Calculate Exponent Y for Molar Solubility of Lead Sulfate

To find the value of $Y$, we first determine the equilibrium concentration of $\text{SO}_4^{2-}$ in the given solution. Since $K_{a1}$ of $\text{H}_2\text{SO}_4$ is very large, $\text{H}_2\text{SO}_4$ dissociates completely into $1\text{ M}$ $\text{H}^+$ and $1\text{ M}$ $\text{HSO}_4^-$. Using the second dissociation constant $K_{a2} = \frac{[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]} = 1.2 \times 10^{-2}$, with $[\text{H}^+] \approx 1\text{ M}$ and $[\text{HSO}_4^-] \approx 1\text{ M}$, the concentration of sulphate ion is given by $[\text{SO}_4^{2-}] = K_{a2} = 1.2 \times 10^{-2}\text{ M}$. The molar solubility $S$ of $\text{PbSO}_4$ is then calculated from the solubility product $K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] = 1.6 \times 10^{-8}$, yielding $S = \frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}} = 1.33 \times 10^{-6}\text{ M}$. Expressing this in scientific notation as $X \times 10^{-Y}\text{ M}$, we find $Y = 6$.