Question

In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid...

In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid ( ka =7.5 x 10^-5) was titrated with a 0.0500 M solution of NaOH at 25 C. The system will acquire this pH after addition of 18.75 mL of the titrant:

answer is 4.535

Homework Answers

Answer #1

moles of acetic acid present = 0.05 M x 50 ml = 2.5 mmol

moles of NaOH added = 0.05 M x 18.75 ml = 0.9375 mmol

[CH3COONa] formed = 0.9375 mmol/68.75 ml = 0.014 M

[CH3COOH] remaining = 1.5625 mmol/68.75 ml = 0.028 M

pH = pKa + log([base]/[acid])

      = 4.125 + log(0.014/0.028)

      = 3.823

The pH would be 3.823.

[I don't think the answer you have quoted 4.535 is correct, in order to get this pH you have to add about 36 ml of NaOH]

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