Question

You have 275 mL of an 0.13 M acetic acid solution. What volume (V) of 2.30 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.84? (The pKa of acetic acid is 4.76.)

Answer 1

pH = pKa + log[A-]/[HA]
4.84 = 4.76 + log[A-]/[HA]
[A-]/[HA] = 10^(4.84 - 4.76) = 1.202
initial [HA] = 0.13
initial moles of HA = 0.275*0.13 = 0.03575 mol
after adding a volume V of 2.30 M NaOH, you reduce moles of HA by 2.30*V and increase moles of A- by the same amount. Then the new solution volume is 0.275 + V. so the new concentrations are
[HA] = (0.03575 - 2.30*V)/(0.275 + V)
[A-] = 2.30*V/(0.275 + V)
[A-]/[HA] = 2.30*V/(0.275- V)
1.202 = 2.30*V/(0.275 - V)
V = 0.0943 L = 94.3 mL