QBManager

Ionic Equilibrium

229413 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $1^{- 31}, 1^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$∙ CuS \Rightarrow S^{2} = 10^{- 37}$ $S = 3.16 \times 10^{- 19}$
- ${Ag}_{2} S \Rightarrow 4 S^{3} = 10^{- 44}$
$S = 1.357 \times 10^{- 15}$
- $H_{g} S \Rightarrow S^{2} = 10^{- 54}$
$S = 1 \times 10^{- 27}$
So, order of solubility is-
${Ag}_{2} S > CuS > HgS$

BITSAT-2010

Ionic Equilibrium

229415 If $K_{sp}$ of ${Ag}_{2} S$ is $10^{- 17}$ , the solubility of ${Ag}_{2} S$ in $0.1 M$ solution of ${Na}_{2} S$ will be

1

10^{- 8}

2

5 \times 10^{- 9}

3

10^{- 15}

4

10^{- 16}

Explanation:

$\begin{aligned} {Ag}_{2} \underset{↓}{S} + {Na}_{2} \underset{↓}{S} \\ S^{2 -} ⟶ commonion \\ {Na}_{2} S 2 {Na}^{+} + S^{2 -} \\ S 2 S S \\ 0.1 m 2 \times 0.1 m 0.1 m \end{aligned}$
${Ag}_{2} S$ ? $2 {Ag}^{+} + S^{2 -}$
$x 2 x 0.1$
$K_{sp} = {[{Ag}^{+}]}^{2} \cdot [S^{2 -}]$
$K_{sp} = [2 x]^{2} [x]$
$K_{sp} = [4 x]^{2} [10^{- 1}]$
$10^{- 17} = [4 x^{2}] \cdot [10^{- 1}]$
$4 x^{2} = \frac{10^{- 17}}{10^{- 1}} = 10^{- 16}$
$x = \sqrt{\frac{10^{- 16}}{4}}$
$x = \frac{1}{2} \sqrt{10^{- 16}}$
$x = \frac{1}{2} \times 10^{- 8}$
$x = 5 \times 10^{- 9} m$

CG PET-2009

Ionic Equilibrium

229416 If solubility of $M (OH)_{3}$ is $S$ , its solubility product will be:

1

108 S^{5}

2

27 S^{3}

3

4 S^{4}

4

27 S^{4}

Explanation:

$M (OH)_{3}$ is ${AB}_{3}$ Type salt its solubility product will be-
$\begin{aligned} \begin{array}{r} M (OH)_{3} ?? \\ M^{3 +} + 3 {OH}^{-} \\ S \end{array} \\ K_{sp} = [M^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ K_{sp} = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \end{aligned}$

CG PET-2019

Ionic Equilibrium

229417 The product of $H^{+}$ and ${OH}^{-}$ of water will be

1

K_{w} = 10^{- 12}

2

K_{w} = 10^{- 14}

3

K_{w} = 10^{- 11}

4

K_{w} = 10^{- 10}

Explanation:

The product of the molar concentration of $H^{+}$ and ${OH}^{-}$ ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water.
At $298 K$ , For pure water
$\begin{aligned} [H^{+}] . [{OH}^{-}] = 1 \times 10^{- 7} mol / {dm}^{- 3} \\ K_{w} = [H^{+}] [{OH}^{-}] \\ = [1 \times 10^{- 7}] . [1 \times 10^{- 7}] \\ K_{w} = 1 \times 10^{- 14} \end{aligned}$

CG PET -2006

Ionic Equilibrium

229413 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $1^{- 31}, 1^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$∙ CuS \Rightarrow S^{2} = 10^{- 37}$ $S = 3.16 \times 10^{- 19}$
- ${Ag}_{2} S \Rightarrow 4 S^{3} = 10^{- 44}$
$S = 1.357 \times 10^{- 15}$
- $H_{g} S \Rightarrow S^{2} = 10^{- 54}$
$S = 1 \times 10^{- 27}$
So, order of solubility is-
${Ag}_{2} S > CuS > HgS$

BITSAT-2010

Ionic Equilibrium

229415 If $K_{sp}$ of ${Ag}_{2} S$ is $10^{- 17}$ , the solubility of ${Ag}_{2} S$ in $0.1 M$ solution of ${Na}_{2} S$ will be

1

10^{- 8}

2

5 \times 10^{- 9}

3

10^{- 15}

4

10^{- 16}

Explanation:

$\begin{aligned} {Ag}_{2} \underset{↓}{S} + {Na}_{2} \underset{↓}{S} \\ S^{2 -} ⟶ commonion \\ {Na}_{2} S 2 {Na}^{+} + S^{2 -} \\ S 2 S S \\ 0.1 m 2 \times 0.1 m 0.1 m \end{aligned}$
${Ag}_{2} S$ ? $2 {Ag}^{+} + S^{2 -}$
$x 2 x 0.1$
$K_{sp} = {[{Ag}^{+}]}^{2} \cdot [S^{2 -}]$
$K_{sp} = [2 x]^{2} [x]$
$K_{sp} = [4 x]^{2} [10^{- 1}]$
$10^{- 17} = [4 x^{2}] \cdot [10^{- 1}]$
$4 x^{2} = \frac{10^{- 17}}{10^{- 1}} = 10^{- 16}$
$x = \sqrt{\frac{10^{- 16}}{4}}$
$x = \frac{1}{2} \sqrt{10^{- 16}}$
$x = \frac{1}{2} \times 10^{- 8}$
$x = 5 \times 10^{- 9} m$

CG PET-2009

Ionic Equilibrium

229416 If solubility of $M (OH)_{3}$ is $S$ , its solubility product will be:

1

108 S^{5}

2

27 S^{3}

3

4 S^{4}

4

27 S^{4}

Explanation:

$M (OH)_{3}$ is ${AB}_{3}$ Type salt its solubility product will be-
$\begin{aligned} \begin{array}{r} M (OH)_{3} ?? \\ M^{3 +} + 3 {OH}^{-} \\ S \end{array} \\ K_{sp} = [M^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ K_{sp} = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \end{aligned}$

CG PET-2019

Ionic Equilibrium

229417 The product of $H^{+}$ and ${OH}^{-}$ of water will be

1

K_{w} = 10^{- 12}

2

K_{w} = 10^{- 14}

3

K_{w} = 10^{- 11}

4

K_{w} = 10^{- 10}

Explanation:

The product of the molar concentration of $H^{+}$ and ${OH}^{-}$ ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water.
At $298 K$ , For pure water
$\begin{aligned} [H^{+}] . [{OH}^{-}] = 1 \times 10^{- 7} mol / {dm}^{- 3} \\ K_{w} = [H^{+}] [{OH}^{-}] \\ = [1 \times 10^{- 7}] . [1 \times 10^{- 7}] \\ K_{w} = 1 \times 10^{- 14} \end{aligned}$

CG PET -2006

Ionic Equilibrium

229413 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $1^{- 31}, 1^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$∙ CuS \Rightarrow S^{2} = 10^{- 37}$ $S = 3.16 \times 10^{- 19}$
- ${Ag}_{2} S \Rightarrow 4 S^{3} = 10^{- 44}$
$S = 1.357 \times 10^{- 15}$
- $H_{g} S \Rightarrow S^{2} = 10^{- 54}$
$S = 1 \times 10^{- 27}$
So, order of solubility is-
${Ag}_{2} S > CuS > HgS$

BITSAT-2010

Ionic Equilibrium

229415 If $K_{sp}$ of ${Ag}_{2} S$ is $10^{- 17}$ , the solubility of ${Ag}_{2} S$ in $0.1 M$ solution of ${Na}_{2} S$ will be

1

10^{- 8}

2

5 \times 10^{- 9}

3

10^{- 15}

4

10^{- 16}

Explanation:

$\begin{aligned} {Ag}_{2} \underset{↓}{S} + {Na}_{2} \underset{↓}{S} \\ S^{2 -} ⟶ commonion \\ {Na}_{2} S 2 {Na}^{+} + S^{2 -} \\ S 2 S S \\ 0.1 m 2 \times 0.1 m 0.1 m \end{aligned}$
${Ag}_{2} S$ ? $2 {Ag}^{+} + S^{2 -}$
$x 2 x 0.1$
$K_{sp} = {[{Ag}^{+}]}^{2} \cdot [S^{2 -}]$
$K_{sp} = [2 x]^{2} [x]$
$K_{sp} = [4 x]^{2} [10^{- 1}]$
$10^{- 17} = [4 x^{2}] \cdot [10^{- 1}]$
$4 x^{2} = \frac{10^{- 17}}{10^{- 1}} = 10^{- 16}$
$x = \sqrt{\frac{10^{- 16}}{4}}$
$x = \frac{1}{2} \sqrt{10^{- 16}}$
$x = \frac{1}{2} \times 10^{- 8}$
$x = 5 \times 10^{- 9} m$

CG PET-2009

Ionic Equilibrium

229416 If solubility of $M (OH)_{3}$ is $S$ , its solubility product will be:

1

108 S^{5}

2

27 S^{3}

3

4 S^{4}

4

27 S^{4}

Explanation:

$M (OH)_{3}$ is ${AB}_{3}$ Type salt its solubility product will be-
$\begin{aligned} \begin{array}{r} M (OH)_{3} ?? \\ M^{3 +} + 3 {OH}^{-} \\ S \end{array} \\ K_{sp} = [M^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ K_{sp} = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \end{aligned}$

CG PET-2019

Ionic Equilibrium

229417 The product of $H^{+}$ and ${OH}^{-}$ of water will be

1

K_{w} = 10^{- 12}

2

K_{w} = 10^{- 14}

3

K_{w} = 10^{- 11}

4

K_{w} = 10^{- 10}

Explanation:

The product of the molar concentration of $H^{+}$ and ${OH}^{-}$ ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water.
At $298 K$ , For pure water
$\begin{aligned} [H^{+}] . [{OH}^{-}] = 1 \times 10^{- 7} mol / {dm}^{- 3} \\ K_{w} = [H^{+}] [{OH}^{-}] \\ = [1 \times 10^{- 7}] . [1 \times 10^{- 7}] \\ K_{w} = 1 \times 10^{- 14} \end{aligned}$

CG PET -2006

Ionic Equilibrium

229413 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $1^{- 31}, 1^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$∙ CuS \Rightarrow S^{2} = 10^{- 37}$ $S = 3.16 \times 10^{- 19}$
- ${Ag}_{2} S \Rightarrow 4 S^{3} = 10^{- 44}$
$S = 1.357 \times 10^{- 15}$
- $H_{g} S \Rightarrow S^{2} = 10^{- 54}$
$S = 1 \times 10^{- 27}$
So, order of solubility is-
${Ag}_{2} S > CuS > HgS$

BITSAT-2010

Ionic Equilibrium

229415 If $K_{sp}$ of ${Ag}_{2} S$ is $10^{- 17}$ , the solubility of ${Ag}_{2} S$ in $0.1 M$ solution of ${Na}_{2} S$ will be

1

10^{- 8}

2

5 \times 10^{- 9}

3

10^{- 15}

4

10^{- 16}

Explanation:

$\begin{aligned} {Ag}_{2} \underset{↓}{S} + {Na}_{2} \underset{↓}{S} \\ S^{2 -} ⟶ commonion \\ {Na}_{2} S 2 {Na}^{+} + S^{2 -} \\ S 2 S S \\ 0.1 m 2 \times 0.1 m 0.1 m \end{aligned}$
${Ag}_{2} S$ ? $2 {Ag}^{+} + S^{2 -}$
$x 2 x 0.1$
$K_{sp} = {[{Ag}^{+}]}^{2} \cdot [S^{2 -}]$
$K_{sp} = [2 x]^{2} [x]$
$K_{sp} = [4 x]^{2} [10^{- 1}]$
$10^{- 17} = [4 x^{2}] \cdot [10^{- 1}]$
$4 x^{2} = \frac{10^{- 17}}{10^{- 1}} = 10^{- 16}$
$x = \sqrt{\frac{10^{- 16}}{4}}$
$x = \frac{1}{2} \sqrt{10^{- 16}}$
$x = \frac{1}{2} \times 10^{- 8}$
$x = 5 \times 10^{- 9} m$

CG PET-2009

Ionic Equilibrium

229416 If solubility of $M (OH)_{3}$ is $S$ , its solubility product will be:

1

108 S^{5}

2

27 S^{3}

3

4 S^{4}

4

27 S^{4}

Explanation:

$M (OH)_{3}$ is ${AB}_{3}$ Type salt its solubility product will be-
$\begin{aligned} \begin{array}{r} M (OH)_{3} ?? \\ M^{3 +} + 3 {OH}^{-} \\ S \end{array} \\ K_{sp} = [M^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ K_{sp} = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \end{aligned}$

CG PET-2019

Ionic Equilibrium

229417 The product of $H^{+}$ and ${OH}^{-}$ of water will be

1

K_{w} = 10^{- 12}

2

K_{w} = 10^{- 14}

3

K_{w} = 10^{- 11}

4

K_{w} = 10^{- 10}

Explanation:

The product of the molar concentration of $H^{+}$ and ${OH}^{-}$ ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water.
At $298 K$ , For pure water
$\begin{aligned} [H^{+}] . [{OH}^{-}] = 1 \times 10^{- 7} mol / {dm}^{- 3} \\ K_{w} = [H^{+}] [{OH}^{-}] \\ = [1 \times 10^{- 7}] . [1 \times 10^{- 7}] \\ K_{w} = 1 \times 10^{- 14} \end{aligned}$

CG PET -2006

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