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JEE Main levelSingle correct optionDifficulty 2 of 3ia009

A 0.10 M solution of the sodium salt $NaA$ of a weak monoprotic acid HA has pH 9.00 at 298 K. The value of $K_{a}$ of HA is

A $1.0 \times 1 0^{- 5}$
B $1.0 \times 1 0^{- 9}$
C $1.0 \times 1 0^{- 4}$
D $1.0 \times 1 0^{- 3}$

Answer: A

[OH X^{-}] = 1 0^{- 5}

M.

K_{h} = [OH X^{-}]^{2} / c = 1 0^{- 10} /0.10 = 1.0 e - 09

. Then

K_{a} = K_{w} / K_{h} = 1.0 e - 05

. The choice

K_{h}

is the hydrolysis constant itself, not

K_{a}

.

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