Question

For H₂S ; $K_{a_1}=10^{-7}$ , ${\text{K}}_{{\text{a}}_2}=10^{-14}$ . 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

$K_{sp};MnS=2.5\times 10^{-10},CoS=4\times 10^{-21},A{g}_{2}S=6.3\times 10^{-50}$

Calculate difference in pH for precipitation of Mns and CoS.

Answer 1

Answer: 5.4

For H₂S,

${\text{H}}_2\text{S}\rightleftharpoons 2{\text{H}}^{+}+{\text{S}}^{2-}$

${\text{K}}_{\text{a}}={\text{K}}_{{\text{a}}_1}{\text{K}}_{{\text{a}}_2}=\frac{{\left[{\text{H}}^{+}\right]}^2\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_2\text{S}\right]}\text{;}{\left[{\text{H}}^{+}\right]}^2=\frac{10^{-21}\left[{\text{H}}_2\text{S}\right]}{\left[{\text{S}}^{2-}\right]}$

${\left[{\text{H}}^{+}\right]}^2=\frac{10^{-22}}{\left[{\text{S}}^{2-}\right]}$

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

${\left(\text{K}\right)}_{\text{sp}}/{\left(\left[{\left(\text{Ag}\right)}^{+}\right]\right)}^2=\frac{6\text{.}3\times 10^{-50}}{{\left(10^{-2}\right)}^2}=6\text{.}3\times 10^{-46}\text{M}$

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

$\Rightarrow$ [S^{2 -} ] for precipitation of CoS $\Rightarrow$ $\frac{4\times 10^{-21}}{10^{-2}}=4\times 10^{-19}$

$\therefore$ $\left[{\text{H}}^{+}\right]=\sqrt{\frac{10^{-22}}{4\times 10^{-19}}}=1\text{.}58\times 10^{-2}\text{M}$ ; pH = 1.8

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

$\frac{2\text{.}5\times 10^{-10}}{10^{-2}}=2\text{.}5\times 10^{-8}$

$\left[{\text{H}}^{+}\right]=\sqrt{\frac{10^{-22}}{2\text{.}5\times 10^{-8}}}=6\text{.}3\times 10^{-8}$ ; 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.