A solution is prepared in which 0.1 mole each of HCl, CH3COOH and CHCl2COOH is present in a litre. I

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A solution is prepared in which 0.1 mole each of HCl, CH3COOH and CHCl2COOH is present in a litre. If the ionization constant of CH3COOH is 10–5 and that of Cl2CHCOOH is 0.15, then the pH of solution is (log 2 = 0.3, log 3 = 0.48)

Accepted Answer

$\underset{0.1 - x}{CH_{3}COOH} ightleftharpoons \underset{x}{CH_{3}COO^{-}} + \underset{0.1 + x + y = 0.1 + y}{H^{+}}$$\underset{0.1 - y}{Cl_{2}CHCOOH} ightleftharpoons \underset{y}{Cl_{2}CHCOO^{-}} + \underset{0.1 + x + y = 0.1 + y}{H^{+}} $$${0.15 = \frac{y 	imes (0.1 + y)}{(0.1 - y)} \Rightarrow y = 0.05 }{	herefore\left\lbrack H^{+} ightbrack = 0.1 + x + y \approx 0.1 + y = 0.15	ext{ M} }{	herefore P^{H} = - \log(0.15) = 0.82}$$

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