The $pH$ of a saturated solution of $Ca(OH)_{2}$ is $9$. The solubility product $(K_{sp})$ of $Ca(OH)_{2}$ is:

If the solubility of $AgCl$ (molar mass $= 143 \, g/mol$) in water at $25 \, ^\circ C$ is $1.43 \times 10^{-4} \, g/100 \, mL$ of solution,then the value of $K_{sp}$ is?

The solubility of a sparingly soluble salt $AB_2$ is $18.78 \times 10^{-4} \text{ g/dm}^3$. What is its solubility product? (Molar mass of $AB_2 = 187.8 \text{ g mol}^{-1}$)

The solubility of $CaF_2$ is $2 \times 10^{-4} \, mol/L$. Its solubility product $(K_{sp})$ is:

If the $[F^{-}] = 2.0 \times 10^{-5} \ M$ in water,how many grams of $CaCl_{2}$ must be added to initiate the precipitation of $CaF_{2}$? Given: $K_{sp}$ for $CaF_{2} = 1.7 \times 10^{-10}$ and the molar mass of $CaCl_{2} = 111 \ g \ mol^{-1}$. Assume the volume of the solution is $1 \ L$.

The correct order of solubility of the given salts in water at $298 K$ is:

Salt	$K_{sp}$ at $298 K$
$AgBr$	$5.0 \times 10^{-13}$
$Zn(OH)_2$	$1.0 \times 10^{-15}$
$Hg_2Cl_2$	$1.3 \times 10^{-18}$