Question

The ${\text{K}}_{\text{sp}}$ of $\text{FeS}=4\times 10^{-19}$ at $298\text{K}\text{.}$ The minimum concentration of ${\text{H}}^{\text{+}}$ ions required to prevent the precipitation of $\text{FeS}$ from a $\text{0}\text{.01}\text{M}$ solution ${\text{Fe}}^{2+}$ salt by passing 0.1 M ${\text{H}}_{\text{2}}\text{S}$

(Given ${\text{H}}_{\text{2}}\text{S}$ ${\text{k}}_{{\text{a}}_{\text{1}}}\times {\text{k}}_{{\text{b}}_{\text{1}}}=10^{-21}$ )

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

Answer 1

Answer: (A)

$\left[F{e}^{2+}\right]\left[S^{-2}\right]=4\times 10^{-19}\Rightarrow \left[S^{-2}\right]=\frac{4\times 10^{-19}}{1\times 10^{-2}}=4\times 10^{-17}M$

In order to precipitate $FeS,\left[S^{-2}\right]$ required is $4\times 10^{-17}M$ from $0.01MF{e}^{2+}$ salts.

Now $\frac{{\left[H^{+}\right]}^2\left[4\times 10^{-17}\right]}{0.1}=1\times 10^{-21}$

$\Rightarrow {\left[H^{+}\right]}^2=2.5\times 10^{-6}$

$\Rightarrow \left[H^{+}\right]=1.6\times 10^{-3}$