Homework Answers
pKa of acetic acid = -log 1.8x10-5 = 4.75
CH3COOH + HCl -------------------------> CH3COONa + H2O
20x0.1 0 0 ---- initial mmoles
I) initial pH
pH of acid = 1/2[pKa -log C] = 1/2[4.75 -log 0.1]
=2.875
II)pKa = 4.75
CH3COOH + HCl -------------------------> CH3COONa + H2O
20x0.1=2 0 0 ---- initial mmoles
1.5 0 0.5 ------ at 1/4 equivalence
pH of this buffer = pKa +log [conjugate base]/[acid] Hendersen equatio
= 4.75 + log 0.5/1.5
= 4.27
CH3COOH + HCl -------------------------> CH3COONa + H2O
20x0.1=2 0 0 ---- initial mmoles
0.5 0 1.5 ------ at 3/4 equivalence
pH of this buffer = pKa +log [conjugate base]/[acid] Hendersen equatio
= 4.75 + log 1.5/0.5
= 5.227
III)
At equivalence
CH3COOH + HCl -------------------------> CH3COONa + H2O
20x0.1=2 0 0 ---- initial mmoles
0 0 2 ------ at equivalence
[salt] formed = mmoles/ total volume = 2 /(20+20) = 0.05 M
pH of salt solution of strong base and weak acid = 1/2[pKw + pKa + log C]
= 1/2[14 + 4.75 + log 0.05] = 8.7244
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