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JEE Main levelNumerical answerDifficulty 2 of 3ia138

An acid-base indicator HIn has $p K_{in} = 5.3$ . At what pH is $80%$ of the indicator present in the ionised form $In X^{-}$ ? (Use $lo g 2 = 0.301$ and give two decimals.)

Numerical answer type, tolerance 0.01.

Answer: 5.902

pH = p K_{in} + lo g ([In X^{-}] / [HIn]) = 5.3 + lo g (80/20) = 5.3 + 2 lo g 2 = 5.3 + 0.602 = 5.90

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