Question

Sol. [Key Concept: Solubility order via Ksp / lattice energy]

$AgBr(s) \rightleftharpoons {Ag}^{+}(aq) + .Br(aq$.

$$\begin{matrix} & K_{sp} = 5 \times 10^{- 13} = S^{2} \\ & \text{ }S = \sqrt{5 \times 10^{- 13}},\text{ }S = 7.07 \times 10^{- 7}\text{ }mol/l \\ & Zn(OH)_{2}(\text{ }s) \rightleftharpoons {Zn}^{2 +}\text{~}\text{(aq.)}\text{~} + 2{OH}^{-}\text{(aq.)}\text{~} \\ & K_{sp} = 10^{- 15} = (S)(2\text{ }S)^{2} \\ & 4{\text{ }S}^{3} = 10^{- 15},\text{ }S = 0.63 \times 10^{- 5} = 6.3 \times 10^{- 6}\text{ }mol/l \\ & {Hg}_{2}{Cl}_{2}(\text{ }s) \rightleftharpoons {Hg}_{2}^{2 +}\text{~}\text{(aq.)}\text{~} + 2{Cl}^{-}(\text{aq.}\text{~}) \\ & K_{sp} = 1.3 \times 10^{- 18} = 4{\text{ }S}^{3},\text{ }S = 0.69 \times 10^{- 6} = 6.9 \times 10^{- 7}\text{ }mol/l \end{matrix}$$

The correct order of solubility is:

$$Zn(OH)_{2} > AgBr > {Hg}_{2}{Cl}_{2}$$