Consider the titration of 100.0 mL of 0.200 M acetic acid by 0.100 m KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added: a) 0.0 mL b) 50.0 mL c) 100.0 mL d) 200.0 mL e) 250.0 mL
1 Answer
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Ka of acetic acid = 1.8 x 10^-5
pKa = 4.74
a)
1.8 x 10^-5 = x^2 / 0.200-x
x = [H+]= 0.00190 M
pH = 2.72
b)
moles acetic acid = 0.100 L x 0.200 M = 0.0200
moles OH- added = 0.0500 L x 0.100 M=0.005
moles acetic acid in excess = 0.0200 – 0.005=0.015
moles acetate = 0.005
pH = 4.74 + log 0.005/0.015=5.26
c)
moles OH- = 0.100 L x 0.100 M = 0.0100
moles acetic acid in excess = 0.0200 – 0.0100 = 0.0100
moles acetate = 0.0100
pH = 4.74 + log 0.0100/0.0100= 4.74
d)
moles OH- = 0.200 L x 0.100 M= 0.0200
moles acetate = 0.0200
total volume = 0.300 L
concentration acetate = 0.0200/0.300=0.0667 M
CH3COO- + H2O <=> CH3COOH + OH-
Kb = 5.6 x 10^-10 =x^2/ 0.0667-x
x = [OH-]= 6.11 x 10^-6 M
pOH = 5.21
pH = 8.79
e)
moles OH- = 0.250 L x 0.100 M= 0.0250
moles OH- in excess = 0.0250- 0.0200 = 0.0050
total volume = 0.350 L
[OH-]= 0.0050/0.350=0.0143 M
pOH = 1.85
pH = 12.15
