Bromophenol blue is an indicator with a `K_(a)` value of `5.84 xx 10^(-5)`. What is the percentage of this indicator in its basic form at a `pH` of `4.84`?

Explanation

`HBPh hArr H^(+) + BPh^(-)`
`K_(a) = ([H^(+)][BPh^(-)])/([HBPh]) , ` when `BPh^(-)` = HBPh , indicator will work . Thus
` [H^(+)] = 5.84 xx10^(-5)`
` :. pH = 4.2336`
Also if pH = 4.84
or `[H^(+] = 1.44 xx10^(-5) ` , then
`K_(a) = ([H^(+)] [BPh^(-)])/([HBPh])`
or `5.84 xx10^(-5) = (1.44 xx10^(-5)Calpha)/(C(1-alpha))`
`alpha = 0.8 ` or 80 %