Chemistry - Ionic Equilibrium Question with Solution | TestHub

Let $s$ be the solubility of the salt in water.

For $\text{AgBr}$:
$\text{AgBr}\rightleftharpoons {\text{Ag}}^{+}+{\text{Br}}^{-}$
$K_{sp}=s^2$
$s=\sqrt{5.0\times 10^{-13}}=\sqrt{50\times 10^{-14}}\approx 7.07\times 10^{-7}\text{ M}$

For ${\text{Zn(OH)}}_2$:
${\text{Zn(OH)}}_2\rightleftharpoons {\text{Zn}}^{2+}+2{\text{OH}}^{-}$
$K_{sp}=s(2s{)}^2=4s^3$
$s={\left(\frac{1.0\times 10^{-15}}{4}\right)}^{1/3}={\left(250\times 10^{-18}\right)}^{1/3}\approx 6.3\times 10^{-6}\text{ M}$

For ${\text{Hg}}_2{\text{Cl}}_2$:
${\text{Hg}}_2{\text{Cl}}_2\rightleftharpoons {\text{Hg}}_2^{2+}+2{\text{Cl}}^{-}$
$K_{sp}=s(2s{)}^2=4s^3$
$s={\left(\frac{1.3\times 10^{-18}}{4}\right)}^{1/3}={\left(0.325\times 10^{-18}\right)}^{1/3}\approx 6.87\times 10^{-7}\text{ M}$

Comparing the solubilities:
$6.3\times 10^{-6}>7.07\times 10^{-7}>6.87\times 10^{-7}$

Thus, the correct order of solubility is ${\text{Zn(OH)}}_2>\text{AgBr}>{\text{Hg}}_2{\text{Cl}}_2$.

Answer: ${\text{Zn(OH)}}_2>\text{AgBr}>{\text{Hg}}_2{\text{Cl}}_2$