Question

The $K_{sp}$ of $PbC{O}_3$ and $MgC{O}_3$ are $1.5\times 10^{-15}$ and $1\times 10^{-15}$ , respectively at $298K$ . The concentration of $p{b}^{2+}$ ions in a saturated solution containing $MgC{O}_3$ and $PbC{O}_3$ 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 $PbC{O}_3$ and $MgC{O}_3$ are present in solution
Suppose solubility of $PbC{O}_3$ is $xmol{L}^{-1}$ and that of $MgC{O}_3$ is $ymol{L}^{-1}$ then,
$pbC{O}_3\rightleftharpoons \underset{\text{x}}{p{b}^{2+}}+\underset{\text{x+y}}{C{O}_3^{2-}}$
$MgC{O}_3\rightleftharpoons \underset{\text{y}}{M{g}^{2+}}+\underset{\text{x+y}}{C{O}_3^{2-}}$
$\frac{K_{sp}\left(pbC{O}_3\right)}{K_{sp}\left(MgC{O}_3\right)}=\frac{x\left(x+y\right)}{y\left(x+y\right)}=\frac{x}{y}$
$=\frac{1.5\times 10^{-15}}{1.0\times 10^{-15}}=1.5$
Thus, $x=1.5y$
$K_{sp}\left(pbC{O}_3\right)=x\left(x+y\right)=1.5\times 10^{-15}$
$\therefore 1.5y\left(1.5y+y\right)=1.5\times 10^{-15}$
or $3.75y^2=1.5\times 10^{-15}$
$y={\left(\frac{1.5\times 10^{-15}}{3.75}\right)}^{1/2}=2\times 10^{-8}$
Now, $x=1.5y$
$=1.5\times (2\times 10^{-8})=3\times 10^{-8}M$