Question

You have 875 mL of an 0.49 M acetic acid solution. What volume (V) of 1.30 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.39? (The pKa of acetic acid is 4.76.)

Answer 1

pH = pKa + log[A-]/[HA]
5.39 = 4.76 + log[A-]/[HA]
[A-]/[HA] = 10^(5.39 - 4.76) = 4.265
initial [HA] = 0.49M
initial moles of HA = 0.875*0.49 = 0.428 mol
after adding a volume V of 1.30 M NaOH, you reduce moles of HA by 1.30*V and increase moles of A- by the same amount. The new solution volume is 0.875 + V so the new concentrations are
[HA] = (0.428 - 1.30*V)/(0.875 + V)
[A-] = 1.30*V/(0.875 + V)
[A-]/[HA] = 1.30*V/(0.875 - V)
4.625 = 1.30*V/(0.875 - V)
V = 0.682 L = 682 mL