Question
The solubility of metal sulphide in saturated solution of H2S (concentration = 0.1 M) can be represented as follows. MS(s) + 2H+ (aq) $\rightleftharpoons$ M2+ (aq) + H2S(aq);$K_{eq} = \frac{\left\lbrack M^{2 +} \right\rbrack\left\lbrack H_{2}S \right\rbrack}{\left\lbrack H^{+} \right\rbrack^{2}}$.
The values of Keq for the metal sulphides, MnS, ZnS, CoS and PbS are 3 × 1010, 3 × 10−2, 3 and 3 × 10−7, respectively. If the concentration of each metal ion in a saturated solution of H2S is 0.01 M, then which metal sulphide(s) will precipitate at [H+] = 1.0 M?
Options
Solution
For ppt, Q > Keq
For the metal sulphides,
$Q = \frac{\left\lbrack M^{2 +} \right\rbrack\left\lbrack H_{2}S \right\rbrack}{\left\lbrack H^{+} \right\rbrack^{2}} = \frac{0.01 \times 0.1}{(1.0)^{2}} = 10^{- 3}$> Keq for PbS only Hence, only Pbs will precipitate.
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