IB Chemistry: Buffer Solutions

IB Chemistry: Buffer Solutions Buffer Solutions 1. Describe the composition of a buffer solution and explain its action A buffer solution is essentially a solution which is designed to resist changes in pH when small quantities of an acid or an alkali is added, removing small amounts of H+ or OH- ions that are added to it. There are two types of buffers: • Acidic buffers • Alkaline buffers An acidic buffer is made from a weak acid and a salt of that weak acid. Most common example would be ethanoic acid and sodium ethanoate, CH3COOH and NaCH3COO. An acidic buffer solution has a pH less than 7. The mechanism of an acidic buffer solution is as followed: NaCH3COO(aq) → Na+(aq) + CH3COO-(aq) CH3COOH(aq) ↔ CH3COO- (aq)+ H+(aq) The weak acid only partially dissociates in a solution while the salt completely separates into its ions. When H+ ions are added, the ethanoate ions react with it to form undissociated ethanoic acid, keeping the [H+] and therefore the pH constant. CH3COO- (aq)+ H+(aq) ↔ CH3COOH(aq) If OH- ions are added, the ethanoic acid will react with it to form water, therefore keeping the [H+] ions and the pH constant CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) A key fact is that an acidic buffer solution is most commonly formed from adding a weak acid to a strong base, until a salt and excess weak acid remain: NaOH(aq) + CH3COOH(aq) → NaCH3COO(aq) + H2O(l) + (excess) CH3COOH(aq) An alkaline buffer operates in a similar fashion. However, it is composed of a weak base and one of its salts instead, with a pH greater than 7. A common example would ammonia mixed with ammonium chloride, NH3 and NH4Cl. Again, in a solution, the salt completely dissociates whilst the weak base partially dissociates: NH4Cl(aq) → NH4+(aq) + Cl-(aq) NH3(aq) + H2O(l) –> NH4+(aq) + OH-(aq) When H+ ions are added to a solution, it will react with the OH- from the weak base to form water and the weak base will further dissociate to replace them. If OH- ions are added, the NH4+ ammonium ions will react with it to form NH3 and water. Similar to acidic buffers, an alkaline buffer solution can be made if a strong acid is mixed with a weak base until only the salt and excess weak base remain. 2. Solve problems involving the composition and pH of a specified buffer system. Since a buffer solution contains a weak acid in it, the dissociation equilibrium equation can be used to find the Ka and pH: Ka = ([H+] [CH3COO-])/ [CH3COOH] But wait a minute! Doesn’t a buffer solution also contains the completely dissociated salt! We would therefore have to make the assumption that [CH3COO-] in the solution in equilibrium is provided largely by the dissociated salt and not by the weak acid. A common question they will ask you in a exam regarding buffers is to work out the pH of a buffer solution formed. For example: 25.0 cm3 of 1.00 × 10–2 mol dm–3 hydrochloric acid solution is added to 50.0 cm3 of 1.00 × 10–2 mol dm–3 aqueous ammonia solution. Calculate the concentrations of both ammonia and ammonium ions in the resulting solution and hence determine the pH of the solution. Reaction equation: HCl(aq) + NH3(aq) → NH4+(aq) + Cl-(aq) Number of moles before reaction: HCl: 0.025 x 0.01 = 0.00025 mols NH3: 0.05 x 0.01 = 0.0005 mols 1:1 mol ratio when they react, therefore the NH3 is in excess. Number of moles after reaction: HCl: 0.00025 mols → 0.00 mols (completely reacted) NH3: 0.0005 mols → 0.00025 mols (1:1 ratio) NH4+: 0.00025 mols formed Concentration: Total volume of solution: 0.025 dm3 + 0.05 dm3 = 0.075 dm3 [NH3] = 0.00025 / 0.075 = 0.0033 mol dm-3 [NH4+] = 0.00025 / 0.075 = 0.0033 mol dm-3 Equilibrium equation: pKb of NH3 = 4.75 (from data booklet) ; Kb therefore equals 10-4.75 Kb = ([NH4+] [OH-]) / [NH3] You want to work out the [OH-] so you can work out the pH. Rearranging the equation gives us: [OH-] = (Kb x [NH3]) / [NH4+] Substituting the values in: [OH-] = (10-4.75 x 0.0033) / 0.0033 [NH3] = [NH4+] Therefore [OH-] = 10-4.75 & pOH = 4.75 pH = pKw – pOH pH = 14 – 4.75 pH = 9.25