The degree of ionisation of a 0.1(M) bromoacetic acid solution is 0.132. calculate the pH of the solution and the `pK_(a)` of bromoacetic acid.

Explanation

`BrCH_(2)COOH(aq)hArr BrCH_(2)COO^(-)(aq)+H^(+)(aq)`
the degre of ionisation `(alpha)` of `BrCH_(2)COOH` in its 0.1(M) solution=0.132
So, in the solution,
`[H^(+)]=alphaxx0.1(M)=0.132xxx0.1(M)=0.0132(M)`
`therefore pH` of the solution`=-log_(10)[H^(+)]=-log_(10)(0.0132)=1.88`
FOr a solution of a weak monoprotic acid like
`BrCH_(2)COOH,pH=(1)/(2)pK_(a)-(1)/(2)logc`
`therefore 1.88=(1)/(2)pK_(a)-(1)/(2)log0.1 or, pK_(a)=2.76`