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The indicator constant of an indicator is 3 .0. The percentage of the indicator in the unionized form in a solution of $pH = 2.6$ is

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40%

55%

62%

71 %

answer is D.

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Detailed Solution

$2.6 = 3.0 + \log \frac{[{In}^{-}]}{[HIn]}$ Hence $\frac{[{In}^{-}]}{[HIn]} = antilog (- 0.4)$

or $\frac{[{In}^{-}]}{[HIn]} = 0.40$ or $\frac{[{In}^{-}] + [HIn]}{[HIn]} = 1.40$

Per cent of $HIn = (1 / 1.40) (100) = 71.4 %$

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