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.
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
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