$\begin{array}{lcc} H _2 SO _4 \\ 1 M & HSO _4^{-}+ H ^{+} \\ – & – & – \\ 1 M & 1 M \end{array}$

$\begin{array}{l} Na _2 SO _4 \quad \longrightarrow \quad 2 Na ^{+}+ SO _4^{2-} \\ 3.6 \times 10^{-2} M \quad 1.8 \times 10^{-2} M \\ HSO _4^{-} \rightleftharpoons H ^{+}+ SO _4{ }^{2-} ; K _{ a _2}=1.2 \times 10^{-2} M \\ 1 M \quad 1 M \quad 1.8 \times 10^{-2} M \\ \text { Since } Q _C> K _C \text { it will move in backward direction. } \\ 1+ x \quad 1- x \quad 1.8 \times 10^{-2}- x \\ K _{ a _2}=1.2 \times 10^{-2}=\frac{(1- x )\left(1.8 \times 10^{-2}- x \right)}{(1+ x )} \\\end{array}$

Since $x$ is very small $(1+x) \simeq 1$ and $(1-x) \simeq 1$

$x =\left(1.8 \times 10^{-2}-1.2 \times 10^{-2}\right) M$

${\left[ SO _4^{2-}\right]=\left(1.8 \times 10^{-2}-0.6 \times 10^{-2}\right) M }$

$=1.2 \times 10^{-2} M$

$PbSO _4 \quad \longrightarrow \quad Pb ^{2+}+ SO _4^{2-}$

$\text { – } \quad 1.2 \times 10^{-2} M$

$s \quad\left(s+1.2 \times 10^{-2}\right)$

$K _{ sp }= s \left( s +1.2 \times 10^{-2}\right)=1.6 \times 10^{-8}$

$( PbSO _4)$

Here, $\left(s+1.2 \times 10^{-2}\right) \simeq 1.2 \times 10^{-2}$ (since ' $s$ ' is very small)

$s\left(1.2 \times 10^{-2}\right)=1.6 \times 10^{-8}$

$\Rightarrow s =\frac{1.6}{1.2} \times 10^{-6} M = X \times 10^{-7} M$

$\Rightarrow Y =6$