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 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

[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

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+]

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 [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] for a weak acid [H+] = [A-] Ka ≈ [H+]2 [HA]total

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+]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 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 = -log10 1.34 x 10-3 pH = 2.87

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 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 0.1 Ka ≈ 1.0 x 10-4 mol dm-3