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

`K_(a) = 5.84 xx 10^(-5), pK_(a) = 4.2336`
`pH = pK_(a) + log ["Ind"^(Θ)//"Hin"]`
`4.84 = 4.2336 + log ["Ind"^(Θ)//"Hin"]`
`:. Log ["Ind"^(Θ)//"Hin"] = 0.6064`,
`["Ind"^(Θ)//"Hin"] = 4.04`
`%` of basic form `= (["Ind"^(Θ)])/(["Ind"^(Θ)]+["HIn"]) xx 100`
`["HIn"//"Ind"^(Θ)] = 1.4//04...(i)`
Add 1 to both sides of equaiton (i)
`(["HIn"])/(["Ind"^(Θ)])+1 = (1)/(4.04) +1:. (["HIn"]+["Ind"^(Θ)])/(["Ind"^(Θ)]) =(5.04)/(4.04)`
`(["Ind"^(Θ)])/(["HIn"]+["Ind"^(Θ)]) xx100 = (4.04)/(5.04) xx 100 = 80%`