How would you prepare 2.00 l of a 0.25 m acetate buffer at ph= 4.50 from concentrated acetic acid (17.4 m) and 1.00 m naoh? the pka of acetic acid is 4.76?

The reaction of acetic acid with sodium hydroxide is:

The ratio of A-/HA is calculated as follows:

According to Henderson Hasslebach equation:

The total concentration of HA and A- = 2.0 L * 0.25 M = 0.5 mol.

[ HA] = 0.156 mol and [ A- ]= 0.343 mol

Total of 0.5 moles of acetic acid is required:

HA = 30.0 g acetic acid

Conversion of 0.343 moles of the acetic acid to acetate can be performed by adding NaOH

= 343 mL  

Thus, 343 mL of 1M NaOH is required