QBManager

Ionic Equilibrium

229391 What will be solubility of $KAl {({SO}_{4})}_{2}$ in water if solubility product $(K_{sp})$ for $KAl {({SO}_{4})}_{2}$ is $1.6 \times 10^{- 11}$ (unit)?

1

6 \times 10^{- 5} M

2

3 \times 10^{- 4} M

3

2 \times 10^{- 6} M

4

1.4 \times 10^{- 3} M

Explanation:

Given that, $K_{sp} = 1.6 \times 10^{- 11}$
$\begin{aligned} KAl {({SO}_{4})}_{2} ? \\ K_{sp} = {[K_{S}^{+} + [{Al}^{3 +} S^{3 +} + 2 {SO}_{4}^{2 -}] {SO}_{4}^{2}]}^{2} \\ = S \times S \times 2 S^{2} \\ K_{sp} = 4 S^{4} \\ S = {(\frac{1.6 \times 10^{- 11}}{4})}^{1 / 4} \\ = {(4 \times 10^{- 12})}^{1 / 4} \\ S = 1.4 \times 10^{- 3} \end{aligned}$

AMU-2019.

Ionic Equilibrium

229436 The solubility in water of a sparingly soluble salt ${AB}_{2}$ is $1.0 \times 10^{- 5}$ mol $L^{- 1}$ . Its solubility product will be

1

4 \times 10^{- 15}

2

4 \times 10^{- 10}

3

1 \times 10^{- 15}

4

1 \times 10^{- 10}

Explanation:

Solubility process is :-
${AB}_{2} ⟶ A_{s}^{2 +} + 2 B^{-}$
using the relation of solubility and solubility product as:
$\begin{aligned} K_{sp} = [A^{2 +}] {[B^{-}]}^{2} \\ K_{sp} = [S .] [2 S]^{2} \end{aligned}$
Substituting $. S = 10^{- 5}$
$\begin{aligned} K_{sp} = 10^{- 5} \times {(2 \times 10^{- 5})}^{2} \\ K_{sp} = 4 \times 10^{- 15} \end{aligned}$

AIEEE-2003

Ionic Equilibrium

229392 The $K_{sp}$ of ${PbCO}_{3}$ and ${MgCO}_{3}$ are $1.5 \times 10^{- 15}$ and $1 \times 10^{- 15}$ respectively at $298 K$ . The concentration of ${Pb}^{2 +}$ ions in a saturated solution containing ${MgCO}_{3}$ and ${PbCO}_{3}$ is

1

1.5 \times 10^{- 8} M

2

3 \times 10^{- 8} M

3

2 \times 10^{- 8} M

4

2.5 \times 10^{- 8} M

Explanation:

Exp:

$\begin{aligned} K_{sp} ({MgCO}_{3}) = (y) (x + y) \\ \frac{K_{sp} ({PbCO}_{3})}{K_{sp} ({MgCO}_{3})} = \frac{(x) \cdot (x + y)}{y (x + y)} \\ \frac{1.5 \times 10^{- 15}}{1 \times 10^{- 15}} = \frac{x}{y} \Rightarrow x = 1.5 y \end{aligned}$
Putting the value of $X$ in equation (i), we get-
$\begin{aligned} 1.5 \times 10^{- 15} & = 1.5 y (1.5 y + y) \\ = 1.5 y (2.5 y) \\ = 3.75 y^{2} \\ y & = \sqrt{\frac{1.5 \times 10^{- 15}}{3.75}} \\ y & = 2 \times 10^{- 8} \end{aligned}$
Then, $x = 1.5 y$
$= 1.5 \times 2 \times 10^{- 8}, x = 3 \times 10^{- 8} m$

AMU-2018

Ionic Equilibrium

229393 The solubility of pure oxygen in water at $20^{\circ} C$ and 1.0 atmosphere pressure is $1.38 \times 10^{- 3}$ mole/litre. What will be the concentration of oxygen at $20^{\circ} C$ and partial pressure of 0.21 atmosphere?

1

2.9 \times 10^{- 4} mole / litre

2

5.8 \times 10^{- 4} mole / litre

3

7.6 \times 10^{- 4} mole / litre

4

11.6 \times 10^{- 4} mole / litre

Explanation:

The solubility of gas in liquid at particular temperature is given by Henrys law is -
$m_{a} = K_{H} P_{a}$
Where $m_{a} =$ mass of gas dissolved in a unit volume of solvent.
$Pa =$ pressure of the gas in equilibrium
$K_{H} =$ Henry's Law constant
On putting the given values in above expression.
$\begin{aligned} 1.38 \times 10^{- 3} mol / l = K_{H} \times 1 atm \\ K_{H} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \end{aligned}$
At $0.21 atm m_{a}$ will be calculated as
$\begin{aligned} m_{a} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \times 0.21 atm \\ m_{a} = 2.9 \times 10^{- 4} mol / L . \end{aligned}$

AMU-2018

Ionic Equilibrium

229391 What will be solubility of $KAl {({SO}_{4})}_{2}$ in water if solubility product $(K_{sp})$ for $KAl {({SO}_{4})}_{2}$ is $1.6 \times 10^{- 11}$ (unit)?

1

6 \times 10^{- 5} M

2

3 \times 10^{- 4} M

3

2 \times 10^{- 6} M

4

1.4 \times 10^{- 3} M

Explanation:

Given that, $K_{sp} = 1.6 \times 10^{- 11}$
$\begin{aligned} KAl {({SO}_{4})}_{2} ? \\ K_{sp} = {[K_{S}^{+} + [{Al}^{3 +} S^{3 +} + 2 {SO}_{4}^{2 -}] {SO}_{4}^{2}]}^{2} \\ = S \times S \times 2 S^{2} \\ K_{sp} = 4 S^{4} \\ S = {(\frac{1.6 \times 10^{- 11}}{4})}^{1 / 4} \\ = {(4 \times 10^{- 12})}^{1 / 4} \\ S = 1.4 \times 10^{- 3} \end{aligned}$

AMU-2019.

Ionic Equilibrium

229436 The solubility in water of a sparingly soluble salt ${AB}_{2}$ is $1.0 \times 10^{- 5}$ mol $L^{- 1}$ . Its solubility product will be

1

4 \times 10^{- 15}

2

4 \times 10^{- 10}

3

1 \times 10^{- 15}

4

1 \times 10^{- 10}

Explanation:

Solubility process is :-
${AB}_{2} ⟶ A_{s}^{2 +} + 2 B^{-}$
using the relation of solubility and solubility product as:
$\begin{aligned} K_{sp} = [A^{2 +}] {[B^{-}]}^{2} \\ K_{sp} = [S .] [2 S]^{2} \end{aligned}$
Substituting $. S = 10^{- 5}$
$\begin{aligned} K_{sp} = 10^{- 5} \times {(2 \times 10^{- 5})}^{2} \\ K_{sp} = 4 \times 10^{- 15} \end{aligned}$

AIEEE-2003

Ionic Equilibrium

229392 The $K_{sp}$ of ${PbCO}_{3}$ and ${MgCO}_{3}$ are $1.5 \times 10^{- 15}$ and $1 \times 10^{- 15}$ respectively at $298 K$ . The concentration of ${Pb}^{2 +}$ ions in a saturated solution containing ${MgCO}_{3}$ and ${PbCO}_{3}$ is

1

1.5 \times 10^{- 8} M

2

3 \times 10^{- 8} M

3

2 \times 10^{- 8} M

4

2.5 \times 10^{- 8} M

Explanation:

Exp:

$\begin{aligned} K_{sp} ({MgCO}_{3}) = (y) (x + y) \\ \frac{K_{sp} ({PbCO}_{3})}{K_{sp} ({MgCO}_{3})} = \frac{(x) \cdot (x + y)}{y (x + y)} \\ \frac{1.5 \times 10^{- 15}}{1 \times 10^{- 15}} = \frac{x}{y} \Rightarrow x = 1.5 y \end{aligned}$
Putting the value of $X$ in equation (i), we get-
$\begin{aligned} 1.5 \times 10^{- 15} & = 1.5 y (1.5 y + y) \\ = 1.5 y (2.5 y) \\ = 3.75 y^{2} \\ y & = \sqrt{\frac{1.5 \times 10^{- 15}}{3.75}} \\ y & = 2 \times 10^{- 8} \end{aligned}$
Then, $x = 1.5 y$
$= 1.5 \times 2 \times 10^{- 8}, x = 3 \times 10^{- 8} m$

AMU-2018

Ionic Equilibrium

229393 The solubility of pure oxygen in water at $20^{\circ} C$ and 1.0 atmosphere pressure is $1.38 \times 10^{- 3}$ mole/litre. What will be the concentration of oxygen at $20^{\circ} C$ and partial pressure of 0.21 atmosphere?

1

2.9 \times 10^{- 4} mole / litre

2

5.8 \times 10^{- 4} mole / litre

3

7.6 \times 10^{- 4} mole / litre

4

11.6 \times 10^{- 4} mole / litre

Explanation:

The solubility of gas in liquid at particular temperature is given by Henrys law is -
$m_{a} = K_{H} P_{a}$
Where $m_{a} =$ mass of gas dissolved in a unit volume of solvent.
$Pa =$ pressure of the gas in equilibrium
$K_{H} =$ Henry's Law constant
On putting the given values in above expression.
$\begin{aligned} 1.38 \times 10^{- 3} mol / l = K_{H} \times 1 atm \\ K_{H} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \end{aligned}$
At $0.21 atm m_{a}$ will be calculated as
$\begin{aligned} m_{a} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \times 0.21 atm \\ m_{a} = 2.9 \times 10^{- 4} mol / L . \end{aligned}$

AMU-2018

Ionic Equilibrium

229391 What will be solubility of $KAl {({SO}_{4})}_{2}$ in water if solubility product $(K_{sp})$ for $KAl {({SO}_{4})}_{2}$ is $1.6 \times 10^{- 11}$ (unit)?

1

6 \times 10^{- 5} M

2

3 \times 10^{- 4} M

3

2 \times 10^{- 6} M

4

1.4 \times 10^{- 3} M

Explanation:

Given that, $K_{sp} = 1.6 \times 10^{- 11}$
$\begin{aligned} KAl {({SO}_{4})}_{2} ? \\ K_{sp} = {[K_{S}^{+} + [{Al}^{3 +} S^{3 +} + 2 {SO}_{4}^{2 -}] {SO}_{4}^{2}]}^{2} \\ = S \times S \times 2 S^{2} \\ K_{sp} = 4 S^{4} \\ S = {(\frac{1.6 \times 10^{- 11}}{4})}^{1 / 4} \\ = {(4 \times 10^{- 12})}^{1 / 4} \\ S = 1.4 \times 10^{- 3} \end{aligned}$

AMU-2019.

Ionic Equilibrium

229436 The solubility in water of a sparingly soluble salt ${AB}_{2}$ is $1.0 \times 10^{- 5}$ mol $L^{- 1}$ . Its solubility product will be

1

4 \times 10^{- 15}

2

4 \times 10^{- 10}

3

1 \times 10^{- 15}

4

1 \times 10^{- 10}

Explanation:

Solubility process is :-
${AB}_{2} ⟶ A_{s}^{2 +} + 2 B^{-}$
using the relation of solubility and solubility product as:
$\begin{aligned} K_{sp} = [A^{2 +}] {[B^{-}]}^{2} \\ K_{sp} = [S .] [2 S]^{2} \end{aligned}$
Substituting $. S = 10^{- 5}$
$\begin{aligned} K_{sp} = 10^{- 5} \times {(2 \times 10^{- 5})}^{2} \\ K_{sp} = 4 \times 10^{- 15} \end{aligned}$

AIEEE-2003

Ionic Equilibrium

229392 The $K_{sp}$ of ${PbCO}_{3}$ and ${MgCO}_{3}$ are $1.5 \times 10^{- 15}$ and $1 \times 10^{- 15}$ respectively at $298 K$ . The concentration of ${Pb}^{2 +}$ ions in a saturated solution containing ${MgCO}_{3}$ and ${PbCO}_{3}$ is

1

1.5 \times 10^{- 8} M

2

3 \times 10^{- 8} M

3

2 \times 10^{- 8} M

4

2.5 \times 10^{- 8} M

Explanation:

Exp:

$\begin{aligned} K_{sp} ({MgCO}_{3}) = (y) (x + y) \\ \frac{K_{sp} ({PbCO}_{3})}{K_{sp} ({MgCO}_{3})} = \frac{(x) \cdot (x + y)}{y (x + y)} \\ \frac{1.5 \times 10^{- 15}}{1 \times 10^{- 15}} = \frac{x}{y} \Rightarrow x = 1.5 y \end{aligned}$
Putting the value of $X$ in equation (i), we get-
$\begin{aligned} 1.5 \times 10^{- 15} & = 1.5 y (1.5 y + y) \\ = 1.5 y (2.5 y) \\ = 3.75 y^{2} \\ y & = \sqrt{\frac{1.5 \times 10^{- 15}}{3.75}} \\ y & = 2 \times 10^{- 8} \end{aligned}$
Then, $x = 1.5 y$
$= 1.5 \times 2 \times 10^{- 8}, x = 3 \times 10^{- 8} m$

AMU-2018

Ionic Equilibrium

229393 The solubility of pure oxygen in water at $20^{\circ} C$ and 1.0 atmosphere pressure is $1.38 \times 10^{- 3}$ mole/litre. What will be the concentration of oxygen at $20^{\circ} C$ and partial pressure of 0.21 atmosphere?

1

2.9 \times 10^{- 4} mole / litre

2

5.8 \times 10^{- 4} mole / litre

3

7.6 \times 10^{- 4} mole / litre

4

11.6 \times 10^{- 4} mole / litre

Explanation:

The solubility of gas in liquid at particular temperature is given by Henrys law is -
$m_{a} = K_{H} P_{a}$
Where $m_{a} =$ mass of gas dissolved in a unit volume of solvent.
$Pa =$ pressure of the gas in equilibrium
$K_{H} =$ Henry's Law constant
On putting the given values in above expression.
$\begin{aligned} 1.38 \times 10^{- 3} mol / l = K_{H} \times 1 atm \\ K_{H} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \end{aligned}$
At $0.21 atm m_{a}$ will be calculated as
$\begin{aligned} m_{a} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \times 0.21 atm \\ m_{a} = 2.9 \times 10^{- 4} mol / L . \end{aligned}$

AMU-2018

Ionic Equilibrium

229391 What will be solubility of $KAl {({SO}_{4})}_{2}$ in water if solubility product $(K_{sp})$ for $KAl {({SO}_{4})}_{2}$ is $1.6 \times 10^{- 11}$ (unit)?

1

6 \times 10^{- 5} M

2

3 \times 10^{- 4} M

3

2 \times 10^{- 6} M

4

1.4 \times 10^{- 3} M

Explanation:

Given that, $K_{sp} = 1.6 \times 10^{- 11}$
$\begin{aligned} KAl {({SO}_{4})}_{2} ? \\ K_{sp} = {[K_{S}^{+} + [{Al}^{3 +} S^{3 +} + 2 {SO}_{4}^{2 -}] {SO}_{4}^{2}]}^{2} \\ = S \times S \times 2 S^{2} \\ K_{sp} = 4 S^{4} \\ S = {(\frac{1.6 \times 10^{- 11}}{4})}^{1 / 4} \\ = {(4 \times 10^{- 12})}^{1 / 4} \\ S = 1.4 \times 10^{- 3} \end{aligned}$

AMU-2019.

Ionic Equilibrium

229436 The solubility in water of a sparingly soluble salt ${AB}_{2}$ is $1.0 \times 10^{- 5}$ mol $L^{- 1}$ . Its solubility product will be

1

4 \times 10^{- 15}

2

4 \times 10^{- 10}

3

1 \times 10^{- 15}

4

1 \times 10^{- 10}

Explanation:

Solubility process is :-
${AB}_{2} ⟶ A_{s}^{2 +} + 2 B^{-}$
using the relation of solubility and solubility product as:
$\begin{aligned} K_{sp} = [A^{2 +}] {[B^{-}]}^{2} \\ K_{sp} = [S .] [2 S]^{2} \end{aligned}$
Substituting $. S = 10^{- 5}$
$\begin{aligned} K_{sp} = 10^{- 5} \times {(2 \times 10^{- 5})}^{2} \\ K_{sp} = 4 \times 10^{- 15} \end{aligned}$

AIEEE-2003

Ionic Equilibrium

229392 The $K_{sp}$ of ${PbCO}_{3}$ and ${MgCO}_{3}$ are $1.5 \times 10^{- 15}$ and $1 \times 10^{- 15}$ respectively at $298 K$ . The concentration of ${Pb}^{2 +}$ ions in a saturated solution containing ${MgCO}_{3}$ and ${PbCO}_{3}$ is

1

1.5 \times 10^{- 8} M

2

3 \times 10^{- 8} M

3

2 \times 10^{- 8} M

4

2.5 \times 10^{- 8} M

Explanation:

Exp:

$\begin{aligned} K_{sp} ({MgCO}_{3}) = (y) (x + y) \\ \frac{K_{sp} ({PbCO}_{3})}{K_{sp} ({MgCO}_{3})} = \frac{(x) \cdot (x + y)}{y (x + y)} \\ \frac{1.5 \times 10^{- 15}}{1 \times 10^{- 15}} = \frac{x}{y} \Rightarrow x = 1.5 y \end{aligned}$
Putting the value of $X$ in equation (i), we get-
$\begin{aligned} 1.5 \times 10^{- 15} & = 1.5 y (1.5 y + y) \\ = 1.5 y (2.5 y) \\ = 3.75 y^{2} \\ y & = \sqrt{\frac{1.5 \times 10^{- 15}}{3.75}} \\ y & = 2 \times 10^{- 8} \end{aligned}$
Then, $x = 1.5 y$
$= 1.5 \times 2 \times 10^{- 8}, x = 3 \times 10^{- 8} m$

AMU-2018

Ionic Equilibrium

229393 The solubility of pure oxygen in water at $20^{\circ} C$ and 1.0 atmosphere pressure is $1.38 \times 10^{- 3}$ mole/litre. What will be the concentration of oxygen at $20^{\circ} C$ and partial pressure of 0.21 atmosphere?

1

2.9 \times 10^{- 4} mole / litre

2

5.8 \times 10^{- 4} mole / litre

3

7.6 \times 10^{- 4} mole / litre

4

11.6 \times 10^{- 4} mole / litre

Explanation:

The solubility of gas in liquid at particular temperature is given by Henrys law is -
$m_{a} = K_{H} P_{a}$
Where $m_{a} =$ mass of gas dissolved in a unit volume of solvent.
$Pa =$ pressure of the gas in equilibrium
$K_{H} =$ Henry's Law constant
On putting the given values in above expression.
$\begin{aligned} 1.38 \times 10^{- 3} mol / l = K_{H} \times 1 atm \\ K_{H} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \end{aligned}$
At $0.21 atm m_{a}$ will be calculated as
$\begin{aligned} m_{a} = \frac{1.38 \times 10^{- 3} mol / L}{1 atm} \times 0.21 atm \\ m_{a} = 2.9 \times 10^{- 4} mol / L . \end{aligned}$

AMU-2018

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