Question

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 470 mg/L .

Answer 1

Molar mass of C8H10N4O2 = 8*MM(C) + 10*MM(H) + 4*MM(N) + 2*MM(O)

= 8*12.01 + 10*1.008 + 4*14.01 + 2*16.0

= 194.2 g/mol

mass of C8H10N4O2 = 470 mg

= 0.47 g [using conversion 1 g = 1000 mg]

we have below equation to be used:

number of mol of C8H10N4O2,

n = mass of C8H10N4O2/molar mass of C8H10N4O2

=(0.47 g)/(194.2 g/mol)

= 2.42*10^-3 mol

volume , V = 1.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2.42*10^-3/1

= 2.42*10^-3 M

we have below equation to be used:

pKb = -log Kb

10.4= -log Kb

log Kb = -10.4

K = 10^(-10.4)

Kb = 3.981*10^-11

Lets write the dissociation equation of C8H10N4O2

C8H10N4O2 +H2O -----> C8H10N5O2+ + OH-

2.42*10^-3 0 0

2.42*10^-3-x x x

Kb = [C8H10N5O2+][OH-]/[C8H10N4O2]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((3.981*10^-11)*2.42*10^-3) = 3.104*10^-7

since c is much greater than x, our assumption is correct

so, x = 3.104*10^-7 M

So, [OH-] = x = 3.104*10^-7 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.104*10^-7)

= 6.51

we have below equation to be used:

PH = 14 - pOH

= 14 - 6.51

= 7.49

Answer: 7.49