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.

(N/A) $(i)$ For $MX$ type salts: $MX_{(s)} \rightleftharpoons M_{(aq)}^{+} + X_{(aq)}^{-}$. Let solubility be $S \ mol/L$. $K_{sp} = [M^{+}][X^{-}] = (S)(S) = S^{2}$.
$(ii)$ For $MX_{2}$ type: $MX_{2(s)} \rightleftharpoons M_{(aq)}^{2+} + 2X_{(aq)}^{-}$. $K_{sp} = [M^{2+}][X^{-}]^{2} = (S)(2S)^{2} = 4S^{3}$.
For $M_{2}X$ type: $M_{2}X_{(s)} \rightleftharpoons 2M_{(aq)}^{+} + X_{(aq)}^{2-}$. $K_{sp} = [M^{+}]^{2}[X^{2-}] = (2S)^{2}(S) = 4S^{3}$.
$(iii)$ For $AX_{3}$ type: $AX_{3(s)} \rightleftharpoons A_{(aq)}^{3+} + 3X_{(aq)}^{-}$. $K_{sp} = [A^{3+}][X^{-}]^{3} = (S)(3S)^{3} = 27S^{4}$.
For $A_{3}X$ type: $A_{3}X_{(s)} \rightleftharpoons 3A_{(aq)}^{+} + X_{(aq)}^{3-}$. $K_{sp} = [A^{+}]^{3}[X^{3-}] = (3S)^{3}(S) = 27S^{4}$.
$(iv)$ For $A_{2}X_{3}$ type: $A_{2}X_{3(s)} \rightleftharpoons 2A_{(aq)}^{3+} + 3X_{(aq)}^{2-}$. $K_{sp} = [A^{3+}]^{2}[X^{2-}]^{3} = (2S)^{2}(3S)^{3} = 4S^{2} \times 27S^{3} = 108S^{5}$.
For $A_{3}X_{2}$ type: $A_{3}X_{2(s)} \rightleftharpoons 3A_{(aq)}^{2+} + 2X_{(aq)}^{3-}$. $K_{sp} = [A^{2+}]^{3}[X^{3-}]^{2} = (3S)^{3}(2S)^{2} = 27S^{3} \times 4S^{2} = 108S^{5}$.