Question

Calculate the solubility product of $PbC{l}_2$ at a certain temperature if the solubility of $PbC{l}_2$ is $4.4g/L$ at the same temperature. (Given: molar masses in $g$, $Pb$ = 207, $Cl$ = 35.5)

• A. $1.63\times {10}^{-7}mo{l}^3/L^3$
• B. $1.63\times {10}^{-5}mo{l}^3/L^3$
• C. $1.36\times {10}^{-5}mo{l}^3/L^3$
• D. $1.36\times {10}^{-7}mo{l}^3/L^3$

Answer 1

The correct option is B $1.63\times {10}^{-5}mo{l}^3/L^3$

Molar mass of $PbC{l}_2=278$
Concentration of $PbC{l}_2=4.4g/L$

So, molar Concentration of $PbC{l}_2=4.4g/L=\frac{4.4}{278}mol/L=1.6\times {10}^{-2}mol/L$
Molar concentration of $PbC{l}_2=1.6\times {10}^{-2}mol/L$

$\Rightarrow \text{solubility of}PbC{l}_2=1.6\times {10}^{-2}mol/L$
$\Rightarrow s=1.6\times {10}^{-2}mol/L$

Now Equilibrium established will be,
$\begin{array}{ccccc}PbC{l}_2& \rightleftharpoons & P{b}^{+2}& +& 2C{l}^{-}\\ 1& & 0& & 0\\ 1-s& & s& & 2s\end{array}$
$K_{sp}=\left[P{b}^{+2}\right][C{l}^{-}{]}^2$
$K_{sp}=\left[s\right][2s{]}^2$

putting the values,
$=(1.6\times {10}^{-2}{)}^3\times 4=1.63\times {10}^{-5}mo{l}^3/L^3$