Ionic EquilibriumHard

Question

The concentration of CH3COO ion in a solution prepared by adding 0.1 mole of CH3COOAg(s) in 1 L of 0.1 M-HCl solution is [Given: Ka(CH3COOH) = 10−5; Ksp(AgCl) = 10−10; Ksp(CH3COOAg) = 10−8]

Options

A.10−3 M
B.10−2 M
C.10−1 M
D.1 M

Solution

$\underset{0.1\text{ mole}}{ACOAg(s)} + \underset{0.1\text{ M}}{H^{+}} + \underset{0.1\text{ M}}{Cl^{-}} \rightleftharpoons ACOH + AgCl(s)$

$K_{eq} = \frac{\lbrack ACOH\rbrack}{\left\lbrack H^{+} \right\rbrack\left\lbrack Cl^{-} \right\rbrack} \times \frac{\left\lbrack ACO^{-} \right\rbrack}{\left\lbrack ACO^{-} \right\rbrack} \times \frac{\left\lbrack Ag^{+} \right\rbrack}{\left\lbrack Ag^{+} \right\rbrack} = \frac{K_{sp}\lbrack ACOAg\rbrack}{K_{sp}(AgCl) \times K_{a}} $$${= \frac{10^{- 8}}{10^{- 10} \times 10^{- 5}} = 10^{7} \Rightarrow \text{Almost complete reaction} }{\therefore\lbrack ACOH\rbrack \simeq 0.1\text{ M, }\left\lbrack H^{+} \right\rbrack = \sqrt{\frac{0.1}{10^{7}}} = 10^{- 4}\text{ M and }\left\lbrack ACO^{-} \right\rbrack = \frac{K_{a} \times \lbrack ACOH\rbrack}{\left\lbrack H^{+} \right\rbrack} = 0.01\text{ M}}$$

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