Question

The Kb for an amine is 1.448 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.810? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

Use the Henderson-Hasselbalch equation: pOH = pKb + log[BH⁺]/[B], where BH⁺ represents the protonated amine (conjugate base) and B represents the amine (weak base).

For a monoprotic acid or base the expression: pKa + pKb = pKw, and pH + pOH = pKw are helpful.

pOH = 14.00 - 9.810 = 4.19
pKb = -log(1.448 x 10^-5) = 4.84

4.19 = 4.84 + log[BH⁺]/[B] = 0.224

You need to find [BH⁺]/[B] so take the antilog of both sides.

Antilog(log [BH⁺]/[B]) = [BH⁺]/[B]
antilog 0.224 = 10^0.224 = 1.675

[BH⁺]/[B] = 1.675

[BH⁺] = 1.675[B]

%BH⁺ protonated = {[BH⁺]/([B] + [BH⁺])}*100% = {1.675[B]/([B] + 1.675[B]}*100%
% BH⁺ protonated = (1.675/2.675)*100% = 62.6% protonated