# Consider the titration of 100.0 mL of 0.200 M acetic acid by 0.100 m KOH. Calculate the pH of the resulting…?

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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

• 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

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