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JEE Advanced levelSingle correct optionDifficulty 3 of 3g2040

At pH 8.15, what is the ratio $[anion] / [acid]$ for 4-nitrophenol ( $p K_{a} = 7.15$ ) divided by the same ratio for phenol ( $p K_{a} = 10.0$ )?

A $1.4 \times 1 0^{- 3}$
B $7.1 \times 1 0^{1}$
C $10$
D $7.1 \times 1 0^{2}$

Answer: D

From the Henderson-Hasselbalch equation,

[A X^{-}] / [HA] = 1 0^{pH - p K_{a}}

. For 4-nitrophenol:

1 0^{1.00} = 10

. For phenol:

1 0^{- 1.85} = 0.0141

. The quotient is

10/0.0141 = 708

, i.e.

7.1 \times 1 0^{2}

(equal to

1 0^{2.85}

, the difference in pKa values).

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