Detailed and comprehensive notes on unit 3.1.12 (acids and bases) for AQA A level chemistry. Covers: acids, strong acid calculations, pH, ionisation of water (Kw), bases, strong base calculations, reactions of strong acids and bases, acid dissociation constant (Ka), pKa, reactions of weak acids and...
• Brønsted-Lowry acids are proton donors – they release H+ ions when mixed with warer.
• H+ combines with H2O to form hydroxonium ions, H3O+: HA (aq) + H2O (l) → H3O+ (aq) + A- (aq).
o Monoprotic acids release 1 H+ ion per molecule
E.g. HCl, HNO3, CH3COOH
o Diprotic acids release 2 H+ ions per molecule
E.g. H2SO4, H2C2O4
• Strong acids dissociate almost completely in water, e.g. HCl → H+ + Cl-
• Weak acids only partially dissociate in water, e.g. CH3COOH ⇌ CH3COO- + H+
o An equilibrium is set up which lies to the left
pH of Strong Acids
pH = -log[H+] [H+] = 10-pH
Always give pH to 2 d.p.
Examples:
Calculate the pH of 0.2 M HCl.
• [H+] = 0.2 M
• pH = -log[0.2] = 0.70
Calculate the pH of 0.04 M H2SO4.
• [H+] = 2 x 0.04 = 0.08 M
• pH = -log[0.8] = 1.10
Calculate the concentration of H2SO4 with pH -0.50.
• [H+] = 10-(-0.50) = 3.16 M
• [H2SO4] = 3. = 1.58 M
Dilution of a Strong Acid
[H+] in diluted solution = [H+] in original solution x old volume
new volume
Examples:
Calculate the pH of the solution formed from the addition of 250cm3 of water to 50cm3 of 0.2 mol dm-3
HNO3.
• [H+] in diluted solution = 0.2 x 50/300 = 1/30 M
• pH = -log[1/30] = 1.51
Calculate the pH of the solution formed by adding water to 100cm3 of 2 mol dm-3 H2SO4 to make 500cm3 of
solution.
• [H+] = 2 x 2 x 100/500 = 0.8 M
• pH = -log[0.8] = 0.10
, Ionisation of Water
Ionisation of Water
• Water can act as both an acid and a base.
• Water self-ionises to a small extent to produce H+ (aq) and OH- (aq):
• 2H2O (l) ⇌ H3O+ (aq) + OH- (aq)
o This reaction is a dissociation reaction involving only bond breaking, which is endothermic.
o If the temperature increases, Kw will shift to the right to absorb the added heat.
Kc = [H3O+] [OH-] Kw = [H3O+] [OH-]
[H2O]2
• Since this is pure water, [H2O] is very large and effectively constant.
• Kw is called the ionic product of water and at 25°C is 1.0 x 10-14 mol2 dm-6
o (Equilibrium, so changes with temperature)
Example:
At 30°C, Kw = 1.471 x 10-14 mol2 dm-6. Calculate the pH of pure water at this temperature.
• Pure water, so [H+] = [OH-]
• Kw = [H+]2 and [H+] = √Kw
• [H+] = √(1.471 x 10-14) = 1.213 x 10-7 M
• pH = -log[1.213 x 10-7) = 6.92
• (Note: the water is still neutral, despite the pH not being 7, because the [H +] and [OH-] are equal
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