Question

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

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

Answer 1

Answer: (A)

$[\frac{{10}^{-4}}{1.6\times 1.6}{]}^{1/3}$

$PbB{r}_2$ is a $A{B}_2$ type of salt and will have Ksp,

$ksp=4s^3$

if salt is 80% ionised, solubility is,

$s\times 0.8$

after putting the values

$8\times {10}^{-5}=4\times (s^3\times {0.8}^3)$

$s=[\frac{{10}^{-4}}{1.6\times 1.6}{]}^{1/3}$