Estimate the approximate pK_a of 0.5 M CH_3C | vedclass

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(C) The dissociation of acetic acid is represented as: $CH_3COOH \rightleftharpoons H^{+} + CH_3COO^{-}$Using the expression for the dissociation cons

Vedclass · Accepted Answer

$1.88$

Vedclass · Answer

(C) The dissociation of acetic acid is represented as: $CH_3COOH ightleftharpoons H^{+} + CH_3COO^{-}$Using the expression for the dissociation constant $K_a = \frac{C\alpha^2}{1-\alpha}$,where $C = 0.5 \ M$ and $\alpha = 0.15$.$K_a = \frac{0.5 	imes (0.15)^2}{1 - 0.15} = \frac{0.5 	imes 0.0225}{0.85} = \frac{0.01125}{0.85} \approx 0.0132$.Now,calculate $pK_a$ using $pK_a = -\log K_a$.$pK_a = -\log(0.0132) = -\log(1.32 	imes 10^{-2})$.$pK_a = -(\log 1.32 + \log 10^{-2}) = -(0.12 - 2) = 1.88$.

Estimate the approximate $pK_a$ of $0.5 \ M$ $CH_3COOH$. Degree of dissociation (ionisation) is $0.15$ $(\log 1.32 = 0.12)$.

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