Question

At $298K,$ the solubility product of ${PbCl}_2$ is $1.0\times {10}^{-6}$. What will be the solubility of $PbC{l}_2$ in moles/litre.

• A. $6.3\times {10}^{-3}$
• B. $1.0\times {10}^{-3}$
• C. $3.0\times {10}^{-3}$
• D. $4.6\times {10}^{-14}$

Answer 1

The correct option is A $6.3\times {10}^{-3}$

$PbC{l}_2\to P{b}^{2+}+2C{l}^{-}$

$K_{sp}=4s^3$

$S=\sqrt[3]{3\frac{K_{sp}}{4}}=\sqrt[3]{3\frac{1.0\times {10}^{-6}}{4}}=6.3\times {10}^{-3}$
Option A is correct.