1. Brønsted-Lowry Acids and Bases
Acid – proton donor
Base – proton acceptor
Acids can only behave as acids in the presence of a base – the proton acceptor.
Water often fulfils this role.
Acids do not demonstrate acidic properties (taste sour, turn UI red etc) in the absence of water.
This is because the H+(aq) ion must be formed.
It is the active ingredient in acids.

2. Formation of the hydronium ion from HCl and H2O
Week 19
Formation of the hydronium ion from HCl and H2O
© Pearson Education Ltd 2009
This document may have been altered from the original 2

3. Conjugate acid-base pairs
HCl(g) +H2O(l) D H3O+(aq) + Cl-(aq)
acid base acid base
2 conjugate acid-base pairs.
Each is a set of 2 pairs that can transform into each other by the gain or loss of a proton.

4. Conjugate acid–base pair
Week 19
Conjugate acid–base pair
© Pearson Education Ltd 2009
This document may have been altered from the original 4

5. NH3(g) + H2O(l) D NH4+ (aq) + OH-(aq)
More Examples
NH3(g) + H2O(l) D NH4+ (aq) + OH-(aq)
Base acid acid base
The conjugate acid-base pairs are:
Ammonia D ammonium and Water D hydroxide
This reminds us that water is AMPHOTERIC ie can be both an acid and a base.
i.e. H2O(l) + H2O(l) D H3O+(aq) + OH-(aq)
acid base acid base

6. Identify the acid-base pairs in the examples below:
Insert state symbols as required.
HIO3 + H2O D H3O+ +IO3-
CH3CO2H + H2O D CH3CO2- + H3O+
HF + H2O D H3O+ +F-
H2SO4 + H2O D H3O+ + HSO4-
HSO4- + H2O D H3O+ + SO42-
Write the equation for the reaction of methylamine in water.
Write the equations for the reaction of Alka- Seltzer in water using HA to represent citric acid. (p.139)

7. Polybasic Acids
On the previous slide there were 2 equations used to explain the acidity of sulphuric acid.
Sulphuric acid is an example of a DIBASIC acid since it is able to release 2 protons per molecule.
Phosphoric acid H3PO4 is an example of a TRIBASIC acid.
Write the 3 equations needed for phosphoric acid to release all 3 protons.
Write the overall equation which produces 3 hydronium ions per molecule of phosphoric acid.

8. Acid-Base Reactions
H3O+ -the hydronium or oxonium ion is often simplified to H+(aq).
Both symbols mean the same but H+(aq) is easier to use in equation writing.
Reactions of Acids:
Acid + carbonate → salt + water + carbon dioxide
Acid + base → salt + water
Acid + alkali → salt + water
Acid + metal → salt + hydrogen

9. Ionic Equations – Spectator Ions
Write the equation including state symbols.
Balance it.
Identify spectator ions which remain unchanged in the reaction.
Cancel them out.
Rewrite the ionic equation.
(Remember all acids have ionised (dissociated) in solution.)

10. HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g)
HCl(aq) + Na2CO3(aq) → NaCl(aq) + H2O(l) + CO2(g)

11. HNO3(aq) + MgO(s)→ H2O(l) + Mg(NO3)(aq)
Nitric acid +magnesium oxide → magnesium nitrate + water
Sulphuric acid + potassium hydroxide → potassium sulphate + water
H2SO4(aq) + KOH(aq) → K2SO4(aq) + H2O(l)

12. Reactions of Acids with Metals
These are REDOX reactions.
Hydrochloric acid + magnesium→ magnesium chloride + hydrogen
HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

13. Strong and Weak Acids and Bases
Strong acids and bases are completely ionised /disociated when dissolved in water.
HCl is a strong acid. If 1 mole of HCl is dissolved in water 1 mole of H+ ions and 1 mole of Cl- ions are formed.
Similarly group 1 hydroxides are strong bases. E.g. If 1 mole of the solid lithium hydroxide is dissolved in water 1 mole of Li+ ions and 1 mole of OH- ions are produced.
Weak acids and bases do not fully ionise.

14. General equation
If the acid HA is dissolved in water the position of the equilibrium depends on the relative strength of the conjugate base.
HA + H2O D H3O+ +A-
If HA is strong the equilibrium lies RIGHT and the back reaction is limited.
This means the conjugate base A- is WEAK since it does not easily accept a proton from H3O+.
If HA is weak then the equilibrium lies LEFT and the conjugate base A- is strong and readily accepts a proton from H3O+

15. Acid Dissociation Constant Ka
So the STRONGEST acids have the WEAKEST conjugate bases and vice versa.
Since ‘weak’ and ‘strong’ are relative acid strength is quantified using the acid dissociation constant Ka
Strictly speaking this is an equilibrium constant for the following reaction when an acid dissociates in water.
HA + H2O D H3O+ +A-
Kc = [H3O+][A-] is a constant at given t
[HA][H2O]
Since the concentration of water in dilute solution is very little different from pure water the above can be simplified.

16. (concentration of water = 1kg dm-3.
0.1 moldm-3 HCl contains 3.65g HCl in g water.
55.56 moldm-3 in pure water and mol dm-3 in the acid)
So simplified:
HA(aq) D H+(aq) + A-(aq)
Ka = [H+(aq)][A-(aq)]
[HA(aq)]
Ka values vary from 40 (for nitric acid) to
10-14 for water.

17. pKa Rather than using indexes all the time it is easier to use pKa.
pKa = -log10Ka
(Ka = 10-pKa)
The smaller the pKa the stronger the acid.
For di and tribasic acids each ionisation is quoted separately.
The second ionisationis smaller than the first and the third smaller again due to the difficulty of removing H+ from a negatively charged ion.
Thus pKa values INCREASE for successive ionisations.
Complete questions p.143.

18. The Dissociation of Water
Even very pure water conducts electricity to a very small extent due to self ionisation.
H2O(l) D H+(aq) + OH-(aq)
Applying the equilibrium law at fixed t
[H+(aq)][OH-(aq)] = Kc
[H2O(l)]
Because the degree of ionisation is so small that it is virtually all unionised then [H2O] in water = 1000/18 = 55.6 mol dm-3 and a constant.
Then:
Kc[H2O] = [H+][OH-] = new constant Kw
constxconst

19. Kw – The Ionic Product of Water
The value of Kw ranges from 1.14 x mol2 dm-6 at 0oC to 5.13 x mol2dm-6 at 100oC.
At 25oC Kw = x ≈ 1 x mol2 dm-6.
In pure water because ionisation produces equimolar concentrations of H+ and OH- ions:
[H+] [OH-] =10-14
[H+] = [OH-] =√ =10-7 mol dm-3
This value Kw at a given t controls the [H+] and [OH-] for ALL aqueous solutions, even if they are from different sources because the equilibrium:
H2O D H+ + OH- ALWAYS holds at a given t.

20. Using Kw
In an acid or alkali [H+] and [OH-] are no longer equal but [H+][OH-] will ALWAYS be mol2 dm-6.
There will always be OH- present in an acid and H+ present in an alkali.
If 1 is known the other can be calculated.
E.g mol dm-3 HCl
[H+] = 10-2
Since [H+][OH-] =10-14
Then [OH-] = = = 10-12
[H+]

21. Using Kw to Calculate [H+] in an alkali
10-1 mol dm-3 NaOH
[OH-] = 10-1
Since [H+][OH-] =10-14
[H+] = 10-14
10-1
= mol dm-3
Complete questions bottom p.147

22. The pH Scale
Since [H+] falls between 100(1) and mol dm-3 Sørensen suggested logs to simplify the scale.
BY DEFINITION
pH= -lg [H+] (and [H+(aq)] = 10-pH)
The set of numbers to 14 are the pH Scale.
For pure water at 250C
[H+] = 10-7 mol dm-3.
So pH =-lg[H+]
= - lg 10-7
= 7

23. Calculations pH scale is logarithmic.
This means that a change of 1 pH unit corresponds to a ten fold change in [H+].
Dilute a mol dm-3 solution to moldm-3 and pH will change from 3 to 4.
pH values not always whole numbers.
E.g. HCl concentration 4.3 x 10-4 mol dm-3.
[H+] = 4.3 x 10-4
pH = -lg [H+]
= 3.37
Do questions p.141

24. Calculating pH for strong acids from concentrations
A sample of Hydrochloric acid has a concentration of 1.22 x 10-3 mol dm-3.
What is its pH?
[H+(aq)] = [HCl(aq)] = 1.22 x 10-3 mol dm-3
pH = -log[H+(aq)]
=-log(1.22 x10-3)
= 2.91

25. Calculating concentration for strong acids from pH
A sample of Nitric Acid has a pH of 5.63
What is the concentration of the nitric acid?
[H+(aq)] = 10-pH = = 2.34 x 10-6 mol dm-3
HNO3 is a strong acid
so [HNO3(aq)] = [H+(aq)]
= 2.34 x 10-6 mol dm-3

26. Calculating the pH of Weak Acids
pH of a weak acid may be calculated from the acid dissociation constant, Ka ,the equilibrium constant expression and the concentration of the acid solution.
HA(aq) D H+(aq)+ A-(aq)
Ka = [H+(aq)] [A-(aq)]
[HA(aq)]
We simplify this:
We assume that
[H+(aq)] = [A-(aq)] and that [HA] = the concentration of the acid used since degree of dissociation is so small in dilute solutions.

27. Ka now becomes :
[H+(aq)]2 mol dm-3
[HA(aq)]
Although some molecules of acid will have dissociated the proportion is such that it will not affect the calculated value of the pH until the third decimal place – which is fine at this level.
Since Ka = [H+(aq)]2 mol dm-3
Then [H+(aq)]2 = Ka [HA(aq)]
 [H+(aq)] =√(Ka [HA(aq)])
And pH = -log [H+(aq)] as before

28. Worked Example
Calculate the pH of mol dm-3 ethanoic acid (Ka = 1.7 x 10-5 mol dm-3)
Ka = [H+(aq)]2 mol dm [HA(aq)]
[H+(aq)]2 = Ka [HA(aq)]
= 1.7 x10-5 x = 1.7 x 10-6
Taking square roots:
[H+(aq)] = x 10-3 mol dm-3
pH –log(1.304 x 10-3) = 2.9

29. Calculating Ka for weak acids
Using the worked example from the book:
A sample of mol dm-3 methanoic acid HCOOH has a pH of 2.66.
Calculate Ka
[H+] = 10-pH =
= 2.19 x 10-3 mol dm-3
Ka = [H+(aq)]2 mol dm [HA(aq)]
= (2.19 x 10-3)2 = 1.6 x 10-4 mol dm-3
0.030

30. Some examples check these
Calculate Ka for the following acids:
a) mol dm-3 ethanoic acid with pH 4.50
Ans = about 2.5 x 10-8
b) mol dm-3 methanoic acid with pH 3.25
Ans = about 2.5 x 10-6
c) mol dm-3 HX with pH 5.70
Ans =about 4.5 x 10-12
Complete questions on the bottom of p.145

31. pH values of strong bases
Strong bases are completely dissociated.
NaOH + aq →Na+(aq) + OH-(aq)
In an alkaline solution of a strong base the [OH-(aq)] is the same as the concentration of the alkali.
The [H+(aq)] depends on Kw, the ionic product of water.
Kw= [H+(aq)] [OH-(aq)] = 1.00 x mol2dm-6 at 25oC
So log [H+(aq)] + log [OH-(aq)] =-14
- log [H+(aq)] +- log [OH-(aq)] =14
pH + POH = 14
So pH=14-pOH

32. Examples
Find the pH of the following strong bases given that Kw =1.0 x mol2dm-6 at 25oC.
0.001 mol dm-3 potassium hydroxide solution.
11.0
An aqueous solution containing 0.2g of sodium hydroxide, NaOH , per dm3
11.7
Do the questions on p.149