The $K_{sp}$ of $M(OH)_2$ is $3.2 \times 10^{-11}$. The $pH$ of its saturated solution in water is:

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$Ba(AcO)_2 + K_2CrO_4 \longrightarrow BaCrO_4 \downarrow + 2AcOK$

$K_{sp}$ for $BaSO_4$ at $25 \, ^oC$ 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 $....... \ M$

If the solubility product of $AB_2$ is $3.20 \times 10^{-11} \ mol^3 L^{-3}$,then the solubility of $AB_2$ in pure water is......... $\times 10^{-4} \ mol \ L^{-1}$. [Assuming that neither kind of ion reacts with water]

If the $K_{sp}$ of a saturated solution of $Mg(OH)_2$ is $4 \times 10^{-12}$,then the $pH$ is equal to:

When $150 \ mL$ of $0.0008 \ M$ ammonium sulfate solution is mixed with $50 \ mL$ of $0.04 \ M$ calcium nitrate solution,given $K_{sp} = 2.4 \times 10^{-5}$ for $CaSO_4$,what is the ionic product of $CaSO_4$?