Question

An indicator is a weak acid and pH range of its colour is 3 to 5. If the neutral point of the indicator lies in the centre of the $[H^{+}]$ corresponding to given pH range, then pH at the equivalence point is

Difficult

Solution

The neutral point is at the centre of the $H^{+}$ ion concentration $[H^{+}] at (pH = 3) is 10^{- 3}$
$[H^{+}] at (pH = 5) is 10^{- 5}$
Thus, $[H^{+}] at the centre = \frac{10^{- 3} + 10^{- 5}}{2} = \frac{101 \times 10^{- 5}}{2}$
$∴ pH = - \log (\frac{101 \times 10^{- 5}}{2}) = 3.3$

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