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Consider the titration of 100.0 mL of 0.200M acetic acid (CH3COOH, Ka=1.8 x10-5) by 0.100M KOH....

Consider the titration of 100.0 mL of 0.200M acetic acid (CH3COOH, Ka=1.8 x10-5) by 0.100M KOH. Calculate the pH of the resulting solution after 50.0 mL 0.100M KOH is added.

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Answer #1

Ka of acetic acid = 1.8 x 10^-5
pKa = 4.74

a) when 0.00 ml of KOH is added
1.8 x 10^-5 = x^2 / 0.200-x
x = [H+]= 0.00190 M
pH = 2.72

b) when 50.00 ml of KOH is added
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

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