The degree of ionisation of a `0.1 M` bromoacetic acid solution is `0.13`. Calculate the `pH` of the solution and the `pK_(a)` of bromoacetic acid.

Explanation

`K_(a) =((Calpha)xx(Calpha))/(C(1-alpha))=(Calpha^(2))/(1-alpha)=Calpha^(2) =(0.1) xx (0.132)^(2) =1.74 xx 10^(-3)`
`pK_(a) =- log (1.74 xx 10^(-3)) =(3- log 1.74) =(3-0.2405)=2.76`
`[H^(+)] =Calpha =(0.1) xx(0.132) =1.32 xx 10^(-2)M`
`pH =- log (1.32 xx10^(-2)) =(2 -log 1.32) =(2-0.1206) =1.88`