Buffer solutions in action
The following page will demonstrate how to calulate the pH of a buffer solution after a small amount of acid or base has been added to it.
It will also try to demonstrate how they limit the changes in pH through worked examples:
Example
If we take an example from the previous page; we made a buffer solution from 0.3M CH3COOH and 0.17M CH3COONa and we found
that it buffered at a pH of 4.5.
Lets assume we take 100 cm3 of this buffered solution and add to it 10cm3 of 0.1M HCl.
Note: The pH of HCl alone is equal to -log100.1 = 1
As the acid is added a new equilibrium is formed because the CH3COO- ions of
the conjugate base will react to remove the H+ ions from solution:
CH3COO- + H+ 
CH3COOH
The reaction of this is 1:1:1 so the number of moles of HCl added will completely react with the same number of moles of
ethanoate ions to produce an equal number of moles of Acetic acid.
Steps to calculate change in pH of buffer solution
- First we need to calculate the initial amounts of the acid and base present in the buffer solution and the amount in moles of
HCl added. To do this we use the equation:
Moles = Concentration x Volume
| CH3COOH |
HCl |
CH3COO- |
0.3M x 0.1 moldm-3
= 0.03 moles |
0.1M x 0.01 moldm-3
= 0.001 moles |
0.17M x 0.1 moldm-3
0.017 moles |
-
Then we need to calculate the amounts of each substance that will be present in this new
equilibrium formed after the addition of the acid:
Because of the stochiometry we mentioned before 1:1:1
0.001 moles of
HCl present will react with
0.001 moles of
CH3COO- ions to
produce an extra
0.001 moles of
acetic acid.
| CH3COOH |
HCl |
CH3COO- |
0.03moles + 0.001 moles
= 0.031 moles |
0.001moles - 0.001 moles
= 0 moles |
0.017 moles - 0.001 moles
= 0.016 moles |
- We can now put these values for the acid and base back into the Henderson-Hasselbalch equation
to work out the new pH:
pH = pKa(CH3COOH) - log10([acid]/[base])
pH=4.75 - log10(0.031 / 0.016)
pH = 4.75 - log10 1.9375
pH= 4.75 - 0.287
pH= 4.46
What the values mean
On the previous page we calculated that the buffer solution consisting of
0.3M CH3COOH and 0.17M CH3COONa would buffer at a pH of = 4.5.(heres how we calculated this)
After the addition of 10cm3 of 0.1M HCl, the solution is now at a pH of
4.46.
The pH of the HCl solution on its own was 1.(look here)
From this we can clearly see that the buffer solution has severely limited the changes in pH
that would be otherwise caused by the addition of a very strong acid with a pH of 1.
On the next page we will investigate the uses of buffer
solutions in a variety of different environments.
Author: Christopher Saywell (document modification date: 7th April 2004)
