A solution is 0.10 M in CH3COOH (K_a = 1.8x10^-5) and 0.10 M in HCN (K_a = 6.2x10^-10). The concentration of CN- is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia215

A solution is $0.10 M$ in $CH X_{3} COOH$ ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) and $0.10 M$ in $HCN$ ( $K_{a} = 6.2 \times 1 0^{- 10}$ ). The concentration of $CN X^{-}$ is

Acetic acid fixes

[H X^{+}] = 1.8 \times 1 0^{- 5} \times 0.10 = 1.34 \times 1 0^{- 3} M

(the contribution of

HCN

is negligible). Then

[CN X^{-}] = K_{a} [HCN] / [H X^{+}] = 6.2 \times 1 0^{- 10} \times 0.10/1.34 \times 1 0^{- 3} = 4.62 \times 1 0^{- 8} M

. Treating

HCN

alone would give

7.87 \times 1 0^{- 6}

, which ignores the common ion.