Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 1 of 47 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 5: Introduction to Reactions in Aqueous Solutions

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 2 of 47 Contents 5-1The Nature of Aqueous Solutions 5-2Precipitation Reactions 5-3Acid-Base Reactions 5-4Oxidation-Reduction: Some General Principles 5-5Balancing Oxidation-Reduction Equations 5-6Oxidizing and Reducing Agents 5-7Stoichiometry of Reactions in Aqueous Solutions: Titrations  Focus On Water Treatment

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 3 of The Nature of Aqueous Solutions  Water  Inexpensive  Can dissolve a vast number of substances  Many substances dissociate into ions  Aqueous solutions are found everywhere ◦Seawater ◦Living systems

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 4 of 47 Electrolytes  Some solutes can dissociate into ions.  Electric charge can be carried.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 5 of 47 Types of Electrolytes  Weak electrolyte partially dissociates.  Fair conductor of electricity.  Non-electrolyte does not dissociate.  Poor conductor of electricity.  Strong electrolyte dissociates completely.  Good electrical conduction.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 6 of 47 Representation of Electrolytes using Chemical Equations MgCl 2 (s) → Mg 2+ (aq) + 2 Cl - (aq) A strong electrolyte: A weak electrolyte: CH 3 CO 2 H(aq) ← CH 3 CO 2 - (aq) + H + (aq) → CH 3 OH(aq) A non-electrolyte:

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 7 of 47 Three Types of Electrolytes

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 8 of 47 Solvation: The Hydrated Proton

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 9 of 47 MgCl 2 (s) → Mg 2+ (aq) + 2 Cl - (aq) [Mg 2+ ] = M [Cl - ] = M [MgCl 2 ] = 0 M Relative Concentrations in Solution In M MgCl 2 : Stoichiometry is important.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 10 of 47 Calculating Ion concentrations in a Solution of a Strong Electolyte. What are the aluminum and sulfate ion concentrations in M Al 2 (SO 4 ) 3 ?. Al 2 (SO 4 ) 3 (s) → 2 Al 3+ (aq) + 3 SO 4 2- (aq) Write a Balanced Chemical Equation: EXAMPLE 5-1 Identify the Stoichiometric Factors : 2 mol Al 3+ 1 mol Al 2 (SO 4 ) 3 3 mol SO mol Al 2 (SO 4 ) 3

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 11 of 47 [Al] =  = 1 L 2 mol Al 3+ 1 mol Al 2 (SO 4 ) mol Al 2 (SO 4 ) M SO 4 2- [SO 4 2- ] =  = 1 mol Al 2 (SO 4 ) 3 Sulfate Concentration: 1 L 3 mol SO mol Al 2 (SO 4 ) 3 Aluminum Concentration: EXAMPLE 7-3 mol Al 3+ 1 L

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 12 of Precipitation Reactions  Soluble ions can combine to form an insoluble compound.  Precipitation occurs.  A test for the presence of chloride ion in water. Ag + (aq) + Cl - (aq) → AgCl(s)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 13 of 47 Silver Nitrate and Sodium Iodide AgNO 3 (aq)AgI(s) Na + (aq) NO 3 - (aq) NaI(aq)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 14 of 47 Ag + (aq) + NO 3 - (aq) + Na + (aq) + I - (aq) → AgI(s) + Na + (aq) + NO 3 - (aq) Spectator ions Ag + (aq) + NO 3 - (aq) + Na + (aq) + I - (aq) → AgI(s) + Na + (aq) + NO 3 - (aq) Net Ionic Equation AgNO 3 (aq) +NaI(aq) → AgI(s) + NaNO 3 (aq) Overall Precipitation Reaction: Complete ionic equation: Ag + (aq) + I - (aq) → AgI(s) Net ionic equation:

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 15 of 47 Solubility Rules  Compounds that are soluble: Li +, Na +, K +, Rb +, Cs + NH 4 + NO 3 - ClO 4 - CH 3 CO 2 -  Alkali metal ion and ammonium ion salts  Nitrates, perchlorates and acetates

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 16 of 47 Solubility Rules  Chlorides, bromides and iodides Cl - Br - I - ◦Except those of Pb 2+, Ag +, and Hg  Sulfates SO 4 2- ◦Except those of Sr 2+, Ba 2+, Pb 2+ and Hg ◦Ca(SO 4 ) is slightly soluble.  Compounds that are mostly soluble:

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 17 of 47  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  Silver, Lead and Mercury Ag + Pb 2+ Hg 2 2+ ◦Except nitrates, acetates and perchlorates  Compounds that are insoluble: Solubility Rules

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 18 of Acid-Base Reactions  Latin acidus (sour)  Sour taste  Arabic al-qali (ashes of certain plants)  Bitter taste  Svante Arrhenius 1884 Acid-Base theory.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 19 of 47 Acids  Acids provide H + in aqueous solution.  Strong acids completely ionize:  Weak acid ionization is not complete: HCl(aq) H + (aq) + Cl - (aq) → → ← CH 3 CO 2 H(aq) H + (aq) + CH 3 CO 2 - (aq)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 20 of 47 Bases  Bases provide OH - in aqueous solution.  Strong bases:  Weak bases: → ← NH 3 (aq) + H 2 O(l) OH - (aq) + NH 4 + (aq) NaOH(aq) Na + (aq) + OH - (aq) → H2OH2O

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 21 of 47 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 can be identified by chemical equations Na 2 CO 3 (s) + H 2 O(l)→ HCO 3 - (aq) + 2 Na + (aq) + OH - (aq)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 22 of 47 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)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 23 of 47 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) CaCO 3 (s) + 2 H + (aq) → Ca 2+ (aq) + H 2 CO 3 (aq) CaCO 3 (s) + 2 H + (aq) → Ca 2+ (aq) + H 2 O(l) + CO 2 (g)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 24 of 47

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 25 of 47 Fe 2 O 3 (s) + 3 CO(g) → 2 Fe(l) + 3 CO 2 (g)   Hematite is converted to iron in a blast furnace. Fe 2 O 3 (s) + 3 CO(g) → 2 Fe(l) + 3 CO 2 (g)   CO(g) is oxidized to carbon dioxide. Fe 3+ is reduced to metallic iron. 5-4 Oxidation-Reduction: Some General Principles  Oxidation and reduction always occur together.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 26 of 47 Oxidation State Changes Fe 2 O 3 (s) + 3 CO(g) → 2 Fe(l) + 3 CO 2 (g)   Assign oxidation states: CO(g) is oxidized to carbon dioxide. Fe 3+ is reduced to metallic iron.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 27 of 47 Thermite Reaction Fe 2 O 3 (s) + 2 Al(s)  Al 2 O 3 (s) + 2 Fe(l)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 28 of 47 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.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 29 of 47 An Oxidation Reduction Reaction Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 30 of 47 Oxidation and Reduction Half-Reactions  A reaction represented by two half-reactions. Oxidation: Reduction: Overall: Zn(s) → Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e- → Cu(s) Cu 2+ (aq) + Zn(s) → Cu(s) + Zn 2+ (aq)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 31 of 47 Balancing Oxidation-Reduction Equations  Few can be balanced by inspection.  Systematic approach required.  The Half-Reaction (Ion-Electron) Method

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 32 of 47 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.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 33 of 47 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) EXAMPLE 5-6

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 34 of 47 SO 3 2- (aq) + MnO 4 - (aq) → SO 4 2- (aq) + Mn 2+ (aq) Determine the oxidation states: SO 3 2- (aq) → SO 4 2- (aq) + 2 e - (aq) Write the half-reactions: 5 e - (aq) +MnO 4 - (aq) → Mn 2+ (aq) Balance atoms other than H and O: Already balanced for elements. EXAMPLE 5-6

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 35 of 47 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. EXAMPLE 5-6

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 36 of 47 Multiply the half-reactions to balance all e - : 5 H 2 O(l) + 5 SO 3 2- (aq) → 5 SO 4 2- (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 3 2- (aq) + 2 MnO 4 - (aq) + 6H + (aq) → 5 SO 4 2- (aq) + 2 Mn 2+ (aq) + 3 H 2 O(l) Check the balance! EXAMPLE

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 37 of 47 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.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 38 of 47 Disproportionation Reactions  The same substance is both oxidized and reduced.  Some have practical significance  Hydrogen peroxide  Sodium thiosulphate 2 H 2 O 2 (aq)  H 2 O(l) + O 2 (g) 2 S 2 O 3 (aq) + H + (aq)  S(s) + SO 2 (g) + H 2 O(l)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 39 of 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.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 40 of 47 Oxidation States of Nitrogen

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 41 of 47 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. EXAMPLE 5-8

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 42 of 47 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) 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. Manganese is reduced and peroxide is oxidized. EXAMPLE

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 43 of 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.

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 44 of 47 Indicators 5.0 mL CH 3 CO 2 H A few drops phenolpthalein Add M NaOHThe “endpoint” (close to the equivalence point)

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 45 of 47 Standardizing a Solution for Use in Redox Titrations. A piece of iron wire weighing g is converted to Fe 2+ (aq) and requires 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) → 5 Fe 3+ (aq) + Mn 2+ (aq) + 4 H 2 O(l) EXAMPLE 5-10

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 46 of 47 Determine KMnO 4 consumed in the reaction: Determine the concentration: 5 Fe 2+ (aq) + MnO 4 - (aq) + 8 H + (aq) → 4 H 2 O(l) + 5 Fe 3+ (aq) + Mn 2+ (aq) EXAMPLE 5-10 n MnO = g Fe 1 mol Fe1 mol Fe g Fe 1 mol MnO 4 1 mol Fe 1 mol KMnO 4 5 mol Fe     - - =  mol KMnO M KMnO 4 c KMnO = L  mol KMnO 4 =

Prentice-Hall © 2007 General Chemistry: Chapter 5 Slide 47 of 47 d? Answer End of Chapter Questions  Try a different line of reasoning if you are stumped on a problem.  Practice Lateral Thinking. d e f a b c ~ ~ ???