The solubility product of a sparingly soluble salt $BA$ is $4 \times 10^{-13}$. Calculate the $[A^-]$ if $[B^+]$ is $1 \times 10^{-6} \text{ M}$.

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 $K_{a2}=1.2 \times 10^{-2}$]

Derive the solubility product $(K_{sp})$ expression for the following sparingly soluble salts:
$(i)$ Two ions having $MX$ formula
$(ii)$ Three ions having $MX_{2}$ or $M_{2}X$ types
$(iii)$ Four ions having $AX_{3}$ or $A_{3}X$ type salts
$(iv)$ Five ions $A_{2}X_{3}$ or $A_{3}X_{2}$ type salts.

The solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$. The value of $K_{sp}$ of the electrolyte is:

The solubility product of a sparingly soluble salt $A_2B$ is $3.2 \times 10^{-11}$. Its solubility in $mol \cdot L^{-1}$ is

$A$ $2.901 \ L$ saturated solution is formed by $0.08 \ g$ of $CaF_2$ at $298 \ K$. Calculate the $K_{sp}$ of $CaF_2$. (Molar mass of $CaF_2$ is $78 \ g \ mol^{-1}$)