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...

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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?

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Answer 1

Answer #1

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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Answer 2

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