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

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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 $K_{a} = 2 \times 10^{-5}$,so $pK_{a} = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.30 = 4.7$.Concentration $C = 0.2 \ M = 2 \times 10^{-1} \ M$.$pH = \frac{1}{2} (4.7 - \log(2 \times 10^{-1}))$.$pH = \frac{1}{2} (4.7 - (\log 2 + \log 10^{-1}))$.$pH = \frac{1}{2} (4.7 - (0.30 - 1)) = \frac{1}{2} (4.7 - (-0.7)) = \frac{1}{2} (5.4) = 2.7$.$pH = 2.7 = 27 \times 10^{-1}$.Thus,the value is $27$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of weak Acids and weak Bases

Medium

$K_{a}$ 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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A
$26$
B
$25$
C
$27$
D
$29$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of weak Acids and weak Bases

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$K_{a}$ 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 $\log 2 = 0.30$]

Similar Questions

Nicotinic acid $(K_a = 10^{-5})$ is represented by the formula $HNiC$. Calculate the percentage of dissociation in a solution containing $0.1 \ mol$ of nicotinic acid in $2 \ L$ of solution.

The $pK_a$ of a weak acid is $5.85$. The concentrations of the acid and its conjugate base are equal at a $pH$ of (in $.85$)

The percent dissociation of a weak monobasic acid is $3 \%$ in its $0.02 \ M$ solution. What is the dissociation constant of the acid?

For the ionization of a weak acid $HA$ as $HA \rightleftharpoons H^{+} + A^{-}$,the $pH$ of a $1.0 \ M$ solution is $5$. The dissociation constant $K_a$ is:

Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^{+}$ ions is $3.4 \times 10^{-4} \ M$. Find the initial concentration of $CH_3COOH$ molecules.

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