Question

For H₂S ; $K_{a_{1}} = 10^{- 7}$ , $\text{K}_{\text{a}_{2}} = 1 0^{- 1 4}$ . A Saturated solution of H₂ S in 0.1 M H₂S contains Mn^{2 +} , Co^{2 +} and Ag⁺ at an original concentration of 0.01 M each. Determine the pH range for selective precipitation of these metal ions

$K_{s p} \, ;MnS=2.5\times 10^{- 10} \, , \, CoS=4\times 10^{- 21} \, , \, Ag_{2}S=6.3\times 10^{- 50}$

Calculate difference in pH for precipitation of Mns and CoS.

Answer 1

For H₂S,

$\text{H}_{2} \text{S} \rightleftharpoons 2 \text{H}^{+} + \text{S}^{2 -}$

$\text{K}_{\text{a}} = \text{K}_{\text{a}_{1}} \text{K}_{\text{a}_{2}} = \frac{\left[\text{H}^{+}\right]^{2} \left[\text{S}^{2 -}\right]}{\left[\text{H}_{2} \text{S}\right]} \text{;} \left[\text{H}^{+}\right]^{2} = \frac{1 0^{- 2 1} \left[\text{H}_{2} \text{S}\right]}{\left[\text{S}^{2 -}\right]}$

$\left[\text{H}^{+}\right]^{2} = \frac{1 0^{- 2 2}}{\left[\text{S}^{2 -}\right]}$

$\Rightarrow $ [S^{2 -} ] for precipitation of Ag₂S is

$\left(\text{K}\right)_{\text{sp}} / \left(\left[\left(\text{Ag}\right)^{+}\right]\right)^{2} = \frac{6 \text{.} 3 \times 1 0^{- 5 0}}{\left(1 0^{- 2}\right)^{2}} = 6 \text{.} 3 \times 1 0^{- 4 6} \text{M}$

i.e., too low value it will precipitate at any pH.

$\Rightarrow $ [S^{2 -} ] for precipitation of CoS $\Rightarrow $ $\frac{4 \times 1 0^{- 2 1}}{1 0^{- 2}} = 4 \times 1 0^{- 1 9}$

$\therefore $ $\left[\text{H}^{+}\right] = \sqrt{\frac{1 0^{- 2 2}}{4 \times 1 0^{- 1 9}}} = 1 \text{.} 5 8 \times 1 0^{- 2} \text{M}$ ; pH = 1.8

$\Rightarrow $ [S^{2 -} ] for precipitation of MnS is

$\frac{2 \text{.} 5 \times 1 0^{- 1 0}}{1 0^{- 2}} = 2 \text{.} 5 \times 1 0^{- 8}$

$\left[\text{H}^{+}\right] = \sqrt{\frac{1 0^{- 2 2}}{2 \text{.} 5 \times 1 0^{- 8}}} = 6 \text{.} 3 \times 1 0^{- 8}$ ; pH = 7.2

pH < 1.8 Ag₂S will precipitate out.

pH - 1.8 to 7.2 only CoS will precipitate out.

pH > 7.2 only MnS will precipitate out.