Calculate the molar solubility of calcium hydroxide $Ca(OH)_2$ in $0.10 \ M \ NaOH$ solution. The solubility product constant $(K_{sp})$ of calcium hydroxide is $5.5 \times 10^{-6}$.

The solubility of a salt of weak acid $(AB)$ at $pH = 3$ is $Y \times 10^{-3} \ mol \ L^{-1}$. The value of $Y$ is (Given that the value of solubility product of $AB$ $(K_{sp}) = 2 \times 10^{-10}$ and the value of ionization constant of $HB$ $(K_{a}) = 1 \times 10^{-8}$)

$A$ saturated solution of $Ag_2SO_4$ is $2.5 \times 10^{-2} \ M$. The value of its solubility product is:

The solubility product of $PbS$ is $3.4 \times 10^{-28}$. If $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$,what concentration of $[S^{2-}]$ is required to precipitate $PbS$?

Solubility of a salt $M_2X_3$ is $y \ mol \ dm^{-3}$. The solubility product of the salt will be

Increasing order of solubility of $AgCl$ in $(i) H_2O$,$(ii) 1 \ M \ NaCl \ (aq.)$,$(iii) 1 \ M \ CaCl_2 \ (aq.)$ and $(iv) 1 \ M \ NaNO_3 \ (aq.)$ solution.