Making 100-mL of buffer from 0.100 M NaCH3CO2 and 0.100 M HCH3CO2 with a pH of 4.80

Question

Use the Henderson-Hasselbalch equation for this calculation. Pka of acetic acid = 4.74. What is the amount of acid and base required to make a solution with a pH of 4.8.

Answer 1

Hi Lilly,

The Henderson Hasselbalch equation is pH = pKa + log [conj,base] / [ acid]

In this problem we are asked to mix the NaCH₃CO2 (conj.base) and the HCH₃CO₂ (acid) in the correct amounts to produce a buffer with a pH of 4.80. We are also given the pKa of the HCH₃CO₂ = 474.

So, essentially we are being asked to solve the Henderson Hasselbalch equation for [CB] / [ A]

pH = pKa + log [CB] / [ A]

4.80 = 4.74 + log [CB] / [A]

log [CB] / [ A] = 0.06

[CB] / [A] = 1.15

[CB] = 1.15[A]

Also, [CB] + [A] = 0.100 M

solving the simultaneous equations we arrive at

1.15[A] + [A] = 0.100

2.15[A] = 0.100

[A] = 0.0465

[CB] = 0.1 - 0.0465 = 0.0535

So, to make 100 mls, MIX 46.5 mls HCH₃CO₂ AND 53.5 mls NaCH₃CO₂