QBManager

Ionic Equilibrium

229381 At $25^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 10^{- 11}$ . At which $pH$ , will ${Mg}^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M {Mg}^{2 +}$ ions?

1 9

2 10

3 11

4 8

Explanation:

Given that, $K_{sp}$ of $Mg (OH)_{2} = 1.0 \times 10^{- 11}$ Solution molarity is $= 0.001$
$\begin{aligned} Mg (OH)_{2} ⟶ {Mg}^{2 +} + 2 {OH}^{-} \\ Moles of Mg (OH)_{2} = Moles of [{Mg}^{2 +}] \\ ∵ K_{sp} = [{Mg}^{2 +}] \times {[{OH}^{-}]}^{2} \\ 1.0 \times 10^{- 11} = 0.001 \times {[{OH}^{- 1}]}^{2} \\ {[{OH}^{-}]}^{2} = \frac{1.0 \times 10^{- 11}}{0.001} = 1.0 \times 10^{- 8} M \\ [{OH}^{-}] = 10^{- 4} M \\ H^{+} = \frac{10^{- 14}}{[{OH}^{- 1}]} = \frac{10^{- 14}}{10^{- 4}} = 10^{- 10} M \\ pH = - \log [H]^{+} \\ pH = - \log [10]^{- 10} \\ pH = 10 \end{aligned}$

2010

Ionic Equilibrium

229382 The solubility product $(K_{sp})$ of the following compounds are given at $25^{\circ} C$ .
$\begin{array}{ll} Compound & K_{sp} \\ AgCl & 1.1 \times 10^{- 10} \\ AgI & 1.0 \times 10^{- 16} \\ {PbCrO}_{4} & 4.0 \times 10^{- 14} \\ {Ag}_{2} {CO}_{3} & 8.0 \times 10^{- 12} \end{array}$
The most soluble and least soluble compounds are respectively.

1

AgCl

and

{PbCrO}_{4}

2

AgI

and

{Ag}_{2} {CO}_{3}

3

AgCl

and

{Ag}_{2} {CO}_{3}

4

{Ag}_{2} {CO}_{3}

and

AgI

KEAM-2011

Explanation:

(i) $AgCl ⟶ {Ag}^{+} (aq) + {Cl}^{-}$ (aq)
$\begin{aligned} K_{Sp} & = [{Ag}^{+}] [{Cl}^{-}] \\ = S \times S \end{aligned}$
Given, $K_{sp} = 1.1 \times 10^{- 10}$
$\begin{aligned} 1.1 \times 10^{- 10} = S^{2} \\ S = 1.04 \times 10^{- 5} mol L^{- 1} \end{aligned}$
(ii) $AgI ⟶ {Ag}^{+} + I^{-}$
$\begin{aligned} K_{sp} & = [{Ag}^{+}] [I^{-}] \\ = S \times S \\ K_{sp} & = S^{2} = 1 \times 10^{- 16} \\ S & = 1 \times 10^{- 8} mol L^{- 1} \end{aligned}$
(iii)
$\begin{aligned} {PbCrO}_{4} ⟶ Pb \\ S \\ K_{Sp} = [{Pb}^{2 +}] [{CrO}^{2 -}] \\ = S \\ K_{sp} = S^{2} \\ 4 \times 10^{- 14} = S^{2} \\ S = 2 \times 10^{- 7} {molL}^{- 1} \end{aligned}$
(iv) ${Ag}_{2} {CO}_{3} ⟶ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] [{CO}_{3}^{2 -}] \\ = [2 S]^{2} \cdot [S] \\ K_{sp} = 4 S^{5} \\ 8 \times 10^{- 12} = 4 S^{5} \\ S = 1.26 \times 10^{- 4} {molL}^{- 1} \end{aligned}$
Hence, ${Ag}_{2} {CO}_{3}$ are most soluble end $AgI$ are least Soluble).

Ionic Equilibrium

229384 Solubility product of a salt $A B$ is $1 \times 10^{- 8} M^{2}$ in a solution in which the concentration of $A^{+}$ ions is $10^{- 3}$ M. The salt will precipitate when the concentration of $B^{-}$ ions is kept

1 Between

10^{- 8} M

to

10^{- 7} M

2 Between

10^{- 7} M

to

10^{- 8} M

3

> 10^{- 5} M

4

< 10^{- 8} M

KCET-2006

Explanation:

Given that, $K_{sp}$ of $AB$ concentration of $A^{+} =$ $10^{- 3} M$
$\begin{aligned} K_{sp} = 1 \times 10^{- 8} \\ AB AB [A^{+}] [B^{-}] \\ [10^{- 3}] [B^{-}] \end{aligned}$
The salt will be precipitated, if ionic product $> K_{sp}$ $10^{- 3} [B^{-}] > 1 \times 10^{- 8}$
$\begin{aligned} [B^{-}] > \frac{10^{- 8}}{10^{- 3}} \\ [B^{-}] > 10^{- 5} M \end{aligned}$

Ionic Equilibrium

229385 The solubility of ${PbF}_{2}$ in water at $25^{\circ} C$ is $\approx 10^{- 3}$ M. What is its solubility in $0.05 M$ NaF solution? Assume the later to be fully ionised.

1

1.6 \times 10^{- 6} M

2

1.2 \times 10^{- 6} M

3

1.2 \times 10^{- 5} M

4

1.6 \times 10^{- 4} M

Explanation:

Solubility of ${PbF}_{2} \approx 10^{- 3} M$
$\begin{aligned} {PbF}_{2} ◻ {Pb}^{2 +} + 2 F^{-} \\ K_{sp} = [Pb]^{2 +} \cdot {[F^{-}]}^{2} \\ S \times 2 S^{2} \\ K_{sp} = 4 S^{3} \\ = 4 \times {(10^{- 3})}^{3} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
In $0.05 NaF$ We have 0.05 of $F^{-}$ ion contributed by $NaF$ If the solubility of ${PbF}_{2}$ in this solution is $S$ then total $[F^{-}] = [2 S + 0.05] M$
$K_{sp} = S [2 S + 0.05]^{2}$
Assuming $2 S << 0.05$
$\begin{aligned} 4 \times 10^{- 9} = S \times 25 \times 10^{- 4} \\ S = 0.16 \times 10^{- 5} M = 1.6 \times 10^{- 6} M \end{aligned}$

AIPMT-2009

Ionic Equilibrium

229381 At $25^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 10^{- 11}$ . At which $pH$ , will ${Mg}^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M {Mg}^{2 +}$ ions?

1 9

2 10

3 11

4 8

Explanation:

Given that, $K_{sp}$ of $Mg (OH)_{2} = 1.0 \times 10^{- 11}$ Solution molarity is $= 0.001$
$\begin{aligned} Mg (OH)_{2} ⟶ {Mg}^{2 +} + 2 {OH}^{-} \\ Moles of Mg (OH)_{2} = Moles of [{Mg}^{2 +}] \\ ∵ K_{sp} = [{Mg}^{2 +}] \times {[{OH}^{-}]}^{2} \\ 1.0 \times 10^{- 11} = 0.001 \times {[{OH}^{- 1}]}^{2} \\ {[{OH}^{-}]}^{2} = \frac{1.0 \times 10^{- 11}}{0.001} = 1.0 \times 10^{- 8} M \\ [{OH}^{-}] = 10^{- 4} M \\ H^{+} = \frac{10^{- 14}}{[{OH}^{- 1}]} = \frac{10^{- 14}}{10^{- 4}} = 10^{- 10} M \\ pH = - \log [H]^{+} \\ pH = - \log [10]^{- 10} \\ pH = 10 \end{aligned}$

2010

Ionic Equilibrium

229382 The solubility product $(K_{sp})$ of the following compounds are given at $25^{\circ} C$ .
$\begin{array}{ll} Compound & K_{sp} \\ AgCl & 1.1 \times 10^{- 10} \\ AgI & 1.0 \times 10^{- 16} \\ {PbCrO}_{4} & 4.0 \times 10^{- 14} \\ {Ag}_{2} {CO}_{3} & 8.0 \times 10^{- 12} \end{array}$
The most soluble and least soluble compounds are respectively.

1

AgCl

and

{PbCrO}_{4}

2

AgI

and

{Ag}_{2} {CO}_{3}

3

AgCl

and

{Ag}_{2} {CO}_{3}

4

{Ag}_{2} {CO}_{3}

and

AgI

KEAM-2011

Explanation:

(i) $AgCl ⟶ {Ag}^{+} (aq) + {Cl}^{-}$ (aq)
$\begin{aligned} K_{Sp} & = [{Ag}^{+}] [{Cl}^{-}] \\ = S \times S \end{aligned}$
Given, $K_{sp} = 1.1 \times 10^{- 10}$
$\begin{aligned} 1.1 \times 10^{- 10} = S^{2} \\ S = 1.04 \times 10^{- 5} mol L^{- 1} \end{aligned}$
(ii) $AgI ⟶ {Ag}^{+} + I^{-}$
$\begin{aligned} K_{sp} & = [{Ag}^{+}] [I^{-}] \\ = S \times S \\ K_{sp} & = S^{2} = 1 \times 10^{- 16} \\ S & = 1 \times 10^{- 8} mol L^{- 1} \end{aligned}$
(iii)
$\begin{aligned} {PbCrO}_{4} ⟶ Pb \\ S \\ K_{Sp} = [{Pb}^{2 +}] [{CrO}^{2 -}] \\ = S \\ K_{sp} = S^{2} \\ 4 \times 10^{- 14} = S^{2} \\ S = 2 \times 10^{- 7} {molL}^{- 1} \end{aligned}$
(iv) ${Ag}_{2} {CO}_{3} ⟶ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] [{CO}_{3}^{2 -}] \\ = [2 S]^{2} \cdot [S] \\ K_{sp} = 4 S^{5} \\ 8 \times 10^{- 12} = 4 S^{5} \\ S = 1.26 \times 10^{- 4} {molL}^{- 1} \end{aligned}$
Hence, ${Ag}_{2} {CO}_{3}$ are most soluble end $AgI$ are least Soluble).

Ionic Equilibrium

229384 Solubility product of a salt $A B$ is $1 \times 10^{- 8} M^{2}$ in a solution in which the concentration of $A^{+}$ ions is $10^{- 3}$ M. The salt will precipitate when the concentration of $B^{-}$ ions is kept

1 Between

10^{- 8} M

to

10^{- 7} M

2 Between

10^{- 7} M

to

10^{- 8} M

3

> 10^{- 5} M

4

< 10^{- 8} M

KCET-2006

Explanation:

Given that, $K_{sp}$ of $AB$ concentration of $A^{+} =$ $10^{- 3} M$
$\begin{aligned} K_{sp} = 1 \times 10^{- 8} \\ AB AB [A^{+}] [B^{-}] \\ [10^{- 3}] [B^{-}] \end{aligned}$
The salt will be precipitated, if ionic product $> K_{sp}$ $10^{- 3} [B^{-}] > 1 \times 10^{- 8}$
$\begin{aligned} [B^{-}] > \frac{10^{- 8}}{10^{- 3}} \\ [B^{-}] > 10^{- 5} M \end{aligned}$

Ionic Equilibrium

229385 The solubility of ${PbF}_{2}$ in water at $25^{\circ} C$ is $\approx 10^{- 3}$ M. What is its solubility in $0.05 M$ NaF solution? Assume the later to be fully ionised.

1

1.6 \times 10^{- 6} M

2

1.2 \times 10^{- 6} M

3

1.2 \times 10^{- 5} M

4

1.6 \times 10^{- 4} M

Explanation:

Solubility of ${PbF}_{2} \approx 10^{- 3} M$
$\begin{aligned} {PbF}_{2} ◻ {Pb}^{2 +} + 2 F^{-} \\ K_{sp} = [Pb]^{2 +} \cdot {[F^{-}]}^{2} \\ S \times 2 S^{2} \\ K_{sp} = 4 S^{3} \\ = 4 \times {(10^{- 3})}^{3} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
In $0.05 NaF$ We have 0.05 of $F^{-}$ ion contributed by $NaF$ If the solubility of ${PbF}_{2}$ in this solution is $S$ then total $[F^{-}] = [2 S + 0.05] M$
$K_{sp} = S [2 S + 0.05]^{2}$
Assuming $2 S << 0.05$
$\begin{aligned} 4 \times 10^{- 9} = S \times 25 \times 10^{- 4} \\ S = 0.16 \times 10^{- 5} M = 1.6 \times 10^{- 6} M \end{aligned}$

AIPMT-2009

Ionic Equilibrium

229381 At $25^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 10^{- 11}$ . At which $pH$ , will ${Mg}^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M {Mg}^{2 +}$ ions?

1 9

2 10

3 11

4 8

Explanation:

Given that, $K_{sp}$ of $Mg (OH)_{2} = 1.0 \times 10^{- 11}$ Solution molarity is $= 0.001$
$\begin{aligned} Mg (OH)_{2} ⟶ {Mg}^{2 +} + 2 {OH}^{-} \\ Moles of Mg (OH)_{2} = Moles of [{Mg}^{2 +}] \\ ∵ K_{sp} = [{Mg}^{2 +}] \times {[{OH}^{-}]}^{2} \\ 1.0 \times 10^{- 11} = 0.001 \times {[{OH}^{- 1}]}^{2} \\ {[{OH}^{-}]}^{2} = \frac{1.0 \times 10^{- 11}}{0.001} = 1.0 \times 10^{- 8} M \\ [{OH}^{-}] = 10^{- 4} M \\ H^{+} = \frac{10^{- 14}}{[{OH}^{- 1}]} = \frac{10^{- 14}}{10^{- 4}} = 10^{- 10} M \\ pH = - \log [H]^{+} \\ pH = - \log [10]^{- 10} \\ pH = 10 \end{aligned}$

2010

Ionic Equilibrium

229382 The solubility product $(K_{sp})$ of the following compounds are given at $25^{\circ} C$ .
$\begin{array}{ll} Compound & K_{sp} \\ AgCl & 1.1 \times 10^{- 10} \\ AgI & 1.0 \times 10^{- 16} \\ {PbCrO}_{4} & 4.0 \times 10^{- 14} \\ {Ag}_{2} {CO}_{3} & 8.0 \times 10^{- 12} \end{array}$
The most soluble and least soluble compounds are respectively.

1

AgCl

and

{PbCrO}_{4}

2

AgI

and

{Ag}_{2} {CO}_{3}

3

AgCl

and

{Ag}_{2} {CO}_{3}

4

{Ag}_{2} {CO}_{3}

and

AgI

KEAM-2011

Explanation:

(i) $AgCl ⟶ {Ag}^{+} (aq) + {Cl}^{-}$ (aq)
$\begin{aligned} K_{Sp} & = [{Ag}^{+}] [{Cl}^{-}] \\ = S \times S \end{aligned}$
Given, $K_{sp} = 1.1 \times 10^{- 10}$
$\begin{aligned} 1.1 \times 10^{- 10} = S^{2} \\ S = 1.04 \times 10^{- 5} mol L^{- 1} \end{aligned}$
(ii) $AgI ⟶ {Ag}^{+} + I^{-}$
$\begin{aligned} K_{sp} & = [{Ag}^{+}] [I^{-}] \\ = S \times S \\ K_{sp} & = S^{2} = 1 \times 10^{- 16} \\ S & = 1 \times 10^{- 8} mol L^{- 1} \end{aligned}$
(iii)
$\begin{aligned} {PbCrO}_{4} ⟶ Pb \\ S \\ K_{Sp} = [{Pb}^{2 +}] [{CrO}^{2 -}] \\ = S \\ K_{sp} = S^{2} \\ 4 \times 10^{- 14} = S^{2} \\ S = 2 \times 10^{- 7} {molL}^{- 1} \end{aligned}$
(iv) ${Ag}_{2} {CO}_{3} ⟶ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] [{CO}_{3}^{2 -}] \\ = [2 S]^{2} \cdot [S] \\ K_{sp} = 4 S^{5} \\ 8 \times 10^{- 12} = 4 S^{5} \\ S = 1.26 \times 10^{- 4} {molL}^{- 1} \end{aligned}$
Hence, ${Ag}_{2} {CO}_{3}$ are most soluble end $AgI$ are least Soluble).

Ionic Equilibrium

229384 Solubility product of a salt $A B$ is $1 \times 10^{- 8} M^{2}$ in a solution in which the concentration of $A^{+}$ ions is $10^{- 3}$ M. The salt will precipitate when the concentration of $B^{-}$ ions is kept

1 Between

10^{- 8} M

to

10^{- 7} M

2 Between

10^{- 7} M

to

10^{- 8} M

3

> 10^{- 5} M

4

< 10^{- 8} M

KCET-2006

Explanation:

Given that, $K_{sp}$ of $AB$ concentration of $A^{+} =$ $10^{- 3} M$
$\begin{aligned} K_{sp} = 1 \times 10^{- 8} \\ AB AB [A^{+}] [B^{-}] \\ [10^{- 3}] [B^{-}] \end{aligned}$
The salt will be precipitated, if ionic product $> K_{sp}$ $10^{- 3} [B^{-}] > 1 \times 10^{- 8}$
$\begin{aligned} [B^{-}] > \frac{10^{- 8}}{10^{- 3}} \\ [B^{-}] > 10^{- 5} M \end{aligned}$

Ionic Equilibrium

229385 The solubility of ${PbF}_{2}$ in water at $25^{\circ} C$ is $\approx 10^{- 3}$ M. What is its solubility in $0.05 M$ NaF solution? Assume the later to be fully ionised.

1

1.6 \times 10^{- 6} M

2

1.2 \times 10^{- 6} M

3

1.2 \times 10^{- 5} M

4

1.6 \times 10^{- 4} M

Explanation:

Solubility of ${PbF}_{2} \approx 10^{- 3} M$
$\begin{aligned} {PbF}_{2} ◻ {Pb}^{2 +} + 2 F^{-} \\ K_{sp} = [Pb]^{2 +} \cdot {[F^{-}]}^{2} \\ S \times 2 S^{2} \\ K_{sp} = 4 S^{3} \\ = 4 \times {(10^{- 3})}^{3} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
In $0.05 NaF$ We have 0.05 of $F^{-}$ ion contributed by $NaF$ If the solubility of ${PbF}_{2}$ in this solution is $S$ then total $[F^{-}] = [2 S + 0.05] M$
$K_{sp} = S [2 S + 0.05]^{2}$
Assuming $2 S << 0.05$
$\begin{aligned} 4 \times 10^{- 9} = S \times 25 \times 10^{- 4} \\ S = 0.16 \times 10^{- 5} M = 1.6 \times 10^{- 6} M \end{aligned}$

AIPMT-2009

Ionic Equilibrium

229381 At $25^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 10^{- 11}$ . At which $pH$ , will ${Mg}^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M {Mg}^{2 +}$ ions?

1 9

2 10

3 11

4 8

Explanation:

Given that, $K_{sp}$ of $Mg (OH)_{2} = 1.0 \times 10^{- 11}$ Solution molarity is $= 0.001$
$\begin{aligned} Mg (OH)_{2} ⟶ {Mg}^{2 +} + 2 {OH}^{-} \\ Moles of Mg (OH)_{2} = Moles of [{Mg}^{2 +}] \\ ∵ K_{sp} = [{Mg}^{2 +}] \times {[{OH}^{-}]}^{2} \\ 1.0 \times 10^{- 11} = 0.001 \times {[{OH}^{- 1}]}^{2} \\ {[{OH}^{-}]}^{2} = \frac{1.0 \times 10^{- 11}}{0.001} = 1.0 \times 10^{- 8} M \\ [{OH}^{-}] = 10^{- 4} M \\ H^{+} = \frac{10^{- 14}}{[{OH}^{- 1}]} = \frac{10^{- 14}}{10^{- 4}} = 10^{- 10} M \\ pH = - \log [H]^{+} \\ pH = - \log [10]^{- 10} \\ pH = 10 \end{aligned}$

2010

Ionic Equilibrium

229382 The solubility product $(K_{sp})$ of the following compounds are given at $25^{\circ} C$ .
$\begin{array}{ll} Compound & K_{sp} \\ AgCl & 1.1 \times 10^{- 10} \\ AgI & 1.0 \times 10^{- 16} \\ {PbCrO}_{4} & 4.0 \times 10^{- 14} \\ {Ag}_{2} {CO}_{3} & 8.0 \times 10^{- 12} \end{array}$
The most soluble and least soluble compounds are respectively.

1

AgCl

and

{PbCrO}_{4}

2

AgI

and

{Ag}_{2} {CO}_{3}

3

AgCl

and

{Ag}_{2} {CO}_{3}

4

{Ag}_{2} {CO}_{3}

and

AgI

KEAM-2011

Explanation:

(i) $AgCl ⟶ {Ag}^{+} (aq) + {Cl}^{-}$ (aq)
$\begin{aligned} K_{Sp} & = [{Ag}^{+}] [{Cl}^{-}] \\ = S \times S \end{aligned}$
Given, $K_{sp} = 1.1 \times 10^{- 10}$
$\begin{aligned} 1.1 \times 10^{- 10} = S^{2} \\ S = 1.04 \times 10^{- 5} mol L^{- 1} \end{aligned}$
(ii) $AgI ⟶ {Ag}^{+} + I^{-}$
$\begin{aligned} K_{sp} & = [{Ag}^{+}] [I^{-}] \\ = S \times S \\ K_{sp} & = S^{2} = 1 \times 10^{- 16} \\ S & = 1 \times 10^{- 8} mol L^{- 1} \end{aligned}$
(iii)
$\begin{aligned} {PbCrO}_{4} ⟶ Pb \\ S \\ K_{Sp} = [{Pb}^{2 +}] [{CrO}^{2 -}] \\ = S \\ K_{sp} = S^{2} \\ 4 \times 10^{- 14} = S^{2} \\ S = 2 \times 10^{- 7} {molL}^{- 1} \end{aligned}$
(iv) ${Ag}_{2} {CO}_{3} ⟶ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] [{CO}_{3}^{2 -}] \\ = [2 S]^{2} \cdot [S] \\ K_{sp} = 4 S^{5} \\ 8 \times 10^{- 12} = 4 S^{5} \\ S = 1.26 \times 10^{- 4} {molL}^{- 1} \end{aligned}$
Hence, ${Ag}_{2} {CO}_{3}$ are most soluble end $AgI$ are least Soluble).

Ionic Equilibrium

229384 Solubility product of a salt $A B$ is $1 \times 10^{- 8} M^{2}$ in a solution in which the concentration of $A^{+}$ ions is $10^{- 3}$ M. The salt will precipitate when the concentration of $B^{-}$ ions is kept

1 Between

10^{- 8} M

to

10^{- 7} M

2 Between

10^{- 7} M

to

10^{- 8} M

3

> 10^{- 5} M

4

< 10^{- 8} M

KCET-2006

Explanation:

Given that, $K_{sp}$ of $AB$ concentration of $A^{+} =$ $10^{- 3} M$
$\begin{aligned} K_{sp} = 1 \times 10^{- 8} \\ AB AB [A^{+}] [B^{-}] \\ [10^{- 3}] [B^{-}] \end{aligned}$
The salt will be precipitated, if ionic product $> K_{sp}$ $10^{- 3} [B^{-}] > 1 \times 10^{- 8}$
$\begin{aligned} [B^{-}] > \frac{10^{- 8}}{10^{- 3}} \\ [B^{-}] > 10^{- 5} M \end{aligned}$

Ionic Equilibrium

229385 The solubility of ${PbF}_{2}$ in water at $25^{\circ} C$ is $\approx 10^{- 3}$ M. What is its solubility in $0.05 M$ NaF solution? Assume the later to be fully ionised.

1

1.6 \times 10^{- 6} M

2

1.2 \times 10^{- 6} M

3

1.2 \times 10^{- 5} M

4

1.6 \times 10^{- 4} M

Explanation:

Solubility of ${PbF}_{2} \approx 10^{- 3} M$
$\begin{aligned} {PbF}_{2} ◻ {Pb}^{2 +} + 2 F^{-} \\ K_{sp} = [Pb]^{2 +} \cdot {[F^{-}]}^{2} \\ S \times 2 S^{2} \\ K_{sp} = 4 S^{3} \\ = 4 \times {(10^{- 3})}^{3} \\ K_{sp} = 4 \times 10^{- 9} \end{aligned}$
In $0.05 NaF$ We have 0.05 of $F^{-}$ ion contributed by $NaF$ If the solubility of ${PbF}_{2}$ in this solution is $S$ then total $[F^{-}] = [2 S + 0.05] M$
$K_{sp} = S [2 S + 0.05]^{2}$
Assuming $2 S << 0.05$
$\begin{aligned} 4 \times 10^{- 9} = S \times 25 \times 10^{- 4} \\ S = 0.16 \times 10^{- 5} M = 1.6 \times 10^{- 6} M \end{aligned}$

AIPMT-2009

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