Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH

pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

Explanation:

pKa acetic acid, CH3COOH = 4.7

It is possible to determine pH of a buffer using H-H equation:

pH = pka + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH3COONa] / [CH3COOH]

As you want a pH 4.7 buffer:

4.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = [CH3COONa] / [CH3COOH]

That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:

For pH 5.7:

5.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = log [CH3COONa] / [CH3COOH]

10 = [CH3COONa] / [CH3COOH] (1)

That means you need 10 times [CH3COONa] over [CH3COOH]

And as you know:

10mL=  [CH3COONa] + [CH3COOH] (2)

Replacing (1) in (2):

10 = 10mL + [CH3COOH] / [CH3COOH]

10[CH3COOH] = 10mL + [CH3COOH]

11[CH3COOH] = 10mL

[CH3COOH] = 0.91mL

And [CH3COONa] = 10mL - 0.91mL =

[CH3COONa] = 9.09mL

That is:

c.

explanation: