General Chemistry. Chapter 5: Introduction to Reactions in Aqueous Solutions. Principles and Modern Applications Petrucci Harwood Herring 8 th Edition

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General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 5: Introduction to Reactions in Aqueous Solutions Philip Dutton University of Windsor, Canada N9B 3P4 Slide 1 of 43

Contents 5-1 The Nature of Aqueous Solutions 5-2 Precipitation Reactions 5-3 Acid-Base Reactions 5-4 Oxidation-Reduction: Some General Principles 5-5 Balancing Oxidation-Reduction Equations 5-6 Oxidizing and Reducing Agents 5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations Focus on Water Treatment Slide 2 of 43

5.1 The Nature of Aqueous Solutions Slide 3 of 43

Electrolytes Some solutes can dissociate into ions. Electric charge can be carried. Slide 4 of 43

Types of Electrolytes Strong electrolyte dissociates completely. Good electrical conduction. Weak electrolyte partially dissociates. Fair conductor of electricity. Non-electrolyte does not dissociate. Poor conductor of electricity. Slide 5 of 43

Representation of Electrolytes using Chemical Equations A strong electrolyte: MgCl 2 (s) Mg 2+ (aq) + 2 Cl - (aq) A weak electrolyte: CH 3 CO 2 H(aq) CH 3CO 2- (aq) + H + (aq) A non-electrolyte: CH 3 OH(aq) Slide 6 of 43

Notation for Concentration MgCl 2 (s) Mg 2+ (aq) + 2 Cl - (aq) In 0.0050 M MgCl 2: Stoichiometry is important. [Mg 2+ ] = 0.0050 M [Cl - ] = 0.0100 M [MgCl 2 ] = 0 M Slide 7 of 43

Example 5-1 Calculating Ion concentrations in a Solution of a Strong Electolyte. What are the aluminum and sulfate ion concentrations in 0.0165 M Al 2 (SO 4 ) 3?. Balanced Chemical Equation: Al 2 (SO 4 ) 3 (s) 2 Al 3+ (aq) + 3 SO 4 2- (aq) Slide 8 of 43

Example 5-1 Aluminum Concentration: 0.0165 mol Al 2 mol Al [Al] = 3+ 2 (SO 4 ) 3 = 0.0330 M Al 3+ 1 L 1 mol Al 2 (SO 4 ) 3 Sulfate Concentration: [SO 2-0.0165 mol Al 3 mol SO 2-4 ] = 2 (SO 4 ) 3 4 = 1 L 1 mol Al 2 (SO 4 ) 3 0.0495 M SO 4 2- Slide 9 of 43

5-2 Precipitation Reactions Soluble ions can combine to form an insoluble compound. Precipitation occurs. Ag + (aq) + Cl - (aq) AgCl(s) Slide 10 of 43

Net Ionic Equation Overall Precipitation Reaction: AgNO 3 (aq) +NaI (aq) AgI(s) + NaNO 3 (aq) Complete ionic equation: Spectator ions Ag + (aq) + NO 3- (aq) + Na + (aq) + I - (aq) AgI(s) + Na + (aq) + NO 3- (aq) Net ionic equation: Ag + (aq) + I - (aq) AgI(s) Slide 11 of 43

Solubility Rules Compounds that are soluble: Alkali metal ion and ammonium ion salts Li +, Na +, K +, Rb +, Cs + NH 4 + Nitrates, perchlorates and acetates NO 3 - ClO 4 - CH 3 CO 2 - Slide 12 of 43

Solubility Rules Compounds that are mostly soluble: Chlorides, bromides and iodides Cl -, Br -, I - Except those of Pb 2+, Ag +, and Hg 2 2+. Sulfates SO 4 2- Except those of Sr 2+, Ba 2+, Pb 2+ and Hg 2+ 2. Ca(SO 4 ) is slightly soluble. Slide 13 of 43

Solubility Rules Compounds that are insoluble: Hydroxides and sulfides HO -, S 2- Except alkali metal and ammonium salts Sulfides of alkaline earths are soluble Hydroxides of Sr 2+ and Ca 2 + are slightly soluble. Carbonates and phosphates CO 3 2-, PO 4 3- Except alkali metal and ammonium salts Slide 14 of 43

5-3 Acid-Base Reactions Latin acidus (sour) Sour taste Arabic al-qali (ashes of certain plants) Bitter taste Svante Arrhenius 1884 Acid-Base theory. Slide 15 of 43

Acids Acids provide H + in aqueous solution. Strong acids: HCl(aq) Weak acids: CH 3 CO 2 H(aq) H + (aq) + Cl - (aq) H + (aq) + CH 3 CO 2- (aq) Slide 16 of 43

Bases Bases provide OH - in aqueous solution. Strong bases: NaOH(aq) Weak bases: H 2 O Na + (aq) + OH - (aq) NH 3 (aq) + H 2 O(l) OH - (aq) + NH 4+ (aq) Slide 17 of 43

Recognizing Acids and Bases. Acids have ionizable hydrogen ions. CH 3 CO 2 H or HC 2 H 3 O 2 Bases have OH - combined with a metal ion. KOH or are identified by chemical equations Na 2 CO 3 (s) + H 2 O(l) HCO 3- (aq) + 2 Na + (aq) + OH - (aq) Slide 18 of 43

More Acid-Base Reactions Milk of magnesia Mg(OH) 2 Mg(OH) 2 (s) + 2 H + (aq) Mg 2+ (aq) + 2 H 2 O(l) Mg(OH) 2 (s) + 2 CH 3 CO 2 H(aq) Mg 2+ (aq) + 2 CH 3 CO 2- (aq) + 2 H 2 O(l) Slide 19 of 43

More Acid-Base Reactions Limestone and marble. CaCO 3 (s) + 2 H + (aq) Ca 2+ (aq) + H 2 CO 3 (aq) But: H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) CaCO 3 (s) + 2 H + (aq) Ca 2+ (aq) + H 2 O(l) + CO 2 (g) Slide 20 of 43

Limestone and Marble Slide 21 of 43

Gas Forming Reactions Slide 22 of 43

5-4 Oxidation-Reduction: Some General Principles Hematite is converted to iron in a blast furnace. Fe 2 O 3 (s) + 3 CO(g) 2 Fe(l) + 3 CO 2 (g) Oxidation and reduction always occur together. Fe 3+ is reduced to metallic iron. CO(g) is oxidized to carbon dioxide. Slide 23 of 43

Oxidation State Changes Assign oxidation states: 3+ 2-2+ 2-0 4+ 2- Fe 2 O 3 (s) + 3 CO(g) 2 Fe(l) + 3 CO 2 (g) Fe 3+ is reduced to metallic iron. CO(g) is oxidized to carbon dioxide. Slide 24 of 43

Zinc in Copper Sulfate Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Slide 25 of 43

Half-Reactions Represent a reaction by two half-reactions. Oxidation: Reduction: Zn(s) Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e- Cu(s) Overall: Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) Slide 26 of 43

Oxidation and Reduction Oxidation O.S. of some element increases in the reaction. Electrons are on the right of the equation Reduction O.S. of some element decreases in the reaction. Electrons are on the left of the equation. Slide 27 of 43

Balancing Oxidation-Reduction Equations Few can be balanced by inspection. Systematic approach required. The Half-Reaction (Ion-Electron) Method Slide 28 of 43

Example 5-6 Balancing the Equation for a Redox Reaction in Acidic Solution. The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution.. SO 3 2- (aq) + MnO 4- (aq) SO 4 2- (aq) + Mn 2+ (aq) Slide 29 of 43

Example 5-6 Determine the oxidation states: 4+ 7+ 6+ 2+ SO 2-3 (aq) + MnO 4- (aq) SO 2-4 (aq) + Mn 2+ (aq) Write the half-reactions: SO 3 2- (aq) SO 4 2- (aq) + 2 e - (aq) 5 e - (aq) +MnO 4- (aq) Mn 2+ (aq) Balance atoms other than H and O: Slide 30 of 43

Example 5-6 Balance O by adding H 2 O: H 2 O(l) + SO 3 2- (aq) SO 4 2- (aq) + 2 e - (aq) 5 e - (aq) +MnO 4- (aq) Mn 2+ (aq) + 4 H 2 O(l) Balance hydrogen by adding H + : H 2 O(l) + SO 3 2- (aq) SO 4 2- (aq) + 2 e - (aq) + 2 H + (aq) 8 H + (aq) + 5 e - (aq) +MnO 4- (aq) Mn 2+ (aq) + 4 H 2 O(l) Check that the charges are balanced: Add e - if necessary. Slide 31 of 43

Example 5-6 Multiply the half-reactions to balance all e - : 5 H 2 O(l) + 5 SO 2-3 (aq) 5 SO 2-4 (aq) + 10 e - (aq) + 10 H + (aq) 16 H + (aq) + 10 e - (aq) + 2 MnO 4- (aq) 2 Mn 2+ (aq) + 8 H 2 O(l) Add both equations and simplify: 5 SO 2-3 (aq) + 2 MnO 4- (aq) + 6H + (aq) 5 SO 2-4 (aq) + 2 Mn 2+ (aq) + 3 H 2 O(l) Check the balance! Slide 32 of 43

Balancing in Acid Write the equations for the half-reactions. Balance all atoms except H and O. Balance oxygen using H 2 O. Balance hydrogen using H +. Balance charge using e -. Equalize the number of electrons. Add the half reactions. Check the balance. Slide 33 of 43

Balancing in Basic Solution OH - appears instead of H +. Treat the equation as if it were in acid. Then add OH - to each side to neutralize H +. Remove H 2 O appearing on both sides of equation. Check the balance. Slide 34 of 43

5-6 Oxidizing and Reducing Agents. An oxidizing agent (oxidant ): Contains an element whose oxidation state decreases in a redox reaction A reducing agent (reductant): Contains an element whose oxidation state increases in a redox reaction. Slide 35 of 43

Redox Slide 36 of 43

Example 5-8 Identifying Oxidizing and Reducing Agents. Hydrogen peroxide, H 2 O 2, is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or a reducing agent. For the following reactions, identify whether hydrogen peroxide is an oxidizing or reducing agent. 5 H 2 O 2 (aq) + 2 MnO 4- (aq) + 6H + 8 H 2 O(l) + 2 Mn 2+ (aq) + 5 O 2 (g) Slide 37 of 43

Example 5-8 H 2 O 2 (aq) + 2 Fe 2+ (aq) + 2 H + 2 H 2 O(l) + 2 Fe 3+ (aq) Iron is oxidized and peroxide is reduced. 5 H 2 O 2 (aq) + 2 MnO 4- (aq) + 6 H + 8 H 2 O(l) + 2 Mn 2+ (aq) + 5 O 2 (g) Manganese is reduced and peroxide is oxidized. Slide 38 of 43

5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations. Titration Carefully controlled addition of one solution to another. Equivalence Point Both reactants have reacted completely. Indicators Substances which change colour near an equivalence point. Slide 39 of 43

Indicators Slide 40 of 43

Example 5-10 Standardizing a Solution for Use in Redox Titrations. A piece of iron wire weighing 0.1568 g is converted to Fe 2+ (aq) and requires 26.42 ml of a KMnO 4 (aq) solution for its titration. What is the molarity of the KMnO 4 (aq)? 5 Fe 2+ (aq) + MnO 4- (aq) + 8 H + (aq) 4 H 2 O(l) + 5 Fe 3+ (aq) + Mn 2+ (aq) Slide 41 of 43

Example 5-6 5 Fe 2+ (aq) + MnO 4- (aq) + 8 H + (aq) 4 H 2 O(l) + 5 Fe 3+ (aq) + Mn 2+ (aq) Determine KMnO 4 consumed in the reaction: n H O 2 = 0.1568gFe 1molFe 55.847gFe 1molFe 1molFe 2+ 4 2+ 1molMnO 5molFe 1molKMnO 1molMnO 4 4 = 5.615 10 4 molkmno 4 Determine the concentration: 4 5.615 10 mol KMnO4 [ KMnO 4] = = 0. 2140 M KMnO 0.2624 L 4 Slide 42 of 43

Chapter 5 Questions 1, 2, 3, 5, 6, 8, 14, 17, 19, 24, 27, 33, 37, 41, 43, 51, 53, 59, 68, 71, 82, 96. Slide 43 of 43