Question

The $ K_{sp} $ of $ PbCO_{3} $ and $ MgCO_{3} $ are $ 1.5 \times 10^{-15} $ and $ 1 \times 10^{-15} $ , respectively at $ 298\, K $ . The concentration of $ pb^{2+} $ ions in a saturated solution containing $ MgCO_{3} $ and $ PbCO_{3} $ is

• A. $ 1.5 \times 10^{-8}\, M $
• B. $ 3 \times 10^{-8}\, M $
• C. $ 2 \times 10^{-8}\, M $
• D. $ 2.5 \times 10^{-8}\, M $

Answer 1

Answer: (B)

When both $PbCO_{3}$ and $MgCO_{3}$ are present in solution
Suppose solubility of $PbCO_{3}$ is $x\, mol \,L^{-1}$ and that of $MgCO_{3}$ is $y \, mol\,L^{-1}$ then,
$pbCO_{3} \rightleftharpoons \underset{\text{x}}{{pb^{2+}}}+\underset{\text{x+y}}{{CO^{2-}_{3}}}$
$MgCO_{3} \rightleftharpoons \underset{\text{y}}{{Mg^{2+}}}+\underset{\text{x+y}}{{CO^{2-}_{3}}}$
$\frac{K_{sp}\left(pbCO_{3}\right)}{K_{sp}\left(MgCO_{3}\right)}=\frac{x\left(x+y\right)}{y\left(x+y\right)}=\frac{x}{y}$
$=\frac{1.5\times10^{-15}}{1.0\times10^{-15}}=1.5$
Thus, $x=1.5y$
$K_{sp}\left(pbCO_{3}\right)=x\left(x+y\right)=1.5\times10^{-15}$
$\therefore 1.5y\left(1.5y+y\right)=1.5\times10^{-15}$
or $3.75y^{2}=1.5\times10^{-15}$
$y=\left(\frac{1.5\times10^{-15}}{3.75}\right)^{1 /2}=2\times10^{-8}$
Now, $x=1.5y$
$ = 1.5 \times (2 \times 10^{-8}) = 3 \times 10^{-8}M$