How much 5.80 M NaOH must be added to 550.0 mL of a buffer that is 0.0215 M acetic acid and 0.0255 M sodium acetate to raise the pH to 5.75

1.64mL of 5.80M are added

Explanation:

First, we need to calculate total moles of acetic buffer. Then, with H-H equation, we need to find ratio of moles of acetate and acetic acid.

The moles of acetate increased are due the addition of moles of NaOH. Thus, we can find the volume of 5.80M NaOH that must be added:

Moles acetic acid and sodium acetate:

Acetic acid: 0.550L * (0.0215mol/L) = 0.011825 moles

Acetate: 0.550L * (0.0255mol/L) = 0.014025 moles

Total moles = 0.02585 moles = [Acetate] + [Acetic acid] (1)

H-H equation:

pH = pKa + log [Acetate] / [Acetic acid]

Where pH is 5.75;

pKa of acetic acid is 4.74;

And [ ] could be taken as moles of each species.

5.75 = 4.74 + log [Acetate] / [Acetic acid]

10.2329 = [Acetate] / [Acetic acid] (2)

Replacing (1) in (2):

10.2329 = 0.02585 - [Acetic acid] / [Acetic acid]

10.2329[Acetic acid] = 0.02585 - [Acetic acid]

11.2329[Acetic acid] = 0.02585

[Acetic acid] = 0.002301 moles.

Moles acetate:

[Acetate] = 0.02585 moles - 0.002301 moles = 0.023549 moles

Initial moles were 0.014025 moles, moles of acetate added (Due the addition of NaOH) are:

0.023549 moles - 0.014025 moles = 0.009524 moles of NaOH

Volume is:

0.009524 moles of NaOH * (1L / 5.80mol) = 1.64x10⁻³L = 1.64mL of 5.80M are added