Q2 / 102 · NTA Abhyas (JEE Main) 2020 · Chemistry · Ionic Equilibrium

Calculate pH at which an acid indicator HIn with concentration 0.1 M changes its colour ( $K_{a}$ for HIn $= 1 \times 1 0^{- 5}$ ).

Correct answer: 5

Solution

Indicator equation is

p H = p K_{I n} + \log \frac{[I n^{-}]}{[H I n]}

Colour change takes place when

[I n^{-}] = [H I n]

i.e., when

p H = p K_{I n} = 5

.

Hence, the correct answer is 5.

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