Equal volumes of 0.20 M NH3 and 0.20 M HCN are mixed at 298 K. K_b(NH3) = 1.8x10^-5, K_a(HCN) = 4.0x10^-10. Treat the resulting salt solution with pH = 7 + tfrac12(pK_a - pK_b). The pH is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia013

Equal volumes of 0.20 M $NH X_{3}$ and 0.20 M HCN are mixed at 298 K. $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{a} (HCN) = 4.0 \times 1 0^{- 10}$ . Treat the resulting salt solution with $pH = 7 + \frac{1}{2} (p K_{a} - p K_{b})$ . The pH is closest to

The mixture contains the salt

NH X_{4} CN

(0.10 M after mixing) with no excess acid or base.

p K_{a} = 9.398

p K_{b} = 4.745

pH = 7 + \frac{1}{2} (9.398 - 4.745) = 9.33

. The answer is independent of concentration.