The degree of ionization of a $0.1 \, M$ bromoacetic acid solution is $0.132$. Calculate the $pH$ of the solution and the $pK_{a}$ of bromoacetic acid.
(N/A) Degree of ionization,$\alpha = 0.132$
Concentration,$c = 0.1 \, M$
Concentration of $H_{3}O^{+} = c \times \alpha = 0.1 \times 0.132 = 0.0132 \, M$
$pH = -\log[H_{3}O^{+}] = -\log(0.0132) \approx 1.88$
For a weak acid,$K_{a} = c \alpha^{2} / (1 - \alpha)$. Since $\alpha$ is small,$K_{a} \approx c \alpha^{2} = 0.1 \times (0.132)^{2} = 0.1 \times 0.017424 = 0.0017424$
$pK_{a} = -\log(K_{a}) = -\log(0.0017424) \approx 2.76$
Concentration,$c = 0.1 \, M$
Concentration of $H_{3}O^{+} = c \times \alpha = 0.1 \times 0.132 = 0.0132 \, M$
$pH = -\log[H_{3}O^{+}] = -\log(0.0132) \approx 1.88$
For a weak acid,$K_{a} = c \alpha^{2} / (1 - \alpha)$. Since $\alpha$ is small,$K_{a} \approx c \alpha^{2} = 0.1 \times (0.132)^{2} = 0.1 \times 0.017424 = 0.0017424$
$pK_{a} = -\log(K_{a}) = -\log(0.0017424) \approx 2.76$
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