Question

he Kb for an amine is 4.150 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.180? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

pH = 9.180

pOH = 14 - pH = 4.82

pKb = -logKb = -log(4.150 x 10^-5) = 4.382

Using Hendersen-Hasselbalck equation,

pOH = pKb + log([BH+]/[B])

4.82 = 4.382 + log([BH+]/[B])

[BH+] = 2.74[B]

Total [B] = [BH+] + [B]

percentage of [BH+] = [2.74[B]/([2.74[B] + [B]) ] x 100

                                 = (2.74[B]/3.74[B]) x 100

                                 = 73.26%

therefore, 73.26% of amine is in protonated form.