Question

A 210.0 mL buffer solution is 0.250 M in acetic acid and 0.250 M in sodium acetate.

1) What is the pH after addition of 0.0050 mol of HCL

2) What is the pH after addition of 0.0050 mol of NaOH

Answer 1

a)
Number of moles of CH3COOH = M*V = 0.250 M * 210 mL = 52.5 mmol
Number of moles of CH3COONa = M*V = 0.250 M * 210 mL = 52.5 mmol
Number of moles of HCl added = 0.005 mol= 5 mmol

The reaction that takes place is:
CH3COONa+   HCl    <----------------------> CH3COOH   + NaCl
5 mmol of CH3COONa and HCl will react to form 1 mmol of CH3COOH

After reaction,
number of moles of CH3COOH = 52.5 + 5 = 57.5 mmol
number of moles of CH3COONa = 52.5 - 5 = 47.5 mmol

CH3COOH and CH3COONa forms buffer
pKa of CH3COOH = 4.75
[CH3COONa ] = number of moles / total volume
                         = 47.5 mmol / 210 mL
                          = 0.226 M

[CH3COOH ] = number of moles / total volume
                         = 57.5 mmol / 210 mL
                          = 0.274 M

use:
pH = pKa + log {[CH3COONa/CH3COOH]}
       = 4.75 + log {0.226/0.274}
       = 4.67
Answer: 4.67
b)
Number of moles of CH3COOH = M*V = 0.250 M * 210 mL = 52.5 mmol
Number of moles of CH3COONa = M*V = 0.250 M * 210 mL = 52.5 mmol
Number of moles of NaOH added = 0.005 mol= 5 mmol
The reaction that takes place is:
CH3COOH   +   NaOH    <----------------------> CH3COONa

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

After reaction,
number of moles of CH3COONa = 52.5 + 5 = 57.5 mmol
number of moles of CH3COOH = 52.5 - 5 = 47.5 mmol

CH3COOH and CH3COONa forms buffer
pKa of CH3COOH = 4.75
[CH3COOH] = number of moles / total volume
                         = 47.5 mmol / 210 mL
                          = 0.226 M

[CH3COONa] = number of moles / total volume
                         = 57.5 mmol / 210 mL
                          = 0.274 M

use:
pH = pKa + log {[CH3COONa/CH3COOH]}
       = 4.75 + log {0.274/0.226}
       = 4.83
Answer: 4.83