Question

The solubility product of $PbB{r}_2$ is $K_{sp}=6.3\times {10}^{-6}$ at room temperature. If 50 ml of $0.02MPb(N{O}_3{)}_2$ are mixed with 50 ml of $0.01MCaB{r}_2$, then

• A. $PbB{r}_{2}willprecipitateandexcessB{r}^{-}willremaininsolution.$
• B. $Ca(N{O}_3{)}_{2}willprecipitate.$
• C. $PbB{r}_{2}willprecipitateandexcessP{b}^{2+}willremaininsolution.$
• D. $noprecipitatewillform$.

Answer 1

Answer: (D)

$noprecipitatewillform$.

$\left[{Pb}^{+2}\right]=\frac{50}{1000}\times 0.02=1\times {10}^{-3}$

$\left[{Br}^{-}\right]=\frac{50}{1000}\times 0.01=5\times 0^{-4}$

Ionic product $=\left[{Pb}^{+2}\right]{\left[2{Br}^{-}\right]}^2=(1\times {10}^{-3})\times {(2\times 5\times {10}^{-4})}^2={10}^{-9}$

$K_{sp}=6.3\times {10}^{-6}$

$\because K_{sp}>$ Ionic product

Hence no precipitate ill form.