1. Buffers
Year 12 Chemistry

2. What is a buffer?
A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it
There are 2 types:
Acidic
Alkaline

3. Acidic buffers An acidic buffer has a pH less than 7
It is made from equal molar concentrations of a weak acid and it’s conjugate base
Example: ethanoic acid and ethanoate ion
CH3COOH  CH3COO H+
(weak acid) (conj. base)

4. Acidic buffers Take a 0.1M solution of ethanoic acid:
CH3COOH  CH3COO H+
At equil: ( M) ( M) ( M)
What’s the pH of this solution?
Now, say you add HCl to this equilibrium. The acetate ion is the only species available to reduce the added H+ and these are very low in concentration, so the pH will drop dramatically.
What can you do to reduce this effect?
-log [H+] = 2.38
Add acetate ions (sodium acetate)

5. Acidic buffers – adding acid
CH3COOH  CH3COO H+
If we buffer the solution by adding 1M sodium acetate, what will happen?
This will have the following effect:
Increase the amount of acetate ions, shifting the equilibrium to the left
Increase the pH to 4.76
When adding H+ ions, the extra acetate ions will react to reduce the [H+] moving the equilibrium to the left.
The solution resists changes to pH, meaning it is buffered

6. Acidic buffers – adding base
Add OH-
CH3COOH  CH3COO H+
Adding base results in more OH- being added to the equilibrium. This extra amount of ions is removed by 2 processes. What are these?
CH3COOH + OH-  CH3COO- + H2O (hydroxide ions are removed by reaction with undissociated ethanoic acid.
The small amount of H+ ions in the above reaction will also remove OH- ions, shifting the equilibrium to the right to make more H+ ions that further remove OH-.

7. pH of buffers depends on concentration of conjugate pair
vol. of 0.1M acetic acid (ml)
vol. of 0.1M sodium acetate (ml) 3
982.3
17.7 4
847.0
153.0 5
357.0
653.0 6
52.2
947.8
Note: you can also get various pH buffers by changing the acid/base pair.

8. Alkaline buffers An alkaline buffer has a pH greater than 7
It is made from a equal molar concentrations of a weak base and it’s conjugate acid
Example: ammonia and ammonium ion
NH3 + H2O   NH OH-
(weak base) (conj. acid)

9. Alkaline buffers NH3 + H2O   NH4+ + OH- (weak base) (conj. acid)
Due to the weak nature of ammonia, this equilibrium will be well to the left. We can create a buffered solution by adding ammonium chloride.
What will this do to the equilibrium?
Addition of ammonium ions will shift the equilibrium even further to the left. The pH of this solution would be 9.25

10. Alkaline buffers – adding acid
NH3 + H2O   NH OH-
What will happen if you add acid to this solution?
Two processes:
NH3 + H+   NH4+ (removal by reaction with ammonia to produce more ammonium ion)
NH3 + H2O   NH OH- (removal by reaction with OH- to produce water
Combines with H+ to form water

11. Alkaline buffers – adding base
NH3 + H2O   NH OH-
What will happen to the equilibrium if you add base?
Adding base effectively adds OH-. This means:
The ammonium ion reacts with the OH- to shift the equilibrium to the left, consuming most of the OH- ions.
NH OH-   NH3 + H2O

12. Summary
Buffer solutions resist changes in pH when acids and alkalis are added
Buffers generally contain:
Sufficient concentrations of a weak acid and it’s conjugate base OR weak base and it’s conjugate acid
The pH of buffer solutions depend on the concentrations and type of conjugate acid/base pairs that are used.
Go here  for good animations of this (chapter 8)

13. pH of a buffer solution Consider the dissociation of a weak acid
Commercially available buffer solutions are used to calibrate pH meters
Consider the dissociation of a weak acid
HA(aq) + H2O(l)  A-(aq) + H3O+
The acid dissociation constant expression is
𝐾𝑎= 𝐴− [𝐻3𝑂+] [𝐻𝐴]
This can be rearranged to find [𝐻3𝑂+]
𝐻3𝑂+ =𝐾𝑎 [𝐻𝐴] 𝐴−

14. HA(aq) + H2O(l)  A-(aq) + H3O+
pH of a buffer solution
HA(aq) + H2O(l)  A-(aq) + H3O+
Assumptions:
As a large quantity of conjugate base (A-) has been added, the equilibrium shifts far left, so that equilibrium concentration of the acid is approximately equal to the initial concentration of the weak acid :
[HA]eq ≈ [HA]I = [acid]
The equilibrium concentration of the conjugate base ion (A- ) is approximately equal to the concentration of the salt that was added to the equilibrium.
[A-]eq ≈ [A-]i = [salt]

15. Recall from a previous slide that
pH of a buffer solution
Recall from a previous slide that
𝐻3𝑂+ =𝐾𝑎 [𝐻𝐴] 𝐴−
So, considering our 2 assumptions:
[HA]eq ≈ [HA]I = [acid]
[A-]eq ≈ [A-]i = [salt]
Note: some texts will show “ + log [salt]/[acid]”. This is the same value as “-log [acid]/[salt]”
𝐻3𝑂+ =𝐾𝑎 x [𝑎𝑐𝑖𝑑] 𝑠𝑎𝑙𝑡
Alternatively, you may solve for [H+] first and then solve for pH using
pH=-log[H+]
𝑝𝐻=𝑝𝐾𝑎−𝑙𝑜𝑔 [𝑎𝑐𝑖𝑑] 𝑠𝑎𝑙𝑡
When [acid] = [salt]
pH= pKa

16. pOH of a buffer solution
Similarly for a base equilibrium
B(aq) + H2O(l)  HB+(aq) + OH-(aq)
𝑂𝐻− =𝐾𝑏 [𝐵] 𝐻𝐵+ =𝐾𝑏 [𝑏𝑎𝑠𝑒] 𝑠𝑎𝑙𝑡
Alternatively, you may solve for [OH-] first and then solve for pOH using
pOH=-log[OH-]
𝑝𝑂𝐻=𝑝𝐾𝑏−𝑙𝑜𝑔 [𝑏𝑎𝑠𝑒] 𝑠𝑎𝑙𝑡
When [base] = [salt]
pOH= pKb

17. Exercise
An aqueous solution of 0.1M ammonia and 0.1M ammonium chloride has a pH of 9.3. The reaction is: NH3 + H2O  NH4+ + OH-
Calculate the Kb for ammonia
If a pH of 9.0 is needed, what should be added? Explain
Calculate the new concentration of the substance added in b)
Answers:
𝐾𝑏= 𝑂𝐻− [𝑁𝐻4+] [𝑁𝐻3] = 10−4.7 [0.1] [0.1] = = 2.00 x 10-5
Ammonium chloride should be added as this will shift the equil to the left(towards ammonia), reducing the [OH-] and decreasing the pH
pH = 9.0 means pOH = 5, so [OH-] = 10-5
[𝑁𝐻4+]= 𝐾𝑏[𝑁𝐻3] [𝑂𝐻−] = (2𝑥10−5 )(0.1) [10−5 ] = 0.2 mol dm-3
Notice the additional volume has a negligible effect on [NH3] and has been ignored

18. Exercise
If you wanted to make a buffer solution of pH=4.46 using ethanoic acid and sodium ethanoate, what concentrations could you use? pKa (ethanoic acid) = The reaction is: CH3COOH+ H2O  CH3COO- + H3O+
Answer:
pH = pKa – log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡]
4.5 = 4.76 – log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡]
log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] = 4.76 – 4.46 = 0.30
[𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] =
[𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡] = 2.0
So, we need a solution with twice as concentrated acid as ethanoate salt
Note: it should be expected that the acid concentration will be higher than the salt.
At equal concentrations, the pH = pKa = 4.76
At pH 4.46 (a lower pH), we need to add acid to shift equil right, increasing [H3O+], decreasing pH

19. Exercise
An aqueous solution of 0.025M ethanoic acid and 0.050M sodium ethanoate is prepared.
Calculate the pH of this solution given Ka = 1.74x10-5 mol dm-3
If 1.0cm-3 of 1.0M NaOH is added to 250cm3 of buffer, what will happen to the pH?
Answers:
𝑝𝐻=𝑝𝐾𝑎−𝑙𝑜𝑔 [𝑎𝑐𝑖𝑑] 𝑠𝑎𝑙𝑡 = -log (1.74x10-4.7) – 𝑙𝑜𝑔 [.025] = =4.96
CH3COOH + OH- 
CH3COO- + H2O
ni (mol)
0.001
0.0125 -
nc(mol)
-0.001
+0.001
ne(mol)
0.0135
[ ] (mol dm-3)
0.021
0.054
So,
pH = pKa – log [𝑎𝑐𝑖𝑑] [𝑠𝑎𝑙𝑡]
pH = 4.76 – log [0.021] [0.054]
pH =
pH = 5.17
Notice the additional volume has again been ignored as it is assumed to have little effect on the overall pH

20. Buffer applications H2CO3  H+ + HCO3-
Blood
The pH of human blood must be maintained at 7.4 or serious health consequences can result. The buffering system that maintains this pH is H2CO3/HCO3-:
H2CO3  H+ + HCO3-
If blood increases in acidity, the additional H+ ions will react with the bicarbonate ions If blood increases in alkalinity, the H+ ions in the equilibrium will react and shift to the right Other buffers in the blood also help to maintain the required pH. Haemoglobin is itself a weak base that helps in this process

21. Buffer applications Body cells
The phosphate buffer system operates in the internal fluid of all cells. This buffer system consists of dihydrogen phosphate ions (H2PO4-) as hydrogen- ion donor (acid) and hydrogen phosphate ions (HPO42-) as hydrogen-ion acceptor (base).
These two ions are in equilibrium with each other as indicated by the chemical equation below.
H2PO4-(aq) H+(aq) + HPO42-(aq)

22. Buffer applications Swimming pools
The H2CO3/HCO3- buffer system that is used in the blood is also used in swimming pools.
Sodium hydrogen carbonate is often added to swimming pools if it is proving difficult to maintain the pH between 7.2 – 7.4.
The measurement “total alkalinity” in pools is a measure of OH- and HCO3-. This is effectively a measure of the buffering capacity of the water. If it is too low, then bicarbonate must be added.