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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`? |
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Answer» `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%` |
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