Question

You have 625 mL of an 0.47 M acetic acid solution. What volume (V) of 1.40 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.00? (The pKa of acetic acid is 4.76.)

Answer 1

Using Henderson equation,

pH = pKa + log [ A ^- ] / [ HA ]

5 = 4.76 + log [ CH₃COO^- ] / [CH3COOH ]

[ CH₃COO^- ] / [CH3COOH ] = 10 ^{(5 - 4.76)} = 1.73

Initial [Ch₃COOH ] = 0.47 M

Therefore Initial moles of CH3COOH =0 .625 x 0.47 = 0.29 mol

After adding a volume V of 1.40 M NaOH, you reduce moles of CH₃COOH by 1.40 x V and increase moles of CH₃COO^- by the same amount. The new solution volume is 0.625 + V so the new concentrations are

[ CH₃COOH ] = (0.29 - 1.40 x V) / (0.625 + V)

[ CH3COO- ] = 1.40 x V / (0.625 + V)

[ CH3COO-] / [CH₃COOH] = 1.40 x V / (0.625 - V)

1.73 = 1.40 x V / (0.625 - V )

V = 0.316 L = 316 mL