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)

$\left[F e^{2 +}\right] \, \left[S^{- 2}\right]=4 \, \times \, 10^{- 19}\Rightarrow \, \left[S^{- 2}\right]=\frac{4 \times 1 0^{- 19}}{1 \times 1 0^{- 2}}= \, 4\times 10^{- 17} \, M$

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

Now $\frac{\left[H^{+}\right]^{2} \left[4 \times 1 0^{- 17}\right]}{0.1}= \, 1 \, \times \, 10^{- 21}$

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

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