Question

Calculate the pH if 10mL of 0.1 M NaOH are added to the buffer made by mixing 25 mL of 1.0 M CH3COOH and 25 mL of 0.5 M CH3COONa.

Answer 1

Number of moles of CH3COOH added = M*V = 1 M * 25 mL = 25 mmol
Number of moles of CH3COONa added = M*V = 0.5 M * 25 mL = 12.5 mmol
Number of moles of NaOH added = M*V = 0.1 M * 10 mL = 1 mmol

The reaction that takes place is:
CH3COOH   +   NaOH    <----------------------> CH3COONa

1 mmol of CH3COOH and NaOH will react to form 1 mmol of CH3COONa

After reaction,
number of moles of CH3COOH = 25 -1 = 24 mmol
number of moles of CH3COONa = 12.5 + 1 = 13.5 mmol
number of moles of NaOH = 1 -1 = 0 mmol
Total volume = 10 mL + 25 mL + 25 mL = 60 mL

CH3COOH and CH3COONa forms buffer
pKa of CH3COOH = 4.75
[CH3COOH] = number of moles / total volume
                         = 24 mmol / 60 mL
                          = 0.4 M

[CH3COONa] = number of moles / total volume
                         = 13.5 mmol / 60 mL
                          = 0.225 M

use:
pH = pKa + log {[CH3COONa/CH3COOH]}
       = 4.75 + log {0.225/0.4}
       = 4.5
Answer: 4.5