Question

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?

Answer 1

The reaction of acetic acid with sodium hydroxide is:

$CH_3COOH + NaOH CH_3COONa + H_2O$

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

According to Henderson Hasslebach equation:

$[A-]/[HA] = 10^p^H^-^p^K^_a$

$= 10^4^.^7^6^-^4^.^5^ = ^2^.^1^8$

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

$[ A- ]+ [ HA ]= 0.5$

$[ A^- ] = 0.5 – [ HA]$

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

Total of 0.5 moles of acetic acid is required:

$0.5 mol HA * 60.0 g$

HA = 30.0 g acetic acid

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

$0.343 mol HA * (1 mol NaOH/1 mol HA) * (1000 mL/1 mol NaOH)$

= 343 mL  

Thus, 343 mL of 1M NaOH is required