What is the volume (in mL) of glacial acetic acid (17.4 M) that would have to be added to 500 mL of a solution of 0.20 M sodium acetate in order to achieve a pH of 5.0

3.16mL of glacial acetic acid must be added

Explanation:

The mixture of acetic acid with sodium acetate produce a buffer. The buffer follows the H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the buffer: 5.0

pKa is pKa of the buffer = 4.74

[A⁻] could be taken as moles of acetate ion = 0.500L * (0.20mol/L) = 0.100mol

And [HA] are moles of acetic acid.

Replacing:

5.0 = 4.74 + log [0.100mol] / [HA]

0.26 = log [0.100mol] / [HA]

1.8197 = [0.100mol] / [HA]

[HA] = 0.100mol / 1.8197

[HA] = 0.05495 moles of acetic acid

As the glacial acetic acid is 17.4M, the volume we must add to prepare the buffer is:

0.05495 mol * (1L / 17.4mol) = 3.16x10⁻³L =

2b2t

explanation:

2b2t