1. Modified from Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chapter 4 Chemical Quantities and Aqueous Reactions

2. Global Warming Scientists have measured an average 0.6 °C rise in atmospheric temperature since 1860 During the same period atmospheric CO 2 levels have risen 25% Are the two trends causal?

3. The Sources of Increased CO 2 One source of CO 2 is the combustion reactions of fossil fuels we use to get energy Another source of CO 2 is volcanic action How can we judge whether global warming is natural or due to our use of fossil fuels?

4. Quantities in Chemical Reactions The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction – Law of Conservation of Mass – Balancing equations by balancing atoms The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry

5. Reaction Stoichiometry The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 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 2 mol C 8 H 18 : 25 mol O 2 : 16 mol CO 2 : 18 mol H 2 O

6. Making Pizza The number of pizzas you can make depends on the amount of the ingredients you use This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza 1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza If you want to make more or less than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make assuming you have enough crusts and tomato sauce

7. 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

8. 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

9. Practice  How many moles of water are made in the combustion of 0.10 moles of glucose? 0.6 mol H 2 O = 0.60 mol H 2 O because 6x moles of H 2 O as C 6 H 12 O 6, the number makes sense C 6 H 12 O 6 + 6 O 2 → 6 CO 2 + 6 H 2 O  1 mol C 6 H 12 O 6 : 6 mol H 2 O 0.10 moles C 6 H 12 O 6 moles H 2 O Check: Solution: Conceptual Plan: Relationships: Given: Find: mol H 2 Omol C 6 H 12 O 6

10. Example: Estimate the mass of CO 2 produced in 2007 by the combustion of 3.5 x 10 15 g gasolne Assuming that gasoline is octane, C 8 H 18, the equation for the reaction is 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) The equation for the reaction gives the mole relationship between amount of C 8 H 18 and CO 2, but we need to know the mass relationship, so the conceptual plan will be g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18

11. Example: Estimate the mass of CO 2 produced in 2007 by the combustion of 3.5 x 10 15 g gasoline because 8x moles of CO 2 as C 8 H 18, but the molar mass of C 8 H 18 is 3x CO 2, the number makes sense 1 mol C 8 H 18 = 114.22g, 1 mol CO 2 = 44.01g, 2 mol C 8 H 18 :16 mol CO 2 3.4 x 10 15 g C 8 H 18 g CO 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18

12. Which Produces More CO 2 ; Volcanoes or Fossil Fuel Combustion? Our calculation just showed that the world produced 1.1 x 10 16 g of CO 2 just from petroleum combustion in 2007 – 1.1 x 10 13 kg CO 2 Estimates of volcanic CO 2 production are 2 x 10 11 kg/year This means that volcanoes produce less than 2% of the CO 2 added to the air annually

13. Example 4.1: How many grams of glucose can be synthesized from 37.8 g of CO 2 in photosynthesis? because 6x moles of CO 2 as C 6 H 12 O 6, but the molar mass of C 6 H 12 O 6 is 4x CO 2, the number makes sense 1 mol C 6 H 12 O 6 = 180.2g, 1 mol CO 2 = 44.01g, 1 mol C 6 H 12 O 6 : 6 mol CO 2 37.8 g CO 2, 6 CO 2 + 6 H 2 O  C 6 H 12 O 6 + 6 O 2 g C 6 H 12 O 6 Check: Solution: Conceptual Plan: Relationships: Given: Find: g 44.01 mol 1 6126 6 6 6 6 2 6 6 2 2 2 OHC g 5.82 OHC mol 1 OHC g 180.2 CO mol 6 OHC 1 CO g 44.01 CO mol 1 CO g 7.83   2 6126 CO mol 6 OHC 1 g C 6 H 12 O 6 mol CO 2 g CO 2 mol C 6 H 12 O 6 mol 1 g 180.2

14. Practice — How many grams of O 2 can be made from the decomposition of 100.0 g of PbO 2 ? 2 PbO 2 (s) → 2 PbO(s) + O 2 (g) (PbO 2 = 239.2, O 2 = 32.00)

15. Practice — How many grams of O 2 can be made from the decomposition of 100.0 g of PbO 2 ? 2 PbO 2 (s) → 2 PbO(s) + O 2 (g) because ½ moles of O 2 as PbO 2, and the molar mass of PbO 2 is 7x O 2, the number makes sense 1 mol O 2 = 32.00g, 1 mol PbO 2 = 239.2g, 1 mol O 2 : 2 mol PbO 2 100.0 g PbO 2, 2 PbO 2 → 2 PbO + O 2 g O 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: g O 2 mol PbO 2 g PbO 2 mol O 2

16. Stoichiometry Road Map a A  b B Moles A Moles B mass A mass B volume A (l) volume B(l) Pure Substance Solution Volume A(g) Volume B(g) % A(aq) ppm A(aq) % B(aq) ppm B(aq) M A(aq) M B(aq) MM density equation 22.4 L M = moles L

17. More Making Pizzas We know that 1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza But what would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese?

18. More Making Pizzas, Continued Each ingredient could potentially make a different number of pizzas But all the ingredients have to work together! We only have enough tomato sauce to make three pizzas, so once we make three pizzas, the tomato sauce runs out no matter how much of the other ingredients we have.

19. More Making Pizzas, Continued The tomato sauce limits the amount of pizzas we can make. In chemical reactions we call this the limiting reactant. – also known as the limiting reagent The maximum number of pizzas we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield. – it also determines the amounts of the other ingredients we will use!

20. The Limiting Reactant For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others When this reactant is used up, the reaction stops and no more product is made The reactant that limits the amount of product is called the limiting reactant – sometimes called the limiting reagent – the limiting reactant gets completely consumed Reactants not completely consumed are called excess reactants The amount of product that can be made from the limiting reactant is called the theoretical yield

21. Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) Our balanced equation for the combustion of methane implies that every one molecule of CH 4 reacts with two molecules of O 2

22. Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) since less CO 2 can be made from the O 2 than the CH 4, so the O 2 is the limiting reactant If we have five molecules of CH 4 and eight molecules of O 2, which is the limiting reactant?

23. Practice — How many moles of Si 3 N 4 can be made from 1.20 moles of Si and 1.00 moles of N 2 in the reaction 3 Si + 2 N 2  Si 3 N 4 ?

24. Limiting reactant Practice — How many moles of Si 3 N 4 can be made from 1.20 moles of Si and 1.00 moles of N 2 in the reaction 3 Si + 2 N 2  Si 3 N 4 ? 2 mol N 2 : 1 Si 3 N 4 ; 3 mol Si : 1 Si 3 N 4 1.20 mol Si, 1.00 mol N 2 mol Si 3 N 4 Solution: Conceptual Plan: Relationships: Given: Find: mol N 2 mol Si 3 N 4 mol Simol Si 3 N 4 Pick least amount Limiting reactant and theoretical yield Theoretical yield

25. More Making Pizzas Let’s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.

26. Theoretical and Actual Yield As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make The theoretical yield will always be the least possible amount of product – the theoretical yield will always come from the limiting reactant Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield

27. Example 4.4: Finding limiting reactant, theoretical yield, and percent yield

28. Example: When 28.6 kg of C are allowed to react with 88.2 kg of TiO 2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.

29. Example: When 28.6 kg of C reacts with 88.2 kg of TiO 2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write down the given quantity and its units Given:28.6 kg C 88.2 kg TiO 2 42.8 kg Ti produced

30. Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write down the quantity to find and/or its units Find: limiting reactant theoretical yield percent yield Information Given: 28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti

31. Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write a conceptual plan Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. kg TiO 2 kg C } smallest amount is from limiting reactant smallest mol Ti

32. Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Collect needed relationships 1000 g = 1 kg Molar Mass TiO 2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO 2 : 1 mol Ti (from the chem. equation) 2 mole C : 1 mol Ti (from the chem. equation) Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti

33. Apply the conceptual plan Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) smallest moles of Ti limiting reactant

34. Apply the conceptual plan theoretical yield Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

35. Apply the conceptual plan Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

36. Check the solutions limiting reactant = TiO 2 theoretical yield = 52.9 kg percent yield = 80.9% Because Ti has lower molar mass than TiO 2, the T.Y. makes sense and the percent yield makes sense as it is less than 100% Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the limiting reactant, theoretical yield, and percent yield TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

37. Practice — How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) If 4.61 g of N 2 are made, what is the percent yield?

38. 1 mol NH 3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N 2 = 28.02 g 2 mol NH 3 : 1 mol N 2, 3 mol CuO : 1 mol N 2 9.05 g NH 3, 45.2 g CuO g N 2 Conceptual Plan: Relationships: Given: Find: g NH 3 mol N 2 mol NH 3 g N 2 g CuOmol N 2 mol CuO Choose smallest

39. because the percent yield is less than 100, the answer makes sense Check: Solution: Theoretical yield Practice — How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) If 4.61 g of N 2 are made, what is the percent yield?

40. Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous – the salt is still there, as you can tell from the taste, or simply boiling away the water 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

41. Describing Solutions Because solutions are mixtures, the composition can vary from one sample to another – pure substances have constant composition – saltwater samples from different seas or lakes have different amounts of salt So to describe solutions accurately, we must describe how much of each component is present – we saw that with pure substances, we can describe them with a single name because all samples are identical

42. Solution Concentration 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

43. 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 – occasionally amount of solvent

44. 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

45. Preparing 1 L of a 1.00 M NaCl Solution

46. Example 4.5: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution because most solutions are between 0 and 18 M, the answer makes sense 1 mol KBr = 119.00 g, M = moles/L 25.5 g KBr, 1.75 L solution molarity, M Check: Solution: Conceptual Plan: Relationships: Given: Find: g KBrmol KBr L sol’n M

47. Practice — What Is the molarity of a solution containing 3.4 g of NH 3 (MM 17.03) in 200.0 mL of solution?

48. the unit is correct, the number is reasonable because the fraction of moles is less than the fraction of liters Check: Solve: M = mol/L, 1 mol NH 3 = 17.03 g, 1 mL = 0.001 L Conceptual Plan: Relationships: 3.4 g NH 3, 200.0 mL solution M Given: Find: g NH 3 mol NH 3 mL sol’nL sol’n M 0.20 mol NH 3, 0.2000 L solution M

49. Using Molarity in Calculations Molarity shows the relationship between the moles of solute and liters of solution If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar – 2 liters = 4.0 moles sugar – 0.5 liters = 1.0 mole sugar 1 L solution : 2 moles sugar

50. Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH? because each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L 0.125 mol NaOH = 1 L solution 0.125 M NaOH, 0.255 mol NaOH liters, L Check: Solution: Conceptual Plan: Relationships : Given: Find: mol NaOHL sol’n

51. Practice — Determine the mass of CaCl 2 (MM = 110.98) in 1.75 L of 1.50 M solution

52. because each L has 1.50 mol CaCl 2, it makes sense that 1.75 L should have almost 3 moles 1.50 mol CaCl 2 = 1 L solution; 110.98 g CaCl 2 = 1 mol 1.50 M CaCl 2, 1.75 L mass CaCl 2, g Check: Solution: Conceptual Plan: Relationships : Given: Find: mol CaCl 2 L sol’ng CaCl 2

53. Example: How would you prepare 250.0 mL of a 1.00 M solution CuSO 4  5 H 2 O(MM 249.69)? the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter Check: Solution: 1.00 L sol’n = 1.00 mol; 1 mL = 0.001 L; 1 mol = 249.69 g Conceptual Plan: Relationships: 250.0 mL solution mass CuSO 4  5 H 2 O, g Given: Find: mL sol’nL sol’n g CuSO 4 mol CuSO 4 Dissolve 62.4 g of CuSO 4 ∙5H 2 O in enough water to total 250.0 mL

54. Practice – How would you prepare 250.0 mL of 0.150 M CaCl 2 (MM = 110.98)?

55. Practice – How would you prepare 250.0 mL of 0.150 M CaCl 2 ? the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter Check: Solution: 1.00 L sol’n = 0.150 mol; 1 mL = 0.001L; 1 mol = 110.98 g Conceptual Plan: Relationships: 250.0 mL solution mass CaCl 2, g Given: Find: mL sol’nL sol’n g CaCl 2 mol CaCl 2 Dissolve 4.16 g of CaCl 2 in enough water to total 250.0 mL

56. Dilution Often, solutions are stored as concentrated stock solutions To make solutions of lower concentrations from these stock solutions, more solvent is added – the amount of solute doesn’t change, just the volume of solution moles solute in solution 1 = moles solute in solution 2 The concentrations and volumes of the stock and new solutions are inversely proportional M 1 ∙V 1 = M 2 ∙V 2

57. Example 4.7: To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH? because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does M 1 V 1 = M 2 V 2 V 1 = 0.200L, M 1 = 15.0 M, M 2 = 3.00 M V 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, M 1, M 2 V2V2

58. Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO 3 to 135.0 mL?

59. because the solution is diluted by a factor of 3, the molarity should decrease by a factor of 3, and it does M 1 V 1 = M 2 V 2 V 1 = 45.0 mL, M 1 = 8.25 M, V 2 = 135.0 mL M 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, M 1, V 2 M2M2

60. Practice – How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution?

61. because the solution is diluted by a factor of 8, the volume should increase by a factor of 8, and it does M 1 V 1 = M 2 V 2 M 1 = 2.0 M, M 2 = 0.25 M, V 2 = 200.0 mL V 1, L Check: Solution: Conceptual Plan: Relationships: Given: Find: M 1, M 2, V 2 V1V1 Dilute 25 mL of 2.0 M solution up to 200.0 mL

62. Solution Stoichiometry Because molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

63. Example 4.8: What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO 3 ) 2 in the reaction 2 KCl(aq) + Pb(NO 3 ) 2 (aq)  PbCl 2 (s) + 2 KNO 3 (aq)? because you need 2x moles of KCl as Pb(NO 3 ) 2, and the molarity of Pb(NO 3 ) 2 > KCl, the volume of KCl should be more than 2x the volume of Pb(NO 3 ) 2 1 L Pb(NO 3 ) 2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO 3 ) 2 : 2 mol KCl 0.150 M KCl, 0.150 L of 0.175 M Pb(NO 3 ) 2 L KCl Check: Solution: Conceptual Plan: Relationships: Given: Find: L Pb(NO 3 ) 2 mol KClL KClmol Pb(NO 3 ) 2

64. Practice — Solution stoichiometry 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH) 2 solution. What is the molarity of the base? 2 HCl + Ba(OH) 2  BaCl 2 + 2 H 2 O

65. Practice – 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH) 2 solution. What is the molarity of the base? 2 HCl(aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2 H 2 O(aq) the units are correct, the number makes sense because there are fewer moles than liters 1 mL= 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH) 2 : 2 mol HCl 43.8 mL of 0.107 M HCl, 37.6 mL Ba(OH) 2 M Ba(OH) 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: L HClmol Ba(OH) 2 mol HCl L Ba(OH) 2 M Ba(OH) 2 0.0438 L of 0.107 M HCl, 0.0376 L Ba(OH) 2 M Ba(OH) 2

66. 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

67. Table Salt Dissolving in Water Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity

68. Electrolytes and Nonelectrolytes Materials that dissolve in water to form a solution that will conduct electricity are called electrolytes Materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes

69. Molecular View of Electrolytes and Nonelectrolytes To conduct electricity, a material must have charged particles that are able to flow Electrolyte solutions all contain ions dissolved in the water – ionic compounds are electrolytes because they dissociate into their ions when they dissolve Nonelectrolyte solutions contain whole molecules dissolved in the water – generally, molecular compounds do not ionize when they dissolve in water the notable exception being molecular acids

70. Salt vs. Sugar Dissolved in Water ionic compounds dissociate into ions when they dissolve molecular compounds do not dissociate when they dissolve

71. Acids Acids are molecular compounds that ionize when they dissolve in water – the molecules are pulled apart by their attraction for the water – when acids ionize, they form H + cations and also anions The percentage of molecules that ionize varies from one acid to another Acids that ionize virtually 100% are called strong acids HCl(aq)  H + (aq) + Cl − (aq) Acids that only ionize a small percentage are called weak acids HF(aq)  H + (aq) + F − (aq)

72. Strong and Weak Electrolytes Strong electrolytes are materials that dissolve completely as ions – ionic compounds and strong acids – their solutions conduct electricity well Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions – weak acids – their solutions conduct electricity, but not well When compounds containing a polyatomic ion dissolve, the polyatomic ion stays together HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 − (aq)

73. Classes of Dissolved Materials

74. Dissociation and Ionization When ionic compounds dissolve in water, the anions and cations are separated from each other. This is called dissociation. Na 2 S(aq)  2 Na + (aq) + S 2- (aq) When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion Na 2 SO 4 (aq)  2 Na + (aq) + SO 4 2− (aq) When strong acids dissolve in water, the molecule ionizes into H + and anions H 2 SO 4 (aq)  2 H + (aq) + SO 4 2− (aq)

75. Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water CaCl 2 HNO 3 (NH 4 ) 2 CO 3 CaCl 2 (aq)  Ca 2+ (aq) + 2 Cl − (aq) HNO 3 (aq)  H + (aq) + NO 3 − (aq) (NH 4 ) 2 CO 3 (aq)  2 NH 4 + (aq) + CO 3 2− (aq)

76. Solubility of Ionic Compounds Some ionic compounds, such as NaCl, dissolve very well in water at room temperature Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature Compounds that dissolve in a solvent are said to be soluble, where as those that do not are said to be insoluble – NaCl is soluble in water, AgCl is insoluble in water – the degree of solubility depends on the temperature – even insoluble compounds dissolve, just not enough to be meaningful

77. 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

78. Solubility Rules Compounds that Are Generally Soluble in Water

79. Solubility Rules Compounds that Are Generally Insoluble in Water

80. Practice – 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

81. Precipitation Reactions 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

82. 2 KI(aq) + Pb(NO 3 ) 2 (aq)  PbI 2 (s) + 2 KNO 3 (aq)

83. No Precipitate Formation = No Reaction KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) all ions still present,  no reaction

84. 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

85. 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

86. Example 4.10: 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

87. Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 3.Determine the solubility of each product KCl is soluble NiCO 3 is insoluble 4.If both products are soluble, write no reaction does not apply because NiCO 3 is insoluble

88. Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 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)

89. Practice – Predict the products and balance the equation KCl(aq) + AgNO 3 (aq)  KNO 3 (aq) + AgCl(s) Na 2 S(aq) + CaCl 2 (aq)  NaCl(aq) + CaS(aq) No reaction (K + + Cl − ) + (Ag + + NO 3 − ) → (K + + NO 3 − ) + (Ag + + Cl − ) KCl(aq) + AgNO 3 (aq) → KNO 3 + AgCl (Na + + S 2− ) + (Ca 2+ + Cl − ) → (Na + + Cl − ) + (Ca 2+ + S 2− ) Na 2 S(aq) + CaCl 2 (aq) → NaCl + CaS KCl(aq) + AgNO 3 (aq)  Na 2 S(aq) + CaCl 2 (aq) 

90. Practice – Write an equation for the reaction that takes place when an aqueous solution of (NH 4 ) 2 SO 4 is mixed with an aqueous solution of Pb(C 2 H 3 O 2 ) 2. (NH 4 ) 2 SO 4 (aq) + Pb(C 2 H 3 O 2 ) 2 (aq)  NH 4 C 2 H 3 O 2 (aq) + PbSO 4 (s) (NH 4 + + SO 4 2− ) + (Pb 2+ + C 2 H 3 O 2 − ) → (NH 4 + + C 2 H 3 O 2 − ) + (Pb 2+ + SO 4 2− ) (NH 4 ) 2 SO 4 (aq) + Pb(C 2 H 3 O 2 ) 2 (aq)  (NH 4 ) 2 SO 4 (aq) + Pb(C 2 H 3 O 2 ) 2 (aq)  NH 4 C 2 H 3 O 2 + PbSO 4 (NH 4 ) 2 SO 4 (aq) + Pb(C 2 H 3 O 2 ) 2 (aq)  2 NH 4 C 2 H 3 O 2 (aq) + PbSO 4 (s)

91. Ionic Equations Equations that describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO 3 ) 2 (aq)  2 KNO 3 (aq) + Mg(OH) 2 (s) Equations that describe the material’s structure when dissolved are called complete ionic equations – aqueous strong electrolytes are written as ions soluble salts, strong acids, strong bases – insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form solids, liquids, and gases are not dissolved, therefore molecule form 2K + (aq) + 2OH − (aq) + Mg 2+ (aq) + 2NO 3 − (aq)  K + (aq) + 2NO 3 − (aq) + Mg(OH) 2(s)

92. Ionic Equations 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)

93. Practice – Write the ionic and net ionic equation for each K 2 SO 4 (aq) + 2 AgNO 3 (aq)  2 KNO 3 (aq) + Ag 2 SO 4 (s) 2K + (aq) + SO 4 2− (aq) + 2Ag + (aq) + 2NO 3 − (aq)  2K + (aq) + 2NO 3 − (aq) + Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2− (aq)  Ag 2 SO 4 (s) Na 2 CO 3 (aq) + 2 HCl(aq)  2 NaCl(aq) + CO 2 (g) + H 2 O(l) 2Na + (aq) + CO 3 2− (aq) + 2H + (aq) + 2Cl − (aq)  2Na + (aq) + 2Cl − (aq) + CO 2 (g) + H 2 O(l) CO 3 2− (aq) + 2 H + (aq)  CO 2 (g) + H 2 O(l) K 2 SO 4 (aq) + 2 AgNO 3 (aq)  2 KNO 3 (aq) + Ag 2 SO 4 (s) Na 2 CO 3 (aq) + 2 HCl(aq)  2 NaCl(aq) + CO 2 (g) + H 2 O(l)

94. Acid-Base Reactions Also called neutralization reactions because the acid and base neutralize each other’s properties 2 HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) The net ionic equation for an acid-base reaction is H + (aq) + OH  (aq)  H 2 O(l) – as long as the salt that forms is soluble in water

95. Acids and Bases in Solution Acids ionize in water to form H + ions – more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H 3 O + most chemists use H + and H 3 O + interchangeably Bases dissociate in water to form OH  ions – bases, such as NH 3, that do not contain OH  ions, produce OH  by pulling H off water molecules In the reaction of an acid with a base, the H + from the acid combines with the OH  from the base to make water The cation from the base combines with the anion from the acid to make the salt acid + base  salt + water

96. Common Acids

97. Common Bases

98. HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)

99. Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 1.Write the formulas of the reactants HNO 3 (aq) + Ca(OH) 2 (aq)  2.Determine the possible products a)determine the ions present when each reactant dissociates or ionizes (H + + NO 3 − ) + (Ca 2+ + OH − )  b)exchange the ions, H + combines with OH − to make H 2 O(l) (H + + NO 3 − ) + (Ca 2+ + OH − )  (Ca 2+ + NO 3 − ) + H 2 O(l) cwrite the formula of the salt (H + + NO 3 − ) + (Ca 2+ + OH − )  Ca(NO 3 ) 2 + H 2 O(l)

100. Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 3.Determine the solubility of the salt Ca(NO 3 ) 2 is soluble 4.Write an (s) after the insoluble products and an (aq) after the soluble products HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + H 2 O(l) 5.Balance the equation 2 HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 H 2 O(l)

101. Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 6.Dissociate all aqueous strong electrolytes to get complete ionic equation – not H 2 O 2 H + (aq) + 2 NO 3 − (aq) + Ca 2+ (aq) + 2 OH − (aq)   Ca 2+ (aq) + 2 NO 3 − (aq) + H 2 O(l) 7.Eliminate spectator ions to get net-ionic equation 2 H + (aq) + 2 OH − (aq)  H 2 O(l) H + (aq) + OH − (aq)  H 2 O(l)

102. Practice – Predict the products and balance the equation 2 HCl(aq) + Ba(OH) 2 (aq)  2 H 2 O(l) + BaCl 2 (aq) H 2 SO 4 (aq) + Sr(OH) 2 (aq)  2 H 2 O(l) + SrSO 4 (s) (H + + Cl − ) + (Ba 2+ + OH − ) → (H + + OH − ) + (Ba 2+ + Cl − ) HCl(aq) + Ba(OH) 2 (aq) → H 2 O(l) + BaCl 2 (H + + SO 4 2− ) + (Sr 2+ + OH − ) → (H + + OH − ) + (Sr 2+ + SO 4 2− ) H 2 SO 4 (aq) + Sr(OH) 2 (aq) → H 2 O(l) + SrSO 4 H 2 SO 4 (aq) + Sr(OH) 2 (aq) → 2 H 2 O(l) + SrSO 4 HCl(aq) + Ba(OH) 2 (aq)  H 2 SO 4 (aq) + Sr(OH) 2 (aq) 

103. Titration Often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio. – the unknown solution is added slowly from an instrument called a burette a long glass tube with precise volume markings that allows small additions of solution

104. Acid-Base Titrations The difficulty is determining when there has been just enough titrant added to complete the reaction – the titrant is the solution in the burette In acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity – the chemical is called an indicator At the endpoint of an acid-base titration, the number of moles of H + equals the number of moles of OH  – also known as the equivalence point

105. Titration

106. The titrant is the base solution in the burette. As the titrant is added to the flask, the H + reacts with the OH – to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color.

107. Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Write down the given quantity and its units Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH

108. Write down the quantity to find, and/or its units Find: concentration HCl, M Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH

109. Collect needed equations and conversion factors HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)  1 mole HCl = 1 mole NaOH 0.100 M NaOH  0.100 mol NaOH  1 L sol’n Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl

110. Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Write a conceptual plan Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Find:M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L mL NaOH L NaOH mol NaOH mol HCl mL HCl L HCl

111. Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Apply the conceptual plan Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Find:M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M = 1.25 x 10  3 mol HCl

112. Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Apply the conceptual plan Information Given:10.00 mL HCl 12.54 mL NaOH Find:M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M

113. Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given:10.00 mL HCl 12.54 mL NaOH Find:M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Check the solution HCl solution = 0.125 M The units of the answer, M, are correct. The magnitude of the answer makes sense because the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated.

114. Practice − What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H 2 SO 4 ? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O 50.0 mL27.5 mL 0.1015 M ? M 114

115. 1 mL= 0.001L, 1 LH 2 SO 4 = 0.1015mol, 2mol NaOH : 1mol H 2 SO 4 Practice — What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H 2 SO 4 ? 2 NaOH(aq) + H 2 SO 4 (aq)  Na 2 SO 4 (aq) + 2 H 2 O(aq) the units are correct, the number makes because the volume of NaOH is about ½ the H 2 SO 4, but the stoichiometry says you need twice the moles of NaOH as H 2 SO 4 50.0 mL of 0.1015 M H 2 SO 4, 27.5 mL NaOH M NaOH Check: Solution: Conceptual Plan: Relationships: Given: Find: 0.0500 L of 0.1015 M H 2 SO 4, 0.0275 L NaOH M NaOH L H 2 SO 4 mol H 2 SO 4 mol NaOH L NaOH M NaOH

116. Gas-Evolving Reactions Some reactions form a gas directly from the ion exchange K 2 S(aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + H 2 S(g) Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K 2 SO 3 (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + H 2 SO 3 (aq) H 2 SO 3  H 2 O(l) + SO 2 (g)

117. NaHCO 3 (aq) + HCl(aq)  NaCl(aq) + CO 2 (g) + H 2 O(l)

118. Compounds that Undergo Gas-Evolving Reactions

119. Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 1.Write the formulas of the reactants Na 2 CO 3 (aq) + HNO 3 (aq)  2.Determine the possible products a)determine the ions present when each reactant dissociates or ionizes (Na + + CO 3 2− ) + (H + + NO 3 − )  b)exchange the anions (Na + + CO 3 2 − ) + (H + + NO 3 − )  (Na + + NO 3 − ) + (H + + CO 3 2 − ) c)write the formula of compounds balance the charges Na 2 CO 3 (aq) + HNO 3 (aq)  NaNO 3 + H 2 CO 3

120. Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 3.Check to see if either product is H 2 S - No 4.Check to see if either product decomposes – Yes – H 2 CO 3 decomposes into CO 2 (g) + H 2 O(l) Na 2 CO 3 (aq) + HNO 3 (aq)  NaNO 3 + CO 2 (g) + H 2 O(l)

121. Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 5.Determine the solubility of other product NaNO 3 is soluble 6.Write an (s) after the insoluble products and an (aq) after the soluble products Na 2 CO 3 (aq) + 2 HNO 3 (aq)  NaNO 3 (aq) + CO 2 (g) + H 2 O(l) 7.Balance the equation Na 2 CO 3 (aq) + 2 HNO 3 (aq)  NaNO 3 (aq) + CO 2 (g) + H 2 O(l)

122. Practice – Predict the products and balance the equation 2 HCl(aq) + Na 2 SO 3 (aq)  H 2 O(l) + SO 2 (g) + 2 NaCl(aq) (H + + Cl − ) + (Na + + SO 3 2− ) → (H + + SO 3 2− ) + (Na + + Cl − ) HCl(aq) + Na 2 SO 3 (aq) → H 2 SO 3 + NaCl (H + + SO 4 2− ) + (Ca 2+ + S 2− ) → (H + + S 2− ) + (Ca 2+ + SO 4 2− ) H 2 SO 4 (aq) + CaS(aq) → H 2 S(g) + CaSO 4 H 2 SO 4 (aq) + CaS(aq) → H 2 S(g) + CaSO 4 (s) HCl(aq) + Na 2 SO 3 (aq)  H 2 SO 4 (aq) + CaS(aq)  HCl(aq) + Na 2 SO 3 (aq) → H 2 O(l) + SO 2 (g) + NaCl

123. Other Patterns in Reactions The precipitation, acid-base, and gas-evolving reactions all involve exchanging the ions in the solution Other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions – also known as redox reactions – many involve the reaction of a substance with O 2 (g) 4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s)

124. Combustion as Redox 2 H 2 (g) + O 2 (g)  2 H 2 O(g)

125. Redox without Combustion 2 Na(s) + Cl 2 (g)  2 NaCl(s) 2 Na  2 Na + + 2 e  Cl 2 + 2 e   2 Cl 

126. Reactions of Metals with Nonmetals Consider the following reactions: 4 Na(s) + O 2 (g) → 2 Na 2 O(s) 2 Na(s) + Cl 2 (g) → 2 NaCl(s) The reactions involve a metal reacting with a nonmetal In addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O 2 (g) → 2 Na + 2 O 2– (s) 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s)

127. Oxidation and Reduction 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 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na → Na + + 1 e – oxidation Cl 2 + 2 e – → 2 Cl – reduction Leo Ger

128. Electron Bookkeeping For reactions that are not metal + nonmetal, or do not involve O 2, we need a method for determining how the electrons are transferred Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction – even though they look like them, oxidation states are not ion charges! oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges

129. Rules for Assigning Oxidation States 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

130. Rules for Assigning Oxidation States 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

131. Rules for Assigning Oxidation States 5.in their compounds, nonmetals have oxidation states according to the table below – nonmetals higher on the table take priority

132. Example: Determine the oxidation states of all the atoms in a propanoate ion, C 3 H 5 O 2 – There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b (C 3 ) + (H 5 ) + (O 2 ) = −1 Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order: – H = +1 – O = −2 (C 3 ) + 5(+1) + 2(−2) = −1 (C 3 ) = −2 C = −⅔ Note: unlike charges, oxidation states can be fractions!

133. Practice – Assign an oxidation state to each element in the following Br 2 K + LiF CO 2 SO 4 2− Na 2 O 2 Br = 0, (Rule 1) K = +1, (Rule 2) Li = +1, (Rule 4a) & F = −1, (Rule 5) O = −2, (Rule 5) & C = +4, (Rule 3a) O = −2, (Rule 5) & S = +6, (Rule 3b) Na = +1, (Rule 4a) & O = −1, (Rule 3a)

134. Oxidation and Reduction Another Definition Oxidation occurs when an atom’s oxidation state increases during a reaction Reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O −4 +1 0 +4 –2 +1 −2 oxidation reduction

135. Oxidation–Reduction Oxidation and reduction must occur simultaneously – if an atom loses electrons another atom must take them The reactant that reduces an element in another reactant is called the reducing agent – the reducing agent contains the element that is oxidized The reactant that oxidizes an element in another reactant is called the oxidizing agent – the oxidizing agent contains the element that is reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl 2 is the oxidizing agent

136. Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions: Fe + MnO 4 − + 4 H + → Fe 3+ + MnO 2 + 2 H 2 O 0−2+7+1+3−2+4+1−2 Oxidation Reduction Oxidizing agent Reducing agent

137. Practice – Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions: Sn 4+ + Ca → Sn 2+ + Ca 2+ F 2 + S → SF 4 +4 0 +2 +2 Ca is oxidized, Sn 4+ is reduced Ca is the reducing agent, Sn 4+ is the oxidizing agent 0 0 +4−1 S is oxidized, F is reduced S is the reducing agent, F 2 is the oxidizing agent