The pH Scale

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+]

[H+] pH Substance1.0 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

[H+] pH Substance2.5 x 10-7 6.6 Milk

1.0 x 10-7 7.0 Pure water

4.0 x 10-8 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

Calculating pH

• 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+]

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 = -log10 0.05 = 1.3

Calculating pH

• pH = -log10 [H+]

• Calculate the pH of the following solutions

• [H+] = 0.025 moldm-3

• [H+] = 0.125 moldm-3

• [H+] = 0.005 moldm-3

pH 1.6

pH 0.9

pH 2.3

As [H+] increases pH decreases

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

Calculating [H+]

• [H+] = 10-pH

• Calculate [H+] from the following pH

• pH = 1

• pH = 2

• pH = 2.5

• pH = 6.95

• pH = -0.5

[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

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]

Ka ≈ [H+]2 [HA]total

for a weak acid [H+] = [A-]

• 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+]2 [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

• 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+]

pH = -log10 1.34 x 10-3

[H+] = 1.34 x 10-3 mol dm-3

pH = 2.87

• A 0.1M solution of a weak acid, HA has a pH value of 2.50. Calculate the value of Ka

pH = -log10[H+]

2.50 = -log10 [H+]

[H+] = 3.16 x 10-3 mol dm-3

Ka ≈ [H+]2 [HA]total

Ka ≈ (3.16 x 10-3) 2

0.1Ka ≈ 1.0 x 10-4 mol dm-3