Question

At $298K$, the solubility product of $Pb{Cl}_2$ is $1.0\times {10}^{-6}$. The solubility of $Pb{Cl}_2$ in $mol{L}^{-1}$ would be:

• 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 $6.3\times {10}^{-3}$

Given:

Solubility product of $PbC{l}_2$ = 1.0× ${10}^{-6}$

Solution:

$PbC{l}_2\to P{b}^{+2}+2C{l}^{-1}$

Let Solubility be S

$\text{Let K\_sp is the solubility product constant}$

$K_{s}p=\left[P{b}^{+2}\right][C{l}^{-1}{]}^2$

$Ksp=S\times (2S{)}^2$

$\Rightarrow K_{s}p=4(S{)}^3$

$\Rightarrow 1.0\times {10}^{-6}=4(S{)}^3$

$\Rightarrow S^3=0.25\times {10}^{-6}$

$\Rightarrow S=(0.25\times {10}^{-6}{)}^{1/3}$

$\Rightarrow S=0.63\times {10}^{-2}$

$\Rightarrow S=6.3\times {10}^{-3}M$

Hence, option A is correct.