QBManager

Ionic Equilibrium

229430 The $K_{sp}$ for $Cr (OH)_{3}$ is $1.6 \times 10^{- 30}$ . The molar solubility of this compound in water is

1

\sqrt[2]{1.6 \times 10^{- 30}}

2

\sqrt[4]{1.6 \times 10^{- 30}}

3

\sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}

4

\frac{1.6 \times 10^{- 30}}{27}

Explanation:

Let's be the molar solubility of $Cr (OH)_{3}$ then,
$\begin{aligned} Cr (OH)_{3} ?? {Cr}_{S}^{3 +} + 3 {OH}_{3}^{-} \\ K_{sp} = [{Cr}^{3 +}] . [{OH}^{-}] \\ K_{sp} = S \times (3 S)^{3} \\ = 27 S^{4} \\ S = \sqrt[4]{\frac{K_{sp}}{27}} \\ S = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}} \end{aligned}$

AIEEE-2011

Ionic Equilibrium

229431 Solubility product of silver bromide is $5.0 \times 10^{- 13}$ - The quantity of potassium bromide (molar mass taken as $120 g {mol}^{- 1}$ ) to be added to $1 L$ of $0.05 M$ solution of silver nitrate of start the precipitation of $A g B r$ is

1

1.2 \times 10^{- 10} g

2

1.2 \times 10^{- 9} g

3

6.2 \times 10^{- 5} g

4

5.0 \times 10^{- 8} g

Explanation:

Given $(AgBr) = 5 \times 10^{- 13}, [{Ag}^{+}] = 0.05 m$ Molar mass of $KBr = 120 {gmol}^{- 1}$
${Ag}^{+} (aq) + {Br}^{-} (aq) = AgBr$
Precipitation starts when ionic product just exceeds solubility product.
$K_{sp} = [{Ag}^{+}] [{Br}^{-}]$ putting the valves -
$\Rightarrow [{Br}^{-}] = \frac{K_{sp}}{[{Ag}^{+}]} = \frac{5 \times 10^{- 13}}{0.05} = 10^{- 11}$
Precipitation starts when $10^{- 11}$ moles of $KBr$ is added to ${AgNO}_{3}$ of ${AgNO}_{3}$ solution.
No. of moles of $KBr$ to be added $= 10^{- 11}$
Weight of $KBr =$ no. of moles $\times$ molar mass.
Putting the valves.
$= 10^{- 11} \times 120 = 1.2 \times 10^{- 9} g$

AIEEE-2010

Ionic Equilibrium

229432 Solid $Ba ({NO}_{3})$ is gradually dissolved in a $1.0 \times 10^{- 4} {MNa}_{2} {CO}_{3}$ solution. At what concentration of ${Ba}^{2 +}$ will a precipitate begin to form? $(K_{sp}$ for ${BaCO}_{3} = 5.1 \times 10^{- 9}$ )

1

4.1 \times 10^{- 5} M

2

5.1 \times 10^{- 5} M

3

8.1 \times 10^{- 8} M

4

8.1 \times 10^{- 7} M

Explanation:

according to situation -
${Na}_{2} {CO}_{3} \underset{1 \times 10^{- 4} m}{2} 2 {Na}^{+} + {CO}_{3}^{2 -}$
For
$\begin{aligned} Ba CO {CO}_{3} \to {Ba}^{2 +} + {CO}_{3}^{2 -} \\ K_{sp} = [{Ba}^{2 +}] \cdot [{CO}_{3}^{2 -}] \end{aligned}$
Hence. $5.1 \times 10^{- 9} = [{Ba}^{2 +}] [1 \times 10^{- 4}]$
$\begin{aligned} [{Ba}^{2 +}] = \frac{5.1 \times 10^{- 9}}{1 \times 10^{- 4}} \\ {Ba}^{2 +} = 5.1 \times 10^{- 3} m \end{aligned}$

AIEEE-2009

Ionic Equilibrium

229433 In a saturated solution of the spatingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets in is ${AgIO}_{3}$ (s) $◻ A g + (aq) + {IO}_{3}^{-} (aq)$
If the solubility product constant $K_{sp}$ of ${AgIO}_{3}$ at a given temperature is $1.0 \times 10^{- 8}$ , what is the mass of ${AgIO}_{3}$ contained in $100 mL$ of its saturated solution?

1

28.3 \times 10^{- 2} g

2

28.3 \times 10^{- 4} g

3

1.0 \times 10^{- 7} g

4

1.0 \times 10^{- 4} g

Explanation:

Exp:
$Ag {IO}_{3}$ ??
(s = Solubility)
$K_{sp} = x^{x} \times y^{y} \times s^{x + y}$
$x =$ no. of ${Ag}^{+}$ ions and
$y =$ no of ${IO}_{3}^{-}$ ions
$K_{sp} = S^{2} \Rightarrow S = \sqrt{K_{sp}} = \sqrt{10^{- 8}} = 10^{- 4} M$
$1000 mL \to 10^{- 4} mL = 10^{- 4} \times 283 g$
$100 mL \to \frac{10^{- 4} \times 283 \times 100}{1000} = 28.3 \times 10^{- 4} g$

AIEEE-2007

Ionic Equilibrium

229434 The solubility product of a salt having general formula $M_{2}$ , in water is $4 \times 10^{- 12}$ . The concentration of $M^{2 +}$ ions in the aqueous solution of the salt is

1

4.0 \times 10^{- 10} M

2

1.6 \times 10^{- 4} M

3

1.0 \times 10^{- 4} M

4

2.0 \times 10^{- 6} M

Explanation:

Solubility produt $= 4 \times 10^{- 12}$
$\begin{aligned} {MX}_{2} ◻ \frac{M^{2 +} + \underset{[S]}{2 S]}}{} \\ K_{sp} = [M^{2 +}] \times {[M^{-}]}^{2} \\ K_{sp} = [S] \cdot [2 S]^{2} \\ K_{sp} = 4 S^{3} \end{aligned}$
The solubility of $M^{2 +}$ ion $= S$
$\begin{aligned} S = {(\frac{K_{sp}}{4})}^{1 / 3} = {(\frac{4 \times 10^{- 12}}{4})}^{1 / 3} \\ S = 1 \times 10^{- 4} M \end{aligned}$

AIEEE-2004

Ionic Equilibrium

229430 The $K_{sp}$ for $Cr (OH)_{3}$ is $1.6 \times 10^{- 30}$ . The molar solubility of this compound in water is

1

\sqrt[2]{1.6 \times 10^{- 30}}

2

\sqrt[4]{1.6 \times 10^{- 30}}

3

\sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}

4

\frac{1.6 \times 10^{- 30}}{27}

Explanation:

Let's be the molar solubility of $Cr (OH)_{3}$ then,
$\begin{aligned} Cr (OH)_{3} ?? {Cr}_{S}^{3 +} + 3 {OH}_{3}^{-} \\ K_{sp} = [{Cr}^{3 +}] . [{OH}^{-}] \\ K_{sp} = S \times (3 S)^{3} \\ = 27 S^{4} \\ S = \sqrt[4]{\frac{K_{sp}}{27}} \\ S = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}} \end{aligned}$

AIEEE-2011

Ionic Equilibrium

229431 Solubility product of silver bromide is $5.0 \times 10^{- 13}$ - The quantity of potassium bromide (molar mass taken as $120 g {mol}^{- 1}$ ) to be added to $1 L$ of $0.05 M$ solution of silver nitrate of start the precipitation of $A g B r$ is

1

1.2 \times 10^{- 10} g

2

1.2 \times 10^{- 9} g

3

6.2 \times 10^{- 5} g

4

5.0 \times 10^{- 8} g

Explanation:

Given $(AgBr) = 5 \times 10^{- 13}, [{Ag}^{+}] = 0.05 m$ Molar mass of $KBr = 120 {gmol}^{- 1}$
${Ag}^{+} (aq) + {Br}^{-} (aq) = AgBr$
Precipitation starts when ionic product just exceeds solubility product.
$K_{sp} = [{Ag}^{+}] [{Br}^{-}]$ putting the valves -
$\Rightarrow [{Br}^{-}] = \frac{K_{sp}}{[{Ag}^{+}]} = \frac{5 \times 10^{- 13}}{0.05} = 10^{- 11}$
Precipitation starts when $10^{- 11}$ moles of $KBr$ is added to ${AgNO}_{3}$ of ${AgNO}_{3}$ solution.
No. of moles of $KBr$ to be added $= 10^{- 11}$
Weight of $KBr =$ no. of moles $\times$ molar mass.
Putting the valves.
$= 10^{- 11} \times 120 = 1.2 \times 10^{- 9} g$

AIEEE-2010

Ionic Equilibrium

229432 Solid $Ba ({NO}_{3})$ is gradually dissolved in a $1.0 \times 10^{- 4} {MNa}_{2} {CO}_{3}$ solution. At what concentration of ${Ba}^{2 +}$ will a precipitate begin to form? $(K_{sp}$ for ${BaCO}_{3} = 5.1 \times 10^{- 9}$ )

1

4.1 \times 10^{- 5} M

2

5.1 \times 10^{- 5} M

3

8.1 \times 10^{- 8} M

4

8.1 \times 10^{- 7} M

Explanation:

according to situation -
${Na}_{2} {CO}_{3} \underset{1 \times 10^{- 4} m}{2} 2 {Na}^{+} + {CO}_{3}^{2 -}$
For
$\begin{aligned} Ba CO {CO}_{3} \to {Ba}^{2 +} + {CO}_{3}^{2 -} \\ K_{sp} = [{Ba}^{2 +}] \cdot [{CO}_{3}^{2 -}] \end{aligned}$
Hence. $5.1 \times 10^{- 9} = [{Ba}^{2 +}] [1 \times 10^{- 4}]$
$\begin{aligned} [{Ba}^{2 +}] = \frac{5.1 \times 10^{- 9}}{1 \times 10^{- 4}} \\ {Ba}^{2 +} = 5.1 \times 10^{- 3} m \end{aligned}$

AIEEE-2009

Ionic Equilibrium

229433 In a saturated solution of the spatingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets in is ${AgIO}_{3}$ (s) $◻ A g + (aq) + {IO}_{3}^{-} (aq)$
If the solubility product constant $K_{sp}$ of ${AgIO}_{3}$ at a given temperature is $1.0 \times 10^{- 8}$ , what is the mass of ${AgIO}_{3}$ contained in $100 mL$ of its saturated solution?

1

28.3 \times 10^{- 2} g

2

28.3 \times 10^{- 4} g

3

1.0 \times 10^{- 7} g

4

1.0 \times 10^{- 4} g

Explanation:

Exp:
$Ag {IO}_{3}$ ??
(s = Solubility)
$K_{sp} = x^{x} \times y^{y} \times s^{x + y}$
$x =$ no. of ${Ag}^{+}$ ions and
$y =$ no of ${IO}_{3}^{-}$ ions
$K_{sp} = S^{2} \Rightarrow S = \sqrt{K_{sp}} = \sqrt{10^{- 8}} = 10^{- 4} M$
$1000 mL \to 10^{- 4} mL = 10^{- 4} \times 283 g$
$100 mL \to \frac{10^{- 4} \times 283 \times 100}{1000} = 28.3 \times 10^{- 4} g$

AIEEE-2007

Ionic Equilibrium

229434 The solubility product of a salt having general formula $M_{2}$ , in water is $4 \times 10^{- 12}$ . The concentration of $M^{2 +}$ ions in the aqueous solution of the salt is

1

4.0 \times 10^{- 10} M

2

1.6 \times 10^{- 4} M

3

1.0 \times 10^{- 4} M

4

2.0 \times 10^{- 6} M

Explanation:

Solubility produt $= 4 \times 10^{- 12}$
$\begin{aligned} {MX}_{2} ◻ \frac{M^{2 +} + \underset{[S]}{2 S]}}{} \\ K_{sp} = [M^{2 +}] \times {[M^{-}]}^{2} \\ K_{sp} = [S] \cdot [2 S]^{2} \\ K_{sp} = 4 S^{3} \end{aligned}$
The solubility of $M^{2 +}$ ion $= S$
$\begin{aligned} S = {(\frac{K_{sp}}{4})}^{1 / 3} = {(\frac{4 \times 10^{- 12}}{4})}^{1 / 3} \\ S = 1 \times 10^{- 4} M \end{aligned}$

AIEEE-2004

Ionic Equilibrium

229430 The $K_{sp}$ for $Cr (OH)_{3}$ is $1.6 \times 10^{- 30}$ . The molar solubility of this compound in water is

1

\sqrt[2]{1.6 \times 10^{- 30}}

2

\sqrt[4]{1.6 \times 10^{- 30}}

3

\sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}

4

\frac{1.6 \times 10^{- 30}}{27}

Explanation:

Let's be the molar solubility of $Cr (OH)_{3}$ then,
$\begin{aligned} Cr (OH)_{3} ?? {Cr}_{S}^{3 +} + 3 {OH}_{3}^{-} \\ K_{sp} = [{Cr}^{3 +}] . [{OH}^{-}] \\ K_{sp} = S \times (3 S)^{3} \\ = 27 S^{4} \\ S = \sqrt[4]{\frac{K_{sp}}{27}} \\ S = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}} \end{aligned}$

AIEEE-2011

Ionic Equilibrium

229431 Solubility product of silver bromide is $5.0 \times 10^{- 13}$ - The quantity of potassium bromide (molar mass taken as $120 g {mol}^{- 1}$ ) to be added to $1 L$ of $0.05 M$ solution of silver nitrate of start the precipitation of $A g B r$ is

1

1.2 \times 10^{- 10} g

2

1.2 \times 10^{- 9} g

3

6.2 \times 10^{- 5} g

4

5.0 \times 10^{- 8} g

Explanation:

Given $(AgBr) = 5 \times 10^{- 13}, [{Ag}^{+}] = 0.05 m$ Molar mass of $KBr = 120 {gmol}^{- 1}$
${Ag}^{+} (aq) + {Br}^{-} (aq) = AgBr$
Precipitation starts when ionic product just exceeds solubility product.
$K_{sp} = [{Ag}^{+}] [{Br}^{-}]$ putting the valves -
$\Rightarrow [{Br}^{-}] = \frac{K_{sp}}{[{Ag}^{+}]} = \frac{5 \times 10^{- 13}}{0.05} = 10^{- 11}$
Precipitation starts when $10^{- 11}$ moles of $KBr$ is added to ${AgNO}_{3}$ of ${AgNO}_{3}$ solution.
No. of moles of $KBr$ to be added $= 10^{- 11}$
Weight of $KBr =$ no. of moles $\times$ molar mass.
Putting the valves.
$= 10^{- 11} \times 120 = 1.2 \times 10^{- 9} g$

AIEEE-2010

Ionic Equilibrium

229432 Solid $Ba ({NO}_{3})$ is gradually dissolved in a $1.0 \times 10^{- 4} {MNa}_{2} {CO}_{3}$ solution. At what concentration of ${Ba}^{2 +}$ will a precipitate begin to form? $(K_{sp}$ for ${BaCO}_{3} = 5.1 \times 10^{- 9}$ )

1

4.1 \times 10^{- 5} M

2

5.1 \times 10^{- 5} M

3

8.1 \times 10^{- 8} M

4

8.1 \times 10^{- 7} M

Explanation:

according to situation -
${Na}_{2} {CO}_{3} \underset{1 \times 10^{- 4} m}{2} 2 {Na}^{+} + {CO}_{3}^{2 -}$
For
$\begin{aligned} Ba CO {CO}_{3} \to {Ba}^{2 +} + {CO}_{3}^{2 -} \\ K_{sp} = [{Ba}^{2 +}] \cdot [{CO}_{3}^{2 -}] \end{aligned}$
Hence. $5.1 \times 10^{- 9} = [{Ba}^{2 +}] [1 \times 10^{- 4}]$
$\begin{aligned} [{Ba}^{2 +}] = \frac{5.1 \times 10^{- 9}}{1 \times 10^{- 4}} \\ {Ba}^{2 +} = 5.1 \times 10^{- 3} m \end{aligned}$

AIEEE-2009

Ionic Equilibrium

229433 In a saturated solution of the spatingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets in is ${AgIO}_{3}$ (s) $◻ A g + (aq) + {IO}_{3}^{-} (aq)$
If the solubility product constant $K_{sp}$ of ${AgIO}_{3}$ at a given temperature is $1.0 \times 10^{- 8}$ , what is the mass of ${AgIO}_{3}$ contained in $100 mL$ of its saturated solution?

1

28.3 \times 10^{- 2} g

2

28.3 \times 10^{- 4} g

3

1.0 \times 10^{- 7} g

4

1.0 \times 10^{- 4} g

Explanation:

Exp:
$Ag {IO}_{3}$ ??
(s = Solubility)
$K_{sp} = x^{x} \times y^{y} \times s^{x + y}$
$x =$ no. of ${Ag}^{+}$ ions and
$y =$ no of ${IO}_{3}^{-}$ ions
$K_{sp} = S^{2} \Rightarrow S = \sqrt{K_{sp}} = \sqrt{10^{- 8}} = 10^{- 4} M$
$1000 mL \to 10^{- 4} mL = 10^{- 4} \times 283 g$
$100 mL \to \frac{10^{- 4} \times 283 \times 100}{1000} = 28.3 \times 10^{- 4} g$

AIEEE-2007

Ionic Equilibrium

229434 The solubility product of a salt having general formula $M_{2}$ , in water is $4 \times 10^{- 12}$ . The concentration of $M^{2 +}$ ions in the aqueous solution of the salt is

1

4.0 \times 10^{- 10} M

2

1.6 \times 10^{- 4} M

3

1.0 \times 10^{- 4} M

4

2.0 \times 10^{- 6} M

Explanation:

Solubility produt $= 4 \times 10^{- 12}$
$\begin{aligned} {MX}_{2} ◻ \frac{M^{2 +} + \underset{[S]}{2 S]}}{} \\ K_{sp} = [M^{2 +}] \times {[M^{-}]}^{2} \\ K_{sp} = [S] \cdot [2 S]^{2} \\ K_{sp} = 4 S^{3} \end{aligned}$
The solubility of $M^{2 +}$ ion $= S$
$\begin{aligned} S = {(\frac{K_{sp}}{4})}^{1 / 3} = {(\frac{4 \times 10^{- 12}}{4})}^{1 / 3} \\ S = 1 \times 10^{- 4} M \end{aligned}$

AIEEE-2004

Ionic Equilibrium

229430 The $K_{sp}$ for $Cr (OH)_{3}$ is $1.6 \times 10^{- 30}$ . The molar solubility of this compound in water is

1

\sqrt[2]{1.6 \times 10^{- 30}}

2

\sqrt[4]{1.6 \times 10^{- 30}}

3

\sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}

4

\frac{1.6 \times 10^{- 30}}{27}

Explanation:

Let's be the molar solubility of $Cr (OH)_{3}$ then,
$\begin{aligned} Cr (OH)_{3} ?? {Cr}_{S}^{3 +} + 3 {OH}_{3}^{-} \\ K_{sp} = [{Cr}^{3 +}] . [{OH}^{-}] \\ K_{sp} = S \times (3 S)^{3} \\ = 27 S^{4} \\ S = \sqrt[4]{\frac{K_{sp}}{27}} \\ S = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}} \end{aligned}$

AIEEE-2011

Ionic Equilibrium

229431 Solubility product of silver bromide is $5.0 \times 10^{- 13}$ - The quantity of potassium bromide (molar mass taken as $120 g {mol}^{- 1}$ ) to be added to $1 L$ of $0.05 M$ solution of silver nitrate of start the precipitation of $A g B r$ is

1

1.2 \times 10^{- 10} g

2

1.2 \times 10^{- 9} g

3

6.2 \times 10^{- 5} g

4

5.0 \times 10^{- 8} g

Explanation:

Given $(AgBr) = 5 \times 10^{- 13}, [{Ag}^{+}] = 0.05 m$ Molar mass of $KBr = 120 {gmol}^{- 1}$
${Ag}^{+} (aq) + {Br}^{-} (aq) = AgBr$
Precipitation starts when ionic product just exceeds solubility product.
$K_{sp} = [{Ag}^{+}] [{Br}^{-}]$ putting the valves -
$\Rightarrow [{Br}^{-}] = \frac{K_{sp}}{[{Ag}^{+}]} = \frac{5 \times 10^{- 13}}{0.05} = 10^{- 11}$
Precipitation starts when $10^{- 11}$ moles of $KBr$ is added to ${AgNO}_{3}$ of ${AgNO}_{3}$ solution.
No. of moles of $KBr$ to be added $= 10^{- 11}$
Weight of $KBr =$ no. of moles $\times$ molar mass.
Putting the valves.
$= 10^{- 11} \times 120 = 1.2 \times 10^{- 9} g$

AIEEE-2010

Ionic Equilibrium

229432 Solid $Ba ({NO}_{3})$ is gradually dissolved in a $1.0 \times 10^{- 4} {MNa}_{2} {CO}_{3}$ solution. At what concentration of ${Ba}^{2 +}$ will a precipitate begin to form? $(K_{sp}$ for ${BaCO}_{3} = 5.1 \times 10^{- 9}$ )

1

4.1 \times 10^{- 5} M

2

5.1 \times 10^{- 5} M

3

8.1 \times 10^{- 8} M

4

8.1 \times 10^{- 7} M

Explanation:

according to situation -
${Na}_{2} {CO}_{3} \underset{1 \times 10^{- 4} m}{2} 2 {Na}^{+} + {CO}_{3}^{2 -}$
For
$\begin{aligned} Ba CO {CO}_{3} \to {Ba}^{2 +} + {CO}_{3}^{2 -} \\ K_{sp} = [{Ba}^{2 +}] \cdot [{CO}_{3}^{2 -}] \end{aligned}$
Hence. $5.1 \times 10^{- 9} = [{Ba}^{2 +}] [1 \times 10^{- 4}]$
$\begin{aligned} [{Ba}^{2 +}] = \frac{5.1 \times 10^{- 9}}{1 \times 10^{- 4}} \\ {Ba}^{2 +} = 5.1 \times 10^{- 3} m \end{aligned}$

AIEEE-2009

Ionic Equilibrium

229433 In a saturated solution of the spatingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets in is ${AgIO}_{3}$ (s) $◻ A g + (aq) + {IO}_{3}^{-} (aq)$
If the solubility product constant $K_{sp}$ of ${AgIO}_{3}$ at a given temperature is $1.0 \times 10^{- 8}$ , what is the mass of ${AgIO}_{3}$ contained in $100 mL$ of its saturated solution?

1

28.3 \times 10^{- 2} g

2

28.3 \times 10^{- 4} g

3

1.0 \times 10^{- 7} g

4

1.0 \times 10^{- 4} g

Explanation:

Exp:
$Ag {IO}_{3}$ ??
(s = Solubility)
$K_{sp} = x^{x} \times y^{y} \times s^{x + y}$
$x =$ no. of ${Ag}^{+}$ ions and
$y =$ no of ${IO}_{3}^{-}$ ions
$K_{sp} = S^{2} \Rightarrow S = \sqrt{K_{sp}} = \sqrt{10^{- 8}} = 10^{- 4} M$
$1000 mL \to 10^{- 4} mL = 10^{- 4} \times 283 g$
$100 mL \to \frac{10^{- 4} \times 283 \times 100}{1000} = 28.3 \times 10^{- 4} g$

AIEEE-2007

Ionic Equilibrium

229434 The solubility product of a salt having general formula $M_{2}$ , in water is $4 \times 10^{- 12}$ . The concentration of $M^{2 +}$ ions in the aqueous solution of the salt is

1

4.0 \times 10^{- 10} M

2

1.6 \times 10^{- 4} M

3

1.0 \times 10^{- 4} M

4

2.0 \times 10^{- 6} M

Explanation:

Solubility produt $= 4 \times 10^{- 12}$
$\begin{aligned} {MX}_{2} ◻ \frac{M^{2 +} + \underset{[S]}{2 S]}}{} \\ K_{sp} = [M^{2 +}] \times {[M^{-}]}^{2} \\ K_{sp} = [S] \cdot [2 S]^{2} \\ K_{sp} = 4 S^{3} \end{aligned}$
The solubility of $M^{2 +}$ ion $= S$
$\begin{aligned} S = {(\frac{K_{sp}}{4})}^{1 / 3} = {(\frac{4 \times 10^{- 12}}{4})}^{1 / 3} \\ S = 1 \times 10^{- 4} M \end{aligned}$

AIEEE-2004

Ionic Equilibrium

229430 The $K_{sp}$ for $Cr (OH)_{3}$ is $1.6 \times 10^{- 30}$ . The molar solubility of this compound in water is

1

\sqrt[2]{1.6 \times 10^{- 30}}

2

\sqrt[4]{1.6 \times 10^{- 30}}

3

\sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}

4

\frac{1.6 \times 10^{- 30}}{27}

Explanation:

Let's be the molar solubility of $Cr (OH)_{3}$ then,
$\begin{aligned} Cr (OH)_{3} ?? {Cr}_{S}^{3 +} + 3 {OH}_{3}^{-} \\ K_{sp} = [{Cr}^{3 +}] . [{OH}^{-}] \\ K_{sp} = S \times (3 S)^{3} \\ = 27 S^{4} \\ S = \sqrt[4]{\frac{K_{sp}}{27}} \\ S = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}} \end{aligned}$

AIEEE-2011

Ionic Equilibrium

229431 Solubility product of silver bromide is $5.0 \times 10^{- 13}$ - The quantity of potassium bromide (molar mass taken as $120 g {mol}^{- 1}$ ) to be added to $1 L$ of $0.05 M$ solution of silver nitrate of start the precipitation of $A g B r$ is

1

1.2 \times 10^{- 10} g

2

1.2 \times 10^{- 9} g

3

6.2 \times 10^{- 5} g

4

5.0 \times 10^{- 8} g

Explanation:

Given $(AgBr) = 5 \times 10^{- 13}, [{Ag}^{+}] = 0.05 m$ Molar mass of $KBr = 120 {gmol}^{- 1}$
${Ag}^{+} (aq) + {Br}^{-} (aq) = AgBr$
Precipitation starts when ionic product just exceeds solubility product.
$K_{sp} = [{Ag}^{+}] [{Br}^{-}]$ putting the valves -
$\Rightarrow [{Br}^{-}] = \frac{K_{sp}}{[{Ag}^{+}]} = \frac{5 \times 10^{- 13}}{0.05} = 10^{- 11}$
Precipitation starts when $10^{- 11}$ moles of $KBr$ is added to ${AgNO}_{3}$ of ${AgNO}_{3}$ solution.
No. of moles of $KBr$ to be added $= 10^{- 11}$
Weight of $KBr =$ no. of moles $\times$ molar mass.
Putting the valves.
$= 10^{- 11} \times 120 = 1.2 \times 10^{- 9} g$

AIEEE-2010

Ionic Equilibrium

229432 Solid $Ba ({NO}_{3})$ is gradually dissolved in a $1.0 \times 10^{- 4} {MNa}_{2} {CO}_{3}$ solution. At what concentration of ${Ba}^{2 +}$ will a precipitate begin to form? $(K_{sp}$ for ${BaCO}_{3} = 5.1 \times 10^{- 9}$ )

1

4.1 \times 10^{- 5} M

2

5.1 \times 10^{- 5} M

3

8.1 \times 10^{- 8} M

4

8.1 \times 10^{- 7} M

Explanation:

according to situation -
${Na}_{2} {CO}_{3} \underset{1 \times 10^{- 4} m}{2} 2 {Na}^{+} + {CO}_{3}^{2 -}$
For
$\begin{aligned} Ba CO {CO}_{3} \to {Ba}^{2 +} + {CO}_{3}^{2 -} \\ K_{sp} = [{Ba}^{2 +}] \cdot [{CO}_{3}^{2 -}] \end{aligned}$
Hence. $5.1 \times 10^{- 9} = [{Ba}^{2 +}] [1 \times 10^{- 4}]$
$\begin{aligned} [{Ba}^{2 +}] = \frac{5.1 \times 10^{- 9}}{1 \times 10^{- 4}} \\ {Ba}^{2 +} = 5.1 \times 10^{- 3} m \end{aligned}$

AIEEE-2009

Ionic Equilibrium

229433 In a saturated solution of the spatingly soluble strong electrolyte ${AgIO}_{3}$ (molecular mass = 283) the equilibrium which sets in is ${AgIO}_{3}$ (s) $◻ A g + (aq) + {IO}_{3}^{-} (aq)$
If the solubility product constant $K_{sp}$ of ${AgIO}_{3}$ at a given temperature is $1.0 \times 10^{- 8}$ , what is the mass of ${AgIO}_{3}$ contained in $100 mL$ of its saturated solution?

1

28.3 \times 10^{- 2} g

2

28.3 \times 10^{- 4} g

3

1.0 \times 10^{- 7} g

4

1.0 \times 10^{- 4} g

Explanation:

Exp:
$Ag {IO}_{3}$ ??
(s = Solubility)
$K_{sp} = x^{x} \times y^{y} \times s^{x + y}$
$x =$ no. of ${Ag}^{+}$ ions and
$y =$ no of ${IO}_{3}^{-}$ ions
$K_{sp} = S^{2} \Rightarrow S = \sqrt{K_{sp}} = \sqrt{10^{- 8}} = 10^{- 4} M$
$1000 mL \to 10^{- 4} mL = 10^{- 4} \times 283 g$
$100 mL \to \frac{10^{- 4} \times 283 \times 100}{1000} = 28.3 \times 10^{- 4} g$

AIEEE-2007

Ionic Equilibrium

229434 The solubility product of a salt having general formula $M_{2}$ , in water is $4 \times 10^{- 12}$ . The concentration of $M^{2 +}$ ions in the aqueous solution of the salt is

1

4.0 \times 10^{- 10} M

2

1.6 \times 10^{- 4} M

3

1.0 \times 10^{- 4} M

4

2.0 \times 10^{- 6} M

Explanation:

Solubility produt $= 4 \times 10^{- 12}$
$\begin{aligned} {MX}_{2} ◻ \frac{M^{2 +} + \underset{[S]}{2 S]}}{} \\ K_{sp} = [M^{2 +}] \times {[M^{-}]}^{2} \\ K_{sp} = [S] \cdot [2 S]^{2} \\ K_{sp} = 4 S^{3} \end{aligned}$
The solubility of $M^{2 +}$ ion $= S$
$\begin{aligned} S = {(\frac{K_{sp}}{4})}^{1 / 3} = {(\frac{4 \times 10^{- 12}}{4})}^{1 / 3} \\ S = 1 \times 10^{- 4} M \end{aligned}$

AIEEE-2004

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