Question
Dissociation constants of acids HA and HB are 2.0 × 10−4 and 5 × 10−5, respectively. The [H+] in the resulting solution obtained by mixing 20 ml of 0.5 M-HA solution and 30 ml of 0.2 M-HB solution is
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Solution
$\lbrack HA\rbrack_{O} = \frac{20 \times 0.5}{50} = 0.2\text{ M}$
$\lbrack HB\rbrack_{O} = \frac{30 \times 0.2}{50} = 0.12\text{ M}$
$\underset{0.2 - x}{HA} \rightleftharpoons \underset{x + y}{H^{+}} + \underset{x}{A^{-}} $$$\underset{0.12 - y}{HA} \rightleftharpoons \underset{x + y}{H^{+}} + \underset{y}{B^{-}}$$
Now, $2 \times 10^{- 4} = \frac{(x + y).x}{0.2 - x} \approx \frac{(x + y).x}{0.2}$
$\therefore(x + y).x = 4 \times 10^{- 5}(1)$
and $5 \times 10^{- 5} = \frac{(x + y).y}{(0.12 - y)} \approx \frac{(x + y).y}{0.12}$
$\therefore(x + y).y = 6 \times 10^{- 6}(2)$
From (1) and (2), [H+] = x + y = 6.78 × 10–3 M
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