The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

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Solution

Degree of ionization, $α = 0.132$
Concentration, c = 0.1 M
Thus, the concentration of $H_{3} O^{+} = c . α$
$= 0.1 \times 0.132 = 0.0132 p H = - l o g [H^{+}] = - l o g (0.0132) = 1.879 (o r) 1.88$
Now,
$K_{a} = C α^{2} = 0.1 \times (0.132)^{2} K_{a} = .0017 p K_{a} = 2.75$

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The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Degree of ionization, α=0.132 Concentration, c = 0.1 M Thus, the concentration of H3O+=c.α =0.1×0.132=0.0132pH=−log[H+]=−log(0.0132)=1.879(or)1.88 Now, Ka=Cα2=0.1×(0.132)2Ka=.0017pKa=2.75

Degree of ionization, $α = 0.132$
Concentration, c = 0.1 M
Thus, the concentration of $H_{3} O^{+} = c . α$
$= 0.1 \times 0.132 = 0.0132 p H = - l o g [H^{+}] = - l o g (0.0132) = 1.879 (o r) 1.88$
Now,
$K_{a} = C α^{2} = 0.1 \times (0.132)^{2} K_{a} = .0017 p K_{a} = 2.75$