Question

The solubility of $ PbCl_{2}$ is:

• A. $ \sqrt{{{K}_{sp}}} $
• B. $ {{({{K}_{sp}})}^{1/3}} $
• C. $ {{({{K}_{sp/4}})}^{1/3}} $
• D. $ {{(8{{K}_{sp}})}^{1/2}} $

Answer 1

Answer: (C)

Let the solubility of $PbCl _{2}=x\, mol / L$
$\underset{x\, mol / L}{PbCl _{2}} \rightleftharpoons \underset{x}{Pb ^{2+}}+\underset{2x}{2 Cl ^{-}}$
$K_{s p}=\left[ Pb ^{2+}\right]\left[ Cl ^{-}\right]^{2}$
or $=(x) \times(2 x)^{2}$
or $=4 x^{3}$
$\therefore x=\sqrt[3]{\frac{K_{s p}}{4}}$
$=\left(\frac{K_{s p}}{4}\right)^{1 / 3}$