MAKING BUFFER SOLUTIONS
Introduction
Buffers are solutions that resist changes to their pH when small amounts of acid, water or base are added.
The key components of a buffer solution are a weak acid and a salt of its conjugate base (or a weak base and a salt of
its conjugate acid).
Buffer solution: CH3CO2H(aq) + CH3CO2Na(aq)  CH3CO2H(aq) + CH3CO2-(aq) + H+(aq) + Na+(aq)
partially dissociates fully dissociates
lots
lots
v. little
In this example, the sodium ethanoate fully dissociates, which increases the concentration of ethanoate (CH3CO2-)
ions, forcing the acid disociation equilibrium towards the left due to Le Chatelier’s principle. This creates reservoir of
undissociated weak acid so that:
 Adding base – H+ ions used up, equilibrium shifts right, dissociating acid to create more [H+]
 Adding acid – H+ ions added, equilibrium shifts to left, H+ combines with A- to reduce [H+]
 Adding water – concentration of HA and A- changes by same proportion so [H+] unaffected
If we consider the expression for Ka of the ethanoic acid used:
[𝐻 + ][CH3 CO2 − ]
𝐾𝑎 =
[CH3 CO2 H]
We can then calculate the pH by taking logs and rearranging to give:
[𝐶𝐻3 𝐶𝑂2− ]
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔10 (
)
[𝐶𝐻3 𝐶𝑂2 𝐻]
Because very little of our acid is actually dissociated in a buffer, we can simplify some otherwise quite complicated
maths by assuming that the concentration of CH3CO2- ions is equal to the concentration of sodium ethanoate used
and the concentration of ethanoic acid is equal to the original used.
Working backwards to determine the quantities to use to make a buffer of a given strength is a little more involved.
For example, what quanity (moles) of sodium ethanoate should be dissolved in 1.00 dm3 of 0.10 M ethanoic acid to
produce a buffer solution with a pH of 5?
1. Calculate the concentration of H+ ions:
[𝐻 + ] = 10−𝑝𝐻 = 10−5 = 1𝑥10−5
2. Write out the expression for Ka
𝐾𝑎 =
[𝐻 + ][CH3 CO2 − ]
= 1.7𝑥10−5
[CH3 CO2 H]
3. Assuming that the concentration of ethanoic acid at
equilibrium is equal to the starting amount. We can
rearrange the above to:
𝐾𝑎 [CH3 CO2 H]
[𝐻 + ]
1.7𝑥10−5 𝑥 0.10
[CH3 CO2 − ] =
1𝑥10−5
−
[CH3 CO2 ] = 0.17 𝑚𝑜𝑙
[CH3 CO2 − ] =
In this experiment you will be producing two buffer solutions, measuring their pH and then adding small amounts of
acid, alkali and water to verify that the pH does not change (much).
Materials and Equipment
Standard laboratory glassware
Data-logger with pH probe
Analytical balance
0.10 M Ethanoic acid solution
0.10 M Ammonia solution
0.10 M sodium hydroxide
0.10 M hydrochloric acid
Ammonium chloride
Sodium ethanoate (a.k.a sodium acetate)
Task


Follow the procedure below to produce an acidic buffer.
Adapt the procedure below to produce an alkaline buffer of pH 10.
Procedure to make a buffer
1. Use a measuring cylinder to measure 200 cm3 0.10 M ethanoic acid into a 250 cm3 beaker.
2. Add 2.78 g of sodium acetate and stir until fully dissolved.
3. Calculate the pH you expect this buffer to have.
Testing the buffer
Do this with each of your two buffers and with distilled water.
4.
5.
6.
7.
8.
Transfer 50 cm3 of buffer solution to a clean dry beaker.
Record the pH
Add 5 successive 5 cm3 portions of 0.10 M sodium hydroxide solution
Repeat steps 3-5 adding 0.10 M hydrochloric acid instead of sodium hydroxide
Repeat steps 3-5 adding distilled water instead of sodium hydroxide.
Analysis





Calculate the pH of your acidic buffer. How does this compare to the measured pH?
How does the pH of alkaline buffer compare to the intended pH?
Looking at your testing data, how successfully were your buffers able to control the pH?
What effect did diluting the buffer have on your pH? Why should this be the case and is it what you
would expect?
Explain how the equilibrium in the alkaline buffer responds to and minimise changes and thus control
pH.