Question

The solubility of $lead\left(II\right)iodide$ is $0.064g/100mL$ at ${25}^{o}C.$ What is its solubility product?

• A. $2.08\times {10}^{-8}mo{l}^{-3}{L}^3$
• B. $1.07\times {10}^{-8}mo{l}^3{L}^{-3}$
• C. $0.17\times {10}^{-8}mo{l}^{-3}{L}^3$
• D. $3.5\times {10}^{-8}mo{l}^3{L}^{-3}$

Answer 1

The correct option is B $1.07\times {10}^{-8}mo{l}^3{L}^{-3}$

Given:
Solubility$=0.064g/100mL$
$\Rightarrow \text{Solubility}=0.64g/L$
So, Calculating Solubility in $mol/L$,
$\Rightarrow s=\frac{0.64}{461}mol/L,\text{as molar mass of}Pb{I}_2=461g$
$\left[Pb{I}_2\right]=1.39\times {10}^{-3}M$

The equilibrium established will be,
$\begin{array}{cccccc}& Pb{l}_2\left(s\right)& \rightleftharpoons & P{b}^{2+}\left(aq\right)& +& 2I^{-}\left(aq\right)\\ [c{]}_i& 1.39\times {10}^{-3}& & 0& & 0\\ [c{]}_e& & & 1.39\times {10}^{-3}& & 2.78\times {10}^{-3}\end{array}$

$K_{sp}=\left[P{b}^{2+}\right]\times [C{l}^{-}{]}^2$

Putting the values,
$K_{sp}=1.39\times {10}^{-3}\times (2.78\times {10}^{-3}{)}^2=1.07\times {10}^{-8}mo{l}^3{L}^{-3}$