K_{a} for butyric acid (C_{3}H_{7}COOH) is 2 | vedclass

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(C) For a weak acid,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_{a} - \log C)$.Given $K_{a} = 2 	imes 10^{-5}$,so $pK_{a} = -\log(2 	ime

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) For a weak acid,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_{a} - \log C)$.Given $K_{a} = 2 	imes 10^{-5}$,so $pK_{a} = -\log(2 	imes 10^{-5}) = 5 - \log 2 = 5 - 0.30 = 4.7$.Concentration $C = 0.2 \ M = 2 	imes 10^{-1} \ M$.$pH = \frac{1}{2} (4.7 - \log(2 	imes 10^{-1}))$.$pH = \frac{1}{2} (4.7 - (\log 2 + \log 10^{-1}))$.$pH = \frac{1}{2} (4.7 - (0.30 - 1)) = \frac{1}{2} (4.7 - (-0.7)) = \frac{1}{2} (5.4) = 2.7$.$pH = 2.7 = 27 	imes 10^{-1}$.Thus,the value is $27$.

$K_{a}$ for butyric acid $(C_{3}H_{7}COOH)$ is $2 \times 10^{-5}$. The $pH$ of $0.2 \ M$ solution of butyric acid is $........... \times 10^{-1}$. (Nearest integer) [Given $\log 2 = 0.30$]

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