Question
A buffer solution is 0.25 M – CH3COOH + 0.15 M – CH3COONa, saturated in H2S (0.1 M) and has [Mn2+] = 0.04 M, Ka(CH3COOH) = 2.0 × 10–5,Ka(H2S) = 1.0 × 10–21 and Ksp(MnS) = 2.5 × 10–13. Which buffer component should be increased in concentration and to which minimum value to just start precipitation of MnS?
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Solution
Mn2+ (aq) + H2S(aq) $\rightleftharpoons$ MnS(s) + 2H+(aq)
To just start precipitation of MnS, Q < Keq
Or, $\frac{\left\lbrack H^{+} \right\rbrack^{2}}{\left\lbrack Mn^{2 +} \right\rbrack\left\lbrack H_{2}S \right\rbrack} < \frac{K_{a}\left( H_{2}S \right)}{K_{sp}(MnS)}$
Or, $\frac{\left\lbrack H^{+} \right\rbrack^{2}}{0.04 \times 0.1} < \frac{1.0 \times 10^{- 21}}{2.5 \times 10^{- 13}} \Rightarrow \left\lbrack H^{+} \right\rbrack < 4 \times 10^{- 6}\text{ M}$
Now, in the given buffer, $\left\lbrack H^{+} \right\rbrack = \frac{K_{a}.\left\lbrack CH_{3}COOH \right\rbrack_{O}}{\left\lbrack CH_{3}COO^{-} \right\rbrack_{O}}$
$= \frac{2 \times 10^{- 5} \times 0.25}{0.15} = 3.33 \times 10^{- 5} > 4 \times 10^{- 6}\text{ M}$
Hence, no precipitation. To start precipitation [H+] should decrease and hence, CH3COONa should be added.
Now, $4 \times 10^{- 6} = \frac{2 \times 10^{- 5} \times 0.25}{\left\lbrack CH_{3}COONa \right\rbrack_{O}}$
∴ [CH3COONa]O = 1.25 M
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