At 298 K, K_b of ammonia is 1.74 x 10^-5 and the pK_a of the anilinium ion C6H5NH3+ is 4.60. Take pK_w = 14.00 and log_10 1.74 = 0.2406. Calculate pK_b(aniline) - pK_b(ammonia), correct to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3nc030

At 298 K, $K_{b}$ of ammonia is $1.74 \times 1 0^{- 5}$ and the $p K_{a}$ of the anilinium ion $C X_{6} H X_{5} NH X_{3} X^{+}$ is 4.60. Take $p K_{w} = 14.00$ and $lo g_{10} 1.74 = 0.2406$ . Calculate $p K_{b} (aniline) - p K_{b} (ammonia)$ , correct to two decimal places.

Numerical answer type, tolerance 0.02.

For a base and its conjugate acid

p K_{a} + p K_{b} = p K_{w}

, so

p K_{b} (aniline) = 14.00 - 4.60 = 9.40

. For ammonia

p K_{b} = - lo g (1.74 \times 1 0^{- 5}) = 5 - 0.2406 = 4.76

. The difference is

9.40 - 4.76 = 4.64

(positive, since aniline is the weaker base).