Question

The solubility in water of a sparingly soluble salt $PbB{r}_2$ is $6.6\times {10}^{-6}$ mol $L^{-1}$. What is the concentration of $B{r}^{-}\left(aq\right)$ in a saturated solution of $PbB{r}_2\left(s\right)$?
$(1.65{)}^{\frac{1}{3}}=1.18$

• A. $4\times {10}^{-10}$
• B. $2.36\times {10}^{-2}$
• C. $1.18\times {10}^{-2}$
• D. $4\times {10}^{-15}$

Answer 1

The correct option is B $2.36\times {10}^{-2}$

Given $K_{sp}=6.6\times {10}^{-6}$
The dissociation equation is$PbB{r}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2B{r}^{-}\left(aq\right)t=0100ateqb1-ss2s$

$K_{sp}=\left[P{b}^{2+}\right][B{r}^{-}{]}^2$
$K_{sp}=s\times (2s{)}^2=4s^3$
$K_{sp}=4s^3=6.6\times {10}^{-6}$
$s^3=1.65\times {10}^{-6}$
$s=1.18\times {10}^{-2}$
$\left[B{r}^{-}\right]\left(aq\right)=2s=2.36\times {10}^{-2}$