Question

The molar solubility of $PbB{r}_2$ is $2.17\times {10}^{-3}$M at a certain temperature. Calculate $K_{sp}$ for $PbB{r}_2$.

• A. $6.2\times {10}^{-6}$
• B. $6.4\times {10}^{-7}$
• C. $4.1\times {10}^{-8}$
• D. $3.4\times {10}^{-6}$
• E. $1.4\times {10}^{-5}$

Answer 1

The correct option is C $4.1\times {10}^{-8}$

For $PbB{r}_2$ the expression of solubility product is $K_{SP}=\left[P{b}^{2+}\right][B{r}^{-}{]}^2=(S\left)\right(2S{)}^2=4S^3$
Substitute $S=2.17\times {10}^{-3}$ in the above expression.
$K_{SP}=4S^3=4(2.17\times {10}^{-3}{)}^3=4.1\times {10}^{-8}$