01. Solubility and Solubility Product Constant
Ionic Equilibrium

229386 On adding $0.1 \mathrm{M}$ solution each of $\left[\mathrm{Ag}^{+}\right],\left[\mathrm{Ba}^{2+}\right]$, $\left[\mathrm{Ca}^{2+}\right]$ in a $\mathrm{Na}_2 \mathrm{SO}_4$ solution, species first precipitated is $\left[\mathrm{K}_{\mathrm{sp}} \mathrm{BaSO}_4^{-5}=10^{-11}, \mathrm{~K}_{\mathrm{sp}} \mathrm{CaSO}_4=\mathbf{1 0}^{-6}, \mathrm{~K}_{\mathrm{sp}} \mathrm{Ag}_2 \mathrm{SO}_4\right.$

1 $\mathrm{Ag}_2 \mathrm{SO}_4$
2 $\mathrm{BaSO}_4$
3 $\mathrm{CaSO}_4$
4 All of these
Ionic Equilibrium

229388 If the ionic product of $\mathrm{Ni}(\mathrm{OH})_2$ is $1.9 \times 10^{-15}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $1.0 \mathrm{M} \mathrm{NaOH}$ is

1 $1.9 \times 10^{-18} \mathrm{M}$
2 $1.9 \times 10^{-13} \mathrm{M}$
3 $1.9 \times 10^{-15} \mathrm{M}$
4 $1.9 \times 10^{-14} \mathrm{M}$
Ionic Equilibrium

229389 A saturated solution of $\mathrm{H}_2 \mathrm{~S}$ in $0.1 \mathrm{M} \mathrm{HCl}$ at $25^{\circ} \mathrm{C}$ contains $\mathrm{S}^{2-}$ ion concentration of $10^{-23} \mathrm{~mol}$ $L^{-1}$. The solubility product of some sulphides are $\mathrm{CuS}=10^{-44}, \mathrm{FeS}=10^{-14}, \mathrm{MnS}=10^{-15}, \mathrm{CdS}=$ $10^{-25}$. If $0.01 \mathrm{M}$ solution of these salts in $1 \mathrm{M}$ $\mathrm{HCl}$ are saturated with $\mathrm{H}_2 \mathrm{~S}$, which of these will be precipitated?

1 All
2 All except $\mathrm{MnS}$
3 All except $\mathrm{MnS}$ and FeS
4 Only-CuS
Ionic Equilibrium

229390 The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{-2}$ ions in that solution?

1 $2 \times 10^{-4} \mathrm{M}$
2 $16 \times 10^{-4} \mathrm{M}$
3 $8 \times 10^{-4} \mathrm{M}$
4 $8 \times 10^{-8} \mathrm{M}$
VIEEE-2014
Ionic Equilibrium

229386 On adding $0.1 \mathrm{M}$ solution each of $\left[\mathrm{Ag}^{+}\right],\left[\mathrm{Ba}^{2+}\right]$, $\left[\mathrm{Ca}^{2+}\right]$ in a $\mathrm{Na}_2 \mathrm{SO}_4$ solution, species first precipitated is $\left[\mathrm{K}_{\mathrm{sp}} \mathrm{BaSO}_4^{-5}=10^{-11}, \mathrm{~K}_{\mathrm{sp}} \mathrm{CaSO}_4=\mathbf{1 0}^{-6}, \mathrm{~K}_{\mathrm{sp}} \mathrm{Ag}_2 \mathrm{SO}_4\right.$

1 $\mathrm{Ag}_2 \mathrm{SO}_4$
2 $\mathrm{BaSO}_4$
3 $\mathrm{CaSO}_4$
4 All of these
Ionic Equilibrium

229388 If the ionic product of $\mathrm{Ni}(\mathrm{OH})_2$ is $1.9 \times 10^{-15}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $1.0 \mathrm{M} \mathrm{NaOH}$ is

1 $1.9 \times 10^{-18} \mathrm{M}$
2 $1.9 \times 10^{-13} \mathrm{M}$
3 $1.9 \times 10^{-15} \mathrm{M}$
4 $1.9 \times 10^{-14} \mathrm{M}$
Ionic Equilibrium

229389 A saturated solution of $\mathrm{H}_2 \mathrm{~S}$ in $0.1 \mathrm{M} \mathrm{HCl}$ at $25^{\circ} \mathrm{C}$ contains $\mathrm{S}^{2-}$ ion concentration of $10^{-23} \mathrm{~mol}$ $L^{-1}$. The solubility product of some sulphides are $\mathrm{CuS}=10^{-44}, \mathrm{FeS}=10^{-14}, \mathrm{MnS}=10^{-15}, \mathrm{CdS}=$ $10^{-25}$. If $0.01 \mathrm{M}$ solution of these salts in $1 \mathrm{M}$ $\mathrm{HCl}$ are saturated with $\mathrm{H}_2 \mathrm{~S}$, which of these will be precipitated?

1 All
2 All except $\mathrm{MnS}$
3 All except $\mathrm{MnS}$ and FeS
4 Only-CuS
Ionic Equilibrium

229390 The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{-2}$ ions in that solution?

1 $2 \times 10^{-4} \mathrm{M}$
2 $16 \times 10^{-4} \mathrm{M}$
3 $8 \times 10^{-4} \mathrm{M}$
4 $8 \times 10^{-8} \mathrm{M}$
VIEEE-2014
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Ionic Equilibrium

229386 On adding $0.1 \mathrm{M}$ solution each of $\left[\mathrm{Ag}^{+}\right],\left[\mathrm{Ba}^{2+}\right]$, $\left[\mathrm{Ca}^{2+}\right]$ in a $\mathrm{Na}_2 \mathrm{SO}_4$ solution, species first precipitated is $\left[\mathrm{K}_{\mathrm{sp}} \mathrm{BaSO}_4^{-5}=10^{-11}, \mathrm{~K}_{\mathrm{sp}} \mathrm{CaSO}_4=\mathbf{1 0}^{-6}, \mathrm{~K}_{\mathrm{sp}} \mathrm{Ag}_2 \mathrm{SO}_4\right.$

1 $\mathrm{Ag}_2 \mathrm{SO}_4$
2 $\mathrm{BaSO}_4$
3 $\mathrm{CaSO}_4$
4 All of these
Ionic Equilibrium

229388 If the ionic product of $\mathrm{Ni}(\mathrm{OH})_2$ is $1.9 \times 10^{-15}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $1.0 \mathrm{M} \mathrm{NaOH}$ is

1 $1.9 \times 10^{-18} \mathrm{M}$
2 $1.9 \times 10^{-13} \mathrm{M}$
3 $1.9 \times 10^{-15} \mathrm{M}$
4 $1.9 \times 10^{-14} \mathrm{M}$
Ionic Equilibrium

229389 A saturated solution of $\mathrm{H}_2 \mathrm{~S}$ in $0.1 \mathrm{M} \mathrm{HCl}$ at $25^{\circ} \mathrm{C}$ contains $\mathrm{S}^{2-}$ ion concentration of $10^{-23} \mathrm{~mol}$ $L^{-1}$. The solubility product of some sulphides are $\mathrm{CuS}=10^{-44}, \mathrm{FeS}=10^{-14}, \mathrm{MnS}=10^{-15}, \mathrm{CdS}=$ $10^{-25}$. If $0.01 \mathrm{M}$ solution of these salts in $1 \mathrm{M}$ $\mathrm{HCl}$ are saturated with $\mathrm{H}_2 \mathrm{~S}$, which of these will be precipitated?

1 All
2 All except $\mathrm{MnS}$
3 All except $\mathrm{MnS}$ and FeS
4 Only-CuS
Ionic Equilibrium

229390 The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{-2}$ ions in that solution?

1 $2 \times 10^{-4} \mathrm{M}$
2 $16 \times 10^{-4} \mathrm{M}$
3 $8 \times 10^{-4} \mathrm{M}$
4 $8 \times 10^{-8} \mathrm{M}$
VIEEE-2014
Ionic Equilibrium

229386 On adding $0.1 \mathrm{M}$ solution each of $\left[\mathrm{Ag}^{+}\right],\left[\mathrm{Ba}^{2+}\right]$, $\left[\mathrm{Ca}^{2+}\right]$ in a $\mathrm{Na}_2 \mathrm{SO}_4$ solution, species first precipitated is $\left[\mathrm{K}_{\mathrm{sp}} \mathrm{BaSO}_4^{-5}=10^{-11}, \mathrm{~K}_{\mathrm{sp}} \mathrm{CaSO}_4=\mathbf{1 0}^{-6}, \mathrm{~K}_{\mathrm{sp}} \mathrm{Ag}_2 \mathrm{SO}_4\right.$

1 $\mathrm{Ag}_2 \mathrm{SO}_4$
2 $\mathrm{BaSO}_4$
3 $\mathrm{CaSO}_4$
4 All of these
Ionic Equilibrium

229388 If the ionic product of $\mathrm{Ni}(\mathrm{OH})_2$ is $1.9 \times 10^{-15}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $1.0 \mathrm{M} \mathrm{NaOH}$ is

1 $1.9 \times 10^{-18} \mathrm{M}$
2 $1.9 \times 10^{-13} \mathrm{M}$
3 $1.9 \times 10^{-15} \mathrm{M}$
4 $1.9 \times 10^{-14} \mathrm{M}$
Ionic Equilibrium

229389 A saturated solution of $\mathrm{H}_2 \mathrm{~S}$ in $0.1 \mathrm{M} \mathrm{HCl}$ at $25^{\circ} \mathrm{C}$ contains $\mathrm{S}^{2-}$ ion concentration of $10^{-23} \mathrm{~mol}$ $L^{-1}$. The solubility product of some sulphides are $\mathrm{CuS}=10^{-44}, \mathrm{FeS}=10^{-14}, \mathrm{MnS}=10^{-15}, \mathrm{CdS}=$ $10^{-25}$. If $0.01 \mathrm{M}$ solution of these salts in $1 \mathrm{M}$ $\mathrm{HCl}$ are saturated with $\mathrm{H}_2 \mathrm{~S}$, which of these will be precipitated?

1 All
2 All except $\mathrm{MnS}$
3 All except $\mathrm{MnS}$ and FeS
4 Only-CuS
Ionic Equilibrium

229390 The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{-2}$ ions in that solution?

1 $2 \times 10^{-4} \mathrm{M}$
2 $16 \times 10^{-4} \mathrm{M}$
3 $8 \times 10^{-4} \mathrm{M}$
4 $8 \times 10^{-8} \mathrm{M}$
VIEEE-2014