Correct Answer - D
Consider the equilibrium expression
`HIn hArr H^(+)+In^(-)`
`K_(In)=([H^(+)][In^(-)])/([HIn])=[H^(+)] ([In^(-)])/([HIn])`
Taking negative log of both sides, we have
`-log K_(In)= -log[H^(+)]- log.([In^(-)])/([H In])`
`pK_(In)=pH-log. ([In^(-)])/([H In])`
or `log. ([In^(-)])/([H In]) = pH-pK_(In)`