Question

The pOH of an aqueous solution of 0.520 M caffeine (a weak base with the formula C8H10N4O2) is

Answer 1

Kb value of caffeine = 3.98 * 10^-11

Assume formula of caffeine is B:

Consider a hydrolysis reaction of caffeine

:B(aq) + H2O (l) ----> OH^-(aq) + BH^+(aq)

IC. : 0.520 0 0

CIC : -x +x +x

FC : 0.520 - x x. x

Kb = [OH^-][BH^+] / [B:] = (x)(x) / (0.520 - x) = x^2 / (0.520 - x)

Since Kb is very small hence 0.520 - x ~ 0.520

x^2 = 0.520 * Kb = 0.520 * 3.98 * 10^-11 = 2.07 * 10^-11 = 20.7 * 10^-12

x = √20.7 * 10^-12 = 4.55 * 10^-6

Since [OH^-] = x M hence [OH^-] = 4.55 * 10^-6 M

[OH^-] = 4.55 * 10^-6 M

pOH = - log [OH^-] = - log (4.55 * 10^-6) = - log 4.55 + 6 log 10 = - 0.66 + 6 * 1 = 5.34

pOH = 5.34