An indicator has `pK_(In)=5.3`. In a certain titration, this indicator is found to be `80%` ionized in its acid form. Thus, `pH` of the solution is

Options

A. `4.7`
B. `5.3`
C. `5.9`
D. `6.2`

Answer

A

Explanation

`underset(Acid form)(HIn)hArrunderset(Basic form)(H^(+))+In^(-)`
Thus, acid form `[HIn] =0.80 M`
`[In^(-)]=0.20 M`
`pH=pK_(In)+log``([In^(-)])/([HIn])=5.3+log``(0.2)/(0.8)`
`=5.3+log(1)/(4)=5.3-0.60=4.7`