0.05 M NH4CN is prepared. Given pK_a(HCN) = 9.3 and pK_b(NH3) = 4.76, and using pH = 7 + tfrac12(pK_a - pK_b) for the salt of a weak acid and a weak base, the pH is

JEE Main levelSingle correct optionDifficulty 2 of 3ia116

$0.05 M$ $NH X_{4} CN$ is prepared. Given $p K_{a} (HCN) = 9.3$ and $p K_{b} (NH X_{3}) = 4.76$ , and using $pH = 7 + \frac{1}{2} (p K_{a} - p K_{b})$ for the salt of a weak acid and a weak base, the pH is

For a salt of a weak acid and a weak base the pH is independent of concentration:

pH = 7 + \frac{1}{2} (9.3 - 4.76) = 9.27

. Since

K_{a} (HCN) < K_{b} (NH X_{3})

the solution is basic.