Preparation of buffer

Question

Calculate the volumes of 1M sodium acetate, 1M acetic acid and water to prepare 100mL of a pH 4 buffer assuming that the final concentration of acetic acid in the buffer is 0.25M. Ka = 1.8 x 10^-4

Answer 1

We can use the Henderson Hasselbalch equation for this problem.

The HH equation is pH = pKa + log [acetate]/[acetic acid]

From the Ka we find the pKa, and pKa = - log Ka = 3.74 (note; the actual Ka = 1.8x10^-5 but I'm using the value given in the problem)

We know the pH = 4 and we have the pKa value so we can solve for [acetate]/[acetic acid]

4 = 3.74 + log [acetate]/[acetic acid]

log [acetate]/[acetic acid] = 0.26

[acetate]/[acetic acid] = 1.82 This represents the ratio of the conj.base to the weak acid

We also know that the final concentration of acetic acid is to be 0.25 M

We therefore can find the final [acetate].

[acetate]/[acetic acid] = 1.82 = x/0.25

x = 0.45 M = [acetate]

Since we want 100 ml (0.1L) of buffer, the moles are...

0.25 mol/L x 0.1 L = 0.025 moles acetic acid

0.45 mol/L x 0.1 L = 0.045 moles acetate

Volumes...

(x L)(1 mol/L) = 0.025 moles and x = 0.025 L = 25 ml acetic acid

(x l)(1 mol/L) = 0.045 moles and x = 0.045 L = 45 ml sodium acetate

Volume of water = 100 ml - 25 ml - 45 ml = 30 ml H2O