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An acid-base indicator HIn (colourless in its un-ionised form) shows its pink ionised colour only when at least 25% of the indicator exists in the ionised form In⁻. Given pKa(HIn) = 9.0 and log 2 = 0.3, what happens when this indicator is added to a solution of pH = 9.6?

1. Pink colour will be visible
2. Pink colour will not be visible
3. Percentage of ionised form is less than 25%
4. Percentage of ionised form is more than 25%

Correct answer: Pink colour will be visible

Solution

Using Henderson-Hasselbalch: [In⁻]/[HIn] = 10^(pH - pKa) = 10⁰.6 = 4. The percentage of ionised form = 4/(4+1) * 100 = 80%, which exceeds 25%, so pink colour is visible.

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