Concentration of H _ 2 SO _ 4 and Na _ 2 SO _ 4 in a solution is 1M...

$\begin{cases}{H}_{2}{SO}_{4} & \mathop{\to}_{} & {H}^{+} & + & {HSO}_{4}^{2 -} \\ & & 1M & & 1M\end{cases}$ (${K}_{{a}_{1}}$ is very large)

${HSO}_{4}^{-}\overset{}{⇌}{H}^{+} + {SO}_{4}^{2 -}{K}_{{a}_{2}} = 1 . 2 \times {10}^{- 2}$

$\left[{SO}_{4}^{2 -}\right]$ coming from ${Na}_{2}{SO}_{4} = 1 . 8 \times {10}^{- 2}$

$\dfrac{\left[{SO}_{4}^{2 -}\right]\left[{H}^{+}\right]}{\left[{HSO}_{4}^{-}\right]} = \dfrac{1 . 8 \times {10}^{- 2} \times 1}{1} > {K}_{{a}_{2}}$

$∴$ Rather than dissociation of ${HSO}_{4}^{-}$ into ${H}^{+}$and ${SO}_{4}^{2 -}$ ions, association between already present ${H}^{+}$ and ${SO}_{4}^{2 -}$ will take place.

Assuming '$x$' $mol / L$ of ${SO}_{4}^{2 -}$ and ${H}^{+}$combines to form ${HSO}_{4}^{-}$

$∴ \left[{SO}_{4}^{2 -}\right] = 1 . 8 \times {10}^{- 2} - x$

\left$$\begin{array}{ll}\left[{H}^{+}\right] = 1 - x \approx 1 \\ \left[{HSO}_{4}^{-}\right] = 1 + x \approx 1\end{array}$$\right\}$(assuming$x < < 1$)</p><p>$\dfrac{\left(1 . 8 \times {10}^{- 2} - x\right)1}{1} = 1 . 2 \times {10}^{- 2}$</p><p>$⇒ x = 0 . 6 \times {10}^{- 2}$</p><p>\left[{SO}_{4}^{2 -}\right] = 1 . 2 \times {10}^{- 2}M$

${PbSO}_{4}\left(s\right) ⇌ {Pb}^{2 +}\left(aq\right) + {SO}_{4}^{2 -}\left(aq\right)$

If solubility of ${PbSO}_{4} = sM$

$∴ \left[{Pb}^{2 +}\right] = s$

\left[{SO}_{4}^{2 -}\right] = s + 1 . 2 \times {10}^{- 2} \approx 1 . 2 \times {10}^{- 2}$</p><p>(assuming$s ≪ 1 . 2 \times {10}^{- 2}$)</p><p>$∴ s \times 1 . 2 \times {10}^{- 2} = 1 . 6 \times {10}^{- 8}$</p><p>$s = \dfrac{1 . 6}{1 . 2} \times {10}^{- 6} = 1 . 33 \times {10}^{- 6}$</p><p>On comparing with$X \times {10}^{- Y}$</p><p>$Y = 6$