100 mL of 0.20 M NH4Cl is mixed with 100 mL of 0.20 M NaCN. K_a(HCN) = 4.9x 10^-10, K_b(NH3) = 1.8x 10^-5. Assume that the equilibrium NH4+ + CN- NH3 + HCN is the only significant one, so that [NH3] = [HCN] and [NH4+] approx [CN-]. The pH of the mixture is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia122

$100 mL$ of $0.20 M$ $NH X_{4} Cl$ is mixed with $100 mL$ of $0.20 M$ $NaCN$ . $K_{a} (HCN) = 4.9 \times 1 0^{- 10}$ , $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ . Assume that the equilibrium $NH X_{4} X^{+} + CN X^{-} NH X_{3} + HCN$ is the only significant one, so that $[NH X_{3}] = [HCN]$ and $[NH X_{4} X^{+}] \approx [CN X^{-}]$ . The pH of the mixture is

With

[NH X_{3}] = [HCN]

and

[NH X_{4} X^{+}] = [CN X^{-}]

K_{a} = [H X^{+}] [CN X^{-}] / [HCN]

and

K_{b} = [NH X_{4} X^{+}] [OH X^{-}] / [NH X_{3}]

give

[H X^{+}]^{2} = K_{a} K_{w} / K_{b}

pH = 7 + \frac{1}{2} (9.310 - 4.745) = 9.28

; it does not depend on concentration.