The solubility of $AgCl$ is $1.435 \times 10^{-5} \ g \ L^{-1}$ at $30^{\circ} C$. Calculate its solubility product $(K_{sp})$. (Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)

The solubility product of $Mg(OH)_2$ is $1 \times 10^{-11}$. At what $pH$ will $Mg(OH)_2$ start to precipitate from a $0.1 \, M \, Mg^{2+}$ solution?

The $K_{SP}$ of $AgI$ is $1.5 \times 10^{-16}$. On mixing equal volumes of the following solutions,precipitation will occur only with:

The $K_{sp}$ of $Mg(OH)_2$ is $1.0 \times 10^{-12}$. At which $pH$ does $0.01 \ M$ $Mg^{2+}$ solution begin to precipitate? Calculate the solubility of $Mg(OH)_2$ in pure water.

If the concentration of $Ag^{+}$ ions in the saturated solution of $Ag_2CO_3$ is $1.20 \times 10^{-4} \ mol \ L^{-1}$,then find the solubility product of $Ag_2CO_3$.

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,$K_{a1} \times K_{a2}(H_2S)=1.0 \times 10^{-21}$,$\log 2=0.30, \log 3=0.48, \log 5=0.70$)