QBManager

Ionic Equilibrium

229374 If the concentration of ${Ag}^{+}$ ions in the saturated solution of ${Ag}_{2} {CO}_{3}$ is $1.20 \times 10^{- 4} mol . L^{- 1}$ . then find the solubility product of ${Ag}_{2} {CO}_{3}$ .

1

5.30 \times 10^{- 12}

2

4.50 \times 10^{- 11}

3

2.66 \times 10^{- 12}

4

6.90 \times 10^{- 12}

Explanation:

Given that, $[{Ag}^{+}] = 1.20 \times 10^{- 4} mol . L^{- 1}$ Silver carbonate dissociates as follows -
${Ag}_{2} {CO}_{3} \to 2 {Ag}^{+} + {CO}_{3}^{2 -}$
Let the solubility is $S$ . The solubility of silver will be $2 s$ as two moles ion are dissociated -
$\begin{aligned} K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}] \\ K_{sp} = (2 S)^{2} (S) \end{aligned}$
$\begin{aligned} K_{sp} = 4 S^{3} \\ K_{sp} = 4 \times {(1.20 \times 10^{- 4})}^{3} \\ K_{sp} = 6.91 \times 10^{- 12} \end{aligned}$

Shift-I

Ionic Equilibrium

229376 The solubility of $AgCl$ at $20^{\circ} C$ is $1.435 \times 10^{- 5} {gL}^{- 1}$ . The solubility product of $AgCl$ is

1

2 \times 10^{- 16}

2

108 \times 10^{- 3}

3

1.0 \times 10^{- 14}

4

1.035 \times 10^{- 5}

Explanation:

Solubility of $AgCl = 1.435 \times 10^{- 5} {gL}^{- 1}$
[Molecular weight of $AgCl = 143.5$ ]
We know that solubility of $AgCl$ in moles
$\begin{aligned} AgCl ◻ {Ag}^{+} + {Cl}^{-} \\ K_{sp} = [{Ag}^{+}] [{Cl}^{-}] \\ K_{sp} = (s)^{2} \\ ∵ & S = \frac{(1.435 \times 10^{- 5})}{143.5} = 1 \times 10^{- 7} \end{aligned}$
Therefore solubility product of $AgCl = S^{2}$

COMEDK-2018

Ionic Equilibrium

229377 The solubility of $AgCN$ in a buffer solution of $p H = 3$ is $x$ . The value of $x$ is :
[Assume : No cyano complex is found; $K_{sp} (AgCN) = 2.2 \times 10^{- 16}$ and $K_{a} (HCN) = 6.2 \times$ $10^{- 10}]$

1

0.625 \times 10^{- 6}

2

1.6 \times 10^{- 6}

3

2.2 \times 10^{- 16}

4

1.9 \times 10^{- 5}

Explanation:

Given $: K_{sp} (AgCN) = 2.2 \times 10^{- 16}$
$K_{a} (HCN) = 6.2 \times 10^{- 10}$
The solubility of $AgCN$ in a buffer solution of $pH = 3$ Let, solubility is $S$ then,
$AgCN ? {Ag}^{+} + {CN}^{-}$
$K_{sp} = S \times S = S^{2}$
$H^{+} + {CN}^{-}$ ?? $HCN, K = \frac{1}{K_{a}} = \frac{1}{6.2 \times 10^{- 10}}$
Therefore, $K_{sp} \times \frac{1}{K_{a}} = [{Ag}^{+}] [{CN}^{-}] \frac{[HCN]}{[H^{+}] [{CN}^{-}]}$ or $2.2 \times 10^{- 16} \times \frac{1}{6.2 \times 10^{- 10}} = \frac{S \times S}{10^{- 3}}$
$\begin{array}{ll} S^{2} = \frac{2.2 \times 10^{- 16} \times 10^{- 3}}{6.2 \times 10^{- 10}} = 0.354 \times 10^{- 9} \\ or & S^{2} = 3.54 \times 10^{- 10} \\ or & S = 1.9 \times 10^{- 5} \end{array}$

Shift-I

Ionic Equilibrium

229378 From the graph, the value of Henry's constant for the solubility of $HCl$ gas in cyclohexane is

1

10 k

torr

2 100 torr

3 50 torr

4

2.4 \times 10^{2}

torr

Explanation:

According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(X)$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law.
$P_{solute} \propto X_{solute}$ in solution
$P_{solute} = k_{H} X_{solute}$ in solution
Where, $k_{H} =$ Henry's constant
The slope of the straight line gives the value of $k_{H}$ .
$\begin{aligned} ∴ \tan θ & = k_{H} = \frac{50}{0.005} \\ = \frac{50000}{5} = 10000 \\ k_{H} & = 10 k torr \end{aligned}$

Shift-I

Ionic Equilibrium

229379 The solubility of $Ca (OH)_{2}$ in water is :
[Given : The solubility product of $Ca (OH)_{2}$ in water $= 5.5 \times 10^{- 6}]$

1

1.11 \times 10^{- 2}

2

1.11 \times 10^{- 6}

3

1.77 \times 10^{- 2}

4

1.77 \times 10^{- 6}

Explanation:

Given: $K_{sp} = 5.5 \times 10^{- 6}$
Let, the solubility of $Ca (OH)_{2}$ in water is $S$ ,
$\begin{aligned} Ca (OH)_{2} ◻ {Ca}^{2 +} + 2 {OH}^{-} \\ (S) (2 S) \\ K_{sp} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2} \\ K_{sp} = (S)^{1} \times (2 S)^{2} \\ 5.5 \times 10^{- 6} = 4 S^{3} \\ S = {(\frac{5.5 \times 10^{- 6}}{4})}^{1 / 3} \\ S = 1.11 \times 10^{- 2} \end{aligned}$

Shift-II

Ionic Equilibrium

229374 If the concentration of ${Ag}^{+}$ ions in the saturated solution of ${Ag}_{2} {CO}_{3}$ is $1.20 \times 10^{- 4} mol . L^{- 1}$ . then find the solubility product of ${Ag}_{2} {CO}_{3}$ .

1

5.30 \times 10^{- 12}

2

4.50 \times 10^{- 11}

3

2.66 \times 10^{- 12}

4

6.90 \times 10^{- 12}

Explanation:

Given that, $[{Ag}^{+}] = 1.20 \times 10^{- 4} mol . L^{- 1}$ Silver carbonate dissociates as follows -
${Ag}_{2} {CO}_{3} \to 2 {Ag}^{+} + {CO}_{3}^{2 -}$
Let the solubility is $S$ . The solubility of silver will be $2 s$ as two moles ion are dissociated -
$\begin{aligned} K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}] \\ K_{sp} = (2 S)^{2} (S) \end{aligned}$
$\begin{aligned} K_{sp} = 4 S^{3} \\ K_{sp} = 4 \times {(1.20 \times 10^{- 4})}^{3} \\ K_{sp} = 6.91 \times 10^{- 12} \end{aligned}$

Shift-I

Ionic Equilibrium

229376 The solubility of $AgCl$ at $20^{\circ} C$ is $1.435 \times 10^{- 5} {gL}^{- 1}$ . The solubility product of $AgCl$ is

1

2 \times 10^{- 16}

2

108 \times 10^{- 3}

3

1.0 \times 10^{- 14}

4

1.035 \times 10^{- 5}

Explanation:

Solubility of $AgCl = 1.435 \times 10^{- 5} {gL}^{- 1}$
[Molecular weight of $AgCl = 143.5$ ]
We know that solubility of $AgCl$ in moles
$\begin{aligned} AgCl ◻ {Ag}^{+} + {Cl}^{-} \\ K_{sp} = [{Ag}^{+}] [{Cl}^{-}] \\ K_{sp} = (s)^{2} \\ ∵ & S = \frac{(1.435 \times 10^{- 5})}{143.5} = 1 \times 10^{- 7} \end{aligned}$
Therefore solubility product of $AgCl = S^{2}$

COMEDK-2018

Ionic Equilibrium

229377 The solubility of $AgCN$ in a buffer solution of $p H = 3$ is $x$ . The value of $x$ is :
[Assume : No cyano complex is found; $K_{sp} (AgCN) = 2.2 \times 10^{- 16}$ and $K_{a} (HCN) = 6.2 \times$ $10^{- 10}]$

1

0.625 \times 10^{- 6}

2

1.6 \times 10^{- 6}

3

2.2 \times 10^{- 16}

4

1.9 \times 10^{- 5}

Explanation:

Given $: K_{sp} (AgCN) = 2.2 \times 10^{- 16}$
$K_{a} (HCN) = 6.2 \times 10^{- 10}$
The solubility of $AgCN$ in a buffer solution of $pH = 3$ Let, solubility is $S$ then,
$AgCN ? {Ag}^{+} + {CN}^{-}$
$K_{sp} = S \times S = S^{2}$
$H^{+} + {CN}^{-}$ ?? $HCN, K = \frac{1}{K_{a}} = \frac{1}{6.2 \times 10^{- 10}}$
Therefore, $K_{sp} \times \frac{1}{K_{a}} = [{Ag}^{+}] [{CN}^{-}] \frac{[HCN]}{[H^{+}] [{CN}^{-}]}$ or $2.2 \times 10^{- 16} \times \frac{1}{6.2 \times 10^{- 10}} = \frac{S \times S}{10^{- 3}}$
$\begin{array}{ll} S^{2} = \frac{2.2 \times 10^{- 16} \times 10^{- 3}}{6.2 \times 10^{- 10}} = 0.354 \times 10^{- 9} \\ or & S^{2} = 3.54 \times 10^{- 10} \\ or & S = 1.9 \times 10^{- 5} \end{array}$

Shift-I

Ionic Equilibrium

229378 From the graph, the value of Henry's constant for the solubility of $HCl$ gas in cyclohexane is

1

10 k

torr

2 100 torr

3 50 torr

4

2.4 \times 10^{2}

torr

Explanation:

According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(X)$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law.
$P_{solute} \propto X_{solute}$ in solution
$P_{solute} = k_{H} X_{solute}$ in solution
Where, $k_{H} =$ Henry's constant
The slope of the straight line gives the value of $k_{H}$ .
$\begin{aligned} ∴ \tan θ & = k_{H} = \frac{50}{0.005} \\ = \frac{50000}{5} = 10000 \\ k_{H} & = 10 k torr \end{aligned}$

Shift-I

Ionic Equilibrium

229379 The solubility of $Ca (OH)_{2}$ in water is :
[Given : The solubility product of $Ca (OH)_{2}$ in water $= 5.5 \times 10^{- 6}]$

1

1.11 \times 10^{- 2}

2

1.11 \times 10^{- 6}

3

1.77 \times 10^{- 2}

4

1.77 \times 10^{- 6}

Explanation:

Given: $K_{sp} = 5.5 \times 10^{- 6}$
Let, the solubility of $Ca (OH)_{2}$ in water is $S$ ,
$\begin{aligned} Ca (OH)_{2} ◻ {Ca}^{2 +} + 2 {OH}^{-} \\ (S) (2 S) \\ K_{sp} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2} \\ K_{sp} = (S)^{1} \times (2 S)^{2} \\ 5.5 \times 10^{- 6} = 4 S^{3} \\ S = {(\frac{5.5 \times 10^{- 6}}{4})}^{1 / 3} \\ S = 1.11 \times 10^{- 2} \end{aligned}$

Shift-II

Ionic Equilibrium

229374 If the concentration of ${Ag}^{+}$ ions in the saturated solution of ${Ag}_{2} {CO}_{3}$ is $1.20 \times 10^{- 4} mol . L^{- 1}$ . then find the solubility product of ${Ag}_{2} {CO}_{3}$ .

1

5.30 \times 10^{- 12}

2

4.50 \times 10^{- 11}

3

2.66 \times 10^{- 12}

4

6.90 \times 10^{- 12}

Explanation:

Given that, $[{Ag}^{+}] = 1.20 \times 10^{- 4} mol . L^{- 1}$ Silver carbonate dissociates as follows -
${Ag}_{2} {CO}_{3} \to 2 {Ag}^{+} + {CO}_{3}^{2 -}$
Let the solubility is $S$ . The solubility of silver will be $2 s$ as two moles ion are dissociated -
$\begin{aligned} K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}] \\ K_{sp} = (2 S)^{2} (S) \end{aligned}$
$\begin{aligned} K_{sp} = 4 S^{3} \\ K_{sp} = 4 \times {(1.20 \times 10^{- 4})}^{3} \\ K_{sp} = 6.91 \times 10^{- 12} \end{aligned}$

Shift-I

Ionic Equilibrium

229376 The solubility of $AgCl$ at $20^{\circ} C$ is $1.435 \times 10^{- 5} {gL}^{- 1}$ . The solubility product of $AgCl$ is

1

2 \times 10^{- 16}

2

108 \times 10^{- 3}

3

1.0 \times 10^{- 14}

4

1.035 \times 10^{- 5}

Explanation:

Solubility of $AgCl = 1.435 \times 10^{- 5} {gL}^{- 1}$
[Molecular weight of $AgCl = 143.5$ ]
We know that solubility of $AgCl$ in moles
$\begin{aligned} AgCl ◻ {Ag}^{+} + {Cl}^{-} \\ K_{sp} = [{Ag}^{+}] [{Cl}^{-}] \\ K_{sp} = (s)^{2} \\ ∵ & S = \frac{(1.435 \times 10^{- 5})}{143.5} = 1 \times 10^{- 7} \end{aligned}$
Therefore solubility product of $AgCl = S^{2}$

COMEDK-2018

Ionic Equilibrium

229377 The solubility of $AgCN$ in a buffer solution of $p H = 3$ is $x$ . The value of $x$ is :
[Assume : No cyano complex is found; $K_{sp} (AgCN) = 2.2 \times 10^{- 16}$ and $K_{a} (HCN) = 6.2 \times$ $10^{- 10}]$

1

0.625 \times 10^{- 6}

2

1.6 \times 10^{- 6}

3

2.2 \times 10^{- 16}

4

1.9 \times 10^{- 5}

Explanation:

Given $: K_{sp} (AgCN) = 2.2 \times 10^{- 16}$
$K_{a} (HCN) = 6.2 \times 10^{- 10}$
The solubility of $AgCN$ in a buffer solution of $pH = 3$ Let, solubility is $S$ then,
$AgCN ? {Ag}^{+} + {CN}^{-}$
$K_{sp} = S \times S = S^{2}$
$H^{+} + {CN}^{-}$ ?? $HCN, K = \frac{1}{K_{a}} = \frac{1}{6.2 \times 10^{- 10}}$
Therefore, $K_{sp} \times \frac{1}{K_{a}} = [{Ag}^{+}] [{CN}^{-}] \frac{[HCN]}{[H^{+}] [{CN}^{-}]}$ or $2.2 \times 10^{- 16} \times \frac{1}{6.2 \times 10^{- 10}} = \frac{S \times S}{10^{- 3}}$
$\begin{array}{ll} S^{2} = \frac{2.2 \times 10^{- 16} \times 10^{- 3}}{6.2 \times 10^{- 10}} = 0.354 \times 10^{- 9} \\ or & S^{2} = 3.54 \times 10^{- 10} \\ or & S = 1.9 \times 10^{- 5} \end{array}$

Shift-I

Ionic Equilibrium

229378 From the graph, the value of Henry's constant for the solubility of $HCl$ gas in cyclohexane is

1

10 k

torr

2 100 torr

3 50 torr

4

2.4 \times 10^{2}

torr

Explanation:

According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(X)$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law.
$P_{solute} \propto X_{solute}$ in solution
$P_{solute} = k_{H} X_{solute}$ in solution
Where, $k_{H} =$ Henry's constant
The slope of the straight line gives the value of $k_{H}$ .
$\begin{aligned} ∴ \tan θ & = k_{H} = \frac{50}{0.005} \\ = \frac{50000}{5} = 10000 \\ k_{H} & = 10 k torr \end{aligned}$

Shift-I

Ionic Equilibrium

229379 The solubility of $Ca (OH)_{2}$ in water is :
[Given : The solubility product of $Ca (OH)_{2}$ in water $= 5.5 \times 10^{- 6}]$

1

1.11 \times 10^{- 2}

2

1.11 \times 10^{- 6}

3

1.77 \times 10^{- 2}

4

1.77 \times 10^{- 6}

Explanation:

Given: $K_{sp} = 5.5 \times 10^{- 6}$
Let, the solubility of $Ca (OH)_{2}$ in water is $S$ ,
$\begin{aligned} Ca (OH)_{2} ◻ {Ca}^{2 +} + 2 {OH}^{-} \\ (S) (2 S) \\ K_{sp} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2} \\ K_{sp} = (S)^{1} \times (2 S)^{2} \\ 5.5 \times 10^{- 6} = 4 S^{3} \\ S = {(\frac{5.5 \times 10^{- 6}}{4})}^{1 / 3} \\ S = 1.11 \times 10^{- 2} \end{aligned}$

Shift-II

Ionic Equilibrium

229374 If the concentration of ${Ag}^{+}$ ions in the saturated solution of ${Ag}_{2} {CO}_{3}$ is $1.20 \times 10^{- 4} mol . L^{- 1}$ . then find the solubility product of ${Ag}_{2} {CO}_{3}$ .

1

5.30 \times 10^{- 12}

2

4.50 \times 10^{- 11}

3

2.66 \times 10^{- 12}

4

6.90 \times 10^{- 12}

Explanation:

Given that, $[{Ag}^{+}] = 1.20 \times 10^{- 4} mol . L^{- 1}$ Silver carbonate dissociates as follows -
${Ag}_{2} {CO}_{3} \to 2 {Ag}^{+} + {CO}_{3}^{2 -}$
Let the solubility is $S$ . The solubility of silver will be $2 s$ as two moles ion are dissociated -
$\begin{aligned} K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}] \\ K_{sp} = (2 S)^{2} (S) \end{aligned}$
$\begin{aligned} K_{sp} = 4 S^{3} \\ K_{sp} = 4 \times {(1.20 \times 10^{- 4})}^{3} \\ K_{sp} = 6.91 \times 10^{- 12} \end{aligned}$

Shift-I

Ionic Equilibrium

229376 The solubility of $AgCl$ at $20^{\circ} C$ is $1.435 \times 10^{- 5} {gL}^{- 1}$ . The solubility product of $AgCl$ is

1

2 \times 10^{- 16}

2

108 \times 10^{- 3}

3

1.0 \times 10^{- 14}

4

1.035 \times 10^{- 5}

Explanation:

Solubility of $AgCl = 1.435 \times 10^{- 5} {gL}^{- 1}$
[Molecular weight of $AgCl = 143.5$ ]
We know that solubility of $AgCl$ in moles
$\begin{aligned} AgCl ◻ {Ag}^{+} + {Cl}^{-} \\ K_{sp} = [{Ag}^{+}] [{Cl}^{-}] \\ K_{sp} = (s)^{2} \\ ∵ & S = \frac{(1.435 \times 10^{- 5})}{143.5} = 1 \times 10^{- 7} \end{aligned}$
Therefore solubility product of $AgCl = S^{2}$

COMEDK-2018

Ionic Equilibrium

229377 The solubility of $AgCN$ in a buffer solution of $p H = 3$ is $x$ . The value of $x$ is :
[Assume : No cyano complex is found; $K_{sp} (AgCN) = 2.2 \times 10^{- 16}$ and $K_{a} (HCN) = 6.2 \times$ $10^{- 10}]$

1

0.625 \times 10^{- 6}

2

1.6 \times 10^{- 6}

3

2.2 \times 10^{- 16}

4

1.9 \times 10^{- 5}

Explanation:

Given $: K_{sp} (AgCN) = 2.2 \times 10^{- 16}$
$K_{a} (HCN) = 6.2 \times 10^{- 10}$
The solubility of $AgCN$ in a buffer solution of $pH = 3$ Let, solubility is $S$ then,
$AgCN ? {Ag}^{+} + {CN}^{-}$
$K_{sp} = S \times S = S^{2}$
$H^{+} + {CN}^{-}$ ?? $HCN, K = \frac{1}{K_{a}} = \frac{1}{6.2 \times 10^{- 10}}$
Therefore, $K_{sp} \times \frac{1}{K_{a}} = [{Ag}^{+}] [{CN}^{-}] \frac{[HCN]}{[H^{+}] [{CN}^{-}]}$ or $2.2 \times 10^{- 16} \times \frac{1}{6.2 \times 10^{- 10}} = \frac{S \times S}{10^{- 3}}$
$\begin{array}{ll} S^{2} = \frac{2.2 \times 10^{- 16} \times 10^{- 3}}{6.2 \times 10^{- 10}} = 0.354 \times 10^{- 9} \\ or & S^{2} = 3.54 \times 10^{- 10} \\ or & S = 1.9 \times 10^{- 5} \end{array}$

Shift-I

Ionic Equilibrium

229378 From the graph, the value of Henry's constant for the solubility of $HCl$ gas in cyclohexane is

1

10 k

torr

2 100 torr

3 50 torr

4

2.4 \times 10^{2}

torr

Explanation:

According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(X)$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law.
$P_{solute} \propto X_{solute}$ in solution
$P_{solute} = k_{H} X_{solute}$ in solution
Where, $k_{H} =$ Henry's constant
The slope of the straight line gives the value of $k_{H}$ .
$\begin{aligned} ∴ \tan θ & = k_{H} = \frac{50}{0.005} \\ = \frac{50000}{5} = 10000 \\ k_{H} & = 10 k torr \end{aligned}$

Shift-I

Ionic Equilibrium

229379 The solubility of $Ca (OH)_{2}$ in water is :
[Given : The solubility product of $Ca (OH)_{2}$ in water $= 5.5 \times 10^{- 6}]$

1

1.11 \times 10^{- 2}

2

1.11 \times 10^{- 6}

3

1.77 \times 10^{- 2}

4

1.77 \times 10^{- 6}

Explanation:

Given: $K_{sp} = 5.5 \times 10^{- 6}$
Let, the solubility of $Ca (OH)_{2}$ in water is $S$ ,
$\begin{aligned} Ca (OH)_{2} ◻ {Ca}^{2 +} + 2 {OH}^{-} \\ (S) (2 S) \\ K_{sp} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2} \\ K_{sp} = (S)^{1} \times (2 S)^{2} \\ 5.5 \times 10^{- 6} = 4 S^{3} \\ S = {(\frac{5.5 \times 10^{- 6}}{4})}^{1 / 3} \\ S = 1.11 \times 10^{- 2} \end{aligned}$

Shift-II

Ionic Equilibrium

229374 If the concentration of ${Ag}^{+}$ ions in the saturated solution of ${Ag}_{2} {CO}_{3}$ is $1.20 \times 10^{- 4} mol . L^{- 1}$ . then find the solubility product of ${Ag}_{2} {CO}_{3}$ .

1

5.30 \times 10^{- 12}

2

4.50 \times 10^{- 11}

3

2.66 \times 10^{- 12}

4

6.90 \times 10^{- 12}

Explanation:

Given that, $[{Ag}^{+}] = 1.20 \times 10^{- 4} mol . L^{- 1}$ Silver carbonate dissociates as follows -
${Ag}_{2} {CO}_{3} \to 2 {Ag}^{+} + {CO}_{3}^{2 -}$
Let the solubility is $S$ . The solubility of silver will be $2 s$ as two moles ion are dissociated -
$\begin{aligned} K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}] \\ K_{sp} = (2 S)^{2} (S) \end{aligned}$
$\begin{aligned} K_{sp} = 4 S^{3} \\ K_{sp} = 4 \times {(1.20 \times 10^{- 4})}^{3} \\ K_{sp} = 6.91 \times 10^{- 12} \end{aligned}$

Shift-I

Ionic Equilibrium

229376 The solubility of $AgCl$ at $20^{\circ} C$ is $1.435 \times 10^{- 5} {gL}^{- 1}$ . The solubility product of $AgCl$ is

1

2 \times 10^{- 16}

2

108 \times 10^{- 3}

3

1.0 \times 10^{- 14}

4

1.035 \times 10^{- 5}

Explanation:

Solubility of $AgCl = 1.435 \times 10^{- 5} {gL}^{- 1}$
[Molecular weight of $AgCl = 143.5$ ]
We know that solubility of $AgCl$ in moles
$\begin{aligned} AgCl ◻ {Ag}^{+} + {Cl}^{-} \\ K_{sp} = [{Ag}^{+}] [{Cl}^{-}] \\ K_{sp} = (s)^{2} \\ ∵ & S = \frac{(1.435 \times 10^{- 5})}{143.5} = 1 \times 10^{- 7} \end{aligned}$
Therefore solubility product of $AgCl = S^{2}$

COMEDK-2018

Ionic Equilibrium

229377 The solubility of $AgCN$ in a buffer solution of $p H = 3$ is $x$ . The value of $x$ is :
[Assume : No cyano complex is found; $K_{sp} (AgCN) = 2.2 \times 10^{- 16}$ and $K_{a} (HCN) = 6.2 \times$ $10^{- 10}]$

1

0.625 \times 10^{- 6}

2

1.6 \times 10^{- 6}

3

2.2 \times 10^{- 16}

4

1.9 \times 10^{- 5}

Explanation:

Given $: K_{sp} (AgCN) = 2.2 \times 10^{- 16}$
$K_{a} (HCN) = 6.2 \times 10^{- 10}$
The solubility of $AgCN$ in a buffer solution of $pH = 3$ Let, solubility is $S$ then,
$AgCN ? {Ag}^{+} + {CN}^{-}$
$K_{sp} = S \times S = S^{2}$
$H^{+} + {CN}^{-}$ ?? $HCN, K = \frac{1}{K_{a}} = \frac{1}{6.2 \times 10^{- 10}}$
Therefore, $K_{sp} \times \frac{1}{K_{a}} = [{Ag}^{+}] [{CN}^{-}] \frac{[HCN]}{[H^{+}] [{CN}^{-}]}$ or $2.2 \times 10^{- 16} \times \frac{1}{6.2 \times 10^{- 10}} = \frac{S \times S}{10^{- 3}}$
$\begin{array}{ll} S^{2} = \frac{2.2 \times 10^{- 16} \times 10^{- 3}}{6.2 \times 10^{- 10}} = 0.354 \times 10^{- 9} \\ or & S^{2} = 3.54 \times 10^{- 10} \\ or & S = 1.9 \times 10^{- 5} \end{array}$

Shift-I

Ionic Equilibrium

229378 From the graph, the value of Henry's constant for the solubility of $HCl$ gas in cyclohexane is

1

10 k

torr

2 100 torr

3 50 torr

4

2.4 \times 10^{2}

torr

Explanation:

According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(X)$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law.
$P_{solute} \propto X_{solute}$ in solution
$P_{solute} = k_{H} X_{solute}$ in solution
Where, $k_{H} =$ Henry's constant
The slope of the straight line gives the value of $k_{H}$ .
$\begin{aligned} ∴ \tan θ & = k_{H} = \frac{50}{0.005} \\ = \frac{50000}{5} = 10000 \\ k_{H} & = 10 k torr \end{aligned}$

Shift-I

Ionic Equilibrium

229379 The solubility of $Ca (OH)_{2}$ in water is :
[Given : The solubility product of $Ca (OH)_{2}$ in water $= 5.5 \times 10^{- 6}]$

1

1.11 \times 10^{- 2}

2

1.11 \times 10^{- 6}

3

1.77 \times 10^{- 2}

4

1.77 \times 10^{- 6}

Explanation:

Given: $K_{sp} = 5.5 \times 10^{- 6}$
Let, the solubility of $Ca (OH)_{2}$ in water is $S$ ,
$\begin{aligned} Ca (OH)_{2} ◻ {Ca}^{2 +} + 2 {OH}^{-} \\ (S) (2 S) \\ K_{sp} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2} \\ K_{sp} = (S)^{1} \times (2 S)^{2} \\ 5.5 \times 10^{- 6} = 4 S^{3} \\ S = {(\frac{5.5 \times 10^{- 6}}{4})}^{1 / 3} \\ S = 1.11 \times 10^{- 2} \end{aligned}$

Shift-II