If very small amount of phenolphthalein is added to $0.15 m o l {l i t r e}^{- 1}$ solution of sodium benzoate, what fraction of the indicator will exist in the coloured form? State any assumption that you make.
$K_{a}$ (Benzoic acid) $= 6.2 \times 10^{- 5}, K_{w} (H_{2} O) = 1 \times 10^{- 14}$
$K_{l n}$ (Phenolphthalein) $3.16 \times 10^{- 10}$

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Solution

Formula for pH of salt hydrolysis:
$p H = \frac{1}{2} [p K_{w} + p K_{a} + log C]$
$= \frac{1}{2} [14 + {log}_{10} 6.2 \times 10^{- 5} + log 0.15] = 8.6918$
Formula for pH of indicator:
$p H = p K_{l n} + {log}_{10} \frac{[{I n}^{-}]}{[H I n]}$
$8.6918 = - {log}_{10} 3.16 \times 10^{- 10} + {log}_{10} \frac{[{I n}^{-}]}{[H I n]}$
$0.16 = \frac{[{I n}^{-}]}{[H I n]} =$ Fraction of indicator in coloured form.

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