The degree of ionization of a $0.1 M$ bromoacetic acid solution is $0.132$ . Calculate the $p H$ of the solution and the $p K_{a}$ of bromoacetic acid.

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Solution

$[H^{+}] = c α = 0.1 \times 0.132 = 0.0132 M$
$p H = - l o g [H^{+}] = - l o g 0.0132 = 1.88$
The acid dissociation constant is $K_{a} = \frac{c α^{2}}{1 - α} = \frac{0.1 \times (0.132)^{2}}{1 - 0.132} = 2.01 \times 10^{- 3}$ .
$p K_{a} = - l o g K_{a} = - l o g (2.01 \times 10^{- 3}) \approx 2.7$

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

[H+]=cα=0.1×0.132=0.0132MpH=−log[H+]=−log0.0132=1.88The acid dissociation constant is Ka=cα21−α=0.1×(0.132)21−0.132=2.01×10−3.pKa=−logKa=−log(2.01×10−3)≈2.7

The degree of ionization of a $0.1 M$ bromoacetic acid solution is $0.132$ . Calculate the $p H$ of the solution and the $p K_{a}$ of bromoacetic acid.