[H}}{{\text{S}}^ - }]$ be y M.

 ${{\text{H}}_{\text{2}}}{\text{S}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{H}}{{\text{S}}^ - }$

The final concentration of ${{\text{H}}_{\text{2}}}{\text{S}}$ will be $(0.1 - y)$ 

Now, 

${{\text{K}}_{{\text{a1}}}} =$

$\dfrac{{[HS^{-}][H^{+}]}}{{[H_2S]}}$

${K_{a1}} = \dfrac{{y(0.1 +y)}}{{(0.1 - y)}} $

$9.1 \times {10^{ - 8}} =$

$\dfrac{{y \times 0.1}}{{0.1}}  $

$y = 9.1 \times {10^{- 8}}  $

The concentration of $[{\text{H}}{{\text{S}}^ - }] = 9.1 \times {10^{ - 8}}$ 

To calculate the concentration of $[{{\text{S}}^{2 - }}]$ 

Case I (in the absence of 0.1 M of ${\text{HCl}}$):

 ${\text{H}}{{\text{S}}^ - } \rightleftharpoons {{\text{H}}^ + } + {{\text{S}}^{2 - }}$ 

$[{\text{H}}{{\text{S}}^ - }] = 9.54 \times {10^{ - 5}}{\text{ M}}$ 

Let $[{{\text{S}}^{2 - }}]$ be ${\text{X}}$ .

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

${{K_{{a_2}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{S^{2 - }}} \right]}}{{\left[ {H{S^ - }} \right]}}} $ 

${{K_{{a_2}}} = \dfrac{{\left( {9.54 \times {{10}^{ - 5}}} \right)(X)}}{{9.54 \times {{10}^{ - 5}}}}} $ 

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

Case II (in the presence of 0.1 M of ${\text{HCl}}$):

The concentration of ${{\text{S}}^{2 - }}$ be ${{\text{X}}^{\text{'}}}{\text{ M}}$ 

$[{\text{H}}{{\text{S}}^ - }] = 9.1 \times {10^{ - 8}}{\text{ M}}$ 

$[{{\text{H}}^ + }] = 0.1{\text{ M}}$ 

Then, 

${K_{{a_2}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{S^2}} \right]}}{{\left[ {H{S^ - }} \right]}}$

${1.2 \times {{10}^{ - 13}} = \dfrac{{(0.1)\left( {X'} \right)}}{{9.1 \times {{10}^{ - 8}}}}} $ 

${10.92 \times {{10}^{ - 21}} = 0.1X'} $

${\dfrac{{10.92 \times {{10}^{ - 21}}}}{{0.1}} = X'} $ 

${X' = \dfrac{{1.092 \times {{10}^{ - 20}}}}{{0.1}}} $ 

${ = 1.092 \times {{10}^{ - 19}}{\text{M}}} $ 

${ \Rightarrow {K_{{a_1}}} = 1.74 \times {{10}^{ - 5}}} $

45. The ionization constant of acetic acid is $1.74 \times {10^{ - 5}}$ . Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the Ans:and its pH.

Ans: For the reaction:

${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ 

The final concentration of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }$ and ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$ is $0.05\alpha $ . Then the final concentration of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ will be $(0.05 - 0.05\alpha )$ .

${{K_a} = \dfrac{{(.05\alpha )(.05\alpha )}}{{(.05 - 0.05\alpha )}}} $ 

${ = \dfrac{{(.05\alpha )(0.05\alpha )}}{{.05(1 - \alpha )}}} $ 

${ = \dfrac{{.05{\alpha ^2}}}{{1 - \alpha }}} $

${1.74 \times {{10}^{ - 5}} = \dfrac{{.05{\alpha ^2}}}{{1 - \alpha }}} $ 

${1.74 \times {{10}^{ - 5}} - 1.74 \times {{10}^{ - 5}}\alpha  = 0.05{\alpha ^2}} $ 

${0.05{\alpha ^2} + 1.74 \times {{10}^{ - 5}}\alpha  - 1.74 \times {{10}^{ - 5}}} $ 

${D = {b^2} - 4ac} $

${ = {{\left( {1.74 \times {{10}^{ - 5}}} \right)}^2} - 4(.05)\left( {1.74 \times {{10}^{ - 5}}} \right)} $ 

${ = 3.02 \times {{10}^{ - 25}} + .348 \times {{10}^{ - 5}}} $

${\alpha  = \sqrt {\dfrac{{{K_a}}}{c}} } $ 

${\alpha  = \sqrt {\dfrac{{1.74 \times {{10}^{ - 5}}}}{{.05}}} } $ 

${ = \sqrt {\dfrac{{34.8 \times {{10}^{ - 5}} \times 10}}{{10}}} } $ 

${ = \sqrt {3.48 \times {{10}^{ - 6}}} } $

${\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]}{ = 0.05 \times 1.86 \times {{10}^{ - 3}}} $ 

${ = \dfrac{{0.93 \times {{10}^{ - 3}}}}{{1000}}} $ 

${ = .000093} $

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

${\quad  =  - \log \left( {.093 \times {{10}^{ - 2}}} \right)} $ 

${\therefore pH = 3.03} $

 Hence, the concentration of acetate ion in the Ans:is 0.00093 M and its pH is 3.03.

46. It has been found that the pH of 0.01M of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its ${\text{p}}{{\text{K}}_{\text{a}}}$ .

Ans: Let the organic compound be ${\text{HA}}$.

${\text{HA}} \rightleftharpoons {{\text{H}}^ + } + {{\text{A}}^ - }$ 

Concentration of  ${\text{HA}}$ is 0.01 M and its pH is 4.15.

${ - \log \left[ {{H^ + }} \right] = 4.15} $ 

${\left[ {{H^ + }} \right] = 7.08 \times {{10}^{ - 5}}} $ 

${{\text{Now,}}{K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{[HA]}}} $ 

${\left[ {{H^ + }} \right] = \left[ {{A^ - }} \right] = 7.08 \times {{10}^{ - 5}}} $ 

${[HA] = 0.01} $

Now, 

${{\text{K}}_{\text{a}}} = \dfrac{{[{{\text{H}}^ + }][{{\text{A}}^ - }]}}{{[{\text{HA}}]}}[{{\text{H}}^ + }] = [{{\text{A}}^ - }] = 7.08 \times {10^{ - 5}}[{\text{HA}}] = 0.01{\text{}}$

Then,

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

${{K_a} = 5.01 \times {{10}^{ - 7}}} $ 

${p{K_a} =  - \log {K_a}} $ 

${\quad  =  - \log {K_a}} $ 

${\quad  = \left( {5.01 \times {{10}^{ - 7}}} \right)} $ 

${p{K_a} = 6.3001}$ 

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

1. 0.003 M ${\text{HCl}}$ 

Ans: ${{\text{H}}_{\text{2}}}{\text{O}} + {\text{HCl}} \rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^ + } + {\text{C}}{{\text{l}}^ - }$ 

Since ${\text{HCl}}$ is completely ionized.

$[{{\text{H}}_{\text{3}}}{{\text{O}}^ + }] = [{\text{HCl}}] \Rightarrow [{{\text{H}}_{\text{3}}}{{\text{O}}^ + }] = 0.003{\text{}}$ 

${\text{pH}} = {\text{}} - \log [{{\text{H}}_{\text{3}}}{{\text{O}}^ + }]$

${\text{pH}} = {\text{}} - \log (0.003)$

${\text{pH}} = 2.52$

Hence, the pH of the solution is

2.52.

2. 0.005 M ${\text{NaOH}}$  

Ans: ${\text{NaOH(aq)}} \rightleftharpoons Na^{+} (aq) + OH^{-}(aq)$ 

$[OH^{-}] = [NaOH] \Rightarrow [OH^{-}] =0.005$ 

${\text{pOH}} = {\text{}} - \log [{\text{O}}{{\text{H}}^ - }]$

${\text{pOH}} = {\text{}} - \log (0.005)$

${\text{pOH}} = 2.30$

${\text{pH}} = 14 - {\text{pOH}}$

${\text{pH}} = 14 - 2.30$

${\text{pH}} = 11.70$

Hence, the pH of the solution is

11.70.

3.  0.002 M ${\text{HBr}}$ 

Ans: ${\text{HBr}} +H_{2}O \rightleftharpoons H_{3}O^{+} + Br^{-}$ 

$[H_{3}O^{+}] = [HBr] \Rightarrow [H_{3}O^{+}] =0.002$ 

${\text{pH}} = {\text{}} - \log [{{\text{H}}_{\text{3}}}{{\text{O}}^ + }]$

${\text{pH}} = {\text{}} - \log (0.002)$

${\text{pH}} = 2.69$

Hence, the pH of the solution is

2.69.

4. 0.002 M ${\text{KOH}}$ 

Ans: $KOH (aq) \rightleftharpoons K^{+}(aq) + OH^{ -}(aq)$ 

$[OH^{-}] = [KOH]$

$\Rightarrow [OH^{-}] = 0.002$

${\text{pOH}} = {\text{}} - \log [{\text{O}}{{\text{H}}^ - }]$

${\text{pOH}} = {\text{}} - \log (0.002)$

${\text{pOH}} = 2.69$

${\text{pH}} = 14 - {\text{pOH}}$

${\text{pH}} = 14 - 2.69$

${\text{pH}} = 11.31$

Hence, the pH of the solution is 11.31.

48. Calculate the pH of the following solutions: 

1. 2 g of ${\text{TlOH}}$  dissolved in water to give 2 litre of solution. 

Ans: For 2g of ${\text{TlOH}}$ dissolved in water to give 2 L of solution:

${\left[ {TIO{H_{(aq)}}} \right] = \dfrac{2}{2}g/L} $ 

${ = \dfrac{2}{2} \times \dfrac{1}{{221}}M} $ 

${ = \dfrac{1}{{221}}M} $

${\text{TlOH (aq}}) \to {\text{T}}{{\text{l}}^ + }({\text{aq}}) + {\text{O}}{{\text{H}}^ - }{\text{(aq}})$ 

${TIO{H_{(aq)}} \to Tl_{(aq)}^ +  + OH_{(aq)}^ - } $ 

${\left[ {OH_{(aq)}^ - } \right] = \left[ {TIO{H_{(aq)}}} \right] = \dfrac{1}{{221}}M} $ 

${{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]} $ 

${{{10}^{ - 14}} = \left[ {{H^ + }} \right]\left( {\dfrac{1}{{221}}} \right)} $ 

${221 \times {{10}^{ - 14}} = \left[ {{H^ + }} \right]} $

${ \Rightarrow pH =  - \log \left[ {{H^ + }} \right] =  - \log \left( {221 \times {{10}^{ - 14}}} \right)} $ 

${ =  - \log \left( {2.21 \times {{10}^{ - 12}}} \right)} $ 

${ = 11.65} $

2. 0.3 g of ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$  dissolved in water to give 500 mL of solution. 

Ans: Molecular weight of ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$ is 74g.

So, 0.3g ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$ = $\dfrac{{0.3}}{{74}}{\text{ mole}}$ 

${\text{Molar Concentration  =  }}\dfrac{{{\text{Moles in Solution}}}}{{{\text{Volume in Solution}}}}$ 

Molar concentration of ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$ = $\dfrac{{\dfrac{{0.3}}{{74}}{\text{ mole}}}}{{0.5{\text{ L}}}} = 8.11 \times {10^{ - 3}}{\text{ M}}$ 

${\text{Ca(OH}}{)_2} \to {\text{C}}{{\text{a}}^{ + 2}} + 2{\text{O}}{{\text{H}}^ - }$ 

$[{\text{O}}{{\text{H}}^ - }] = 2[{\text{Ca(OH}}{)_2}]{\text{}}$ 

$[{\text{O}}{{\text{H}}^ - }] = 2 \times (8.11 \times {10^{ - 1}}){\text{ M}}$

$[{\text{O}}{{\text{H}}^ - }] = 16.22 \times {10^{ - 1}}{\text{M}}$

${\text{pOH}} = {\text{}} - \log (16.22 \times {10^{ - 1}})$

${\text{pOH}} = 3 - 1.2101$

${\text{pOH}} = 1.79$

${\text{pH}} = 14 - 1.79$

${\text{pH}} = 12.21$

3. 0.3 g of ${\text{NaOH}}$ 

dissolved in water to give 200 mL of solution.

Ans: 1 mole of  ${\text{NaOH}}$ = 40g

Since, 40 gm of ${\text{NaOH}}$ = 1 mole

of ${\text{NaOH}}$

So, 0.3 gm of $NaOH= $\frac{0.3}{40}$

Molar Concentration=$\dfrac{{{\text{Moles in Solution}}}}{{{\text{Volume in Solution}}}}$

Molar concentration = $\dfrac{{\dfrac{{0.3}}{{40}} Moles}}{{0.2L}} = 3.75 \times 10^{-1} M$ 

$[OH^{-}] = 3.75 \times {10^{ - 2}}{\text{ M}}$ 

${\text{pOH}} = {\text{}} - \log (3.75 \times {10^{ - 2}})$

${\text{pOH}} = 1.43$

${\text{pH}} = 14 - 1.43$

${\text{pH}} = 12.57$

4. 1mL of 13.6 M ${\text{HCl}}$  is diluted with water to give 1 litre of solution.

Ans: For this question:

$M_{1}V_{1}= M_{2}V_{2}$ 

$13.6 \times {\text{1 mL}} = {{\text{M}}_{\text{2}}} \times 100{\text{0 mL}}$

$13.6 \times {10^{ - 3}} = {{\text{M}}_{\text{2}}} \times {\text{1 L}}$

${{\text{M}}_2} = 1.36 \times {10^{ - 2}}$

$[{{\text{H}}^ + }] = 1.36 \times {10^{ - 2}}$

${\text{pH}} = {\text{}} - \log (1.36 \times {10^{ - 2}})$

${\text{pH}} = {\text{}} - 0.1335 + 2$

${\text{pH}} = 1.87$

49. The degree of ionization of a 0.1M bromoacetic acid is 0.132. Calculate the pH of ${\text{p}}{{\text{K}}_{\text{a}}}$  of bromoacetic acid.

Ans: Degree of ionization, $\alpha {\text{}} = 0.132$ 

Concentration, c = 0.1 M

Thus, the concentration of

${H_3}{O^+}$=$c.\alpha$ 

$[{H_3}{O^+}]$ = $0.1\times 0.132$ 

$[{{\text{H}}_{\text{3}}}{{\text{O}}^ + }] = 0.0132$

${\text{pH}} = {\text{}} - \log [{{\text{H}}^ + }]$

${\text{pH}} = {\text{}} - \log (0.0132)$

${\text{pH}} = 1.88$

Now,

${{\text{K}}_{\text{a}}} = C{\alpha ^2}{\text{}}$

${{\text{K}}_{\text{a}}} = 0.1 \times {(0.132)^2}$

${{\text{K}}_{\text{a}}} = 0.0017{\text{p}}$

${{\text{K}}_{\text{a}}} = 2.75$

50. What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Ans: ${{{\text{K}}_{\text{b}}} = 4.27\quad {{10}^{ - 10}}} $ 

${{\text{c}} = 0.001{\text{M}}} $ 

${{\text{pH}} = ?\quad } $ 

${\alpha  = ?} $ 

${{k_a} = C{\alpha ^2}} $ 

${4.27 \times {{10}^{ - 10}} = 0.001 \times {\alpha ^2}} $ 

${4270 \times {{10}^{ - 10}} = {\alpha ^2}} $ 

${65.34 \times {{10}^{ - 5}} = \alpha  = 6.53 \times {{10}^{ - 4}}} $

Then, concentration of anion:

$[{\text{anion}}] = c.\alpha {\text{}} = 0.01 \times 65.34 \times {10^{ - 5}}{\text{}}$

$[{\text{anion}}] = 0.65 \times {10^{ - 5}}$

${pOH =  - \log \left( {.065 \times {{10}^{ - 5}}} \right)} $ 

${ = 6.187} $ 

${pH = 7.813} $

${{K_a} \times {K_b} = {K_w}} $ 

${\therefore 4.27 \times {{10}^{ - 10}} \times {K_a} = {K_w}} $ 

${{K_a} = \dfrac{{{{10}^{ - 14}}}}{{4.27 \times {{10}^{ - 10}}}}} $ 

${ = 2.34 \times {{10}^{ - 5}}} $

Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times {10^{ - 5}}$ .

51. The degree of ionization of a $0.1{\text{M}}$ bromoacetic acid is 0.132. Calculate the ${\text{pH}}$ and the ${\text{p}}{{\text{K}}_{\text{a}}}$ of bromoacetic acid.

Ans: Degree of ionization, $\alpha  = 0.132$

Concentration, ${\text{c}} = 0.1{\text{M}}$

Thus, the concentration of ${{\text{H}}_3}{{\text{O}}^ + } = $c. $\alpha $

$ \Rightarrow 0.1 \times 0.132 = 0.0132{\text{}}$  

$pH= {\text{}} - \log \left[ {{H^ + }} \right]{\text{}} = {\text{}} - \log (0.0132){\text{}} = 1.879:1.88{\text{}}$

Now

${K_a}$ = $C{\alpha ^2}{\text{}}$ = $0.1 \times{(0.132)^2}{\text{}}$

${\text{}}{K_a} = .0017{\text{}}p$

${K_a} = 2.75$

52. What is the pH of $0.001{\text{M}}$ aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Ans:

${{{\text{K}}_{\text{b}}} = 4.27 \times {{10}^{ - 10}}} $ 

${{\text{c}} = 0.001{\text{M}}} $ 

${{\text{pH}} = ?\quad } $ 

${\alpha  = ?} $ 

${{k_a} = C{\alpha ^2}} $ 

${4.27 \times {{10}^{ - 10}} = 0.001 \times {\alpha ^2}} $ 

${4270 \times {{10}^{ - 10}} = {\alpha ^2}} $ 

${65.34 \times {{10}^{ - 5}} = \alpha  = 6.53 \times {{10}^{ - 4}}} $ 

$Then,[anion] =$\text{c}\alpha =0.01\times 65.34\times 10^{-5}$ 

${ = 0.65 \times {{10}^{ - 5}}} $ 

${pOH}{ =  - \log \left( {.065 \times {{10}^{ - 5}}} \right)} $ 

${ = 6.187} $

 $pH = 7.813$

Now,

${{K_a} \times {K_b} = {K_w}} $ 

${\therefore 4.27 \times {{10}^{ - 10}} \times {K_a} = {K_w}} $ 

${{K_a} = \dfrac{{{{10}^{ - 14}}}}{{4.27 \times {{10}^{ - 10}}}}} $ 

${ = 2.34 \times {{10}^{ - 5}}}$ 

Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times {10^{ - 5}}$.

53. Calculate the degree of ionization of $0.05{\text{M}}$ acetic acid if its ${\text{p}}{{\text{K}}_{\text{a}}}$ value is 4.74. How is the degree of dissociation affected when it's also contains 

1. $0.01{\text{M}}$

2. $0.1{\text{M}}$ in ${\text{HCl}}$ ?

Ans:

${{\text{c}} = 0.05{\text{M}}}$

${p{K_a} = 4.74}$ 

$p{K_a} =  - \log {K_a}$ 

${{K_a} = 1.82 \times {{10}^{ - 5}}} $

$ c = 0.05M$  

$p{K_{a} = 4.74} $

$ pK_{a} =  - \log( K_{a} )$  

${K_{a}} = 1.82 \times 10^{-5} $

${{K_a} = c{\alpha ^2}\quad \alpha  = \sqrt {\dfrac{{{K_a}}}{c}} }$

${\alpha  = \sqrt {\dfrac{{1.82 \times {{10}^{ - 5}}}}{{5 \times {{10}^{ - 2}}}}}  = 1.908 \times {{10}^{ - 2}}}$ 

$ K_a = c\alpha ^2 \quad \alpha  = \sqrt {\dfrac{{K_a}}{{c}}} $ 

${\alpha  = \sqrt {\dfrac{{1.82 \times {{10}^{ - 5}}}}{{5 \times {{10}^{ - 2}}}}}  = 1.908 \times {{10}^{ - 2}}} $

When HCl is added to the solution, the concentration of ${{\text{H}}^ + }$ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease. Case I: When $0.01{\text{MHCl}}$ is taken.

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

${{\text{C}}{{\text{H}}_3}{\text{COOH}}}{{{\text{H}}^ + } + }{{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} $ 

${{\text{Initial conc}}{\text{. }}}{0.05{\text{M}}} {{\text{ Afterdissociation }}}{0.05 - x}{0.01 + x}x $

As the dissociation of a very small amount of acetic acid will take place, the values i.e., $0.05 - {\text{x}}$ and $0.01 + {\text{x}}$ can be taken as $0.05$ and $0.01$ respectively.

${{K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}}$

$\therefore {K_{a}} = \dfrac{{(0.01)x}}{{0.05}}$ 

$x = \dfrac{{1.82 \times 10^{-5} \times 0.05}}{{0.01}}$ 

$x = 1.82 \times 10^{-3} \times 0.05M$

$ {{K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}} $ 

${\therefore {K_a} = \dfrac{{(0.01)x}}{{0.05}}} $ 

${x = \dfrac{{1.82 \times {{10}^{ - 5}} \times 0.05}}{{0.01}}} $ 

${x = 1.82 \times {{10}^{ - 3}} \times 0.05M} $

Now,

$\alpha = \dfrac{{{\text{ Amount of acid dissociated }}}}{{{\text{ Amount of acid taken }}}}$

$ \Rightarrow \dfrac{{1.82 \times {{10}^{ - 5}} \times 0.05}}{{0.05}} = 1.82 \times {10^{ - 3}}$

Case II: When $0.1{\text{MHCl}}$ is taken.

Let the amount of acetic acid dissociated in this case be $X$. As we have done in the first case, the concentrations of various species involved in the reaction are:

${\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right] = 0.05 - X;0.05{\text{M}}}$

${\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right] = X}$

${\left[ {{{\text{H}}^ + }} \right] = 0.1 + X;0.1{\text{M}}}$ 

${{K_a} = \dfrac{{\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]\left[ {{{\text{H}}^ + }} \right]}}{{\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right]}}}$

$\therefore K_a = \dfrac{{(0.1)x}}{{0.05}} $

${\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right] = 0.05 - X;0.05{\text{M}}} {\left[{{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right] = X} $ 

${\left[ {{{\text{H}}^ + }} \right] = 0.1 + X;0.1{\text{M}}} $ 

${{K_a} = \dfrac{{\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }}\right]\left[{{{\text{H}}^+}}\right]}}{{\left[{{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right]}}} $$ \therefore K_a = \dfrac{{(0.1)x}}{{0.05}} $

$x = \dfrac{{1.82 \times {{10}^{ - 5}} \times 0.05}}{{0.1}}$ $x = 1.82 \times {10^{ - 4}} \times 0.05M$ Now,

$\alpha = \dfrac{{{\text{ Amount of acid dissociated }}}}{{{\text{ Amount of acid taken }}}}$

$ \Rightarrow \dfrac{{1.82 \times {{10}^{ - 4}} \times 0.05}}{{0.05}} = 1.82 \times {10^{ - 4}}$

54. The ionization constant of dimethylamine is ${5.410^{ - 4}}$. Calculate its degree of ionization in its $0.02{\text{M}}$ solution. What percentage of dimethylamine is ionized if the Ans:is also $0.1{\text{M}}$ in ${\text{NaOH}}?$

Ans:${{K_b} = 5.4 \times {{10}^{ - 4}}} $ 

${{\text{c}} = 0.02{\text{M}}} $

${\text{c}} = 0.02{\text{M}}$ Then, $\alpha  = \sqrt {\dfrac{{{K_b}}}{c}} $ $ = \sqrt {\dfrac{{5.4 \times {{10}^{ - 4}}}}{{0.02}}} $ $ = 0.1643$

Now, if $0.1{\text{M}}$ of ${\text{NaOH}}$ is added to the solution, then ${\text{NaOH}}$ (being a strong base) undergoes complete ionization.

(image will be uploaded soon)

And,

${{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}}_2^ +  + {\text{OH}}} $ 

${(0.02 - x)\quad x} $ 

${;0.02{\text{M}}} $ 

${{\text{ Then, }}\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}}_2^ + } \right] = x} $ 

${\left[ {{\text{O}}{{\text{H}}^ - }} \right] = x + 0.1;0.1} $

${ \Rightarrow {K_b} = \dfrac{{\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}}_2^ + } \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}NH} \right]}}} $ 

${5.4 \times {{10}^{ - 4}} = \dfrac{{x \times 0.1}}{{0.02}}} $ 

${x = 0.0054} $

It means that in the presence of $0.1{\text{MNaOH}},0.54\% $ of dimethylamine will get dissociated.

55.  Calculate the hydrogen ion concentration in the following biological fluids whose ${\text{pH}}$ are given below:

Ans:

1. Human muscle fluid 6.83

${{\text{pH}} = 6.83{\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\therefore 6.83 =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\left[ {{{\text{H}}^ + }} \right] = 1.48\quad {{10}^{ - 7}}{\text{M}}} $

2. Human stomach fluid, 1.2:

${{\text{pH}} = 1.2\quad  \times } $ 

${1.2 =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\therefore \left[ {{{\text{H}}^ + }} \right] = 0.063} $

3. Human blood, 7.38:

${\text{pH}} = 7.38 =  - \log \left[ {{{\text{H}}^ + }} \right]$

$\therefore \left[ {{{\text{H}}^ + }} \right] = 4.17\quad {10^{ - 8}}{\text{M}}$

4. Human saliva, 6.4:

${{\text{pH}} = 6.4\quad  \times } $ 

${6.4 =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\left[ {{{\text{H}}^ + }} \right] = 3.98\quad {{10}^{ - 7}}} $

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

Ans: The hydrogen ion concentration in the given substances can be calculated by using the given relation: ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

1. ${\text{pH}}$ of milk $ = 6.8$

Since, ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

$6.8 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 6.8$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 6.8)$

$ = 1.5\quad {19^{ - 7}}{\text{M}}$

2. ${\text{pH}}$ of black coffee $ = 5.0$

Since, ${\text{pH}} =  \times \log \left[ {{{\text{H}}^ + }} \right]$

$5.0 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 5.0$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 5.0)$

$ = {10^{ - 5}}{\text{M}}$

3. ${\text{pH}}$ of tomato juice $ = 4.2$

Since, ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

$4.2 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 4.2$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 4.2)$

$ = 6.31\quad {10^{ - 5}}{\text{M}}$

4. ${\text{pH}}$ of lemon juice $ = 2.2$ 

Since, ${\text{pH}} =  - $ log $\left[ {{{\text{H}}^ + }} \right]$ 

$2.2 =  - \log \left[ {{{\text{H}}^ + }} \right]$log 

$\left[ {{{\text{H}}^ + }} \right] =  - 2.2$ 

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 2.2)$ 

$\quad  = 6.31\quad {10^{ - 3}}{\text{M}}$

5. ${\text{pH}}$ of egg white $ = 7.8$

Since, ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

$7.8 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 7.8$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 7.8)$

$ = 1.58\quad {10^{ - 8}}{\text{M}}$

57. If $0.561\;{\text{g}}$ of ${\text{KOH}}$ is dissolved in water to give $200\;{\text{mL}}$ at $298\;{\text{K}}$. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its ${\text{pH}}$ ?

Ans: $\left[ {KO{H_{aq}}} \right] = \dfrac{{0.561}}{{\dfrac{1}{5}}}{\text{g/L}}$

${\text{}} = 2.805{\text{g/L}}$

$ = 2.805 \times \dfrac{1}{{56.11}}M{\text{}} =.05M$

$KO{H_{aq}} \to K_{(aq)}^ + {\text{}} + OH_{(aq)}^ - {\text{}}$

$\left[ {O{H^ - }} \right] = .05M$

$ = \left[ {{K^ + }} \right]{\text{}}\left[ {{H^ + }} \right]\left[ {{H^ - }} \right] = {K_w}{\text{}}$

$\left[ {{H^ + }} \right] = \dfrac{{{K_w}}}{{\left[ {O{H^ - }} \right]}}{\text{}}$

$ = \dfrac{{{{10}^{ - 14}}}}{{0.05}}{\text{}}$

$ = 2 \times {10^{ - 11}}M{\text{}}$

$\therefore pH = 12.70$

58. The solubility of ${\text{Sr}}{({\text{OH}})_2}$ at $298\;{\text{K}}$ is $19.23\;{\text{g}}/{\text{L}}$ of solution. Calculate the concentrations of strontium and hydroxyl ions and the ${\text{pH}}$ of the solution.

Ans:Solubility of ${\text{Sr}}{({\text{OH}})_2} = 19.23\;{\text{g}}/{\text{L}}$

Then, concentration of ${\text{Sr}}{({\text{OH}})_2}$

${ = \dfrac{{19.23}}{{121.63}}M} $

${ = 0.1581M} $

${\operatorname{Sr} {{(OH)}_{2(aq)}} \to Sr_{(aq)}^{2 + } + 2{{\left( {O{H^ - }} \right)}_{(aq)}}} $

${\therefore \left[ {S{r^{2 + }}} \right] = 0.1581M} $

${\left[ {O{H^ - }} \right] = 2 \times 0.1581M = 0.3126M} $

Now,

${{K_w} = \left[ {O{H^ - }} \right]\left[ {{H^ + }} \right]} $ 

${\dfrac{{{{10}^{ - 14}}}}{{0.3126}} = \left[ {{H^ + }} \right]} $ 

${ \Rightarrow \left[ {{H^ + }} \right] = 3.2 \times {{10}^{ - 14}}} $ 

${\therefore pH = 13.495;13.50} $

59. The ionization constant of propanoic acid is $1.32 \times {10^{ - 5}}.$ Calculate the degree of ionization of the acid in its $0.05{\text{M}}$ Ans:and also its ${\text{pH}}$. What will be its degree of ionization if the Ans:is $0.01{\text{M}}$ in HCl also?

Ans: Let the degree of ionization of propanoic acid be $\alpha $. Then, representing propionic acid as HA, we have:

${{\text{HA       }} + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{H}}_2}{\text{O}} \Leftrightarrow {{\text{H}}_3}{{\text{O}}^ + }{\text{     +   }}{{\text{A}}^ - }}$ 

${(.05 - 0.0\alpha ) \approx 0.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.05\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,.05\alpha \,\,\,\,\,\,\,\,} $

$K_{a}=\dfrac{\left[H_{3} O^{+}\right]\left[A^{-}\right]}{[H A]} $

$=\dfrac{(.05 \alpha)(.05 \alpha)}{0.05}=.05 \alpha^{2} $

$\alpha=\sqrt{\dfrac{K_{a}}{.05}}=1.63 \times 10^{-2}$

Then, $\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = .05\alpha  = .05 \times 1.63 \times {10^{ - 2}} = {K_b}.15 \times {10^{ - 4}}{\text{M}}$

$\therefore {\text{pH}} = 3.09$

In the presence of $0.1{\text{M}}$ of ${\text{HCl}}$, let $\alpha $ ' be the degree of ionization.

Then, $\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 0.01$

${\left[ {{A^ - }} \right] = 005{\alpha ^\prime }} $ 

${[HA] = .05} $ 

${{K_a} = \dfrac{{0.01 \times .05{\alpha ^\prime }}}{{.05}}} $ 

${1.32 \times {{10}^{ - 5}} = .01 \times {\alpha ^\prime }} $ 

${{\alpha ^\prime } = 1.32 \times {{10}^{ - 5}}} $

60. The ${\text{pH}}$of $0.1{\text{M}}$ of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Ans: ${{\text{c}} = 0.1{\text{M}}} $ 

${{\text{pH}} = 2.34} $ 

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

${ - \log \left[ {{H^ + }} \right] = 2.34} $ 

${\left[ {{H^ + }} \right] = 4.5 \times {{10}^{ - 3}}} $

Also,

$\left[ {{H^ + }} \right] = c\alpha $ $4.5 \times {10^{ - 3}} = 0.1 \times \alpha $ $\dfrac{{4.5 \times {{10}^{ - 3}}}}{{0.1}} = \alpha $ $\alpha  = 4.5 \times {10^{ - 3}} = .045$ Then, 

$=0.1 \times\left(45 \times 10^{-3}\right)^{2} $

$=202.5 \times 10^{-6} $

$=2.02 \times 10^{-4}$

61. The ionization constant of nitrous acid is ${4.510^{ - 4}}$. Calculate the ${\text{pH}}$ of $0.04{\text{M}}$ sodium nitrite and also its degree of hydrolysis.

Ans: ${\text{NaN}}{{\text{O}}_2}$ is the salt of a strong base $({\text{NaOH}})$ and a weak acid $\left( {{\text{HN}}{{\text{O}}_2}} \right)$.

${{\text{NO}}_2^ -  + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{HN}}{{\text{O}}_2} + {\text{O}}{{\text{H}}^ - }} $ 

${{K_b} = \dfrac{{\left[ {{\text{HN}}{{\text{O}}_2}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{NO}}_2^ - } \right]}}} $ 

${ \Rightarrow \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{{{10}^{ - 14}}}}{{4.5 \times {{10}^{ - 14}}}} = .22 \times {{10}^{ - 10}}} $

Now, If $x$ moles of the salt undergo hydrolysis, then the concentration of various species present in the solution, it will be:

${\left[ {NO_2^ - } \right] = .04 - x;0.04} $ 

${\left[ {HN{O_2}} \right] = x} $ 

${\left[ {O{H^ - }} \right] = x} $ 

${{K_b} = \dfrac{{{x^2}}}{{0.04}} = 0.22 \times {{10}^{ - 10}}} $ 

${{x^2} = .0088 \times {{10}^{ - 10}}} $ 

${x = .093 \times {{10}^{ - 5}}} $

${\therefore \left[ {O{H^ - }} \right] = 0.093 \times {{10}^{ - 5}}M} $ 

${\left[ {{H_3}{O^ + }} \right] = \dfrac{{{{10}^{ - 14}}}}{{.093 \times {{10}^{ - 5}}}} = 10.75 \times {{10}^{ - 9}}M} $ 

${ \Rightarrow pH =  - \log \left( {10.75 \times {{10}^{ - 9}}} \right)} $ 

 = 7.96 

Therefore, degree of hydrolysis

${ = \dfrac{x}{{0.04}} = \dfrac{{.093 \times {{10}^{ - 5}}}}{{.04}}} $ 

${ = 2.325 \times {{10}^{ - 5}}} $

62. A $0.02{\text{M}}$ solution of pyridinium hydrochloride has ${\text{pH}} = 3.44$. Calculate the ionization constant of pyridine

Ans: ${\text{pH}} = 3.44$

We know that,

${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$ $\therefore \left[ {{H^ + }} \right] = 3.63 \times {10^{ - 4}}$

Then, ${K_b} = \dfrac{{{{\left( {3.63 \times {{10}^{ - 4}}} \right)}^2}}}{{0.02}}(\because $ concentration $ = 0.02{\text{M}})$

$ \Rightarrow {K_b} = 6.6 \times {10^{ - 6}}$

Now, ${K_b} = \dfrac{{{K_w}}}{{{K_a}}}$

$ \Rightarrow {K_a} = \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{{{10}^{ - 14}}}}{{6.6 \times {{10}^{ - 6}}}} = 1.51 \times {10^{ - 9}}$

63. Predict if the solutions of the following salts are neutral, acidic or basic:

${\text{NaCl}},{\text{KBr}},{\text{NaCN}},{\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3},{\text{NaN}}{{\text{O}}_2}$ and ${\text{KF}}$

Ans:

(i) ${\text{NaCl}}$ :

${\text{NaCl}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{NaOH}} + {\text{HCl}}$

Strong base Strong base

Therefore, it is a neutral solution.

(ii) KBr:

${\text{KBr}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{KOH}} + {\text{HBr}}$

Strong base    Strong base

Therefore, it is a neutral solution.

iii) ${\text{NaCN}}:$

${\text{NaCN}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{HCN}} + {\text{NaOH}}$

Weak acid        Strong base

Therefore, it is a basic solution.

(iv) ${\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}$

${\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{N}}{{\text{H}}_4}{\text{OH}} + {\text{HN}}{{\text{O}}_3}$

Weak acid               Strong base

Therefore, it is an acidic solution.

(v) ${\text{NaN}}{{\text{O}}_{\text{2}}}$

${\text{NaN}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{NaOH}} + {\text{HN}}{{\text{O}}_2}$

Strong base         Weak acid

Therefore, it is a basic solution.

(vi) KF

${\text{KF}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{KOH}} + {\text{HF}}$

Strong base     Weak acid

Therefore, it is a basic solution.

64. The ionization constant of chloro acetic acid is ${1.3510^{ - 3}}.$ What will be the ${\text{pH}}$ of $0.1{\text{M}}$ acid and its $0.1{\text{M}}$ sodium salt solution?

Ans: It is given that ${{\text{K}}_{\text{a}}}$ for ${\text{ClC}}{{\text{H}}_2}{\text{COOH}}$ is $1.35\quad {10^{ - 3}}$

$ \Rightarrow {K_b} = c{\alpha ^2}$ $\therefore \alpha  = \sqrt {\dfrac{{{K_a}}}{c}} $ $ = \sqrt {\dfrac{{1.35 \times {{10}^{ - 3}}}}{{0.1}}} \quad (\therefore $ concentration of acid $ = 0.1{\text{M}})$

$\Rightarrow K_{b}-c \alpha^{2} $

$\therefore \alpha=\sqrt{\frac{K_{a}}{c}} $

$=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}} \quad(\therefore \text { concentration of acid }=0.1 \mathbf{M}) $

$\alpha=\sqrt{1.35 \times 10^{-3}} $

$=0.116 $

$\therefore\left[H^{+}\right]=c \alpha=0.1 \times 0.116 $

$\Rightarrow p H=-\log \left[H^{+}\right]=1.94$

${\text{ClC}}{{\text{H}}_2}{\text{COONa}}$ is the salt of a weak acid i.e., ${\text{ClC}}{{\text{H}}_2}{\text{COOH}}$ and a strong base i.e., ${\text{NaOH}}$.

${{\text{ClC}}{{\text{H}}_2}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{ClC}}{{\text{H}}_2}{\text{COOH}} + {\text{O}}{{\text{H}}^ - }} $ 

${{K_b} = \dfrac{{\left[ {{\text{ClC}}{{\text{H}}_2}{\text{COOH}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{ClC}}{{\text{H}}_2}{\text{CO}}{{\text{O}}^ - }} \right]}}} $ 

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

${{K_b} = \dfrac{{{{10}^{ - 14}}}}{{1.35 \times {{10}^{ - 3}}}}} $ 

${\quad  = 0.740 \times {{10}^{ - 11}}} $

Also, ${K_b} = \dfrac{{{x^2}}}{{0.1}}$ (where ${\text{x}}$ is the concentration of ${\text{O}}{{\text{H}}^ - }$and $\left. {{\text{ClC}}{{\text{H}}_2}{\text{COOH}}} \right)$

${0.740 \times {{10}^{ - 11}} = \dfrac{{{x^2}}}{{0.1}}} $ 

${0.074 \times {{10}^{ - 11}} = {x^2}} $ 

${ \Rightarrow {x^2} = 0.74 \times {{10}^{ - 12}}} $ 

${x = 0.86 \times {{10}^{ - 6}}} $ 

${\left[ {O{H^ - }} \right] = 0.86 \times {{10}^{ - 6}}} $ 

${\therefore \left[ {{H^ + }} \right] = \dfrac{{{K_w}}}{{0.86 \times {{10}^{ - 6}}}}} $ 

${ = \dfrac{{{{10}^{ - 14}}}}{{0.86 \times {{10}^{ - 6}}}}} $ 

${\left[ {{H^ + }} \right] = 1.162 \times {{10}^{ - 8}}} $ 

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

$ = 7.94$

65. Ionic product of water at $310\;{\text{K}}$ is $2.7\quad {10^{ - 14}}$. What is the ${\text{pH}}$ of neutral water at this temperature?

Ans: Ionic product,

${{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]} $ 

${\operatorname{Let} \left[ {{H^ + }} \right] = x} $

${{\text{ Since }}\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right],{K_w} = {x^2}.} $ 

${ \Rightarrow {K_w}{\text{ at }}310\;{\text{K is }}2.7 \times {{10}^{ - 14}}} $ 

${\therefore 2.7 \times {{10}^{ - 14}} = {x^2}} $ 

${ \Rightarrow x = 1.64 \times {{10}^{ - 7}}} $ 

${ \Rightarrow \left[ {{H^ + }} \right] = 1.64 \times {{10}^{ - 7}}} $ 

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

${ =  - \log \left[ {1.64 \times {{10}^{ - 7}}} \right]} $ 

${ = 6.78} $

Hence, the ${\text{pH}}$ of neutral water is 6.78.

66. Calculate the ${\text{pH}}$ of the resultant mixtures:

1. $10\;{\text{mL}}$ of $0.2{\text{MCa}}{({\text{OH}})_2} + 25\;{\text{mL}}$ of $0.1{\text{MHCl}}$

Ans: Moles of ${{\text{H}}_3}{{\text{O}}^ + } = \dfrac{{25 \times 0.1}}{{1000}} = .0025\;{\text{mol}}$

Moles of $O{H^ - } = \dfrac{{10 \times 0.2 \times 2}}{{1000}} = .0040\;{\text{mol}}$

Thus, excess of $O{H^ - } = .0015\;{\text{mol}}$

2. $10\;{\text{mL}}$ of $0.01{\text{M}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 10\;{\text{mL}}$ of $0.01{\text{MCa}}{({\text{OH}})_2}$

Ans: Moles of ${{\text{H}}_3}{{\text{O}}^ + } = \dfrac{{2 \times 10 \times 0.1}}{{1000}} = .0002\;{\text{mol}}$

Moles of $O{H^ - } = \dfrac{{2 \times 10 \times 0.1}}{{1000}} = .0002\;{\text{mol}}$

Since there is neither an excess of ${{\text{H}}_3}{{\text{O}}^ + }$or ${\text{O}}{{\text{H}}^ - }$

The solution is neutral. Hence, ${\text{pH}} = 7$.

3. $10\;{\text{mL}}$ of $0.1{\text{M}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 10\;{\text{mL}}$ of $0.1{\text{MKOH}}$

Ans: Moles of ${{\text{H}}_3}{O^ + } = \dfrac{{2 \times 10 \times 0.1}}{{1000}} = .002\;{\text{mol}}$

Moles of $O{H^ - } = \dfrac{{10 \times 0.1}}{{1000}} = 0.001\;{\text{mol}}$

Excess of ${{\text{H}}_3}{{\text{O}}^ + } = .001\;{\text{mol}}$

Thus, $\left[ {{H_3}{O^ + }} \right] = \dfrac{{.001}}{{20 \times {{10}^{ - 3}}}} = \dfrac{{{{10}^{ - 3}}}}{{20 \times {{10}^{ - 3}}}} = .05$ $\therefore pH =  - \log (0.05)$ $ = 1.30$

67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercuries iodide at $298\;{\text{K}}$ from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

Ans:

1. Silver chromate:${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4} \to 2{\text{A}}{{\text{g}}^ + } + {\text{CrO}}_4^{2 - }$

Then,

${K_{sp}} = {\left[ {A{g^ + }} \right]^2}\left[ {CrO_4^{2 - }} \right]$

Let the solubility of ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4}$ be ${\text{s}}$.

$ \Rightarrow \left[ {{\text{A}}{{\text{g}}^ + }} \right] = 2\;{\text{s and }}\left[ {{\text{CrO}}_4^{2 - }} \right] = s$

Then,

${{K_{sp}} = {{(2s)}^2} \cdot s = 4{s^3}} $ 

${ \Rightarrow 1.1 \times {{10}^{ - 12}} = 4{s^3}} $ 

${.275 \times {{10}^{ - 12}} = {s^3}} $ 

${s = 0.65 \times {{10}^{ - 4}}M} $

$ Molarity of Ag^{+} = 2s = 2x0.65x10^{ - 4} = 1.30x10^{ - 4}M $

Molarity of $CrO_4^{2 - } = {\text{s}} = 0.65 \times {10^{ - 4}}{\text{M}}$

2. Barium Chromate:

${\text{BaCr}}{{\text{O}}_4} \to {\text{B}}{{\text{a}}^{2 + }} + {\text{CrO}}_4^{2 - }$

Then, ${K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {CrO_4^{2 - }} \right]$

Let the solubility of ${\text{BaCr}}{{\text{O}}_4}$ be ${\text{s}}$.

So, $\left[ {B{a^{2 + }}} \right] = {\text{s}}$ and $\left[ {{\text{CrO}}_4^{2 - }} \right] = {\text{s}} \Rightarrow {K_{sp}} = {s^2}$

$ \Rightarrow 1.2 \times {10^{ - 10}} = {s^2}$

$ \Rightarrow s = 1.09 \times {10^{ - 5}}M$

Molarity of $B{a^{2 + }} = $ Molarity of $CrO_4^{2 - } = s = 1.09 \times {10^{ - 5}}M$

3. Ferric Hydroxide:

${{\text{Fe}}{{({\text{OH}})}_3} \to {\text{F}}{{\text{e}}^2} + 3{\text{O}}{{\text{H}}^ - }} $ 

${{K_{sp}} = \left[ {{\text{F}}{{\text{e}}^{2 + }}} \right]{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^3}} $

Let ${\text{s}}$ be the solubility of ${\text{Fe}}{({\text{OH}})_3}$

Molarity of $O{H^ - } = 3s = 4.17 \times {10^{10}}M$

4. Lead Chloride:

${{\text{PbC}}{{\text{l}}_2} \to {\text{P}}{{\text{b}}^{2 + }} + 2{\text{C}}{{\text{l}}^ - }} $ 

${{K_{SP}} = \left[ {{\text{P}}{{\text{b}}^{2 + }}} \right]{{\left[ {{\text{C}}{{\text{l}}^ - }} \right]}^2}} $

Let ${{\text{K}}_{{\text{SP}}}}$ be the solubility of ${\text{PbC}}{{\text{l}}_2}$.

${\left[ {P{B^{2 + }}} \right] = s{\text{ and }}\left[ {C{l^ - }} \right] = 2s} $ 

${{\text{ Thus, }}{K_{sp}} = s.{{(2s)}^2}} $ 

${\quad  = 4{s^3}} $ 

${ \Rightarrow 1.6 \times {{10}^{ - 5}} = 4{s^3}} $ 

${ \Rightarrow 0.4 \times {{10}^{ - 5}} = {s^3}} $ 

${4 \times {{10}^{ - 6}} = {s^3} \Rightarrow 1.58 \times {{10}^{ - 2}}M = S.1}$

Molarity of $P{B^{2 + }} = s = 1.58 \times {10^{ - 2}}M$

Molarity of chloride $ = 2s = 3.16 \times {10^{ - 2}}{\text{M}}$

5. Mercurous Iodide:

${{\text{H}}{{\text{g}}_2}{{\text{I}}_2} \to {\text{H}}{{\text{g}}^{2 + }} + 2{I^ - }} $ 

${{K_{sp}} = \left[ {{\text{Hg}}_2^{2 + }} \right]{{\left[ {{I^ - }} \right]}^2}} $

Let ${\text{s}}$ be the solubility of ${\text{H}}{{\text{g}}_2}{{\text{I}}_2}$.

$ \Rightarrow \left[ {{\text{Hg}}_2^{2 + }} \right] = s{\text{ and }}\left[ {{I^ - }} \right] = 2s$

Thus, ${K_{sp}} = s{(2s)^2} \Rightarrow {K_{sp}} = 4{s^3}$

$4.5 \times {10^{ - 29}} = 4{s^3}$

${1.125 \times {{10}^{ - 29}} = {s^3}} $ 

${ \Rightarrow s = 2.24 \times {{10}^{ - 10}}{\text{M}}} $

Molarity of ${\text{Hg}}_2^{2 + } = s = 2.24 \times {10^{ - 10}}{\text{M}}$

Molarity of ${I^ - } = 2s = 4.48 \times {10^{ - 10}}M$

68. The solubility product constant of ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4}$ and ${\text{AgBr}}$ are $1.1 \times {10^{ - 12}}$ and $5.0 \times {10^{13}}$ respectively. Calculate the ratio of the molarities of their saturated solutions.

Ans: Let ${\text{s}}$ be the solubility of ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4}$

Then, ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4} \to {\text{A}}{{\text{g}}^{2 + }} + 2{\text{CrO}}_4^{2 - }$

${{K_{sp}} = {{(2s)}^2} \cdot s = 4{s^3}} $ 

${1.1 \times {{10}^{ - 12}} = 4{s^3}} $ 

${s = 6.5 \times {{10}^{ - 5}}M} $

Let ${{\text{s}}^\prime }$ be the solubility of ${\text{AgBr}}$.

${AgB{r_{(s)}} \leftrightarrow A{g^ + } + B{r^ - }} $ 

${{K_{sp}} = {s^{\prime 2}} = 5.0 \times {{10}^{ - 13}}} $ 

${\therefore {s^\prime } = 7.07 \times {{10}^{ - 7}}{\text{M}}} $

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

$\dfrac{s}{{{s^\prime }}} = \dfrac{{6.5 \times {{10}^{ - 5}}M}}{{7.07 \times {{10}^{ - 7}}M}} = 91.9$

69. Equal volumes of $0.002{\text{M}}$ solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate?

(For cupric iodate $\left. {{{\text{K}}_{{\text{sp}}}} = 7.4 \times {{10}^{ - 8}}} \right)$

Ans: When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., $0.001{\text{M}}$. Then,

${{\text{Nal}}{{\text{O}}_3} \to {\text{N}}{{\text{a}}^ + } + {\text{lO}}_3^ - } $ 

${0.001{\text{M}}}{0.001{\text{M}}} $ 

${{\text{Cu}}{{\left( {{\text{Cl}}{{\text{O}}_3}} \right)}_2} \to {\text{C}}{{\text{u}}^{2 + }}}{ + 2{\text{ClO}}_3^ - } $ 

${0.001{\text{M}}} {0.001M} $

Now, the solubility equilibrium for copper iodate can be written as:

$Cu(lO_{3})_{2} \to Cu_{(aq)}^{2 + } + 2lO_{3(aq)}^{ -} $

Ionic product of copper iodate:

${ = \left[ {C{u^{2 + }}} \right] \times {{\left[ {lO_3^ - } \right]}^2}} $ 

${ = (0.001){{(0.001)}^2}} $ 

${ = 1 \times {{10}^{ - 9}}} $

Since the ionic product $(1 \times \left. {{{10}^{ - 9}}} \right)$ is less than ${{\text{K}}_{{\text{sp}}}}\left( {7.4 \times {{10}^{ - 8}}} \right)$, precipitation will not occur.

70. The ionization constant of benzoic acid is $6.46 \times {10^{ - 5}}$ and ${{\text{K}}_{{\text{sp}}}}$ for silver benzoate is $2.5 \times {10^{ - 13}}$. How many times is silver benzoate more soluble in a buffer of ${\text{pH}} = 3.19$ compared to its solubility in pure water?

Ans: ${{\text{ Since pH}} = 3.19{\text{, }}} $ 

${\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 6.46 \times {{10}^{ - 4}}{\text{M}}} $ 

${{{\text{C}}_6}{{\text{H}}_5}{\text{COOH}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_3}{\text{O}}} $ 

${{K_a} = \dfrac{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - }} \right]\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]}}{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{COOH}}} \right]}}} $ 

${\dfrac{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{COOH}}} \right]}}{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - }} \right]}} = \dfrac{{\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]}}{{{K_a}}} = \dfrac{{6.46 \times {{10}^{ - 4}}}}{{6.46 \times {{10}^{ - 5}}}} = 10} $

Let the solubility of ${{\text{C}}_6}{{\text{H}}_5}{\text{COOAg}}$ be ${\text{xmol}}/{\text{L}}$.

Then,

${\left[ {A{g^ + }} \right] = x} $ 

${\left[ {{C_6}{H_5}COOH} \right] + \left[ {{C_6}{H_5}CO{O^ - }} \right] = x} $ 

${10\left[ {{C_6}{H_5}CO{O^ - }} \right] + \left[ {{C_6}{H_5}CO{O^ - }} \right] = x} $ 

${\left[ {{C_6}{H_5}CO{O^ - }} \right] = \dfrac{x}{{11}}} $ 

${{K_{sp}}\left[ {A{g^ + }} \right]\left[ {{C_6}{H_5}CO{O^ - }} \right]} $ 

${2.5 \times {{10}^{ - 13}} = x\left( {\dfrac{x}{{11}}} \right)} $ 

${x = 1.66 \times {{10}^{ - 6}}\;{\text{mol}}/{\text{L}}} $

Thus, the solubility of silver benzoate in a ${\text{pH}}\quad 3.19$ solution is $1.66 \times {10^{ - 6}}\;{\text{mol}}/{\text{L}}$. Now, let the solubility of ${{\text{C}}_6}{{\text{H}}_5}{\text{COOAg}}$ be ${x^\prime }{\text{mol}}/{\text{L}}$.

Then, $\left[ {A{g^ + }} \right] = {x^\prime }M$ and $\left[ {{C_6}{H_5}CO{O^ - }} \right] = {x^\prime }M$

${{K_{sp}} = \left[ {{\text{A}}{{\text{g}}^ + }} \right]\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - }} \right]} $ 

${{K_{sp}} = {{\left( {{x^\prime }} \right)}^2}} $ 

${{x^\prime } = \sqrt {{K_{sp}}}  = \sqrt {2.5 \times {{10}^{ - 13}}}  = 5 \times {{10}^{ - 7}}\;{\text{mol}}/{\text{L}}} $ 

${\therefore \dfrac{x}{{{x^\prime }}} = \dfrac{{1.66 \times {{10}^{ - 6}}}}{{5 \times {{10}^{ - 7}}}} = 3.32} $

Hence, ${{\text{C}}_6}{{\text{H}}_5}{\text{COOAg}}$ is approximately 3.317 times more soluble in a low ${\text{pH}}$ solution.

71. 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, $\left. {{{\text{K}}_{{\text{sp}}}} = 6.3\quad {{10}^{ - 18}}} \right)$

Ans:  Let the maximum concentration of each solution be ${\text{xmol}}/{\text{L}}$. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., ${\text{x}}/2$.

$\therefore \left[ {FeS{O^4}} \right] = \left[ {N{a_2}S} \right] = \dfrac{x}{2}M$

Then, $\left[ {F{e^{2 + }}} \right] = \left[ {FeS{O^4}} \right] = \dfrac{x}{2}M$

Also, $\left[ {{S^{2 - }}} \right] = \left[ {N{a_2}S} \right] = \dfrac{x}{2}M$

$Fe{S_{(x)}} \leftrightarrow Fe_{(aq)}^{2 + } + S_{(aq)}^{2 - }$

${{K_{sp}} = \left[ {F{e^{2 + }}} \right]\left[ {{S^{2 - }}} \right]} $ 

${6.3 \times {{10}^{ - 18}} = \left( {\dfrac{x}{2}} \right)\left( {\dfrac{x}{2}} \right)} $ 

${\dfrac{{{x^2}}}{4} = 6.3 \times {{10}^{ - 18}}} $ 

${ \Rightarrow x = 5.02 \times {{10}^{ - 9}}} $

If the concentrations of both solutions are equal to or less than $5.02 \times {10^{ - 9}}{\text{M}}$, then there will be no precipitation of iron sulphide.

72. What is the minimum volume of water required to dissolve $1\;{\text{g}}$ of calcium sulphate at 298 K? (For calcium sulphate, ${K_{{\text{sp }}}}$ is $\left. {9.1\quad {{10}^{ - 6}}} \right)$

Ans:  ${{\text{CaS}}{{\text{O}}_{4(s)}} \leftrightarrow {\text{C}}{{\text{a}}^{2 + }}_{(aq)} + SO_{4(aq)}^{2 - }} $ 

${{K_{sp}} = \left[ {{\text{C}}{{\text{a}}^{2 + }}} \right]\left[ {{\text{SO}}_4^{2 - }} \right]} $

Let the solubility of ${\text{CaS}}{{\text{O}}_4}$ be ${\text{s}}$.

Then, ${K_{sp}} = {s^2}$

${9.1 \times {{10}^{ - 6}} = {s^2}} $ 

${s = 3.02 \times {{10}^{ - 3}}\;{\text{mol}}/{\text{L}}} $

Molecular mass of ${\text{CaS}}{{\text{O}}_4} = 136\;{\text{g}}/{\text{mol}}$

Solubility of ${\text{CaS}}{{\text{O}}_4}$ in gram/L

${ = 3.02 \times {{10}^{ - 3}} \times 136} $ 

${ = 0.41\;{\text{g}}/{\text{L}}} $

This means that we need $1\;{\text{L}}$ of water to dissolve $0.41\;{\text{g}}$ of ${\text{CaS}}{{\text{O}}_4}$

Therefore, to dissolve $1\;{\text{g}}$ of ${\text{CaS}}{{\text{O}}_4}$ we require $ = \dfrac{1}{{0.41}}\;{\text{L}} = 2.44\;{\text{L}}$ of water.

73. The concentration of sulphide ion in $0.1{\text{MHCl}}$ solution saturated with hydrogen sulphide is 1.0 ${10^{ - 19}}{\text{M}}$. If $10\;{\text{mL}}$ of this is added to $5\;{\text{mL}}$ of $0.04{\text{M}}$ solution of the following:

${\text{FeS}}{{\text{O}}_4},{\text{MnC}}{{\text{l}}_2},{\text{ZnC}}{{\text{l}}_2}$ and ${\text{CdC}}{{\text{l}}_2}$. in which of these solutions precipitation will take place?

Givěn ${K_{sp}}$ for ${\text{Fes}} = 6.3 \times {10^{ - 18}},{\text{MnS}} = 2.5 \times {10^{ - 13}},{\text{ZnS}} = 1.6 \times {10^{ - 24}}$

$CdS = 8.0 \times {10^{ - 27}}$

Ans: For precipitation to take place, it is required that the calculated ionic product exceeds the ${{\text{K}}_{{\text{sp}}}}$ value.

Before mixing:

${\left[ {{S^{2 - }}} \right] = 1.0 \times {{10}^{ - 19}}M\left[ {{M^{2 + }}} \right] = 0.04M} $

${{\text{ volume }} = 10\;{\text{mL}}\quad {\text{ volume }} = 5\;{\text{mL}}} $

After mixing:

$\left[ {{S^{2 - }}} \right] = ?\quad \left[ {{M^{2 + }}} \right] = ?$

${{\text{ volume }} = (10 + 5) = 15\;{\text{mL}}\quad {\text{ volume }} = 15\;{\text{mL}}} $

${\left[ {{S^{2 - }}} \right] = \dfrac{{1.0 \times {{10}^{ - 19}} \times 10}}{{15}} = 6.67 \times {{10}^{ - 20}}{\text{M}}} $

${\left[ {{M^{2 + }}} \right] = \dfrac{{0.04 \times 5}}{{15}} = 1.33 \times {{10}^{ - 2}}{\text{M}}} $

${{\text{ Ionic product }} = \left[ {{M^{2 + }}} \right]\left[ {{S^{2 - }}} \right]} $

${ = \left( {1.33 \times {{10}^{ - 2}}} \right)\left( {6.67 \times {{10}^{ - 20}}} \right)} $

${ = 8.87 \times {{10}^{ - 22}}} $

This ionic product exceeds the ${{\text{K}}_{{\text{spof }}}}{\text{Zns}}$ and ${\text{CdS}}$. Therefore, precipitation will occur in ${\text{CdC}}{{\text{l}}_2}$ and ${\text{ZnC}}{{\text{l}}_2}$ solutions.

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