Ionic EquilibriumHard

Question

The concentration of acetate ions in 1 M acetic acid (Ka = 2 × 10−5) solution containing 0.1 M HCl is

Options

A.2 × 10–1 M
B.2 × 10–3 M
C.2 × 10–4 M
D.4.4 × 10–3 M

Solution

CH3COOH $\rightleftharpoons$ CH3COO– + H+

1M 0 0.1 M

Equilibrium 1 – x x 0.1 + x

$\simeq$1M $\simeq$0.1 M

Now, $K_{a} = \frac{\left\lbrack CH_{3}COO^{-} \right\rbrack\left\lbrack H^{+} \right\rbrack}{\left\lbrack CH_{3}COOH \right\rbrack} \Rightarrow 2 \times 10^{- 5} = \frac{x \times 0.1}{1}$

⇒ x = 2 × 10–4 M

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