Solubility and Equilibria of Sparingly Soluble Salts

Solubility products of $\mathrm{CuI}$ and ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ have almost the same value $\left(~4\times {10}^{-12}\right)$. The ratio of solubilities of the two salts $\left(\mathrm{CuI}:{\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)$ is closest to 

The solubility of $\text{B}\text{a}\text{S}{\text{O}}_4$ in pure water (in $\text{g}{\text{L}}^{-1}$ ) is closest to
$\left[\right.$Given; ${\text{K}}_{\text{sp}}$ for $\text{B}\text{a}\text{S}{\text{O}}_4$ is $1.0\times {10}^{-10}$ at $25°\text{C}.$ Molecular weight of $\text{B}\text{a}\text{S}{\text{O}}_4$ is $233 \text{g} \text{m}\text{o}{\text{l}}^{-1}$ ]

If the solubility of a ${\mathrm{M}}_2\mathrm{S}$ salt is $3.6\times {10}^{-5}$, find its solubility product.

By which of the following processes can the common salt be purified?

Which acid is used in the purification of common salt?

A saturated solution of ${\mathrm{CaSO}}_4$ is prepared at $25°\mathrm{C}$. When $200 \mathrm{mL}$ of this solution is allowed to complete evaporation of water, $0.34 \mathrm{g}$ of residue obtained. The ${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{CaSO}}_4$ at $25°\mathrm{C}$ will be:

$\mathrm{pH}$ of a saturated solution of $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ is $12$. Hence, ${\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$

Solubility of thorium phosphate (molar mass $=\mathrm{m}$) is $\mathrm{wg}/100 \mathrm{mL}$ at $25°\mathrm{C}$. The ${\mathrm{K}}_{\mathrm{sp}}$ of thorium phosphate is :

The ${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ is $1.1\times {10}^{-12}$ at $298 \mathrm{K}$. The solubility (in $\mathrm{mol} {\mathrm{L}}^{-1}$) of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ in a $0.1 \mathrm{M} {\mathrm{AgNO}}_3$ solution is :

The solubility of $\mathrm{AgCl}\left(\mathrm{s}\right)$ with solubility product $1.6\times {10}^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ solution would be :

Calculate the molar solubility of $\mathrm{AgCl}$ at $25°\mathrm{C}$ in $3.0{\mathrm{MNH}}_3$

$\left({\mathrm{K}}_{\mathrm{sp}}\right.$ of $\mathrm{AgCl}=2.0\times {10}^{-10}, {\mathrm{K}}_{\mathrm{f}}$ of ${\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}=1.25\times {10}^7)$
 

If solubility of sparingly soluble compound $\mathrm{M}{\left(\mathrm{OH}\right)}_3$ in a solution of $\mathrm{pH}=12$ is $\mathrm{x}\times {10}^{-\left(\mathrm{y}\right)}$ then calculate $\mathrm{y}$. Solubility of $\mathrm{M}{\left(\mathrm{OH}\right)}_3$ in pure water is $2\times {10}^{-7}$.

If ${10}^{-x}$ is the factor by which solubility of $AgBr\left(\mathrm{s}\right) \left[{\mathrm{K}}_{\mathrm{sp}}={10}^{-10}\right]$ changes with respect to initial solubility, when it is dissolved in $0.01\mathrm{M}$ NaBr $\left(\mathrm{aq}\right)$ solution. The value of ${}^{\text{'}}\mathrm{X}^{\text{'}}$ is :

Calculate the solubility of silver phosphate $\left({\mathrm{Ag}}_3{\mathrm{PO}}_4\right)$ in $0.1{\mathrm{MAgNO}}_3$ $\left(\mathrm{Ksp}\right.$ of ${\mathrm{Ag}}_3{\mathrm{PO}}_4=1.1\times {10}^{-16})$:-

Calculate the molar solubility of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ in $0.1 \mathrm{M} \mathrm{NaOH}$. The solubility product of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ is $2\times {10}^{-15}$. (Given: $\sqrt[3]{0.5}=0.79$)

$\mathrm{pH}$ of a saturated solution of $\mathrm{Ba}(\mathrm{OH}{)}_2$ is $12$. Hence, ${\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{Ba}(\mathrm{OH}{)}_2$ is:

At $25°\mathrm{C}$, the solubility product of $\mathrm{Mg} (\mathrm{OH}{)}_2$is $1.0\times {10}^{-11}$. At which $\mathrm{pH}$, will ${\mathrm{Mg}}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH}{)}_2$ from a solution of $0.001{\mathrm{MMg}}^{2+}$ ions ? 

What is the minimum concentration of ${\mathrm{SO}}_4^{2-}$ required to precipitate ${\mathrm{BaSO}}_4$ in a solution containing $1\times {10}^{-4}$ mole of ${\mathrm{Ba}}^{2+}$ ? $\left({\mathrm{K}}_{\mathrm{sp}}\right.$for$\left. {\mathrm{BaSO}}_4=4\times {10}^{-10}\right)$

The solubility of salt in saturated aqueous solution of ${\mathrm{NH}}_4{\mathrm{KHPO}}_3$ is :-