What is the pH at the equivalence point in the titration of a 18.7 mL sample of a 0.405 M aqueous acetic acidsolution wi...
What is the pH at the equivalence point in the titration of a 18.7 mL sample of a 0.405 M aqueous acetic acidsolution with a 0.432 M aqueous sodium hydroxide solution?
Homework Answers
CH3COOH + NaOH ------------> CH3COONa + H2O
initially
millimoles of CH3COOH = 18.7 x 0.405 = 7.5735
7.5735 millimoles NaOH must be added to reach equivalence point
7.5735 = V x 0.432
V = 17.53 mL NaOH must be added
total volume = 17.53 + 18.7 = 36.33 mL
[CH3COONa] = 7.5735 / 36.33 = 0.208 M
for salt of weak acid
pH = 1/2 [pKw + pKa + log C]
pKa of CH3COOH = 4.74
pH = 1/2 [14 + 4.74 + log 0.208]
pH = 1/2 [18.06]
pH = 9.03
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