Question

The Kb for an amine is 1.519 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.764? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibirum Kb:

Kb = [HB+][OH-]/[B]

let "x" be OH- in solution

in equilibrium due to sotichiometry:

[HB+]= x= [OH-]

Acocunt for the dissolved base in solution vs. not in solution:

[OH-] = 10^-(14-pH) = 10^-(14-9.764) = 0.00005807644

[B] = M- 0.00005807644

Substitute in Kb

Kb = [HB+][OH-]/[B]

1.519*10^-5 = (0.00005807644*0.00005807644)/(M-0.00005807644)

solv efor M

M = (0.00005807644^2) /(1.519*10^-5) + 0.00005807644 = 0.00028012205 M

then

% ion = [OH-] / M * 100 = 0.00005807644/0.00028012205 *100 = 20.73 %