Precipitation reactions These are processes in which an insoluble product (precipitate) is formed that drops out of solution, removing material, and therefore driving the reaction along (Le Chatelier again!) To predict whether a reaction will occur we must know the solubilities of the potential products in a reaction. Guidelines: 1.Compounds containing the Group 1 (Li, Na, K etc) or NH 4 + cations are most likely soluble. 2.Compounds containing the Halide anions, Cl -, Br - and I - (except Ag +, Hg 2+ and Pb 2+ compounds), the nitrate, perchlorate, sulphate or acetate anions (except Ba 2+, Hg 2+ and Pb 2+ sulphates) are most likely soluble.

Precipitation reactions Solubility Equilibria In saturated solutions of solutes an equilibrium between the ion concentrations and the solid exists. CaF 2(s.)  Ca 2+ (aq.) + 2F - (aq.) We can write an equilibrium constant for this: K sp = [Ca 2+ ][F - ] 2 Where K sp is the solubility product. 2AgNO 3 (aq.) + Na 2 CO 3(aq.)  Ag 2 CO 3 (s.) + 2NaNO 3 (aq.)

Oxidation-Reduction (REDOX) reactions Originally oxidation was assigned to the combination of an element with oxygen to give an oxide and reduction was the reverse. Today, a much broader definition is given: loss of electron(s) for oxidation gain of electron(s) for reduction Thus redox reactions are electron transfer reactions. Na  Na + + e - Cl 2 + e -  2Cl - 2Na + Cl 2  2Na + + 2Cl - In more complex reactions a bookkeeping system, oxidation numbers, is used to keep track of electron transfers. A redox reaction is therefore a reaction in which changes in oxidation numbers occur.

Oxidation Numbers Rules for assigning oxidation numbers: 1.An atom in its elemental state, 0. 2.An atom in a simple monoatomic ion, charge on the ion. 3.The sum of all the oxidation numbers in a molecule or a polyatomic ion, charge on the particle. 4.In its compounds F, In its compounds H, In its compounds O, -2. In cases of conflicting rules, apply the rule with the lower number first. Try these: MoS 2, ClO 2 -, H 2 SO 4, NH 4 Cl, KMnO 4,Na 2 S 2 O 3

Redox Reactions Oxidation-increase in oxidation numberReduction-decrease in oxidation number example.: rusting of iron. 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Identify substance that is oxidised, then identify substance that is reduced. Identify oxidising and reducing agents

Balancing redox reactions Using the ion-electron (half-reaction) method In acidic solutions Find oxidised and reduced species. Divide chemical equation into two half-reactions. Balance atoms (excluding H and O). Balance O (by adding H 2 O). Balance H (by adding H + ). Balance charge (by adding electrons). Make electron gain equivalent to electron loss, then add the half-reactions. Cancel similar species on both sides of the chemical equation.

Balancing redox reactions In basic solutions 8.Add the same number of hydroxyl ions as there are protons to both sides of the chemical equation. 9.Combine protons and hydroxyls to give water molecules. 10.Cancel H 2 O if you can. Try these: Cr 2 O Fe 2+  Cr 3+ + Fe 3+ in acid SO MnO 4 -  SO MnO 2 in base Using the ion-electron (half-reaction) method

Redox Titrations in Analyses Take the example of 2.00g of iron ore converted with acid to Fe 2+ (aq.). Titrated solution required 27.45mL of potassium permanganate. What is the %Fe in the ore? For the same sample, evaluate the volume of Ce 4+ that would have been required to titrate the Fe 2+ solution