The degree of ionization of a 0.1M bromoacetic acid solution is 0.132

Degree of ionization, α = 0.132

Concentration, c = 0.1 M

Thus, the concentration of $\mathrm{H}_{3} \mathrm{O}^{+}=c . a$

= 0.1 × 0.132

= 0.0132

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$

$=-\log (0.0132)$

$=1.879: 1.88$

Now,

$K_{a}=C \alpha^{2}$

$=0.1 \times(0.132)^{2}$

$K_{a}=.0017$

$p K_{a}=2.75$