Ionic EquilibriumHard
Question
What is the solubility of MnS in pure water, assuming hydrolysis of S2– ions?
Ksp of MnS = 2.5 × 10–10, Ka1 = 1 × 10–7 and Ka2 = 1 × 10–14 for H2S. (0.633 = 0.25)
Options
A.6.3 × 10–4 M
B.2.5 × 10−4 M
C.6.3 × 10−3 M
D.1.58 × 10−5 M
Solution
$MnS(s) \rightleftharpoons \underset{S\text{ M}}{Mn^{2 +}} + \underset{(S - x)\text{ M}}{S^{2 -}}$
$\underset{(S - x)\text{ M}}{S^{2 -}} + H_{2}O \rightleftharpoons \underset{x\text{ M}}{HS^{-}} + \underset{x\text{ M}}{OH^{-}} $$${\frac{10^{- 14}}{10^{- 14}} = \frac{x.x}{(S - x)}\text{ and }2.5 \times 10^{- 10} = S.(S - x) }{\therefore S = 6.3 \times 10^{- 4}\text{ M}}$$
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