Question

If the solubility of sparingly soluble salt $PbC{l}_2$ is $4.41g{L}^{-1}$ at ${25}^{o}C$. Find its solubility product at the given temperature.
(Take Molecular mass $PbC{l}_2\approx 278gmo{l}^{-1}$)

• A. $3.67\times {10}^{-10}{\text{mol}}^3{\text{L}}^{-3}$
• B. $4.64\times {10}^{-8}{\text{mol}}^3{\text{L}}^{-3}$
• C. $1.64\times {10}^{-5}{\text{mol}}^3{\text{L}}^{-3}$
• D. $2.15\times {10}^{-1}{\text{mol}}^3{\text{L}}^{-3}$

Answer 1

The correct option is C $1.64\times {10}^{-5}{\text{mol}}^3{\text{L}}^{-3}$

Given:

$\text{Solubility}=4.41g{L}^{-1}$
So, Calculating Solubility in $mol{L}^{-1}$
$\text{Moles}=\frac{\text{Mass}}{\text{Molecular mass}}$
so,
$\Rightarrow s=\frac{4.41}{278}mol{L}^{-1}s=1.58\times {10}^{-2}\approx 1.6\times {10}^{-2}$

$s=1.6\times {10}^{-2}M$

The equilibrium is established between undissolved $PbC{l}_2$ and its ions. The equilibrium established will be,
$\begin{array}{cccccc}& PbC{l}_2\left(s\right)& \rightleftharpoons & P{b}^{2+}\left(aq\right)& +& 2C{l}^{-}\left(aq\right)\\ Initial& C& & 0& & 0\\ Equilibrium:& C-s& & s& & 2s\end{array}$

$K_{sp}=\left[P{b}^{2+}\right]\times [C{l}^{-}{]}^2$
$K_{sp}=s\times (2s{)}^2=4s^3$
Putting the value of $s$,
$K_{sp}=4\times (1.6\times {10}^{-2}{)}^{3}mo{l}^3{L}^{-3}{K}_{sp}=1.64\times {10}^{-5}mo{l}^3{L}^{-3}$