1. Acids and Bases
pH, Kw, Ka and pKa

2. Weak acids and bases involve partial dissociation in solution – the interaction with water can be shown:
CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
In these equilibria, the equilibrium position lies more to the left i.e. in a solution of ethanoic acid, for example, there are more undissociated ethanoic acid molecules than there are ethanoate and hydronium ions.
Note: generally, a strong acid gives rise to a weak conjugate base whereas a weak acid gives rise to a strong conjugate base. Similarly for bases.

3. Strong acids and bases are much better conductors of electricity than weak acids or bases – why?
Weak acids include Weak bases include
Ethanoic acid ammonia
Phosphoric (V) acid amines
Ethanedioic acid
Methanoic acid
Carbonic acid
What do we use to express how acidic or alkaline a solution is?

4. pH
The pH of a solution is defined as the negative logarithm (to the base 10) of the hydrogen ion concentration in mol dm-3
pH = -log [H+ ] {or pH = -log [H3O+ ] }
e.g. what is the pH with [H+ ] = 1.23 x 10-2 mol dm-3?
Note: always give pH values to two decimal places
-log(1.23 x 10-2) = 1.91

5. Determine the pH of the following
[H+] = 1.0 x 10-4M
[H+] = 2.5 x 10-2M
[H+] = 7.1 x 10-11M
[HCl] = 1.5 M
So, for strong acids, the initial [acid] gives the [H+]
Comment on monoprotic and diprotic etc

6. then [H+] = 10-pH i.e the antilog of the negative pH
If pH = -log [H+ ],
then [H+] = 10-pH i.e the antilog of the negative pH
e.g. find the hydrogen ion concentration of a solution of pH 5.20
[H+ ] = = 6.31 x 10-6 mol dm-3
Try these:
pH = pH = 1.38 pH = 4.00

7. Kw Pure water has a pH of @25OC therefore [H3O+] =
Where does the [H3O+] come from?
Remember that water is amphoteric.
H2O + H2O  H3O+ + OH-
In other words, water dissociates to a slight (but measurable) extent. How can the [H3O+] be measured?
Or H2O  H+ + OH-
Conjugate acids / bases

8. The H+ ion is very reactive and so attaches immediately to a neutral water molecule to form the hydronium ion.

9. An equilibrium expression can thus be written:
K = [H3O+(aq)]eq [OH-(aq)]eq
[H2O(l)]2eq
Combining the constants (K and the [H2O(l)]– why?) gives the dissociation constant (ionic product) of water, Kw.
Kw = [H3O+(aq)]eq [OH-(aq)]eq
= 1.0 x mol2 dm-6 at 25oC
why?
Is pure water always neutral? Will the pH always be 7.00?
A neutral solution is defined as one in which [OH-] = [H3O+]

10. Pure water is neutral
A neutral solution is defined as one in which
So, [H+] = [OH-]
= √ Kw
= √ (1.0 x 10-14) = 1.0 x 10-7 mol
therefore, the pH of a neutral
= -log[1.0 x 10-7] = 7.00
An acidic solution is defined as one in which
therefore, [H+] > 1.0 x 10-7 and pH < 7
An alkaline solution is defined as one in which
therefore, [H+] < 1.0 x 10-7 and pH > 7

11. Using Kw Calculate the [H3O+] and pH for the following solutions:
i) [OH-] = 8.4 x 10-9 M
Kw = [H3O+] [OH-]
[H3O+] = Kw = 1.0 x = 1.2 x 10-6 mol dm-3
[OH-] x 10-9
pH = -log [H3O+] = -log [1.2 x 10-6] = 5.92
ii) [OH-] = 6.1 x 10-2 M
[H3O+] = 1.6 x 10-13
pH = pOH = 1.21
What happens if you sum pH and pOH? = 14
Why? Kw = [H3O+] [OH-] = 1.0 x 10-14
pKw = -log Kw = 14 pKw = -log ( [H3O+] [OH-] ) = pH + pOH as log ab = log a + log b

12. Summary pH = -log [H3O+] [H3O+] = 10-pH Kw = [H3O+] [OH-]
= 1.0 x mol2dm-6 @25oC
pKw = -log Kw = @25oC

13. What is the pH of a solution with [H3O+] = 5.8 x 10-3 M?
What is the pH of a solution with [OH -] = 1.0 x 10-4 M?
What is the [OH-] of a solution of pH = 5.66?
What is the [H3O+] of a solution of pH = 13.86?
Answers:
2.24 10
4.6 x 10-9
1.4 x 10-14

14. What about weak acids (and bases)?
To calculate the pH of a solution of a weak acid (or base), the concentration of the solution is insufficient information. The extent to which the acid (or base) is dissociated (or ionised) must be known.
Remember, the ionisation is a reversible reaction:
HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
The equilibrium expression for this reaction can be given by
But [H2O(l)] is a constant (as it is in huge excess), therefore is absorbed in the constant – Ka – the acid dissociation constant
i.e. Ka = [H+(aq)]eq [A-(aq)]eq
[HA(aq)]eq

15. The higher the value of Ka, the stronger / weaker the acid. Why?
Using Ka
The higher the value of Ka, the stronger / weaker the acid. Why?
e.g HCl Ka = 1 x 107
HNO3 Ka = 40
HF Ka = 5.6 x 10-4
CH3COOH Ka = 1.7 x 10-5
C6H5OH Ka = 1.3 x 10-10
Calculate the pH of a 0.01mol dm-3 solution of hydrofluoric acid..
Ka values for strong acids rarely quoted
Phenol
Do on board

16. When HF ionises,
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
I xs mol dm-3
C -x x x
E x x x
Ka = [H3O+(aq)]eq [F-(aq)]eq
[HF(aq)]eq
Since one F- ion is formed for each H3O+ ion
[H3O+(aq)] = [F-(aq)] and [H3O+(aq)] [F-(aq)] = [H3O+(aq)]2
As the degree of dissociation is small, we assume that the value of x is so small that we can ignore it, so (0.01-x) ≈ 0.01
5.6 x 10-4 = x2
0.01

17. So, (5.6 x 10-4) x 0.01 = x2
x = 2.37 x = [H3O+]
pH = - log [H3O+]
= - log 2.37 x 10-3
= - (-2.63)
= 2.63 (0.01 mol dm-3 HF solution)

18. Calculate the pH of a 0.123 mol dm-3 solution of ethanoic acid. Ka = 1.74 x 10-5 mol dm-3
Ka = [H+]2 / [CH3COOH] = 1.74 x 10-5
[H+] = √(Ka [CH3COOH]) = √(1.74 x 10-5 x 0.123) = 1.46 x 10-3 moldm-3
pH = -log [H+] = 2.83

19. Calculate the pH of a 0. 423 mol dm-3 solution of methanoic acid
Calculate the pH of a mol dm-3 solution of methanoic acid. pKa = 3.75
pKa = -log Ka Ka = 10-pKa
Ka = = 1.78 x 10-4 mol dm-3
[H+] = √(Ka [HCOOH] ) = √ (1.78 x 10-4 x 0.423) = 7.53 x 10-5
pH = -log [H+] = 4.12

20. Calculation of Ka or pKa from pH
Calculate the acid dissociation constant for a weak acid, HA, given that a mol dm-3 solution has a pH of 3.21.
Ka = [H+][A-]
[HA]
[H+] = 10-pH = = x 10-4 mol dm-3 = [A-]
[HA] = – (6.166 x 10-4) = mol dm-3
Ka = (6.166 x 10-4)2 = x 10-6 mol dm-3
0.1054
Note: as x 10-4 mol H+ ions are produced from mol of HA, the amount of HA at equilibrium equals (0.106 – x 10-4) = A less accurate value of Ka can be calculated using [HA] = mol dm-3

21. pKa
pKa = -log Ka

22. Finding pKa from pH of a weak acid
The pH of a weak acid, HA, of concentration mol dm-3 was found to be 4.00.
Calculate the pKa for the acid.
pH = 4.00
[H+] = 1.00 x10-4 mol dm-3

23. Equilibrium concentration [0.100 - 1.00 x10-4] = 0.0999 1.00 x10-4HA H+ A-
Start concentration
mol dm-3
0.100
Equilibrium concentration
[ x10-4] =
1.00 x10-4
A less accurate value of Ka can be calculated using [HA] = mol dm-3

24. Ka = [H+][A-] [HA] = (1.00 x10-4)2 0.0999 = 1.001 x 10-7 mol dm-3
pKa = -log10 Ka
= 7.00
A less accurate value of Ka can be calculated using [HA] = mol dm-3

25. Now try these:
1.) Ethanoic acid mol dm-3 had a pH calculate pKa of acid.
Find Ka and pKa of the following:
2.) mol dm-3 HX has pH = 4.50
3.) mol dm-3 HY has pH = 3.25
4.) mol dm-3 HZ has pH = 5.70

26. Answer to no.1 [H+] = 10-pH = 10-4.56 = A Ka = [H+][CH3COO-] [CH3COOH]=
= A
Ka = [H+][CH3COO-]
[CH3COOH]
= [A]2
[B]
= Ka
= pKa = -log10 Ka
CH3COOH H+
CH3COO-
Start conc
mol dm-3
0.500
Eqm conc A
= B A

27. Quick Quiz! 1.) Calculate pH of 0.056 mol dm-3 H2SO4.
2.) Calculate concentration of HCl pH 3.52.
3.) Write the Ka expression for ethanoic acid in water.
4.) Is ethanoic acid a weak or strong acid?
5.) What is a weak acid?
6.) What is the Ka if pKa value of HNO3 is 4.5 x10-5?
7.) Calculate pKa from Ka value of HCl, 3.2 x10-3 mol dm-3.
8.) What is the pH of 0.56 mol dm-3 nitrous acid (HNO2), which has a Ka of x 10-3 mol dm-3?)