Question

The Kb for an amine is 3.877 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.967? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

we have below equation to be used:

pH = -log [H+]

9.967 = -log [H+]

log [H+] = -9.967

[H+] = 10^(-9.967)

[H+] = 1.079*10^-10 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(1.079*10^-10)

[OH-] = 9.268*10^-5 M

B + H2O —> BH+ + OH-

C       0   0  

C-x       x   x

Here:

x = 9.268*10^-5 M

Kb = x*x/(C-x)

3.877*10^-5 = (9.268*10^-5 * 9.268*10^-5) / (C- 9.268*10^-5)

(C- 9.268*10^-5) = 2.216*10^-4

C = 3.142*10^-4

% protonation = x*100/C

= 9.268*10^-5*100/3.142*10^-4

= 29.49 %

Answer: 29.49 %