Question

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.
Calculate the pH of a solution containing a caffeine concentration of 156 mgL−1 .
Express your answer to one decimal place.

Answer 1

1st find the concentration of caffeine

Let the volume be 1 L

Molar mass of C8H10N4O2,

MM = 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(C8H10N4O2)= 156 mg

= 0.156 g

use:

number of mol of C8H10N4O2,

n = mass of C8H10N4O2/molar mass of C8H10N4O2

=(0.156 g)/(1.942*10^2 g/mol)

= 8.033*10^-4 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 8.033*10^-4/1

= 8.033*10^-4 M

use:

pKb = -log Kb

10.4= -log Kb

Kb = 3.981*10^-11

C8H10N4O2 dissociates as:

C8H10N4O2 +H2O -----> C8H10N4O2H+ + OH-

8.033*10^-4 0 0

8.033*10^-4-x x x

Kb = [C8H10N4O2H+][OH-]/[C8H10N4O2]

Kb = x*x/(c-x)

Assuming 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)*8.033*10^-4) = 1.788*10^-7

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

so, x = 1.788*10^-7 M

So, [OH-] = x = 1.788*10^-7 M

use:

pOH = -log [OH-]

= -log (1.788*10^-7)

= 6.7476

use:

PH = 14 - pOH

= 14 - 6.7476

= 7.2524

Answer: 7.3