0.10 M solutions of the sodium salts $NaA$ and $NaB$ of two weak acids HA and HB have pH 9.00 and 10.00 respectively at 298 K. The ratio $K_{a} (HA) / K_{a} (HB)$ is

Answer: D

K_{h} = [OH X^{-}]^{2} / c

. For NaA,

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

K_{h} = 1 e - 09

; for NaB,

[OH X^{-}] = 1 0^{- 4}

K_{h} = 1 e - 07

. Since

K_{a} = K_{w} / K_{h}

K_{a} (HA) / K_{a} (HB) = K_{h} (B) / K_{h} (A) = 100

. The ratio of [OH-] alone (10) is not the answer because

K_{h}

depends on the square.