Question

You have 225 mL of an 0.43 M acetic acid solution. What volume (V) of 1.10 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.23? (The pKa of acetic acid is 4.76.)

Answer 1

CH3COOH + NaOH ----< CH3COONa + H2O

So finally all NaOH will react

use:
pH= pKa + log ([CH3COONa]/[CH3COOH])
4.23 = 4.76 + log ([CH3COONa]/[CH3COOH])
([CH3COONa]/[CH3COOH]) = 0.295

Let us assume that v ml of NaOH was mixed.

[CH3COOH] =initial number of moles / total volume
                   = 225*0.43 / (225+v)
                   = 96.75 / (225+v)
[NaOH] =initial number of moles / total volume
                   = 1.1*V / (225+v)
                   = 1.1v / (225+v)

So finally all NaOH will react
After reaction:
[CH3COONa] = 1.1v / (225+v)
[CH3COOH] = 96.75 / (225+v) -   1.1v / (225+v)
      = (96.75- 1.1v) / (225+v)

but
([CH3COONa]/[CH3COOH]) = 0.295
[1.1v / (225+v)] / [(96.75- 1.1v) / (225+v)] =0.295
1.1v / (96.75- 1.1v) =0.295
1.1v = 28.54 - 0.32 v
v = 20 mL

Answer: 20 mL