Question

The Kb for an amine is 1.623 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.396? (Assume that all OH– came from the reaction of B with H2O

Answer 1

pkb of amine = -log(kb)   = -log(1.623*10^-5) = 4.8

pH+pOH = 14

pOH of amine = 1/2(pkb-logC)

(14-9.396) = 1/2(4.8-logC)

C = concentration of amine = 3.91*10^-5 M

[OH-] = CX

x = degree of dissociation of amine = ?

[OH-] = 10^-pOH   =   10^-(14-9.396) = 2.5*10^-5 M

x = 2.5*10^-5/(3.91*10^-5) = 0.64

% of dissociation = x*100

                          = 0.64*100 = 64%

percentage of the amine is protonated = 64%