Question

The Kb for an amine is 8.695 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.207? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

pH = 9.207.

pH + pOH = 14

pOH = 14 - pH = 14 - 9.207

pOH = 4.793

[HO^-] = 10^-pOH .

[HO^-] = 10^-4.793.

[HO^-] = 1.61 x 10^-5 M.

Amine (A) undergo rotonation in aqueous medium as,

A + H-OH <--------> AH⁺ (aq) + HO^- (aq)

For which,

Kb = [AH⁺][HO^-] / [A] = 8.695 x 10^-5. ---------- (1)

We have, [HO^-] = 1.61 x 10^-5 M calculated above.

As per reaction stoichiometry,

[AH⁺] = [HO^-] = 1.61 x 10^-5 M

Let initially [A] = X so at equilibrium, [A] = X - (1.61x10^-5)

Using these equilibrium concentration in eq.(1)

(1.61x10^-5)(1.61x10^-5) / [X - (1.61x10^-5)] = 8.695 x 10^-5.

X - (1.61x10^-5) = (1.61x10^-5)(1.61x10^-5) / 8.695 x 10^-5.

X - (1.61x10^-5) = 0.3 x 10^-5.

X = 0.3 x 10^-5 + 1.61 x 10^-5.

X = 1.91 x 10^-5 M

i.e. [A] = X = 1.91 x 10^-5 M

Then,

% protonation of Amine = ([AH⁺] / [A]) x 100

= {(1.61 x 10^-5 )/(1.91 x 10^-5 )} x 100

= (1.61/1.91)x100

= 84.29

84.29 % Amine is protonated.

=======================XXXXXXXXXXXXXXXXXXXXXX=======================