01. Solubility and Solubility Product Constant
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Ionic Equilibrium

229435 The molar solubility (in $\mathrm{Mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is $\mathrm{S}$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ is given in terms of $K_{s p}$ by the relation.

1 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 128\right)^{1 / 4}$
2 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 128\right)^{1 / 4}$
3 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 256\right)^{1 / 5}$
4 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 256\right)^{1 / 5}$
AIEEE-2004
Ionic Equilibrium

229439 A solution is $0.1 \mathrm{M}$ in $\mathrm{Cl}^{-}$and $0.001 \mathrm{M}$ in $\mathrm{CrO}_4^{2-}$. Solid $\mathrm{AgNO}_3$ is gradually added to it.
Assuming that the addition does not change in volume and
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7 \times 10^{-10} \mathrm{M}^2$ and
$\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)=1.9 \times 10^{-12} \mathrm{M}^3$
Select correct statement from the following

1 $\mathrm{AgCl}$ precipitates first because its $\mathrm{K}_{\mathrm{sp}}$ is high.
2 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first as its $\mathrm{K}_{\mathrm{sp}}$ is low.
3 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first because the amount of $\mathrm{Ag}^{+}$needed is low.
4 $\mathrm{AgCl}$ will precipitate first as the amount of $\mathrm{Ag}^{+}$needed to precipitate is low
JEE Main 2021
Ionic Equilibrium

229440 The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is [Given: The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water = $\left.5.5 \times 10^{-6}\right]$

1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
JEE Main 2021
Ionic Equilibrium

229441 The pH of a 0.1 molar solution of the acid $\mathrm{HQ}$ is 3 . The value of the ionisation constant, $K_a$ of the acid is

1 $3 \times 10^{-1}$
2 $1 \times 10^{-3}$
3 $1 \times 10^{-5}$
4 $1 \times 10^{-7}$
AIEEE-2012
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Ionic Equilibrium

229435 The molar solubility (in $\mathrm{Mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is $\mathrm{S}$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ is given in terms of $K_{s p}$ by the relation.

1 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 128\right)^{1 / 4}$
2 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 128\right)^{1 / 4}$
3 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 256\right)^{1 / 5}$
4 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 256\right)^{1 / 5}$
AIEEE-2004
Ionic Equilibrium

229439 A solution is $0.1 \mathrm{M}$ in $\mathrm{Cl}^{-}$and $0.001 \mathrm{M}$ in $\mathrm{CrO}_4^{2-}$. Solid $\mathrm{AgNO}_3$ is gradually added to it.
Assuming that the addition does not change in volume and
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7 \times 10^{-10} \mathrm{M}^2$ and
$\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)=1.9 \times 10^{-12} \mathrm{M}^3$
Select correct statement from the following

1 $\mathrm{AgCl}$ precipitates first because its $\mathrm{K}_{\mathrm{sp}}$ is high.
2 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first as its $\mathrm{K}_{\mathrm{sp}}$ is low.
3 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first because the amount of $\mathrm{Ag}^{+}$needed is low.
4 $\mathrm{AgCl}$ will precipitate first as the amount of $\mathrm{Ag}^{+}$needed to precipitate is low
JEE Main 2021
Ionic Equilibrium

229440 The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is [Given: The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water = $\left.5.5 \times 10^{-6}\right]$

1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
JEE Main 2021
Ionic Equilibrium

229441 The pH of a 0.1 molar solution of the acid $\mathrm{HQ}$ is 3 . The value of the ionisation constant, $K_a$ of the acid is

1 $3 \times 10^{-1}$
2 $1 \times 10^{-3}$
3 $1 \times 10^{-5}$
4 $1 \times 10^{-7}$
AIEEE-2012
NEET DIGITAL LIBRARY
Ionic Equilibrium

229435 The molar solubility (in $\mathrm{Mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is $\mathrm{S}$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ is given in terms of $K_{s p}$ by the relation.

1 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 128\right)^{1 / 4}$
2 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 128\right)^{1 / 4}$
3 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 256\right)^{1 / 5}$
4 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 256\right)^{1 / 5}$
AIEEE-2004
Ionic Equilibrium

229439 A solution is $0.1 \mathrm{M}$ in $\mathrm{Cl}^{-}$and $0.001 \mathrm{M}$ in $\mathrm{CrO}_4^{2-}$. Solid $\mathrm{AgNO}_3$ is gradually added to it.
Assuming that the addition does not change in volume and
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7 \times 10^{-10} \mathrm{M}^2$ and
$\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)=1.9 \times 10^{-12} \mathrm{M}^3$
Select correct statement from the following

1 $\mathrm{AgCl}$ precipitates first because its $\mathrm{K}_{\mathrm{sp}}$ is high.
2 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first as its $\mathrm{K}_{\mathrm{sp}}$ is low.
3 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first because the amount of $\mathrm{Ag}^{+}$needed is low.
4 $\mathrm{AgCl}$ will precipitate first as the amount of $\mathrm{Ag}^{+}$needed to precipitate is low
JEE Main 2021
Ionic Equilibrium

229440 The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is [Given: The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water = $\left.5.5 \times 10^{-6}\right]$

1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
JEE Main 2021
Ionic Equilibrium

229441 The pH of a 0.1 molar solution of the acid $\mathrm{HQ}$ is 3 . The value of the ionisation constant, $K_a$ of the acid is

1 $3 \times 10^{-1}$
2 $1 \times 10^{-3}$
3 $1 \times 10^{-5}$
4 $1 \times 10^{-7}$
AIEEE-2012
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Ionic Equilibrium

229435 The molar solubility (in $\mathrm{Mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is $\mathrm{S}$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ is given in terms of $K_{s p}$ by the relation.

1 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 128\right)^{1 / 4}$
2 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 128\right)^{1 / 4}$
3 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} / 256\right)^{1 / 5}$
4 $\mathrm{S}=\left(\mathrm{K}_{\mathrm{sp}} \times 256\right)^{1 / 5}$
AIEEE-2004
Ionic Equilibrium

229439 A solution is $0.1 \mathrm{M}$ in $\mathrm{Cl}^{-}$and $0.001 \mathrm{M}$ in $\mathrm{CrO}_4^{2-}$. Solid $\mathrm{AgNO}_3$ is gradually added to it.
Assuming that the addition does not change in volume and
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7 \times 10^{-10} \mathrm{M}^2$ and
$\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)=1.9 \times 10^{-12} \mathrm{M}^3$
Select correct statement from the following

1 $\mathrm{AgCl}$ precipitates first because its $\mathrm{K}_{\mathrm{sp}}$ is high.
2 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first as its $\mathrm{K}_{\mathrm{sp}}$ is low.
3 $\mathrm{Ag}_2 \mathrm{CrO}_4$ precipitates first because the amount of $\mathrm{Ag}^{+}$needed is low.
4 $\mathrm{AgCl}$ will precipitate first as the amount of $\mathrm{Ag}^{+}$needed to precipitate is low
JEE Main 2021
Ionic Equilibrium

229440 The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is [Given: The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water = $\left.5.5 \times 10^{-6}\right]$

1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
JEE Main 2021
Ionic Equilibrium

229441 The pH of a 0.1 molar solution of the acid $\mathrm{HQ}$ is 3 . The value of the ionisation constant, $K_a$ of the acid is

1 $3 \times 10^{-1}$
2 $1 \times 10^{-3}$
3 $1 \times 10^{-5}$
4 $1 \times 10^{-7}$
AIEEE-2012
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