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The indicator constant of an indicator is $10^{- 4}$ . If pH of solution in which this indicator is present is 3.5, which of the following facts will be correct?

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Ionized indicator is present in the larger amount

Unionized indicator is present in the larger amount

Both ionized and unionized indicator are present in equal amounts

Percentage of indicator present in ionized form cannot be predicted

answer is B.

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Detailed Solution

$HIn ⇌ H^{+} +$ + In; Since $pH < p K_{Hin}^{\circ},$ the solution contains larger $H^{+}$ than that obtained from ionization of $HIn$ alone. Hence, the above equilibrium is shift to left, thus more of unionized indicator is present.

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