Chemistry - Ionic Equilibrium Question with Solution | TestHub

The solubility of a compound is directly related to its $K_{sp}$ value; the higher the $K_{sp}$, the more soluble the compound. However, we must also consider the stoichiometry of the salts when comparing $K_{sp}$ values. To accurately compare the solubilities, we need to calculate the molar solubility ($s$) for each compound.

A. $\mathrm{MnS}\rightleftharpoons {\mathrm{Mn}}^{2+}+{\mathrm{S}}^{2-}$

$K_{sp}=[{\mathrm{Mn}}^{2+}][{\mathrm{S}}^{2-}]=s^2=8\times 10^{-37}$

$s=\sqrt{8\times 10^{-37}}=2.83\times 10^{-18.5}=2.83\times 10^{-19}\mathrm{M}$

B. $\mathrm{ZnS}\rightleftharpoons {\mathrm{Zn}}^{2+}+{\mathrm{S}}^{2-}$

$K_{sp}=[{\mathrm{Zn}}^{2+}][{\mathrm{S}}^{2-}]=s^2=7\times 10^{-16}$

$s=\sqrt{7\times 10^{-16}}=2.65\times 10^{-8}\mathrm{M}$

C. ${\mathrm{Bi}}_2{\mathrm{S}}_3\rightleftharpoons 2{\mathrm{Bi}}^{3+}+3{\mathrm{S}}^{2-}$

$K_{sp}=[{\mathrm{Bi}}^{3+}{]}^2[{\mathrm{S}}^{2-}{]}^3=(2s{)}^2(3s{)}^3=4s^2\cdot 27s^3=108s^5=1\times 10^{-72}$

$s^5=\frac{1\times 10^{-72}}{108}=9.26\times 10^{-75}$

$s=\sqrt[5]{9.26\times 10^{-75}}=1.56\times 10^{-15}\mathrm{M}$

D. ${\mathrm{Ag}}_3{\mathrm{PO}}_4\rightleftharpoons 3{\mathrm{Ag}}^{+}+{\mathrm{PO}}_4^{3-}$

$K_{sp}=[{\mathrm{Ag}}^{+}{]}^3[{\mathrm{PO}}_4^{3-}]=(3s{)}^3(s)=27s^4=1.8\times 10^{-18}$

$s^4=\frac{1.8\times 10^{-18}}{27}=6.67\times 10^{-20}$

$s=\sqrt[4]{6.67\times 10^{-20}}=9.03\times 10^{-5}\mathrm{M}$

Comparing the molar solubilities:

MnS: $2.83\times 10^{-19}\mathrm{M}$

ZnS: $2.65\times 10^{-8}\mathrm{M}$

Bi₂S₃: $1.56\times 10^{-15}\mathrm{M}$

Ag₃PO₄: $9.03\times 10^{-5}\mathrm{M}$

The largest molar solubility is for Ag₃PO₄. Therefore, Ag₃PO₄ is the most soluble in water.