Which of the following sets of concentrations will cause the precipitation of $ZnCl_2$ $(K_{sp} = 1.2 \times 10^{-12} \ M^3)$?

Precipitation occurs if the ionic product is ........

The solubility product of a salt $AB$ is $1 \times 10^{-8}$ in a solution in which the concentration of $A$ is $10^{-3} \ M$. The salt will precipitate when the concentration of $B$ becomes more than:

If the solubility of $MX_2$ in water is $0.0002 \ mol/L$,then its $K_{sp}$ is:

The solubility product of Lead sulphate,$PbSO_4$ is $1.3 \times 10^{-8}$. Calculate its solubility in pure water. The molecular mass of $PbSO_4 = 303 \ g \ mol^{-1}$.

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