Question

At $25$, $K_{sp}$ for ${PbBr}_2$ is equal to ${8\times 10}^{-5}$. If the salt is 80% dissociated, what is the ${PbBr}_2$ in mol/litre?

• A. ${\left[\frac{{10}^{-4}}{1.6\times 1.6}\right]}^{1/3}$
• B. ${\left[\frac{{10}^{-5}}{1.6\times 1.6}\right]}^{1/2}$
• C. ${\left[\frac{{10}^{-4}}{0.8\times 0.8}\right]}^{1/3}$
• D. None of these

Answer 1

The correct option is D None of these

$PbB{r}_2\to P{b}^{+2}+2B{r}^{-}$

$t=0$ $C$ $0$ $0$

at equilibrium $C(1-0.8)$ $C\times 0.8$ $2C\times 0.8$

$0.2C$ $0.8C$ $1.6C$

$\frac{\left(1.6C{)}^2\right(0.8C)}{\left(0.2C\right)}=8\times {10}^{-5}$

$(1.6C{)}^2=2\times {10}^{-5}$

$1.6C\times 1.6C=2\times {10}^{-5}$

$C^2=\frac{2\times {10}^{-5}}{1.6\times 1.6}$

$C={\left(\frac{2\times {10}^{-5}}{1.6\times 1.6}\right)}^{1/2}$