Question

Solubility product of $PbCl_2$ at $298 K$ is $1 \times 10^{-6}$. At this temperature, solubility of $PbCl_2\, (in\, mol/L)$ is

• A. $(1 \times 10^{-6} )^{1/2}$
• B. $(1 \times 10^{-6} )^{1/3}$
• C. $(0.25 \times 10^{-6} )^{1/3}$
• D. $(2.5 \times 10^{-6} )^{1/2}$

Answer 1

Answer: (C)

$PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-$
$K_{sp} = {[Pb^{2+} ][Cl^-]^2 } = (S)(2S)^2 = 4S^3 $
$1 \times 10^{-6} = 4S^3$
$S = \left( frac{1 \times 10^{-6}}{4} \right)^{1/3} = (0.25 \times 10^{-6} )^{1/3}$