You are asked to prepare 500mL 0.100M acetate buffer using only pure acetic acid (MW= 60.05g/mol and pKa= 4.76), 3.00 M NaOH and water

Question

You are asked to prepare 500mL 0.100M acetate buffer using only pure acetic acid (MW= 60.05g/mol and pKa= 4.76), 3.00 M NaOH and water

What volume of 3.00 M NaOH must be added to the acetic acid to achieve a bigger w a pH of. 5.00 at. A final volume of 500mL?

Answer 1

Let’ HAc represent acetic acid and Ac- is the acetate anion.

Henderson Hasselbalch eq:

pH = pKa + log [Ac-] / [HAc]

5.00 = 4.76 + log [Ac-] / [HAc]

log [Ac-] / [HAc] = 0.24

[Ac-] / [HAc] = 1.74

For 500 ml (0.5 L) we need 0.05 moles HAc in order to have a buffer 0.1 M in acetate.

The ratio of Ac- to HAc is 1.74 to 1

x / 0.05-x = 1.74

x = 0.0318 moles Ac- = moles NaOH

Volume of 3 M NaOH = 0.0328 mol x 1L/3 mol = 0.0109 L = 10.9 mls