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CHXI07:EQUILIBRIUM

314952 The product of ionic concentration in a saturated solution of an electrolyte at a given temperature is constant and is known as

1 Ionic product of the electrolyte

2 Solubility product

3 Ionization constant

4 Dissociation constant

CHXI07:EQUILIBRIUM

314970 Consider the following sparingly soluble salts having same solubility product in their saturated solution in water:
(A) $XY$
(B) $X_{2} Y$
(C) ${XY}_{3}$
The correct order of increasing solubility of these salts in water will be

1 A

<

B

<

C

2 C

<

B

<

A

3 B

<

A

<

C

4 B

<

C

<

A

Explanation:

Let $S_{1}, S_{2}, S_{3}$ be the solubility of $XY$ , $X_{2} Y$ and ${XY}_{3}$ in water, respectively
$Salt Solubality product$
$\begin{matrix} XY & S_{1}^{2} \\ X_{2} Y & {3S}_{2}^{3} \\ X Y_{3} & {27S}_{3}^{4} \end{matrix}$
According to given condition,
$S_{1}^{2} {= 4S}_{2}^{3} {= 27S}_{3}^{4} or S_{1} {= 2S}_{2}^{3/2} = 3 \sqrt{3} S_{3}^{2}$
Since the practical value of solubility product is found to be very small. So, $S_{1} < S_{2} < S_{3}$

CHXI07:EQUILIBRIUM

314953 What is the minimum volume of water required to dissolve $1 g$ of calcium sulphate at $298 K$ ? (For calcium sulphate, $K_{s p}$ is $9.1 \times 10^{- 6}$ )?

1

5.43 L

2

2.43 L

3

8.43 L

4

7.43 L

Explanation:

$\begin{matrix} CaS O_{4} ⇌ C a^{2 +} {+ SO}_{4}^{2 -}; K_{SP} = 9 .1 \times 1 0^{- 6} \\ S S S \end{matrix}$
where $S$ is the solubility of ${CaSO}_{4}$ .
$K_{s p} = [{Ca}^{2 +}] [{SO}_{4}^{2 -}] = S . S = S^{2}$
$S = \sqrt{K_{s p}} = \sqrt{9.1 \times 10^{- 6}}$
$S = 3.017 \times 10^{- 3} M$
${CaSO}_{4}$ solubility $= 3.017 \times 10^{- 3} mol L^{- 1}$
The solubility in $g / L = 3 \times 10^{- 3} \times 136$
$= 0.410 g / L$
Hence, $0.41 g$ dissolve in $1 L$
$1 g$ will dissolve in $\frac{1}{0.41} = 2.43 L$ .

CHXI07:EQUILIBRIUM

314954 The correct representation for solubility product of ${SnS}_{2}$ is

1

[{Sn}^{4 +}] {[S^{2 -}]}^{2}

2

[{Sn}^{4 +}] [S^{2 -}]

3

[{Sn}^{4 +}] [2 S^{2 -}]

4

[{Sn}^{4 +}] {[2 S^{2 -}]}^{2}

Explanation:

$K_{SP}$ of ${SnS}_{2} = [{Sn}^{4 +}] {[S^{2 -}]}^{2}$

CHXI07:EQUILIBRIUM

314952 The product of ionic concentration in a saturated solution of an electrolyte at a given temperature is constant and is known as

1 Ionic product of the electrolyte

2 Solubility product

3 Ionization constant

4 Dissociation constant

CHXI07:EQUILIBRIUM

314970 Consider the following sparingly soluble salts having same solubility product in their saturated solution in water:
(A) $XY$
(B) $X_{2} Y$
(C) ${XY}_{3}$
The correct order of increasing solubility of these salts in water will be

1 A

<

B

<

C

2 C

<

B

<

A

3 B

<

A

<

C

4 B

<

C

<

A

Explanation:

Let $S_{1}, S_{2}, S_{3}$ be the solubility of $XY$ , $X_{2} Y$ and ${XY}_{3}$ in water, respectively
$Salt Solubality product$
$\begin{matrix} XY & S_{1}^{2} \\ X_{2} Y & {3S}_{2}^{3} \\ X Y_{3} & {27S}_{3}^{4} \end{matrix}$
According to given condition,
$S_{1}^{2} {= 4S}_{2}^{3} {= 27S}_{3}^{4} or S_{1} {= 2S}_{2}^{3/2} = 3 \sqrt{3} S_{3}^{2}$
Since the practical value of solubility product is found to be very small. So, $S_{1} < S_{2} < S_{3}$

CHXI07:EQUILIBRIUM

314953 What is the minimum volume of water required to dissolve $1 g$ of calcium sulphate at $298 K$ ? (For calcium sulphate, $K_{s p}$ is $9.1 \times 10^{- 6}$ )?

1

5.43 L

2

2.43 L

3

8.43 L

4

7.43 L

Explanation:

$\begin{matrix} CaS O_{4} ⇌ C a^{2 +} {+ SO}_{4}^{2 -}; K_{SP} = 9 .1 \times 1 0^{- 6} \\ S S S \end{matrix}$
where $S$ is the solubility of ${CaSO}_{4}$ .
$K_{s p} = [{Ca}^{2 +}] [{SO}_{4}^{2 -}] = S . S = S^{2}$
$S = \sqrt{K_{s p}} = \sqrt{9.1 \times 10^{- 6}}$
$S = 3.017 \times 10^{- 3} M$
${CaSO}_{4}$ solubility $= 3.017 \times 10^{- 3} mol L^{- 1}$
The solubility in $g / L = 3 \times 10^{- 3} \times 136$
$= 0.410 g / L$
Hence, $0.41 g$ dissolve in $1 L$
$1 g$ will dissolve in $\frac{1}{0.41} = 2.43 L$ .

CHXI07:EQUILIBRIUM

314954 The correct representation for solubility product of ${SnS}_{2}$ is

1

[{Sn}^{4 +}] {[S^{2 -}]}^{2}

2

[{Sn}^{4 +}] [S^{2 -}]

3

[{Sn}^{4 +}] [2 S^{2 -}]

4

[{Sn}^{4 +}] {[2 S^{2 -}]}^{2}

Explanation:

$K_{SP}$ of ${SnS}_{2} = [{Sn}^{4 +}] {[S^{2 -}]}^{2}$

CHXI07:EQUILIBRIUM

314952 The product of ionic concentration in a saturated solution of an electrolyte at a given temperature is constant and is known as

1 Ionic product of the electrolyte

2 Solubility product

3 Ionization constant

4 Dissociation constant

CHXI07:EQUILIBRIUM

314970 Consider the following sparingly soluble salts having same solubility product in their saturated solution in water:
(A) $XY$
(B) $X_{2} Y$
(C) ${XY}_{3}$
The correct order of increasing solubility of these salts in water will be

1 A

<

B

<

C

2 C

<

B

<

A

3 B

<

A

<

C

4 B

<

C

<

A

Explanation:

Let $S_{1}, S_{2}, S_{3}$ be the solubility of $XY$ , $X_{2} Y$ and ${XY}_{3}$ in water, respectively
$Salt Solubality product$
$\begin{matrix} XY & S_{1}^{2} \\ X_{2} Y & {3S}_{2}^{3} \\ X Y_{3} & {27S}_{3}^{4} \end{matrix}$
According to given condition,
$S_{1}^{2} {= 4S}_{2}^{3} {= 27S}_{3}^{4} or S_{1} {= 2S}_{2}^{3/2} = 3 \sqrt{3} S_{3}^{2}$
Since the practical value of solubility product is found to be very small. So, $S_{1} < S_{2} < S_{3}$

CHXI07:EQUILIBRIUM

314953 What is the minimum volume of water required to dissolve $1 g$ of calcium sulphate at $298 K$ ? (For calcium sulphate, $K_{s p}$ is $9.1 \times 10^{- 6}$ )?

1

5.43 L

2

2.43 L

3

8.43 L

4

7.43 L

Explanation:

$\begin{matrix} CaS O_{4} ⇌ C a^{2 +} {+ SO}_{4}^{2 -}; K_{SP} = 9 .1 \times 1 0^{- 6} \\ S S S \end{matrix}$
where $S$ is the solubility of ${CaSO}_{4}$ .
$K_{s p} = [{Ca}^{2 +}] [{SO}_{4}^{2 -}] = S . S = S^{2}$
$S = \sqrt{K_{s p}} = \sqrt{9.1 \times 10^{- 6}}$
$S = 3.017 \times 10^{- 3} M$
${CaSO}_{4}$ solubility $= 3.017 \times 10^{- 3} mol L^{- 1}$
The solubility in $g / L = 3 \times 10^{- 3} \times 136$
$= 0.410 g / L$
Hence, $0.41 g$ dissolve in $1 L$
$1 g$ will dissolve in $\frac{1}{0.41} = 2.43 L$ .

CHXI07:EQUILIBRIUM

314954 The correct representation for solubility product of ${SnS}_{2}$ is

1

[{Sn}^{4 +}] {[S^{2 -}]}^{2}

2

[{Sn}^{4 +}] [S^{2 -}]

3

[{Sn}^{4 +}] [2 S^{2 -}]

4

[{Sn}^{4 +}] {[2 S^{2 -}]}^{2}

Explanation:

$K_{SP}$ of ${SnS}_{2} = [{Sn}^{4 +}] {[S^{2 -}]}^{2}$

CHXI07:EQUILIBRIUM

314952 The product of ionic concentration in a saturated solution of an electrolyte at a given temperature is constant and is known as

1 Ionic product of the electrolyte

2 Solubility product

3 Ionization constant

4 Dissociation constant

CHXI07:EQUILIBRIUM

314970 Consider the following sparingly soluble salts having same solubility product in their saturated solution in water:
(A) $XY$
(B) $X_{2} Y$
(C) ${XY}_{3}$
The correct order of increasing solubility of these salts in water will be

1 A

<

B

<

C

2 C

<

B

<

A

3 B

<

A

<

C

4 B

<

C

<

A

Explanation:

Let $S_{1}, S_{2}, S_{3}$ be the solubility of $XY$ , $X_{2} Y$ and ${XY}_{3}$ in water, respectively
$Salt Solubality product$
$\begin{matrix} XY & S_{1}^{2} \\ X_{2} Y & {3S}_{2}^{3} \\ X Y_{3} & {27S}_{3}^{4} \end{matrix}$
According to given condition,
$S_{1}^{2} {= 4S}_{2}^{3} {= 27S}_{3}^{4} or S_{1} {= 2S}_{2}^{3/2} = 3 \sqrt{3} S_{3}^{2}$
Since the practical value of solubility product is found to be very small. So, $S_{1} < S_{2} < S_{3}$

CHXI07:EQUILIBRIUM

314953 What is the minimum volume of water required to dissolve $1 g$ of calcium sulphate at $298 K$ ? (For calcium sulphate, $K_{s p}$ is $9.1 \times 10^{- 6}$ )?

1

5.43 L

2

2.43 L

3

8.43 L

4

7.43 L

Explanation:

$\begin{matrix} CaS O_{4} ⇌ C a^{2 +} {+ SO}_{4}^{2 -}; K_{SP} = 9 .1 \times 1 0^{- 6} \\ S S S \end{matrix}$
where $S$ is the solubility of ${CaSO}_{4}$ .
$K_{s p} = [{Ca}^{2 +}] [{SO}_{4}^{2 -}] = S . S = S^{2}$
$S = \sqrt{K_{s p}} = \sqrt{9.1 \times 10^{- 6}}$
$S = 3.017 \times 10^{- 3} M$
${CaSO}_{4}$ solubility $= 3.017 \times 10^{- 3} mol L^{- 1}$
The solubility in $g / L = 3 \times 10^{- 3} \times 136$
$= 0.410 g / L$
Hence, $0.41 g$ dissolve in $1 L$
$1 g$ will dissolve in $\frac{1}{0.41} = 2.43 L$ .

CHXI07:EQUILIBRIUM

314954 The correct representation for solubility product of ${SnS}_{2}$ is

1

[{Sn}^{4 +}] {[S^{2 -}]}^{2}

2

[{Sn}^{4 +}] [S^{2 -}]

3

[{Sn}^{4 +}] [2 S^{2 -}]

4

[{Sn}^{4 +}] {[2 S^{2 -}]}^{2}

Explanation:

$K_{SP}$ of ${SnS}_{2} = [{Sn}^{4 +}] {[S^{2 -}]}^{2}$