The $pH$ of a saturated solution of $Ca(OH)_2$ is $12.25$. Calculate its solubility product $(K_{sp})$.

The $K_{sp}$ for $BaSO_4$ at $25\ ^\circ C$ is $1.1 \times 10^{-10}$. To make the new solubility equal to $1.1 \times 10^{-8} \ M$,it is necessary to use a solution of $Na_2SO_4$ of the following concentration (in $M$):

Which of the following has the maximum solubility? (Given $K_{sp}$ values)

Concentration of $H_2SO_4$ and $Na_2SO_4$ in a solution is $1 \ M$ and $1.8 \times 10^{-2} \ M$,respectively. Molar solubility of $PbSO_4$ in the same solution is $X \times 10^{-Y} \ M$ (expressed in scientific notation). The value of $Y$ is. . . .
[Given: Solubility product of $PbSO_4$ $(K_{sp})=1.6 \times 10^{-8}$. For $H_2SO_4$,$K_{a1}$ is very large and $K_{a2}=1.2 \times 10^{-2}$]

The solubility product of a binary weak electrolyte is $4 \times 10^{-10}$ at $298 \ K$. Its solubility in $\text{mol} \ dm^{-3}$ at the same temperature is

The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately: