Question

Concentration of $H_{2}S{O}_4$ and $N{a}_{2}S{O}_4$ in a solution is $1M$ and $1.8\times 10^{-2}M$, respectively. Molar solubility of $PbS{O}_4$ in the same solution is $X\times 10^{-Y}M$ (expressed in scientific notation). The value of $Y$ is ________.
[Given: Solubility product of $PbS{O}_4\left(K_{sp}\right)=1.6\times 10^{-8}$. For $H_{2}S{O}_4,K_{al}$ is very large and $K_{a2}=1.2\times 10^{-2}]$

Answer 1

Answer: 6

$K_{a_2}=1.2\times 10^{-2}=\frac{(1-x)\left(1.8\times 10^{-2}-x\right)}{(1+x)}$
Since $x$ is very small $(1+x)\simeq 1$ and $(1-x)\simeq 1$
$x=\left(1.8\times 10^{-2}-1.2\times 10^{-2}\right)M$
$\left[S{O}_4^{2-}\right]=\left(1.8\times 10^{-2}-0.6\times 10^{-2}\right)M$
$=1.2\times 10^{-2}M$

$K_{sp}=s\left(s+1.2\times 10^{-2}\right)=1.6\times 10^{-8}$
$\left(PbS{O}_4\right)$
Here, $\left(s+1.2\times 10^{-2}\right)\simeq 1.2\times 10^{-2}($ since ' $s$ ' is very small)
$s\left(1.2\times 10^{-2}\right)=1.6\times 10^{-8}$
$\Rightarrow s=\frac{1.6}{1.2}\times 10^{-6}M=X\times 10^{-Y}M$
$\Rightarrow Y=6$