Question

The Kb for an amine is 4.283 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.350? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

Denote the amine as B Let y % be the percentage at B beins protonated, and C be the concentration of B solution In the solution: p^H = 9.350 p^OH = pk_w - p^H = 14.0 - 9.350 = 4.65 Then [OH^-] = 10^-4.65 M B(aq) + H_2 O rightleftharpoons HB^+ (aq) + OH^-(aq) k_b = 4.283 times 10^-5 At equilibrium: (100-y) % cm - y % cm - 10^-4.65 m At equilibrium: K_b = [HB^+] [OH^-]/[B] 4.283 times 10^-5 = (y % 6) times 10^-4.65/(100-y) % c 4.283 times 10^-5 = y times 10^-4.65/(100-y) (4.283 times 10^-3) - (4.283 times 10^-5)y = 10^-4.65 y [(4.283 times 10^-5) + 10^-4.65]y = 4.283 times 10^-3 [(4.283 times 10^-5) + (2.238 times 10^-5)]y = 4.283 times 10^-3 (6.521 times 10^-5)y = 4.283 times 10^-3 y = 65.68 percentage of the amine protonated = 65.68%