A solution is 0.10 M in CH3COOH (K_a = 1.8 x 10^-5) and 0.10 M in HCN (K_a = 6.2 x 10^-10). Find [CN-] in the solution (in M). Neglect the dissociation of water.

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia105

A solution is $0.10 M$ in $CH X_{3} COOH$ ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) and $0.10 M$ in $HCN$ ( $K_{a} = 6.2 \times 1 0^{- 10}$ ). Find $[CN X^{-}]$ in the solution (in $M$ ). Neglect the dissociation of water.

Acetic acid fixes

[H X^{+}] = K_{a 1} c_{1} + K_{a 2} c_{2} \approx 1.34 \times 1 0^{- 3} M

(HCN contributes negligibly). For HCN,

[CN X^{-}] = K_{a} [HCN] / [H X^{+}] = 6.2 \times 1 0^{- 10} \times 0.10/ (1.34 \times 1 0^{- 3}) = 4.6 \times 1 0^{- 8} M

. The value

K_{a} c

for HCN alone (

7.9 \times 1 0^{- 6} M

) ignores the common

H X^{+}