TUTS Check chem foyer for new tut group lists after 1pm on Monday 22 August. TEST 1 Wed 24 August during the tut Quantitative chemistry, types of reactions 1

TYPES OF REACTIONS Electrolytes Acid- base reactions Solubility rules Precipitation reactions Writing ionic equations Oxidation numbers Redox reactions Concentration 2

ELECTROLYTES “ionic compounds which dissociate in water to produce ion counterparts” -ions conduct electricity -Strong electrolytes: ionic compounds which completely dissociate in water (includes strong acids & bases* and soluble* ionic salts) -Weak electrolytes: ionic compounds which partially dissociate in water 3

ACID-BASE REACTIONS ACID + BASE → SALT + H 2 O One H + from the acid and one OH - from the base form H 2 O e.g. HCl + NaOH → NaCl + H 2 O STRONG ACIDSSTRONG BASES HClLiOH HBrNaOH HIKOH HClO 3 RbOH HClO 4 CsOH HNO 3 Ca(OH) 2 H 2 SO 4 Sr(OH) 2 Ba(OH) 2 4

SOLUBILITY RULES 1.Salts of Group I and ammonium ions are soluble. 2.Nitrates, acetates, and perchlorates are soluble. 3.Salts of Ag, Hg(I), and Pb are insoluble. 4.Chlorides, bromides and iodides are soluble. 5. Carbonates, hydroxides, sulphides, oxides, silicates, and phosphates are insoluble. (Sulphides of Group II ions and hydroxides of Ca 2+, Sr 2+ and Ba 2+ are slightly soluble). 6. Sulfates are soluble, except those of barium, strontium. 5

PRECIPITATION REACTIONS Use solubility rules to predict insoluble products Generally reaction is: AB + CD →AD + CB Try: Pb(NO 3 ) 2 (aq) + 2KI (aq) → PbI 2 ? + 2KNO 3 ? 6

WRITING IONIC EQUATIONS Balanced molecular equation Balanced overall ionic equation (note ppts) Remove spectator ions to give net ionic equation 7

PRACTICE EXAMPLE 8 BaCl 2 + H 2 SO 4 → BaSO 4 + HCl

PRACTICE EXAMPLE 9 H 2 SO 4(aq) + NaOH (aq) →

OXIDATION NUMBERS 1. Elements in their elemental form have an oxidation number of The oxidation number of a monatomic ion is the same as its charge. 3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. – Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1. – Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal. – Fluorine always has an oxidation number of −1. – The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. 4. The sum of the oxidation numbers in a neutral compound is The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. 10

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PRACTICE EXAMPLES S 2 O 3 2- FeO 4 2- Na 2 SO 3 S 8 NH

REDOX REACTIONS OIL RIG Atoms/ions are reduced by oxidising agents or oxidised by reducing agents Balance redox reactions in either acidic or basic medium 13

14 Acidic medium 1.Write reduction and oxidation half -equations 2.Balance half-equations Atoms other than H and O O atoms (add H 2 O) H atoms (adding H + ) 3.Balance charge with electrons 4.Combine half-reactions 5.Simplify 6.Verify

PRACTICE EXAMPLE Fe 2+ (aq) + MnO 4 ¯  Fe 3+ (aq) + Mn 2+ (aq) 15

16 Basic medium 1.Write reduction and oxidation half -equations 2.Balance half-equations Atoms other than H and O O atoms (add H 2 O) H atoms (adding H + ) 3.Balance charge with electrons 4.Combine half-reactions 5.Add number of OH - ion equal to number of H + ions on both sides of overall reaction and combine hydrogen ions and hydroxide ions to form water when they appear on the same side of the equation. 6.Simplify 7.Verify

PRACTICE EXAMPLE S (s) + ClO ¯ (aq)  SO 3 2- (aq) + Cl¯ (aq) 17

CONCENTRATION The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. - can be expressed in various ways e.g. parts per million, mass percent and most commonly molarity. Molarity (M) (molar concentration) - the number of moles of solute in 1 litre of solution. moles of solute (n) volume of solution in litres (V) Molarity = 18 n VM REMEMBER STOICHIOMETRIC RATIOS

PRACTICE EXAMPLE Calculate the molarity of a solution made by dissolving 5.00 g of glucose, C 6 H 12 O 6, in sufficient water to form exactly 100 mL of solution. 19

PRACTICE EXAMPLE How many grams of NaOH are needed to neutralise 20.0 cm 3 of M H 2 SO 4 solution? 20

Dilution of solutions Dilution is a procedure for preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution M conc × V conc = M dil × V dil 21