Calculate the molar solubility of $Ni(OH)_2$ in $0.10 \ M \ NaOH$. The solubility product of $Ni(OH)_2$ is $2 \times 10^{-15}$.

The solubility of $Ba_3(AsO_4)_2$ (molar mass $= 690 \ g/mol$) is $6.9 \times 10^{-2} \ g/100 \ mL$. What is the value of its $K_{sp}$?

If the solubility product of $BaSO_4$ is $1.44 \times 10^{-12}$,then the solubility of $SO_4^{2-}$ is .....

Solubility of $Th_3(PO_4)_4$ $(M_w = m)$ is $w \ g / 100 \ mL$ at $25 \ ^oC$. The $K_{sp}$ is:

The solubility product constants of $Ag_{2}CrO_{4}$ and $AgBr$ are $1.1 \times 10^{-12}$ and $5.0 \times 10^{-13}$ respectively. Calculate the ratio of the molarities of their saturated solutions.

Consider two Group $IV$ metal ions $X^{2+}$ and $Y^{2+}$. $A$ solution containing $0.01 \ M$ $X^{2+}$ and $0.01 \ M$ $Y^{2+}$ is saturated with $H_2S$. The pH at which the metal sulphide $YS$ will form as a precipitate is . . . . . . (Nearest integer). (Given: $K_{sp}(XS)=1 \times 10^{-22}$ at $25^{\circ} C$,$K_{sp}(YS)=4 \times 10^{-16}$ at $25^{\circ} C$,$[H_2S]=0.1 \ M$ in solution,$K_{a1} \times K_{a2}(H_2S)=1.0 \times 10^{-21}$,$\log 2=0.30, \log 3=0.48, \log 5=0.70$)