Question

You have 175 mL of an 0.29 M acetic acid solution. What volume (V) of 1.70 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.44? (The pKa of acetic acid is 4.76.)

Answer 1

pH = pKa + log[A-]/[HA]
4.44 = 4.76 + log[A-]/[HA]
[A-]/[HA] = 10^(4.44 - 4.76) = 0.4786
initial [HA] = 0.29
initial moles of HA = 0.175*0.29 = 0.05075 mol
after adding a volume V of 1.70 M NaOH, you reduce moles of HA by 1.70*V and increase moles of A- by the same amount. The new solution volume is 0.175 + V so the new concentrations are
[HA] = (0.05075 - 1.70*V)/(0.175 + V)
[A-] = 1.70*V/(0.175 + V)
[A-]/[HA] = 1.70*V/(0.175 - V)
0.4786 = 1.70*V/(0.175 - V)
V = 0.038 L = 38 mL