1. Copyright © 2011 Pearson Education, Inc. Chapter 4 Chemical Quantities and Aqueous Reactions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

2. Copyright © 2011 Pearson Education, Inc. 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? 2 Tro: Chemistry: A Molecular Approach, 2/e

3. Copyright © 2011 Pearson Education, Inc. 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? 3 Tro: Chemistry: A Molecular Approach, 2/e

4. Copyright © 2011 Pearson Education, Inc. 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 4 Tro: Chemistry: A Molecular Approach, 2/e

5. Copyright © 2011 Pearson Education, Inc. 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 5 Tro: Chemistry: A Molecular Approach, 2/e

6. Copyright © 2011 Pearson Education, Inc. 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 6 Tro: Chemistry: A Molecular Approach, 2/e

7. Copyright © 2011 Pearson Education, Inc. 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 7 Tro: Chemistry: A Molecular Approach, 2/e

8. Copyright © 2011 Pearson Education, Inc. 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 8 Tro: Chemistry: A Molecular Approach, 2/e

9. Copyright © 2011 Pearson Education, Inc. 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 9 Tro: Chemistry: A Molecular Approach, 2/e

10. Copyright © 2011 Pearson Education, Inc. 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 10 Tro: Chemistry: A Molecular Approach, 2/e

11. Copyright © 2011 Pearson Education, Inc. 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 11 Tro: Chemistry: A Molecular Approach, 2/e

12. Copyright © 2011 Pearson Education, Inc. 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 12 Tro: Chemistry: A Molecular Approach, 2/e

13. Copyright © 2011 Pearson Education, Inc. 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 13 Tro: Chemistry: A Molecular Approach, 2/e

14. Copyright © 2011 Pearson Education, Inc. 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) 14 Tro: Chemistry: A Molecular Approach, 2/e

15. Copyright © 2011 Pearson Education, Inc. 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 15 Tro: Chemistry: A Molecular Approach, 2/e

16. Copyright © 2011 Pearson Education, Inc. 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 16 Tro: Chemistry: A Molecular Approach, 2/e

17. Copyright © 2011 Pearson Education, Inc. More Making Pizzas 17 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? Tro: Chemistry: A Molecular Approach, 2/e

18. Copyright © 2011 Pearson Education, Inc. 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. 18 Tro: Chemistry: A Molecular Approach, 2/e

19. Copyright © 2011 Pearson Education, Inc. 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! 19 Tro: Chemistry: A Molecular Approach, 2/e

20. Copyright © 2011 Pearson Education, Inc. 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 20 Tro: Chemistry: A Molecular Approach, 2/e

21. Copyright © 2011 Pearson Education, Inc. 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 21 Tro: Chemistry: A Molecular Approach, 2/e

22. Copyright © 2011 Pearson Education, Inc. 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 22 If we have five molecules of CH 4 and eight molecules of O 2, which is the limiting reactant? Tro: Chemistry: A Molecular Approach, 2/e

23. Copyright © 2011 Pearson Education, Inc. 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 ? 23 Tro: Chemistry: A Molecular Approach, 2/e

24. Copyright © 2011 Pearson Education, Inc. 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 24 Tro: Chemistry: A Molecular Approach, 2/e

25. Copyright © 2011 Pearson Education, Inc. 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. 25 Tro: Chemistry: A Molecular Approach, 2/e

26. Copyright © 2011 Pearson Education, Inc. 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 26 Tro: Chemistry: A Molecular Approach, 2/e

27. Copyright © 2011 Pearson Education, Inc. Example 4.4: Finding limiting reactant, theoretical yield, and percent yield Tro: Chemistry: A Molecular Approach, 2/e

28. Copyright © 2011 Pearson Education, Inc. 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. 28 Tro: Chemistry: A Molecular Approach, 2/e

29. Copyright © 2011 Pearson Education, Inc. 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 29 Tro: Chemistry: A Molecular Approach, 2/e

30. Copyright © 2011 Pearson Education, Inc. 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 30 Tro: Chemistry: A Molecular Approach, 2/e

31. Copyright © 2011 Pearson Education, Inc. 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 31 Tro: Chemistry: A Molecular Approach, 2/e

32. Copyright © 2011 Pearson Education, Inc. 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 32 Tro: Chemistry: A Molecular Approach, 2/e

33. Copyright © 2011 Pearson Education, Inc. 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 33 Tro: Chemistry: A Molecular Approach, 2/e

34. Copyright © 2011 Pearson Education, Inc. 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) 34 Tro: Chemistry: A Molecular Approach, 2/e

35. Copyright © 2011 Pearson Education, Inc. 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) 35 Tro: Chemistry: A Molecular Approach, 2/e

36. Copyright © 2011 Pearson Education, Inc. 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) 36 Tro: Chemistry: A Molecular Approach, 2/e

37. Copyright © 2011 Pearson Education, Inc. 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? 37 Tro: Chemistry: A Molecular Approach, 2/e

38. Copyright © 2011 Pearson Education, Inc. 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? 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 38 Tro: Chemistry: A Molecular Approach, 2/e

39. Copyright © 2011 Pearson Education, Inc. because the percent yield is less than 100, the answer makes sense Check: Solution: Theoretical yield 39 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? Tro: Chemistry: A Molecular Approach, 2/e

40. Copyright © 2011 Pearson Education, Inc. 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 40 Tro: Chemistry: A Molecular Approach, 2/e

41. Copyright © 2011 Pearson Education, Inc. 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 41 Tro: Chemistry: A Molecular Approach, 2/e

42. Copyright © 2011 Pearson Education, Inc. 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 42 Tro: Chemistry: A Molecular Approach, 2/e

43. Copyright © 2011 Pearson Education, Inc. 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 43 Tro: Chemistry: A Molecular Approach, 2/e

44. Copyright © 2011 Pearson Education, Inc. 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 44 Tro: Chemistry: A Molecular Approach, 2/e

45. Copyright © 2011 Pearson Education, Inc. Preparing 1 L of a 1.00 M NaCl Solution 45 Tro: Chemistry: A Molecular Approach, 2/e

46. Copyright © 2011 Pearson Education, Inc. 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 46 Tro: Chemistry: A Molecular Approach, 2/e

47. Copyright © 2011 Pearson Education, Inc. Practice — What Is the molarity of a solution containing 3.4 g of NH 3 (MM 17.03) in 200.0 mL of solution? 47 Tro: Chemistry: A Molecular Approach, 2/e

48. Copyright © 2011 Pearson Education, Inc. Practice — What Is the molarity of a solution containing 3.4 g of NH 3 (MM 17.03) in 200.0 mL of solution? 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 48 0.20 mol NH 3, 0.2000 L solution M Tro: Chemistry: A Molecular Approach, 2/e

49. Copyright © 2011 Pearson Education, Inc. 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 49 Tro: Chemistry: A Molecular Approach, 2/e

50. Copyright © 2011 Pearson Education, Inc. 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 50 Tro: Chemistry: A Molecular Approach, 2/e

51. Copyright © 2011 Pearson Education, Inc. Practice — Determine the mass of CaCl 2 (MM = 110.98) in 1.75 L of 1.50 M solution 51 Tro: Chemistry: A Molecular Approach, 2/e

52. Copyright © 2011 Pearson Education, Inc. Practice — Determine the mass of CaCl 2 (MM = 110.98) in 1.75 L of 1.50 M solution 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 52 Tro: Chemistry: A Molecular Approach, 2/e

53. Copyright © 2011 Pearson Education, Inc. 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 53 Tro: Chemistry: A Molecular Approach, 2/e

54. Copyright © 2011 Pearson Education, Inc. Practice – How would you prepare 250.0 mL of 0.150 M CaCl 2 (MM = 110.98)? 54 Tro: Chemistry: A Molecular Approach, 2/e

55. Copyright © 2011 Pearson Education, Inc. 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 55 Tro: Chemistry: A Molecular Approach, 2/e

56. Copyright © 2011 Pearson Education, Inc. 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 56 Tro: Chemistry: A Molecular Approach, 2/e

57. Copyright © 2011 Pearson Education, Inc. 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 57 Tro: Chemistry: A Molecular Approach, 2/e

58. Copyright © 2011 Pearson Education, Inc. Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO 3 to 135.0 mL? 58 Tro: Chemistry: A Molecular Approach, 2/e

59. Copyright © 2011 Pearson Education, Inc. Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO 3 to 135.0 mL? 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 59 Tro: Chemistry: A Molecular Approach, 2/e

60. Copyright © 2011 Pearson Education, Inc. Practice – How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? 60 Tro: Chemistry: A Molecular Approach, 2/e

61. Copyright © 2011 Pearson Education, Inc. Practice – How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? 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 61 Tro: Chemistry: A Molecular Approach, 2/e

62. Copyright © 2011 Pearson Education, Inc. 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 62 Tro: Chemistry: A Molecular Approach, 2/e

63. Copyright © 2011 Pearson Education, Inc. 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 63 Tro: Chemistry: A Molecular Approach, 2/e

64. Copyright © 2011 Pearson Education, Inc. 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 64 Tro: Chemistry: A Molecular Approach, 2/e

65. Copyright © 2011 Pearson Education, Inc. 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 65 Tro: Chemistry: A Molecular Approach, 2/e

66. Copyright © 2011 Pearson Education, Inc. 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 66 Tro: Chemistry: A Molecular Approach, 2/e

67. Copyright © 2011 Pearson Education, Inc. 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 67 Tro: Chemistry: A Molecular Approach, 2/e

68. Copyright © 2011 Pearson Education, Inc. 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 68 Tro: Chemistry: A Molecular Approach, 2/e

69. Copyright © 2011 Pearson Education, Inc. 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 69 Tro: Chemistry: A Molecular Approach, 2/e

70. Copyright © 2011 Pearson Education, Inc. Salt vs. Sugar Dissolved in Water 70 Tro: Chemistry: A Molecular Approach, 2/e ionic compounds dissociate into ions when they dissolve molecular compounds do not dissociate when they dissolve

71. Copyright © 2011 Pearson Education, Inc. 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) 71 Tro: Chemistry: A Molecular Approach, 2/e

72. Copyright © 2011 Pearson Education, Inc. 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) 72 Tro: Chemistry: A Molecular Approach, 2/e

73. Copyright © 2011 Pearson Education, Inc. Classes of Dissolved Materials 73 Tro: Chemistry: A Molecular Approach, 2/e

74. Copyright © 2011 Pearson Education, Inc. 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) 74 Tro: Chemistry: A Molecular Approach, 2/e

75. Copyright © 2011 Pearson Education, Inc. 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) 75 Tro: Chemistry: A Molecular Approach, 2/e

76. Copyright © 2011 Pearson Education, Inc. 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 76 Tro: Chemistry: A Molecular Approach, 2/e

77. Copyright © 2011 Pearson Education, Inc. 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 77 Tro: Chemistry: A Molecular Approach, 2/e

78. Copyright © 2011 Pearson Education, Inc. Solubility Rules Compounds that Are Generally Soluble in Water 78 Tro: Chemistry: A Molecular Approach, 2/e

79. Copyright © 2011 Pearson Education, Inc. Solubility Rules Compounds that Are Generally Insoluble in Water 79 Tro: Chemistry: A Molecular Approach, 2/e

80. Copyright © 2011 Pearson Education, Inc. 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 80 Tro: Chemistry: A Molecular Approach, 2/e

81. Copyright © 2011 Pearson Education, Inc. 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 81 Tro: Chemistry: A Molecular Approach, 2/e

82. Copyright © 2011 Pearson Education, Inc. 2 KI(aq) + Pb(NO 3 ) 2 (aq)  PbI 2 (s) + 2 KNO 3 (aq) 82 Tro: Chemistry: A Molecular Approach, 2/e

83. Copyright © 2011 Pearson Education, Inc. No Precipitate Formation = No Reaction KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) all ions still present,  no reaction 83 Tro: Chemistry: A Molecular Approach, 2/e

84. Copyright © 2011 Pearson Education, Inc. 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 84 Tro: Chemistry: A Molecular Approach, 2/e

85. Copyright © 2011 Pearson Education, Inc. 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 85 Tro: Chemistry: A Molecular Approach, 2/e

86. Copyright © 2011 Pearson Education, Inc. 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 86 Tro: Chemistry: A Molecular Approach, 2/e

87. Copyright © 2011 Pearson Education, Inc. 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 87 Tro: Chemistry: A Molecular Approach, 2/e

88. Copyright © 2011 Pearson Education, Inc. 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) 88 Tro: Chemistry: A Molecular Approach, 2/e

89. Copyright © 2011 Pearson Education, Inc. 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)  Tro: Chemistry: A Molecular Approach, 2/e

90. Copyright © 2011 Pearson Education, Inc. 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 90 (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) Tro: Chemistry: A Molecular Approach, 2/e

91. Copyright © 2011 Pearson Education, Inc. 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) 91 Tro: Chemistry: A Molecular Approach, 2/e

92. Copyright © 2011 Pearson Education, Inc. 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) 92 Tro: Chemistry: A Molecular Approach, 2/e

93. Copyright © 2011 Pearson Education, Inc. 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) 93 Tro: Chemistry: A Molecular Approach, 2/e

94. Copyright © 2011 Pearson Education, Inc. 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 94 Tro: Chemistry: A Molecular Approach, 2/e

95. Copyright © 2011 Pearson Education, Inc. 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 95 Tro: Chemistry: A Molecular Approach, 2/e

96. Copyright © 2011 Pearson Education, Inc. Common Acids 96 Tro: Chemistry: A Molecular Approach, 2/e

97. Copyright © 2011 Pearson Education, Inc. 97 Common Bases Tro: Chemistry: A Molecular Approach, 2/e

98. Copyright © 2011 Pearson Education, Inc. HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) 98 Tro: Chemistry: A Molecular Approach, 2/e

99. Copyright © 2011 Pearson Education, Inc. 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) 99 Tro: Chemistry: A Molecular Approach, 2/e

100. Copyright © 2011 Pearson Education, Inc. 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) 100 Tro: Chemistry: A Molecular Approach, 2/e

101. Copyright © 2011 Pearson Education, Inc. 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) 101 Tro: Chemistry: A Molecular Approach, 2/e

102. Copyright © 2011 Pearson Education, Inc. 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)  102 Tro: Chemistry: A Molecular Approach, 2/e

103. Copyright © 2011 Pearson Education, Inc. 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 103 Tro: Chemistry: A Molecular Approach, 2/e

104. Copyright © 2011 Pearson Education, Inc. 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 104 Tro: Chemistry: A Molecular Approach, 2/e

105. Copyright © 2011 Pearson Education, Inc. Titration 105 Tro: Chemistry: A Molecular Approach, 2/e

106. Copyright © 2011 Pearson Education, Inc. Titration 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. 106 Tro: Chemistry: A Molecular Approach, 2/e

107. Copyright © 2011 Pearson Education, Inc. 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 107 Tro: Chemistry: A Molecular Approach, 2/e

108. Copyright © 2011 Pearson Education, Inc. 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 108 Tro: Chemistry: A Molecular Approach, 2/e

109. Copyright © 2011 Pearson Education, Inc. 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 109 Tro: Chemistry: A Molecular Approach, 2/e

110. Copyright © 2011 Pearson Education, Inc. Write a conceptual plan mL NaOH L NaOH mol NaOH mol HCl 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 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? mL HCl L HCl 110 Tro: Chemistry: A Molecular Approach, 2/e

111. Copyright © 2011 Pearson Education, Inc. Apply the conceptual plan = 1.25 x 10  3 mol HCl 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 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? 111 Tro: Chemistry: A Molecular Approach, 2/e

112. Copyright © 2011 Pearson Education, Inc. 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 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? 112 Tro: Chemistry: A Molecular Approach, 2/e

113. Copyright © 2011 Pearson Education, Inc. 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? 113 Tro: Chemistry: A Molecular Approach, 2/e 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. Copyright © 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 114

115. Copyright © 2011 Pearson Education, Inc. 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 115 Tro: Chemistry: A Molecular Approach, 2/e

116. Copyright © 2011 Pearson Education, Inc. 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) 116 Tro: Chemistry: A Molecular Approach, 2/e

117. Copyright © 2011 Pearson Education, Inc. 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 117 Tro: Chemistry: A Molecular Approach, 2/e

118. Copyright © 2011 Pearson Education, Inc. 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) = 0 118 Tro: Chemistry: A Molecular Approach, 2/e

119. Copyright © 2011 Pearson Education, Inc. 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 2 119 Tro: Chemistry: A Molecular Approach, 2/e

120. Copyright © 2011 Pearson Education, Inc. 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 120 Tro: Chemistry: A Molecular Approach, 2/e

121. Copyright © 2011 Pearson Education, Inc. 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! 121 Tro: Chemistry: A Molecular Approach, 2/e

122. Copyright © 2011 Pearson Education, Inc. 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) 122 Tro: Chemistry: A Molecular Approach, 2/e

123. Copyright © 2011 Pearson Education, Inc. 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 123 Tro: Chemistry: A Molecular Approach, 2/e

124. Copyright © 2011 Pearson Education, Inc. 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 124 Tro: Chemistry: A Molecular Approach, 2/e

125. Copyright © 2011 Pearson Education, Inc. 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 125 Tro: Chemistry: A Molecular Approach, 2/e

126. Copyright © 2011 Pearson Education, Inc. 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 126 Tro: Chemistry: A Molecular Approach, 2/e