An acid HA reacts with phenoxide ion: HA + C6H5O- A- + C6H5OH, with equilibrium constant K = 4.0 x 10^5. Given pK_a(C6H5OH) = 10.0, the pK_a of HA is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3g2127

An acid HA reacts with phenoxide ion: $HA + C X_{6} H X_{5} O X^{-} A X^{-} + C X_{6} H X_{5} OH$ , with equilibrium constant $K = 4.0 \times 1 0^{5}$ . Given $p K_{a} (C X_{6} H X_{5} OH) = 10.0$ , the $p K_{a}$ of HA is closest to

lo g K = p K_{a} (C X_{6} H X_{5} OH) - p K_{a} (HA)

lo g (4.0 \times 1 0^{5}) = 5.60

. So

p K_{a} (HA) = 10.0 - 5.60 = 4.40 \approx 4.4

, i.e. a carboxylic-acid-like strength.