Question

If the solubility of $Pb{Br}_2$ is 'S' moles per litre, considering 100% ionisation, its solubility product is:

Answer 1

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

At t=0 : S O O

At equilibrium: O S 2S

Therefore , Solubility product $\left(K_{sp}\right)=\left[P{b}^{2+}\right][B{r}^{-}{]}^2$

$=\left(s\right)(2s{)}^2$

$=\left(s\right)\left(4s^2\right)$

$=4s^3$

Therefore, solubility product is $4s^3$ moles per litre.