Question

The $K_{sp}$ of $FeS=4\times {10}^{-19}\text{at}298\text{K}$. The minimum concentration of $H^{+}$ ions required to prevent the precipitation of $FeS$ from a $0.01\text{M}$ solution of $F{e}^{2+}$ salt by passing $H_{2}S\left(0.1\text{M}\right)$
$(\text{Given}:k_{a_1}\times k_{b_1}={10}^{-21})\text{for}{H}_{2}S$

• A. $1.2\times {10}^{-4}\text{M}$
• B. $1.6\times {10}^{-3}\text{M}$
• C. $2.0\times {10}^{-2}\text{M}$
• D. $2.5\times {10}^{-4}\text{M}$

Answer 1

The correct option is B $1.6\times {10}^{-3}\text{M}$

$\text{Given}:$
$K_{sp}\left(FeS\right)=4\times {10}^{-19},\left[FeS\right]=0.01\text{M},k_{a_1}\times k_{b_1}={10}^{-21}\left(H_{2}S\right)$
$\begin{array}{c}{H}_{2}S\stackrel{k_{a_1}}{\to }{H}^{+}+H{S}^{-}\\ H{S}^{-}\stackrel{k_{b_1}}{\to }{H}^{+}+S^{2-}\\ FeS\to F{e}^{2+}+S^{2-}\\ K_{sp}=\left[F{e}^{2+}\right]\left[S^{2-}\right]=4\times {10}^{-19}\\ \Rightarrow 4\times {10}^{-19}=\frac{\left(0.01\right)\left[S^{2-}\right]}{1}\\ \Rightarrow \left[S^{2-}\right]=4\times {10}^{-17}\text{M}\\ k_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{H_{2}S}\\ k_{b_1}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}\\ k_{a_1}\times k_{b_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}\times \frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}\\ \Rightarrow {10}^{-21}=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}\\ \Rightarrow {10}^{-21}=\frac{[H^{+}{]}^{2}4\times {10}^{-17}}{0.1}\\ \Rightarrow [H^{+}{]}^2=0.25\times {10}^{-5}\\ \Rightarrow [H^{+}{]}^2=2.5\times {10}^{-6}\\ \Rightarrow \left[H^{+}\right]=\sqrt{2.5\times {10}^{-6}}=1.58\times {10}^{-3}\text{M}\approx 1.6\times {10}^{-3}\text{M}\end{array}$