Concentration of H 2 SO 4 and Na 2 SO 4 in a solution is 1 M and 1 . 8 × 10 - 2 M , respectively. Molar…

$\begin{array}{ccccc}{\mathrm{H}}_2{\mathrm{SO}}_4& \underset{}{\to }& {\mathrm{H}}^{+}& +& {\mathrm{HSO}}_4^{2-}\\ & & 1\mathrm{M}& & 1\mathrm{M}\end{array}$ (${\mathrm{K}}_{{\mathrm{a}}_1}$ is very large)

${\mathrm{HSO}}_4^{-}\stackrel{}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-} {\mathrm{K}}_{{\mathrm{a}}_2}=1.2\times {10}^{-2}$

$\left[{\mathrm{SO}}_4^{2-}\right]$ coming from ${\mathrm{Na}}_2{\mathrm{SO}}_4=1.8\times {10}^{-2}$

$\frac{\left[{\mathrm{SO}}_4^{2-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{HSO}}_4^{-}\right]}=\frac{1.8\times {10}^{-2}\times 1}{1}>{\mathrm{K}}_{{\mathrm{a}}_2}$

$\therefore$ Rather than dissociation of ${\mathrm{HSO}}_4^{-}$ into ${\mathrm{H}}^{+}$and ${\mathrm{SO}}_4^{2-}$ ions, association between already present ${\mathrm{H}}^{+}$ and ${\mathrm{SO}}_4^{2-}$ will take place.

Assuming '$\mathrm{x}$' $\mathrm{mol}/\mathrm{L}$ of ${\mathrm{SO}}_4^{2-}$ and ${\mathrm{H}}^{+}$combines to form ${\mathrm{HSO}}_4^{-}$

$\therefore \left[{\mathrm{SO}}_4^{2-}\right]=1.8\times {10}^{-2}-\mathrm{x}$

$\left.\begin{array}{l}\left[{\mathrm{H}}^{+}\right]=1-\mathrm{x}\approx 1\\ \left[{\mathrm{HSO}}_4^{-}\right]=1+\mathrm{x}\approx 1\end{array}\right\}$ (assuming $\mathrm{x}<<1$)

$\frac{\left(1.8\times {10}^{-2}-\mathrm{x}\right)1}{1}=1.2\times {10}^{-2}$

$\Rightarrow \mathrm{x}=0.6\times {10}^{-2}$

$\left[{\mathrm{SO}}_4^{2-}\right]=1.2\times {10}^{-2}\mathrm{M}$

${\mathrm{PbSO}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

If solubility of ${\mathrm{PbSO}}_4=\mathrm{sM}$

$\therefore \left[{\mathrm{Pb}}^{2+}\right]=\mathrm{s}$

$\left[{\mathrm{SO}}_4^{2-}\right]=\mathrm{s}+1.2\times {10}^{-2}\approx 1.2\times {10}^{-2}$

(assuming $\mathrm{s}\ll 1.2\times {10}^{-2}$)

$\therefore \mathrm{s}\times 1.2\times {10}^{-2}=1.6\times {10}^{-8}$

$\mathrm{s}=\frac{1.6}{1.2}\times {10}^{-6}=1.33\times {10}^{-6}$

On comparing with $\mathrm{X}\times {10}^{-\mathrm{Y}}$

$\mathrm{Y}=6$