Question

Concentration of $H _2 SO _4$ and $Na _2 SO _4$ in a solution is $1 M$ and $1.8 \times 10^{-2} M$, respectively. Molar solubility of $PbSO _4$ in the same solution is $X \times 10^{- Y } M$ (expressed in scientific notation). The value of $Y$ is ________.
[Given: Solubility product of $PbSO _4\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $H _2 SO _4, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

Answer 1

$K _{ a _2}=1.2 \times 10^{-2}=\frac{(1- x )\left(1.8 \times 10^{-2}- x \right)}{(1+ x )}$
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[ SO _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} $
$\left( PbSO _4\right)$
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$
$\Rightarrow Y =6$