QBManager

Ionic Equilibrium

229386 On adding $0.1 M$ solution each of $[{Ag}^{+}], [{Ba}^{2 +}]$ , $[{Ca}^{2 +}]$ in a ${Na}_{2} {SO}_{4}$ solution, species first precipitated is $[K_{sp} {BaSO}_{4}^{- 5} = 10^{- 11}, K_{sp} {CaSO}_{4} = 10^{- 6}, K_{sp} {Ag}_{2} {SO}_{4}$

1

{Ag}_{2} {SO}_{4}

2

{BaSO}_{4}

3

{CaSO}_{4}

4 All of these

Explanation:

Given, $K_{sp}$ of ${BaSO}_{4} = 10^{- 11}$
$K_{sp}$ of ${Ag}_{2} {SO}_{4} = 10^{- 5}$
$K_{sp}$ of ${CaSO}_{4} = 10^{- 6}$
Solubility of ${BaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 11}}$
$= 3.16 \times 10^{- 6} mol / L$
Solubility of ${CaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 6}}$
$= 1.0 \times 10^{- 3} mol / L$
Dissociation of ${Ag}_{2} {SO}_{4}$ occurs as follows :
$\begin{aligned} {Ag}_{2} {SO}_{4} ◻ 2 Ag + {SO}_{4}^{2 -} \\ ∴ K_{sp} = (2 S)^{2} (S) = 4 S^{3} \\ ∴ Solubility of {Ag}_{2} {SO}_{4} = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{10^{- 5}}{4}} \\ = 1.4 \times 10^{- 2} mol / L \end{aligned}$
Least solubility is of ${BaSO}_{4}$ , hence it will precipitate first.

AIPMT-2008

Ionic Equilibrium

229388 If the ionic product of $Ni (OH)_{2}$ is $1.9 \times 10^{- 15}$ , the molar solubility of $Ni (OH)_{2}$ in $1.0 M NaOH$ is

1

1.9 \times 10^{- 18} M

2

1.9 \times 10^{- 13} M

3

1.9 \times 10^{- 15} M

4

1.9 \times 10^{- 14} M

Explanation:

Given, $[{OH}^{-}] = 1 M, K_{sp} = 1.9 \times 10^{- 15}$
Let the molar solubility of $Ni (OH)_{2} = S$
$\begin{aligned} [Ni (OH)_{2}] ◻ {Ni}^{- 2 +} + 2 OH \\ ∴ K_{sp} = [{Ni}^{+ 2}] {[{OH}^{-}]}^{2} \\ 1.9 \times 10^{- 15} = S \times (1.0)^{2} \\ S = 1.9 \times 10^{- 15} M \end{aligned}$

AP-EAMCET (Engg.) - 2014

Ionic Equilibrium

229389 A saturated solution of $H_{2} S$ in $0.1 M HCl$ at $25^{\circ} C$ contains $S^{2 -}$ ion concentration of $10^{- 23} mol$ $L^{- 1}$ . The solubility product of some sulphides are $CuS = 10^{- 44}, FeS = 10^{- 14}, MnS = 10^{- 15}, CdS =$ $10^{- 25}$ . If $0.01 M$ solution of these salts in $1 M$ $HCl$ are saturated with $H_{2} S$ , which of these will be precipitated?

1 All

2 All except

MnS

3 All except

MnS

and FeS

4 Only-CuS

Explanation:

Given $[S^{2 -}] = 10^{- 23} mol L^{- 1}$
Ans. C
Exp: Given $[M^{2 +}] = 10^{- 2} M$
Ionic product, $K_{W} = [M^{2 +}] {[S^{2 -}]}_{- 23}$
$\begin{aligned} = 10^{- 2} \times 10^{- 23} \\ = 10^{- 25} \end{aligned}$
So, ionic product is greater than $K_{sp}$ of $CuS$ and $CdS$ .

VITEEE-2015

Ionic Equilibrium

229390 The solubility product of ${Ag}_{2} {CrO}_{4}$ is $32 \times 10^{- 12}$ . What is the concentration of ${CrO}_{4}^{- 2}$ ions in that solution?

1

2 \times 10^{- 4} M

2

16 \times 10^{- 4} M

3

8 \times 10^{- 4} M

4

8 \times 10^{- 8} M

VIEEE-2014

Explanation:

Given, $K_{sp}$ of ${Ag}_{2} {CrO}_{4} = 32 \times 10^{- 12}$
$\begin{aligned} K_{sp} & = [2 {Ag}^{+}] \cdot [{CrO}_{4}^{2 -}] \\ = [2 S^{2} \cdot [S] \\ K_{sp} & = 4 S^{3} \\ S & = {(\frac{K_{sp}}{4})}^{1 / 3} \\ S & = {(\frac{32 \times 10^{- 12}}{4})}^{1 / 3} \\ S & = {(8 \times 10^{- 12})}^{1 / 3} \\ S & = 2 \times 10^{- 4} M . \end{aligned}$

Ionic Equilibrium

229386 On adding $0.1 M$ solution each of $[{Ag}^{+}], [{Ba}^{2 +}]$ , $[{Ca}^{2 +}]$ in a ${Na}_{2} {SO}_{4}$ solution, species first precipitated is $[K_{sp} {BaSO}_{4}^{- 5} = 10^{- 11}, K_{sp} {CaSO}_{4} = 10^{- 6}, K_{sp} {Ag}_{2} {SO}_{4}$

1

{Ag}_{2} {SO}_{4}

2

{BaSO}_{4}

3

{CaSO}_{4}

4 All of these

Explanation:

Given, $K_{sp}$ of ${BaSO}_{4} = 10^{- 11}$
$K_{sp}$ of ${Ag}_{2} {SO}_{4} = 10^{- 5}$
$K_{sp}$ of ${CaSO}_{4} = 10^{- 6}$
Solubility of ${BaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 11}}$
$= 3.16 \times 10^{- 6} mol / L$
Solubility of ${CaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 6}}$
$= 1.0 \times 10^{- 3} mol / L$
Dissociation of ${Ag}_{2} {SO}_{4}$ occurs as follows :
$\begin{aligned} {Ag}_{2} {SO}_{4} ◻ 2 Ag + {SO}_{4}^{2 -} \\ ∴ K_{sp} = (2 S)^{2} (S) = 4 S^{3} \\ ∴ Solubility of {Ag}_{2} {SO}_{4} = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{10^{- 5}}{4}} \\ = 1.4 \times 10^{- 2} mol / L \end{aligned}$
Least solubility is of ${BaSO}_{4}$ , hence it will precipitate first.

AIPMT-2008

Ionic Equilibrium

229388 If the ionic product of $Ni (OH)_{2}$ is $1.9 \times 10^{- 15}$ , the molar solubility of $Ni (OH)_{2}$ in $1.0 M NaOH$ is

1

1.9 \times 10^{- 18} M

2

1.9 \times 10^{- 13} M

3

1.9 \times 10^{- 15} M

4

1.9 \times 10^{- 14} M

Explanation:

Given, $[{OH}^{-}] = 1 M, K_{sp} = 1.9 \times 10^{- 15}$
Let the molar solubility of $Ni (OH)_{2} = S$
$\begin{aligned} [Ni (OH)_{2}] ◻ {Ni}^{- 2 +} + 2 OH \\ ∴ K_{sp} = [{Ni}^{+ 2}] {[{OH}^{-}]}^{2} \\ 1.9 \times 10^{- 15} = S \times (1.0)^{2} \\ S = 1.9 \times 10^{- 15} M \end{aligned}$

AP-EAMCET (Engg.) - 2014

Ionic Equilibrium

229389 A saturated solution of $H_{2} S$ in $0.1 M HCl$ at $25^{\circ} C$ contains $S^{2 -}$ ion concentration of $10^{- 23} mol$ $L^{- 1}$ . The solubility product of some sulphides are $CuS = 10^{- 44}, FeS = 10^{- 14}, MnS = 10^{- 15}, CdS =$ $10^{- 25}$ . If $0.01 M$ solution of these salts in $1 M$ $HCl$ are saturated with $H_{2} S$ , which of these will be precipitated?

1 All

2 All except

MnS

3 All except

MnS

and FeS

4 Only-CuS

Explanation:

Given $[S^{2 -}] = 10^{- 23} mol L^{- 1}$
Ans. C
Exp: Given $[M^{2 +}] = 10^{- 2} M$
Ionic product, $K_{W} = [M^{2 +}] {[S^{2 -}]}_{- 23}$
$\begin{aligned} = 10^{- 2} \times 10^{- 23} \\ = 10^{- 25} \end{aligned}$
So, ionic product is greater than $K_{sp}$ of $CuS$ and $CdS$ .

VITEEE-2015

Ionic Equilibrium

229390 The solubility product of ${Ag}_{2} {CrO}_{4}$ is $32 \times 10^{- 12}$ . What is the concentration of ${CrO}_{4}^{- 2}$ ions in that solution?

1

2 \times 10^{- 4} M

2

16 \times 10^{- 4} M

3

8 \times 10^{- 4} M

4

8 \times 10^{- 8} M

VIEEE-2014

Explanation:

Given, $K_{sp}$ of ${Ag}_{2} {CrO}_{4} = 32 \times 10^{- 12}$
$\begin{aligned} K_{sp} & = [2 {Ag}^{+}] \cdot [{CrO}_{4}^{2 -}] \\ = [2 S^{2} \cdot [S] \\ K_{sp} & = 4 S^{3} \\ S & = {(\frac{K_{sp}}{4})}^{1 / 3} \\ S & = {(\frac{32 \times 10^{- 12}}{4})}^{1 / 3} \\ S & = {(8 \times 10^{- 12})}^{1 / 3} \\ S & = 2 \times 10^{- 4} M . \end{aligned}$

Ionic Equilibrium

229386 On adding $0.1 M$ solution each of $[{Ag}^{+}], [{Ba}^{2 +}]$ , $[{Ca}^{2 +}]$ in a ${Na}_{2} {SO}_{4}$ solution, species first precipitated is $[K_{sp} {BaSO}_{4}^{- 5} = 10^{- 11}, K_{sp} {CaSO}_{4} = 10^{- 6}, K_{sp} {Ag}_{2} {SO}_{4}$

1

{Ag}_{2} {SO}_{4}

2

{BaSO}_{4}

3

{CaSO}_{4}

4 All of these

Explanation:

Given, $K_{sp}$ of ${BaSO}_{4} = 10^{- 11}$
$K_{sp}$ of ${Ag}_{2} {SO}_{4} = 10^{- 5}$
$K_{sp}$ of ${CaSO}_{4} = 10^{- 6}$
Solubility of ${BaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 11}}$
$= 3.16 \times 10^{- 6} mol / L$
Solubility of ${CaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 6}}$
$= 1.0 \times 10^{- 3} mol / L$
Dissociation of ${Ag}_{2} {SO}_{4}$ occurs as follows :
$\begin{aligned} {Ag}_{2} {SO}_{4} ◻ 2 Ag + {SO}_{4}^{2 -} \\ ∴ K_{sp} = (2 S)^{2} (S) = 4 S^{3} \\ ∴ Solubility of {Ag}_{2} {SO}_{4} = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{10^{- 5}}{4}} \\ = 1.4 \times 10^{- 2} mol / L \end{aligned}$
Least solubility is of ${BaSO}_{4}$ , hence it will precipitate first.

AIPMT-2008

Ionic Equilibrium

229388 If the ionic product of $Ni (OH)_{2}$ is $1.9 \times 10^{- 15}$ , the molar solubility of $Ni (OH)_{2}$ in $1.0 M NaOH$ is

1

1.9 \times 10^{- 18} M

2

1.9 \times 10^{- 13} M

3

1.9 \times 10^{- 15} M

4

1.9 \times 10^{- 14} M

Explanation:

Given, $[{OH}^{-}] = 1 M, K_{sp} = 1.9 \times 10^{- 15}$
Let the molar solubility of $Ni (OH)_{2} = S$
$\begin{aligned} [Ni (OH)_{2}] ◻ {Ni}^{- 2 +} + 2 OH \\ ∴ K_{sp} = [{Ni}^{+ 2}] {[{OH}^{-}]}^{2} \\ 1.9 \times 10^{- 15} = S \times (1.0)^{2} \\ S = 1.9 \times 10^{- 15} M \end{aligned}$

AP-EAMCET (Engg.) - 2014

Ionic Equilibrium

229389 A saturated solution of $H_{2} S$ in $0.1 M HCl$ at $25^{\circ} C$ contains $S^{2 -}$ ion concentration of $10^{- 23} mol$ $L^{- 1}$ . The solubility product of some sulphides are $CuS = 10^{- 44}, FeS = 10^{- 14}, MnS = 10^{- 15}, CdS =$ $10^{- 25}$ . If $0.01 M$ solution of these salts in $1 M$ $HCl$ are saturated with $H_{2} S$ , which of these will be precipitated?

1 All

2 All except

MnS

3 All except

MnS

and FeS

4 Only-CuS

Explanation:

Given $[S^{2 -}] = 10^{- 23} mol L^{- 1}$
Ans. C
Exp: Given $[M^{2 +}] = 10^{- 2} M$
Ionic product, $K_{W} = [M^{2 +}] {[S^{2 -}]}_{- 23}$
$\begin{aligned} = 10^{- 2} \times 10^{- 23} \\ = 10^{- 25} \end{aligned}$
So, ionic product is greater than $K_{sp}$ of $CuS$ and $CdS$ .

VITEEE-2015

Ionic Equilibrium

229390 The solubility product of ${Ag}_{2} {CrO}_{4}$ is $32 \times 10^{- 12}$ . What is the concentration of ${CrO}_{4}^{- 2}$ ions in that solution?

1

2 \times 10^{- 4} M

2

16 \times 10^{- 4} M

3

8 \times 10^{- 4} M

4

8 \times 10^{- 8} M

VIEEE-2014

Explanation:

Given, $K_{sp}$ of ${Ag}_{2} {CrO}_{4} = 32 \times 10^{- 12}$
$\begin{aligned} K_{sp} & = [2 {Ag}^{+}] \cdot [{CrO}_{4}^{2 -}] \\ = [2 S^{2} \cdot [S] \\ K_{sp} & = 4 S^{3} \\ S & = {(\frac{K_{sp}}{4})}^{1 / 3} \\ S & = {(\frac{32 \times 10^{- 12}}{4})}^{1 / 3} \\ S & = {(8 \times 10^{- 12})}^{1 / 3} \\ S & = 2 \times 10^{- 4} M . \end{aligned}$

Ionic Equilibrium

229386 On adding $0.1 M$ solution each of $[{Ag}^{+}], [{Ba}^{2 +}]$ , $[{Ca}^{2 +}]$ in a ${Na}_{2} {SO}_{4}$ solution, species first precipitated is $[K_{sp} {BaSO}_{4}^{- 5} = 10^{- 11}, K_{sp} {CaSO}_{4} = 10^{- 6}, K_{sp} {Ag}_{2} {SO}_{4}$

1

{Ag}_{2} {SO}_{4}

2

{BaSO}_{4}

3

{CaSO}_{4}

4 All of these

Explanation:

Given, $K_{sp}$ of ${BaSO}_{4} = 10^{- 11}$
$K_{sp}$ of ${Ag}_{2} {SO}_{4} = 10^{- 5}$
$K_{sp}$ of ${CaSO}_{4} = 10^{- 6}$
Solubility of ${BaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 11}}$
$= 3.16 \times 10^{- 6} mol / L$
Solubility of ${CaSO}_{4} = \sqrt{K_{sp}} = \sqrt{10^{- 6}}$
$= 1.0 \times 10^{- 3} mol / L$
Dissociation of ${Ag}_{2} {SO}_{4}$ occurs as follows :
$\begin{aligned} {Ag}_{2} {SO}_{4} ◻ 2 Ag + {SO}_{4}^{2 -} \\ ∴ K_{sp} = (2 S)^{2} (S) = 4 S^{3} \\ ∴ Solubility of {Ag}_{2} {SO}_{4} = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{10^{- 5}}{4}} \\ = 1.4 \times 10^{- 2} mol / L \end{aligned}$
Least solubility is of ${BaSO}_{4}$ , hence it will precipitate first.

AIPMT-2008

Ionic Equilibrium

229388 If the ionic product of $Ni (OH)_{2}$ is $1.9 \times 10^{- 15}$ , the molar solubility of $Ni (OH)_{2}$ in $1.0 M NaOH$ is

1

1.9 \times 10^{- 18} M

2

1.9 \times 10^{- 13} M

3

1.9 \times 10^{- 15} M

4

1.9 \times 10^{- 14} M

Explanation:

Given, $[{OH}^{-}] = 1 M, K_{sp} = 1.9 \times 10^{- 15}$
Let the molar solubility of $Ni (OH)_{2} = S$
$\begin{aligned} [Ni (OH)_{2}] ◻ {Ni}^{- 2 +} + 2 OH \\ ∴ K_{sp} = [{Ni}^{+ 2}] {[{OH}^{-}]}^{2} \\ 1.9 \times 10^{- 15} = S \times (1.0)^{2} \\ S = 1.9 \times 10^{- 15} M \end{aligned}$

AP-EAMCET (Engg.) - 2014

Ionic Equilibrium

229389 A saturated solution of $H_{2} S$ in $0.1 M HCl$ at $25^{\circ} C$ contains $S^{2 -}$ ion concentration of $10^{- 23} mol$ $L^{- 1}$ . The solubility product of some sulphides are $CuS = 10^{- 44}, FeS = 10^{- 14}, MnS = 10^{- 15}, CdS =$ $10^{- 25}$ . If $0.01 M$ solution of these salts in $1 M$ $HCl$ are saturated with $H_{2} S$ , which of these will be precipitated?

1 All

2 All except

MnS

3 All except

MnS

and FeS

4 Only-CuS

Explanation:

Given $[S^{2 -}] = 10^{- 23} mol L^{- 1}$
Ans. C
Exp: Given $[M^{2 +}] = 10^{- 2} M$
Ionic product, $K_{W} = [M^{2 +}] {[S^{2 -}]}_{- 23}$
$\begin{aligned} = 10^{- 2} \times 10^{- 23} \\ = 10^{- 25} \end{aligned}$
So, ionic product is greater than $K_{sp}$ of $CuS$ and $CdS$ .

VITEEE-2015

Ionic Equilibrium

229390 The solubility product of ${Ag}_{2} {CrO}_{4}$ is $32 \times 10^{- 12}$ . What is the concentration of ${CrO}_{4}^{- 2}$ ions in that solution?

1

2 \times 10^{- 4} M

2

16 \times 10^{- 4} M

3

8 \times 10^{- 4} M

4

8 \times 10^{- 8} M

VIEEE-2014

Explanation:

Given, $K_{sp}$ of ${Ag}_{2} {CrO}_{4} = 32 \times 10^{- 12}$
$\begin{aligned} K_{sp} & = [2 {Ag}^{+}] \cdot [{CrO}_{4}^{2 -}] \\ = [2 S^{2} \cdot [S] \\ K_{sp} & = 4 S^{3} \\ S & = {(\frac{K_{sp}}{4})}^{1 / 3} \\ S & = {(\frac{32 \times 10^{- 12}}{4})}^{1 / 3} \\ S & = {(8 \times 10^{- 12})}^{1 / 3} \\ S & = 2 \times 10^{- 4} M . \end{aligned}$