Question

At ${25}^0$C, $K_{sp}$ for $PbB{r}_2$ is equal to 8 x ${10}^{-5}$. If the salt is 80% dissociated, what is the solubility of $PbB{r}_2$ in mol/litre?

• A. $[\frac{{10}^{-5}}{1.6\times 1.6}{]}^{1/3}$
• B. $[\frac{{10}^{-4}}{1.6\times 1.6}{]}^{1/3}$
• C. $[\frac{{10}^{-4}}{0.8\times 0.8}{]}^{1/3}$
• D. $[\frac{{10}^{-5}}{1.6\times 1.6}{]}^{1/2}$

Answer 1

The correct option is A $[\frac{{10}^{-5}}{1.6\times 1.6}{]}^{1/3}$

$PbB{r}_2⟶\underset{S\alpha }{P{b}^{(2+)}}+\underset{2S\alpha }{2B{r}^{-}}$

$K_{sp}=\left[P{b}^{2+}\right][B{r}^{-}{]}^2$

$\Rightarrow 8\times {10}^{-5}=\left(S\alpha \right)(S\alpha {)}^2$

$\Rightarrow 8\times {10}^{-5}=4(S\alpha {)}^3$

$\Rightarrow 2\times {10}^{-5}=(S\times 0.8{)}^3$

$\Rightarrow S={\left[\frac{2\times {10}^{-5}}{0.8\times 0.8\times 0.8}\right]}^{1/3}={\left[\frac{{10}^{-5}}{1.6\times 1.6}\right]}^{1/3}$