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

Correct Answer: 6

Concentration of $H _2 SO _4$ and $Na _2 SO _4$ in a solution is $1 M$ and $1.8 \times 10^{-2} M$
\(H_2SO_4\rightarrow H^++HSO^-_4\)
\(HSO^-_4\rightleftharpoons H^++SO^{2-}_4 \\ \begin{matrix} 1 && 1 & 1.8\times10^{-2} \\1+x && 1-x & (1.8\times10^{-2}-x) \end{matrix}\)
\(x\approx1 \; x\lt1.8\times10^{-2}\)
\(k_{a_2}=1.2\times10^{-2}\)

\(1.2\times10^{-2}=\frac{1\times[SO^{2-}_4]}{1}\)

\([SO^{2-}_4]=1.2\times10^{-2}M\)
\(PbSO_4 \rightleftharpoons Pb^{2+}+SO^{2-}_4 \\ \begin{matrix} &&&&&s && 1.2\times10^{-2}\end{matrix}\)

\(s=\frac{K_{sp}}{1.2\times10^{-2}}\)

\(s=\frac{1.6\times10^{-8}}{1.2\times10^{-2}}\)

\(s=1.33\times10^{-6}\)
\(Y=6\)

So, the answer is 6.