1. IB Chemistry Power Points
Topic 18
Acids and Bases
Buffers
and Salts

2. Buffer Solutions
DEFINITION: A buffer solution contains a weak acid mixed with its conjugate base (or weak base and conjugate acid)
Buffers resist changes in pH when a small amount of a strong acid or base is added to it.
HA ∏ H+ + A-

3. If a small amount of a strong acid (H+) is added eqm shifts to the left as [H+] increases so system adjusts to increase [HA] and reduce [H+] again.
HA ∏ H+ + A-

4. A small amount of a strong base will react with H+ to form H2O and eqm will shift to the right to increase [H+] again.
HA ∏ H+ + A-

5. Making Buffer Solutions
An example of a weak acid is ethanoic acid. This could be mixed with sodium ethanoate which will provide ethanoate ions (conjugate base).
CH3COOH(aq) ∏ CH3COO-(aq)+H+(aq)
An example of a weak base is ammonia. This could be mixed with ammonium chloride to provide ammonium ions (conjugate acid).
NH3(aq) + H2O(aq) ∏ NH4+(aq) +OH-(aq)
In order for a buffer to work well the concentration of the acid/base and its salt must be much higher than the strong acid/base added.

6. Optimum Buffer
A buffer is most effective when the concentration of weak acid and its salt (the conjugate base) are equal and the pH is equal to pKa.
In practice it will work reasonably well with similar concentrations and the effective buffer range of any weak acid/ base is pKa 1.

7. Blood has buffering capacity
Blood must maintain a pH of 7.4 so its enzymes can work. If 0.01 mol of H+ or OH- is added to blood it only changes pH by 0.1 unit.
The eqm is:
CO2(aq) + H2O(l) ) ∏ H+(aq) + HCO3-(aq)

8. Buffer Calculation #1 pH = 5.6
Calculate the pH of a 1.00 dm3 buffer solution made by dissolving 0.50 mol of sodium ethanoate into a mol dm-3 ethanoic acid solution.
pH = 5.6
Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations.
weak acid [CH3COOH] = mol dm-3
conjugate base [CH3COO-] = 0.50 mol dm-3
Write the equilibrium equation and expression
CH3COOH ∏ H+ + CH3COO-
In this case, assume the equilibrium concentrations of the weak acid and the salt anion are the same as the given information (very little change when equilibrium is established.

9. Buffer Calculation #2 mass = 9.50 g
Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of mol dm-3 NH3 solution to create a buffer with a pH of Assume no change in overall volume.
mass = 9.50 g
In this case, assume the equilibrium concentrations of the weak base is the same as the given information (very little change when equilibrium is established. Calculate [OH-] from desired pH
Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations.
weak base [NH3] = mol dm-3
conjugate acid [NH4+] = ?
Write the equilibrium equation and expression
NH3+ H2O ∏ NH4+ + OH-

10. Buffer Calculation #3 pH= 9.6
A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of mol dm-3 HCl is added to 40 cm3 of a mol dm-3 ammonia solution.
pH= 9.6
In this case, assume the equilibrium concentrations of the weak base and cation are the same as determined by stoichiometry (very little change when equilibrium is established.
Do the stoichiometry to determine what the concentrations are AFTER neutralization.
HCl + NH3  NH4+ + Cl-
Write the equilibrium equation and expression
NH3+ H2O ∏ NH4+ + OH-

11. Salt Hydrolysis
A soluble salt is an ionic compound made of cations (ex. Na+) and anions (ex. Cl-) which completely dissociates into ions in aqueous solution.
Salts can affect the pH of a solution because the cations act as weak acids by bonding with OH- and the anions act as weak bases by accepting H+ ions.

12. Salt Hydrolysis
If the cation comes from a strong base then it will have less acidic activity than one from a weak base. For example Na+ from NaOH is a weaker acid than NH4+ from NH3
If the anion comes from a strong acid then it will have less basic activity than one from a weak acid. For example Cl- from HCl is a weaker base than CH3COO- from CH3COOH.
Salts from a strong acid and strong base ex. NaCl will form a neutral solution.

13. Salt Hydrolysis
Would a solution of sodium ethanoate be acidic, basic or neutral?
Would a solution of ammonium chloride be acidic, basic or neutral?
Would a solution of potassium chloride be acidic, basic or neutral?

14. Salt Hydrolysis
With salts of weak acids and weak bases the pH of the solution formed will reflect the relative strengths of the acid and base. Ex. Ammonium ethanoate is about neutral.
BE CAREFUL THOUGH . . .
Acidity of salts also depends on size and charge of the cation

15. Salt Hydrolysis
Salts with small, highly charged cations are more acidic than large, low charge cations.
recall Period 3 chloride salts : NaCl, MgCl2 and AlCl3  which is most acidic
The aluminum ion and those of transition metals exist in water in hydrated form ie. [Al(H2O)6]3+ , [Fe(H2O)6]3+

16. [Fe(H2O)6]3+  [Fe(OH)(H2O)5]2+(aq) + H+(aq)
Salt Hydrolysis
the e- attracting power of the ion weakens the O-H bond and stabilizes the resulting OH- ion.
As a result these ions are quite acidic in water.
[Fe(H2O)6]3+  [Fe(OH)(H2O)5]2+(aq) + H+(aq)

17. Learning Check – salt hydrolysis
Acid Base Example Salt Solution
Strong Strong NaCl neutral
Weak Weak ammonium ethanoate neutral
Strong Weak ammonium chloride acidic
Weak Strong sodium ethanoate basic

18. IB Chemistry Power Points
Topic 18
Acids and Bases
Buffers
and Salts

19. Acid Base Titrations
If a strong acid is added to a strong base gradually the pH will start off as 1.
Once enough base is added that it is now in excess the pH will change very suddenly to about 13.
The point at which this change is seen is when the amount of acid = amount of base.
This is called the equivalence point.
With this combination it occurs at pH 7 as the acid and base combine to make a neutral solution.

20. Strong Acid and Strong Base
Most indicators will work for this combination.

21. Weak Acid + Strong Base
A weak acid will have a pH of 3-5. When a strong base is added the pH will increase gradually as HA is converted to A-.
This is called the buffering region as it’s acting like a buffer.
When half the amount of base has been added it is called the half-neutralization point and [HA] = [A-]
so [H+] = Ka x [HA]/[A-]
Ka = [H+] and pKa = pH

22. Weak Acid + Strong Base Ka = [H+] and pKa = pH
This is the best way to determine the Ka for a weak acid.
At the equivalence point the pH increases quickly to 13.
The equivalence occurs when pH >7.
Most suitable indicator is phenolphthalein

23. Strong Acid + Weak Base
When a weak base is added to a strong acid the pH will remain around 1 until near the equivalence point when all the base has been converted into its conjugate acid.
B(aq) + H+(aq)  BH+(aq)
At the equivalence point pH is <7 and exact pH can be found from Kb of the base.
[OH-] = Kb x [B]/[BH+]
At equivalence point [B] = [BH+] so Kb = [OH-]
And pKb = pOH = 14 - pH

24. Strong Acid + Weak Base
A suitable indicator here is methyl orange.

25. Weak Acid + Weak Base
When a weak acid is added to a weak base the change in pH is gradual from acidic to basic so it is hard to detect the equivalence point.
It is hard to find a suitable indicator.
The volume of base that will neutralize an acid is not affected by the strength.
It only depends on the stoichiometric amounts reacting.

26. Learning Check
Do exercise 18.4 on packet.

27. Indicators
An indicator is a substance (often an organic dye) that has a different color in acidic and alkaline solutions so it can be used to show the end point of a titration.
The color change is seen because the indicator is a weak acid/base in which the two forms have different colors.
Ex. HIn(aq)H+(aq) + In-(aq)
litmus: red blue
Hin stands for the indicator and In- is the other form when it dissociates. Each of these forms will have a different color.

28. Indicators
In the presence of an acid the eqm is driven to the left to form HIn as the H+ ions combine with In- to reduce [H+].
With a base it shifts to the right to form more In- as the OH- of the base combines with H+ reducing [H+] so more Hin will dissociate.
The usual equation for Ka applies:
Ka = [H+][In-] / [HIn]
[HIn] / [In-] = [H+] / Ka

29. Indicators
The color depends on the pH and [H+] and also on Ka so different indicators change color over different pH ranges.
When pH = pKa the two colored forms will have equal concentrations and the indicator will be in the middle of its color change.
If one form is in excess by 10 fold then the color will be of that form.

30. pH of Indicator If [HIn] is 10x [In-] the pH will be:
[H+] = Ka x [Hin] / [In-]
= Ka x 10/1
= 10Ka
or pH = pKa -1
If [In-] were 10x [HIn] then pH = pKa + 1.
Many indicators change over a region of 2 pH units.
An effective indicator will change color at the equivalence point of the particular titration.

31. Choosing an Indicator Phenolphthalein pKa = 9.6 pH range: 8.3 -10.0
Color in acid: colorless
Color in alkali: pink
Good for titrations with strong bases

32. Choosing an Indicator Methyl Orange pKa = 3.7 pH range: 3.1 - 4.4
Color in acid: red
Color in alkali: yellow
Good for titrations with strong acids

33. Learning Check
Do exercise 18.5 in packet