1. Chapter 4 Types of Chemical Reactions and Solution Stoichiometry

2. Section 4.3 The Composition of Solutions Reaction Stoichiometry  The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry  The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction

3. Section 4.3 The Composition of Solutions Reaction Stoichiometry  2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 molecules of C 8 H 18 react with 25 molecules of O 2 to form 16 molecules of CO 2 and 18 molecules of H 2 O 2 moles of C 8 H 18 react with 25 moles of O 2  to form 16 moles of CO 2 and 18 moles of H 2 O

4. Section 4.3 The Composition of Solutions Predicting Amounts from Stoichiometry  The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance  How much CO 2 can be made from 22.0 moles of C 8 H 18 in the combustion of C 8 H 18 ? 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 moles C 8 H 18 : 16 moles CO 2

5. Section 4.3 The Composition of Solutions Practice  According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

6. Section 4.3 The Composition of Solutions  Homogeneous mixtures are called solutions  The component of the solution that changes state is called the solute  The component that keeps its state is called the solvent  if both components start in the same state, the major component is the solvent

7. Section 4.3 The Composition of Solutions  Qualitatively, solutions are often described as dilute or concentrated  Dilute solutions have a small amount of solute compared to solvent  Concentrated solutions have a large amount of solute compared to solvent

8. Section 4.3 The Composition of Solutions Concentrations—Quantitative Descriptions of Solutions  A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution  Concentration = amount of solute in a given amount of solution

9. Section 4.3 The Composition of Solutions Solution Concentration: MOLARITY  Moles of solute per 1 liter of solution  Used because it describes how many molecules of solute in each liter of solution

10. Section 4.3 The Composition of Solutions Molarity  Molarity (M) = moles of solute per volume of solution in liters: Copyright © Cengage Learning. All rights reserved 10

11. Section 4.3 The Composition of Solutions A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 1.57 M Copyright © Cengage Learning. All rights reserved 11 EXERCISE!

12. Section 4.3 The Composition of Solutions Concentration of Ions  For a 0.25 M CaCl 2 solution: CaCl 2 → Ca 2+ + 2Cl –  Ca 2+ : 1 × 0.25 M = 0.25 M Ca 2+  Cl – : 2 × 0.25 M = 0.50 M Cl –. Copyright © Cengage Learning. All rights reserved 12

13. Section 4.3 The Composition of Solutions Dilution  The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution.  Dilution with water does not alter the numbers of moles of solute present.  Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2 Copyright © Cengage Learning. All rights reserved 13

14. Section 4.3 The Composition of Solutions A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution? a)Add water to the solution. b)Pour some of the solution down the sink drain. c)Add more sodium chloride to the solution. d)Let the solution sit out in the open air for a couple of days. e)At least two of the above would decrease the concentration of the salt solution. Copyright © Cengage Learning. All rights reserved 14 CONCEPT CHECK!

15. Section 4.3 The Composition of Solutions What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? 60.0 mL Copyright © Cengage Learning. All rights reserved 15 EXERCISE!

16. Section 4.4 Types of Chemical Reactions  Precipitation Reactions  Acid–Base Reactions  Oxidation–Reduction Reactions Copyright © Cengage Learning. All rights reserved 16

17. Section 4.4 Types of Chemical Reactions What Happens When a Solute Dissolves?  There are attractive forces between the solute particles holding them together  There are also attractive forces between the solvent molecules  When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules  If the attractions between solute and solvent are strong enough, the solute will dissolve

18. Section 4.4 Types of Chemical Reactions When Will a Salt Dissolve?  Predicting whether a compound will dissolve in water is not easy  The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results  we call this method the empirical method

19. Section 4.4 Types of Chemical Reactions SOLUBILITY RULES Compounds that Are Generally Soluble in Water

20. Section 4.4 Types of Chemical Reactions SOLUBILITY RULES Compounds that Are Generally Insoluble in Water

21. Section 4.4 Types of Chemical Reactions Determine if each of the following is soluble in water  KOH  AgBr  CaCl 2  Pb(NO 3 ) 2  PbSO 4 KOH is soluble because it contains K + AgBr is insoluble; most bromides are soluble, but AgBr is an exception CaCl 2 is soluble; most chlorides are soluble, and CaCl 2 is not an exception Pb(NO 3 ) 2 is soluble because it contains NO 3 − PbSO 4 is insoluble; most sulfates are soluble, but PbSO 4 is an exception

22. Section 4.5 Precipitation Reactions Precipitation Reaction  Precipitation reactions are reactions in which a solid forms when we mix two solutions  reactions between aqueous solutions of ionic compounds  produce an ionic compound that is insoluble in water  the insoluble product is called a precipitate Copyright © Cengage Learning. All rights reserved 22

23. Section 4.5 Precipitation Reactions The Reaction of K 2 CrO 4 (aq) and Ba(NO 3 ) 2 (aq)  Ba 2+ (aq) + CrO 4 2– (aq) → BaCrO 4 (s) Copyright © Cengage Learning. All rights reserved 23

24. Section 4.5 Precipitation Reactions Precipitates  Soluble – solid dissolves in solution; (aq) is used in reaction equation.  Insoluble – solid does not dissolve in solution; (s) is used in reaction equation.  Insoluble and slightly soluble are often used interchangeably. Copyright © Cengage Learning. All rights reserved 24

25. Section 4.5 Precipitation Reactions No Precipitate Formation = No Reaction  KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) all ions still present,  no reaction

26. Section 4.5 Precipitation Reactions Process for Predicting the Products of a Precipitation Reaction 1.Determine what ions each aqueous reactant has 2.Determine formulas of possible products  exchange ions  (+) ion from one reactant with (-) ion from other  balance charges of combined ions to get formula of each product 3.Determine solubility of each product in water  use the solubility rules  if product is insoluble or slightly soluble, it will precipitate 4.If neither product will precipitate, write no reaction after the arrow

27. Section 4.5 Precipitation Reactions Process for Predicting the Products of a Precipitation Reaction 5.If any of the possible products are insoluble, write their formulas as the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate aqueous. 6.Balance the equation  remember to only change coefficients, not subscripts

28. Section 4.5 Precipitation Reactions Which of the following ions form compounds with Pb 2+ that are generally soluble in water? a)S 2– b)Cl – c)NO 3 – d)SO 4 2– e)Na + Copyright © Cengage Learning. All rights reserved 28 CONCEPT CHECK!

29. Section 4.5 Precipitation Reactions Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 1.Write the formulas of the reactants K 2 CO 3 (aq) + NiCl 2 (aq)  2.Determine the possible products a)determine the ions present (K + + CO 3 2− ) + (Ni 2+ + Cl − )  b)exchange the Ions (K + + CO 3 2− ) + (Ni 2+ + Cl − )  (K + + Cl − ) + (Ni 2+ + CO 3 2− ) c)write the formulas of the products  balance charges K 2 CO 3 (aq) + NiCl 2 (aq)  KCl + NiCO 3

30. Section 4.5 Precipitation Reactions 5.Write (aq) next to soluble products and (s) next to insoluble products K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s) 6.Balance the equation K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s)

31. Section 4.5 Precipitation Reactions 5.Write (aq) next to soluble products and (s) next to insoluble products K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s) 6.Balance the equation K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s)

32. Section 4.5 Precipitation Reactions Ionic Equations Equations that describe the material’s structure when dissolved are called complete ionic equations  2K + (aq) + 2OH − (aq) + Mg 2+ (aq) + 2NO 3 − (aq)  K + (aq) + 2NO 3 − (aq) + Mg(OH) 2(s)  Ions that are both reactants and products are called spectator ions 2 K + (aq) + 2 OH − (aq) + Mg 2+ (aq) + 2 NO 3 − (aq)  2 K + (aq) + 2 NO 3 − (aq) + Mg(OH) 2(s)  An ionic equation in which the spectator ions are removed is called a net ionic equation 2 OH − (aq) + Mg 2+ (aq)  Mg(OH) 2(s)

33. Section 4.6 Describing Reactions in Solution Complete Ionic Equation  All substances that are strong electrolytes are represented as ions. Ag + (aq) + NO 3  (aq) + Na + (aq) + Cl  (aq) AgCl(s) + Na + (aq) + NO 3  (aq) Copyright © Cengage Learning. All rights reserved 33

34. Section 4.6 Describing Reactions in Solution Net Ionic Equation  Includes only those solution components undergoing a change.  Show only components that actually react. Ag + (aq) + Cl  (aq)  AgCl(s)  Spectator ions are not included (ions that do not participate directly in the reaction).  Na + and NO 3  are spectator ions. Copyright © Cengage Learning. All rights reserved 34

35. Section 4.8 Acid-Base Reactions Acid–Base Reactions (Brønsted–Lowry)  Acid—proton donor  Base—proton acceptor  For a strong acid and base reaction: H + (aq) + OH – (aq)  H 2 O(l) Copyright © Cengage Learning. All rights reserved 35

36. Section 4.8 Acid-Base Reactions Acid–Base Titrations  Titration – delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).  Equivalence point – enough titrant added to react exactly with the analyte.  Endpoint – the indicator changes color so you can tell the equivalence point has been reached. Copyright © Cengage Learning. All rights reserved 36

37. Section 4.9 Oxidation-Reduction Reactions Redox Reactions  Reactions in which one or more electrons are transferred. Copyright © Cengage Learning. All rights reserved 37  To convert a free element into an ion, the atoms must gain or lose electrons  of course, if one atom loses electrons, another must accept them  Reactions where electrons are transferred from one atom to another are redox reactions  Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced

38. Section 4.9 Oxidation-Reduction Reactions Rules for Assigning Oxidation States Copyright © Cengage Learning. All rights reserved 38  Rules are in order of priority 1.free elements have an oxidation state = 0  Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2.monatomic ions have an oxidation state equal to their charge  Na = +1 and Cl = −1 in NaCl 3.(a) the sum of the oxidation states of all the atoms in a compound is 0  Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0

39. Section 4.9 Oxidation-Reduction Reactions 3.(b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion  N = +5 and O = −2 in NO 3 –, (+5) + 3(−2) = −1 4.(a) Group I metals have an oxidation state of +1 in all their compounds  Na = +1 in NaCl 4.(b) Group II metals have an oxidation state of +2 in all their compounds  Mg = +2 in MgCl 2

40. Section 4.9 Oxidation-Reduction Reactions 5.5.in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority

41. Section 4.9 Oxidation-Reduction Reactions Find the oxidation states for each of the elements in each of the following compounds:  K 2 Cr 2 O 7  CO 3 2-  MnO 2  PCl 5  SF 4 Copyright © Cengage Learning. All rights reserved 41 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1 EXERCISE!

42. Section 4.9 Oxidation-Reduction Reactions Redox Characteristics  Transfer of electrons  Transfer may occur to form ions  Oxidation – increase in oxidation state (loss of electrons); reducing agent  Reduction – decrease in oxidation state (gain of electrons); oxidizing agent Copyright © Cengage Learning. All rights reserved 42

43. Section 4.9 Oxidation-Reduction Reactions Which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent. a)Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) b)Cr 2 O 7 2- (aq) + 2OH - (aq) 2CrO 4 2- (aq) + H 2 O(l) c)2CuCl(aq) CuCl 2 (aq) + Cu(s) Copyright © Cengage Learning. All rights reserved CONCEPT CHECK!

44. Section 4.10 Balancing Oxidation-Reduction Equations Balancing Oxidation–Reduction Reactions by Oxidation States 1.Write the unbalanced equation. 2.Determine the oxidation states of all atoms in the reactants and products. 3.Show electrons gained and lost using “tie lines.” 4.Use coefficients to equalize the electrons gained and lost. 5.Balance the rest of the equation by inspection. 6.Add appropriate states. Copyright © Cengage Learning. All rights reserved 44

45. Section 4.10 Balancing Oxidation-Reduction Equations  Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc(II) chloride and hydrogen gas. Copyright © Cengage Learning. All rights reserved 45

46. Section 4.10 Balancing Oxidation-Reduction Equations 1. What is the unbalanced equation?  Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) Copyright © Cengage Learning. All rights reserved 46

47. Section 4.10 Balancing Oxidation-Reduction Equations 2. What are the oxidation states for each atom?  Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) 0 +1 –1 +2 –1 0 Copyright © Cengage Learning. All rights reserved 47

48. Section 4.10 Balancing Oxidation-Reduction Equations 3. How are electrons gained and lost? 1 e – gained (each atom)  Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) 0 +1 –1 +2 –1 0 2 e – lost  The oxidation state of chlorine remains unchanged. Copyright © Cengage Learning. All rights reserved 48

49. Section 4.10 Balancing Oxidation-Reduction Equations 4. What coefficients are needed to equalize the electrons gained and lost? 1 e – gained (each atom) × 2  Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) 0 +1 –1 +2 –1 0 2 e – lost  Zn(s) + 2HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) Copyright © Cengage Learning. All rights reserved 49

50. Section 4.10 Balancing Oxidation-Reduction Equations 5. What coefficients are needed to balance the remaining elements?  Zn(s) + 2HCl(aq) Zn 2+ (aq) + 2Cl – (aq) + H 2 (g) Copyright © Cengage Learning. All rights reserved 50