Question

The solubility product of $PbB{r}_2$ is $10.8\times 10^{-5}$. It is $70\%$ dissociated in saturated solution. The solubility of salt is:

• A. $4.18\times 10^{-2}$
• B. $6.76\times 10^{-3}$
• C. $3.4\times 10^{-4}$
• D. $5.44\times 10^{-2}$

Answer 1

Answer: (A)

First find relationship between solubility and solubility product.
$PbB{r}_{2}P{b}^2+2B{r}^{-}$
$K_{sp}=\left[P{b}^{2+}\right]{\left[B{r}^{-}\right]}^2$
given $K_{sp}PbB{r}_2=10.8\times 10^{-5}\alpha =70$
Solubility $=xmol/L$
$\therefore \left[P{b}^{2+}\right]=0.7x,\left[B{r}^{-}\right]=2\times 0.7\times x=1.4x$
$\therefore K_{sp}=\left[P{b}^{2+}\right]{\left[B{r}^{-}\right]}^2=(0.7x)(1.4x{)}^2$
or $10.8\times 10^{-5}=1.372x^3$
or $x=\sqrt[3]{\frac{10.8\times 10^{-5}}{1.372}}$
$=4.18\times 10^{-2}$