1. The pH Scale

2. The pH Scale Arrange the substances in order of increasing pH.
Match the pH and [H+] to each substance
Try to work out the relationship between pH and [H+]

3. [H+] pH Substance 1.0 x 10-1 1.0 Battery acid 1.0 x 10-2 2.0
Lemon juice
6.3 x 10-3
2.2
Vinegar
1.0 x 10-3
3.0
Apples
3.2 x 10-4
3.5
Soft drink
1.0 x 10-4
4.0
Wine
3.2 x 10-5
4.5
Tomatoes
2.5 x 10-6
5.6
Unpolluted rainwater

4. [H+] pH Substance 2.5 x 10-7 6.6 Milk 1.0 x 10-7 7.0 Pure water
7.4
Human Blood
5.0 x 10-9
8.3
Baking Soda Solution
4.0 x 10-9
8.4
Sea Water
3.2 x 10-11
10.5
Milk of Magnesia
1.0 x 10-11
11.0
Household Ammonia
4.0 x 10-13
12.4
Lime

5. Calculating pH pH = -log10[H+]
We could use [H+] as a measure of acidity but the numbers would be difficult to work with
So …. We use a log scale
pH = -log10[H+]

6. Calculating pH pH = -log10 [H+] If [H+] is 1 x10-4 moldm-3 then
pH = -log10 (1x 10-4) = 4
If [H+] is 0.05 moldm-3 then
pH = -log = 1.3

7. Calculating pH pH = -log10 [H+]
Calculate the pH of the following solutions
[H+] = moldm-3
[H+] = moldm-3
[H+] = moldm-3
pH 1.6
pH 0.9
pH 2.3
As [H+] increases pH decreases

8. Calculating pH of Strong Acids
pH = -log10 [H+]
Calculate the pH of the following solutions
For strong acids [H+] = [acid], fully dissociated
0.1M HCl
0.25M HNO3
0.1M H2SO4
0.15M H2SO4
pH = 1
pH = 0.6
Diprotic [H+] = 0.2 pH = 0.7
Diprotic [H+] = 0.3 pH = 0.5

9. Calculating [H+] [H+] = 10-pH Calculate [H+] from the following pH
[H+] = 0.1 moldm-3
[H+] = 0.01 moldm-3
[H+] = 3.16 x 10-3 moldm-3
[H+] = 1.12 x 10-7 moldm-3
[H+] = 3.16 moldm-3

10. Calculating pH of Weak Acids
Weak acids are only partially ionised
[H+] is not the same as [acid]
We calculate pH of weak acids using Ka
Ka = [H+][A-] [HA]
for a weak acid [H+] = [A-]
Ka ≈ [H+] [HA]total

11. The weak acid CH3COOH has a Ka value of 1. 8 x 10-5 mol dm-3 at 300K
The weak acid CH3COOH has a Ka value of 1.8 x 10-5 mol dm-3 at 300K. Calculate the pH of a 0.1M solution at this temp.
Ka ≈ [H+] [HA]total
[H+]2 = Ka[HA]total
[H+] = √ Ka[HA]total
[H+] = √ 1.8 x 10-5 x 0.1
[H+] = 1.34 x 10-3 mol dm-3

12. The weak acid CH3COOH has a Ka value of 1. 8 x 10-5 mol dm-3 at 300K
The weak acid CH3COOH has a Ka value of 1.8 x 10-5 mol dm-3 at 300K. Calculate the pH of a 0.1M solution at this temp.
pH = -log10[H+]
[H+] = 1.34 x 10-3 mol dm-3
pH = -log x 10-3
pH = 2.87

13. A 0. 1M solution of a weak acid, HA has a pH value of 2. 50
A 0.1M solution of a weak acid, HA has a pH value of Calculate the value of Ka
pH = -log10[H+]
2.50 = -log10 [H+]
[H+] = 3.16 x 10-3 mol dm-3
Ka ≈ [H+] [HA]total
Ka ≈ (3.16 x 10-3)
Ka ≈ 1.0 x 10-4 mol dm-3