Question

You have 375 mL of an 0.15 M acetic acid solution. What volume (V) of 1.80 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.80? (The pKa of acetic acid is 4.76.)

Answer 1

pH = pKa + log[A-]/[HA]
4.80 = 4.76 + log[A-]/[HA]
[A-]/[HA] = 10^(4.80 - 4.76) = 1.09
initial [HA] = 0.15
initial moles of HA = 0.375*0.15 = 0.05625 mol
after adding a volume V of 1.80 M NaOH, you reduce moles of HA by 1.80*V and increase moles of A- by the same amount. The new solution volume is 0.375 + V so the new concentrations are
[HA] = (0.05625 - 1.80*V)/(0.375 + V)
[A-] = 1.80*V/(0.375 + V)
[A-]/[HA] = 1.80*V/(0.375 - V)
1.09 = 1.80*V/(0.375 - V)
V = 0.141 L = 141 mL