A base type indicator is half in ionised form when pH is 7.2. If the ratio of unionised form to ionised form is 1 : 5, let pH of the solution is pH 1 . With the same pH of solution, indicator is altered (not its type) such that the ratio of unionised form to ionised form is 1 : 4. Let pH 2 be the pH of solution when 50% of new indicator is in ionised form. Then
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(b,d) As indicator (InOH) is base type indicator, so pK InOH = pOH at half ionisation point = 6.8.
pOH 1 = pK InOH + log 10 
= 7.5. ∴ pH 1 = 6.5
pOH 1 = pK In*OH + log 10 
∴ pK In*OH = 6.9 = pOH 2
∴ pH 2 = 7.1.
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