1. Aqueous Reactions © 2009, Prentice-Hall, Inc. Chapter 4 Aqueous Reactions and Solution Stoichiometry

2. Aqueous Reactions © 2009, Prentice-Hall, Inc. Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes.

3. Aqueous Reactions © 2009, Prentice-Hall, Inc. Dissociation When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them. This process is called dissociation.

4. Aqueous Reactions © 2009, Prentice-Hall, Inc. Dissociation An electrolyte is a substances that dissociates into ions when dissolved in water.

5. Aqueous Reactions © 2009, Prentice-Hall, Inc. Electrolytes An electrolyte is a substance that dissociates into ions when dissolved in water. A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

6. Aqueous Reactions © 2009, Prentice-Hall, Inc. Electrolytes and Nonelectrolytes Soluble ionic compounds tend to be electrolytes.

7. Aqueous Reactions © 2009, Prentice-Hall, Inc. Electrolytes and Nonelectrolytes Molecular compounds tend to be nonelectrolytes, except for acids and bases.

8. Aqueous Reactions © 2009, Prentice-Hall, Inc. Electrolytes A strong electrolyte dissociates completely when dissolved in water. A weak electrolyte only dissociates partially when dissolved in water.

9. Aqueous Reactions © 2009, Prentice-Hall, Inc. Strong Electrolytes Are… Strong acids Strong bases

10. Aqueous Reactions © 2009, Prentice-Hall, Inc. Strong Electrolytes Are… Strong acids Strong bases Soluble ionic salts

11. Aqueous Reactions © 2009, Prentice-Hall, Inc. Precipitation Reactions When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

12. Aqueous Reactions

13. Aqueous Reactions © 2009, Prentice-Hall, Inc. Metathesis (Exchange) Reactions Metathesis comes from a Greek word that means “to transpose.” AgNO 3 (aq) + KCl (aq)  AgCl (s) + KNO 3 (aq)

14. Aqueous Reactions © 2009, Prentice-Hall, Inc. Metathesis (Exchange) Reactions Metathesis comes from a Greek word that means “to transpose.” It appears the ions in the reactant compounds exchange, or transpose, ions. AgNO 3 (aq) + KCl (aq)  AgCl (s) + KNO 3 (aq)

15. Aqueous Reactions © 2009, Prentice-Hall, Inc. Solution Chemistry It is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution). If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction.

16. Aqueous Reactions Sample Exercise 4.1 Relating Relative Numbers of Anions and Cations to Chemical Formulas If you were to draw diagrams representing aqueous solutions of each of the following ionic compounds, how many anions would you show if the diagram contained six cations? (a) NiSO 4, (b) Ca(NO 3 ) 2, (c) Na 3 PO 4, (d) Al 2 (SO 4 ) 3 Practice Exercise

17. Aqueous Reactions Answers: (a) 6, (b) 12, (c) 2, (d) 9

18. Aqueous Reactions Sample Exercise 4.2 Using Solubility Rules Classify the following ionic compounds as soluble or insoluble in water: (a) sodium carbonate (Na 2 CO 3 ), (b) lead sulfate (PbSO 4 ). Classify the following compounds as soluble or insoluble in water: (a) cobalt(II) hydroxide, (b) barium nitrate, (c) ammonium phosphate. Practice Exercise

19. Aqueous Reactions Sample Exercise 4.3 Predicting a Metathesis Reaction (a) Predict the identity of the precipitate that forms when solutions of BaCl 2 and K 2 SO 4 are mixed. (b) Write the balanced chemical equation for the reaction. (a) What compound precipitates when solutions of Fe 2 (SO 4 ) 3 and LiOH are mixed? (b) Write a balanced equation for the reaction. (c) Will a precipitate form when solutions of Ba(NO 3 ) 2 and KOH are mixed? Practice Exercise

20. Aqueous Reactions © 2009, Prentice-Hall, Inc. Molecular Equation The molecular equation lists the reactants and products in their molecular form. AgNO 3 (aq) + KCl (aq)  AgCl (s) + KNO 3 (aq)

21. Aqueous Reactions © 2009, Prentice-Hall, Inc. Ionic Equation In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. This more accurately reflects the species that are found in the reaction mixture. Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl (s) + K + (aq) + NO 3 - (aq)

22. Aqueous Reactions © 2009, Prentice-Hall, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl (s) + K + (aq) + NO 3 - (aq)

23. Aqueous Reactions © 2009, Prentice-Hall, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The only things left in the equation are those things that change (i.e., react) during the course of the reaction. Ag + (aq) + Cl - (aq)  AgCl (s)

24. Aqueous Reactions © 2009, Prentice-Hall, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The only things left in the equation are those things that change (i.e., react) during the course of the reaction. Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions. Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl (s) + K + (aq) + NO 3 - (aq)

25. Aqueous Reactions © 2009, Prentice-Hall, Inc. Writing Net Ionic Equations 1.Write a balanced molecular equation. 2.Dissociate all strong electrolytes. 3.Cross out anything that remains unchanged from the left side to the right side of the equation. 4.Write the net ionic equation with the species that remain.

26. Aqueous Reactions Sample Exercise 4.4 Writing a Net Ionic Equation Write the net ionic equation for the precipitation reaction that occurs when solutions of calcium chloride and sodium carbonate are mixed.

27. Aqueous Reactions Sample Exercise 4.4 Writing a Net Ionic Equation Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver nitrate and potassium phosphate are mixed. Practice Exercise

28. Aqueous Reactions Answers: 3 Ag + (aq) + PO 4 3– (aq) → Ag 3 PO 4 (s) Practice Exercise

29. Aqueous Reactions © 2009, Prentice-Hall, Inc. Acids Arrhenius defined acids as substances that increase the concentration of H + when dissolved in water. Brønsted and Lowry defined them as proton donors.

30. Aqueous Reactions © 2009, Prentice-Hall, Inc. Acids There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Chloric (HClO 3 ) Perchloric (HClO 4 )

31. Aqueous Reactions © 2009, Prentice-Hall, Inc. Bases Arrhenius defined bases as substances that increase the concentration of OH − when dissolved in water. Brønsted and Lowry defined them as proton acceptors.

32. Aqueous Reactions © 2009, Prentice-Hall, Inc. Bases The strong bases are the soluble metal salts of hydroxide ion: Alkali metals Calcium Strontium Barium

33. Aqueous Reactions © 2009, Prentice-Hall, Inc. Acid-Base Reactions In an acid-base reaction, the acid donates a proton (H + ) to the base.

34. Aqueous Reactions © 2009, Prentice-Hall, Inc. Neutralization Reactions Generally, when solutions of an acid and a base are combined, the products are a salt and water. CH 3 COOH (aq) + NaOH (aq)  CH 3 COONa (aq) + H 2 O (l)

35. Aqueous Reactions © 2009, Prentice-Hall, Inc. Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l)

36. Aqueous Reactions © 2009, Prentice-Hall, Inc. Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H + ( aq ) + Cl - ( aq ) + Na + ( aq ) + OH - ( aq )  Na + ( aq ) + Cl - ( aq ) + H 2 O ( l )

37. Aqueous Reactions © 2009, Prentice-Hall, Inc. Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H + ( aq ) + Cl - ( aq ) + Na + ( aq ) + OH - ( aq )  Na + ( aq ) + Cl - ( aq ) + H 2 O ( l ) H + (aq) + OH - (aq)  H 2 O (l)

38. Aqueous Reactions Sample Exercise 4.5 Comparing Acid Strengths

39. Aqueous Reactions Sample Exercise 4.5 Writing a Net Ionic Equation Imagine a diagram showing 10 Na + ions and 10 OH – ions. If this solution were mixed with the one pictured on the previous slide for HY, what would the diagram look like that represents the solution after any possible reaction? Practice Exercise

40. Aqueous Reactions Sample Exercise 4.6 Identifying Strong, Weak, and Nonelectrolytes Classify each of the following dissolved substances as a strong electrolyte, weak electrolyte, or nonelectrolyte: CaCl 2, HNO 3, C 2 H 5 OH (ethanol), HCOOH (formic acid), KOH.

41. Aqueous Reactions Consider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L of water: Ca(NO 3 ) 2 (calcium nitrate), C 6 H 12 O 6 (glucose), CH 3 COONa (sodium acetate), and CH 3 COOH (acetic acid). Rank the solutions in order of increasing electrical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity. Practice Exercise

42. Aqueous Reactions Sample Exercise 4.7 Writing Chemical Equations for a Neutralization Reaction (a) Write a balanced molecular equation for the reaction between aqueous solutions of acetic acid (CH 3 COOH) and barium hydroxide, Ba(OH) 2. (b) Write the net ionic equation for this reaction.

43. Aqueous Reactions Solution Plan: neutralization reactions form two products, H 2 O and a salt. We examine the cation of the base and the anion of the acid to determine the composition of the salt. Solve: (a) The salt will contain the cation of the base (Ba 2+ ) and the anion of the acid (CH 3 COO – ). Thus, the formula of the salt is Ba(CH 3 COO) 2. According to the solubility guidelines in Table 4.1, this compound is soluble. The unbalanced molecular equation for the neutralization reaction is To balance this molecular equation, we must provide two molecules of CH 3 COOH to furnish the two CH 3 COO – ions and to supply the two H + ions needed to combine with the two CH – ions of the base. The balanced molecular equation is

44. Aqueous Reactions Sample Exercise 4.7 Writing Chemical Equations for a Neutralization Reaction Write a balanced molecular equation for the reaction of carbonic acid (H 2 CO 3 ) and potassium hydroxide (KOH). (b) Write the net ionic equation for this reaction. Practice Exercise

45. Aqueous Reactions © 2009, Prentice-Hall, Inc. Gas-Forming Reactions Some metathesis reactions do not give the product expected. In this reaction, the expected product (H 2 CO 3 ) decomposes to give a gaseous product (CO 2 ). CaCO 3 (s) + HCl (aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O (l)

46. Aqueous Reactions © 2009, Prentice-Hall, Inc. Gas-Forming Reactions When a carbonate or bicarbonate reacts with an acid, the products are a salt, carbon dioxide, and water. CaCO 3 (s) + HCl (aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O (l) NaHCO 3 (aq) + HBr (aq)  NaBr (aq) + CO 2 (g) + H 2 O (l)

47. Aqueous Reactions © 2009, Prentice-Hall, Inc. Gas-Forming Reactions Similarly, when a sulfite reacts with an acid, the products are a salt, sulfur dioxide, and water. SrSO 3 (s) + 2 HI (aq)  SrI 2 (aq) + SO 2 (g) + H 2 O (l)

48. Aqueous Reactions © 2009, Prentice-Hall, Inc. Gas-Forming Reactions This reaction gives the predicted product, but you had better carry it out in the hood, or you will be very unpopular! But just as in the previous examples, a gas is formed as a product of this reaction. Na 2 S (aq) + H 2 SO 4 (aq)  Na 2 SO 4 (aq) + H 2 S (g)

49. Aqueous Reactions © 2009, Prentice-Hall, Inc. Oxidation-Reduction Reactions An oxidation occurs when an atom or ion loses electrons. A reduction occurs when an atom or ion gains electrons. One cannot occur without the other.

50. Aqueous Reactions © 2009, Prentice-Hall, Inc. Oxidation Numbers To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

51. Aqueous Reactions © 2009, Prentice-Hall, Inc. Oxidation Numbers Elements in their elemental form have an oxidation number of 0. The oxidation number of a monatomic ion is the same as its charge.

52. Aqueous Reactions © 2009, Prentice-Hall, Inc. Oxidation Numbers Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.  Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.  Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

53. Aqueous Reactions © 2009, Prentice-Hall, Inc. Oxidation Numbers Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.  Fluorine always has an oxidation number of −1.  The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

54. Aqueous Reactions © 2009, Prentice-Hall, Inc. Oxidation Numbers The sum of the oxidation numbers in a neutral compound is 0. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

55. Aqueous Reactions Sample Exercise 4.8 Determining Oxidation Numbers Determine the oxidation number of sulfur in each of the following: (a) H 2 S, (b) S 8, (c) SCl 2, (d) Na 2 SO 3, (e) SO 4 2–.

56. Aqueous Reactions Sample Exercise 4.8 Determining Oxidation Numbers What is the oxidation state of the boldfaced element in each of the following: (a) P 2 O 5, (b) NaH, (c) Cr 2 O 7 2–, (d) SnBr 4, (e) BaO 2 ? Practice Exercise

57. Aqueous Reactions Sample Exercise 4.9 Writing Molecular and Net Ionic Equations for Oxidation-Reduction Reactions Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid.

58. Aqueous Reactions Solution Analyze: We must write two equations—molecular and net ionic—for the redox reaction between a metal and an acid. Plan: Metals react with acids to form salts and H 2 gas. To write the balanced equations, we must write the chemical formulas for the two reactants and then determine the formula of the salt. The salt is composed of the cation formed by the metal and the anion of the acid. Solve: The formulas of the given reactants are Al and HBr. The cation formed by Al is Al 3+, and the anion from hydrobromic acid is Br –. Thus, the salt formed in the reaction is AlBr 3. Writing the reactants and products and then balancing the equation gives this molecular equation: 2 Al(s) + 6 HBr(aq) → 2 AlBr 3 (aq) + 3 H 2 (g) Both HBr and AlBr3 are soluble strong electrolytes. Thus, the complete ionic equation is 2 Al(s) + 6 H + (aq) + 6 Br – (aq) → 2 Al 3+ (aq) + 6 Br – (aq) + 3 H 2 (g) Because Br – is a spectator ion, the net ionic equation is 2 Al(s) + 6 H + (aq) → 2 Al 3+ (aq) + 3 H2(g )

59. Aqueous Reactions Sample Exercise 4.9 Writing Molecular and Net Ionic Equations for Oxidation-Reduction Reactions Solution (continued) Comment: The substance oxidized is the aluminum metal because its oxidation state changes from 0 in the metal to +3 in the cation, thereby increasing in oxidation number. The H + is reduced because its oxidation state changes from +1 in the acid to 0 in H 2.

60. Aqueous Reactions Write the balanced molecular and net ionic equations for the reaction between magnesium and cobalt(II) sulfate. What is oxidized and what is reduced in the reaction? Practice Exercise

61. Aqueous Reactions (a)Write the balanced molecular and net ionic equations for the reaction between magnesium and cobalt(II) sulfate. (b)What is oxidized and what is reduced in the reaction? Answers: (a) Mg(s) + CoSO 4 (aq) → MgSO 4 (aq) + Co(s); Mg(s) + Co 2+ (aq) → Mg 2+ (aq) + Co(s) (b) Mg is oxidized and Co 2+ is reduced. Practice Exercise

62. Aqueous Reactions © 2009, Prentice-Hall, Inc. Displacement Reactions In displacement reactions, ions oxidize an element. The ions, then, are reduced.

63. Aqueous Reactions © 2009, Prentice-Hall, Inc. Displacement Reactions In this reaction, silver ions oxidize copper metal. Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag (s)

64. Aqueous Reactions © 2009, Prentice-Hall, Inc. Displacement Reactions The reverse reaction, however, does not occur. Cu 2+ (aq) + 2 Ag (s)  Cu (s) + 2 Ag + (aq) x

65. Aqueous Reactions © 2009, Prentice-Hall, Inc. Activity Series

66. Aqueous Reactions Sample Exercise 4.10 Determining When an Oxidation-Reduction Reaction Can Occur Will an aqueous solution of iron(II) chloride oxidize magnesium metal? If so, write the balanced molecular and net ionic equations for the reaction.

67. Aqueous Reactions Solution Analyze: We are given two substances—an aqueous salt, FeCl 2, and a metal, Mg— and asked if they react with each other. Plan: A reaction will occur if Mg is above Fe in the activity series, Table 4.5. If the reaction occurs, the Fe 2+ ion in FeCl 2 will be reduced to Fe, and the elemental Mg will be oxidized to Mg 2+. Solve: Because Mg is above Fe in the table, the reaction will occur. To write the formula for the salt that is produced in the reaction, we must remember the charges on common ions. Magnesium is always present in compounds as Mg 2+ : the chloride ion is Cl –. The magnesium salt formed in the reaction is MgCl 2, meaning the balanced molecular equation is Mg(s) + FeCl 2 (aq) → MgCl 2 (aq) + Fe(s) Both FeCl 2 and MgCl 2 are soluble strong electrolytes and can be written in ionic form. Cl – then, is a spectator ion in the reaction. The net ionic equation is Mg(s) + Fe 2 +(aq) → Mg 2 +(aq) + Fe(s) The net ionic equation shows that Mg is oxidized and Fe 2+ is reduced in this reaction. Check: Note that the net ionic equation is balanced with respect to both charge and mass.

68. Aqueous Reactions Sample Exercise 4.10 Determining When an Oxidation-Reduction Reaction Can Occur Which of the following metals will be oxidized by Pb(NO 3 ) 2 : Zn, Cu, Fe? Practice Exercise

69. Aqueous Reactions © 2009, Prentice-Hall, Inc. Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution. moles of solute volume of solution in liters Molarity (M) =

70. Aqueous Reactions © 2009, Prentice-Hall, Inc. Mixing a Solution To create a solution of a known molarity, weigh out a known mass (and, therefore, number of moles) of the solute. The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

71. Aqueous Reactions © 2009, Prentice-Hall, Inc. Dilution You can dilute a more concentrated solution by –Using a pipet to deliver a volume of the solution to a new volumetric flask, and –Adding solvent to the line on the neck of the new flask.

72. Aqueous Reactions © 2009, Prentice-Hall, Inc. Dilution The molarity of the new solution can be determined from the equation M c  V c = M d  V d, where M c and M d are the molarity of the concentrated and dilute solutions, respectively, and V c and V d are the volumes of the two solutions.

73. Aqueous Reactions © 2009, Prentice-Hall, Inc. Using Molarities in Stoichiometric Calculations

74. Aqueous Reactions © 2009, Prentice-Hall, Inc. Titration Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.

75. Aqueous Reactions Sample Exercise 4.11 Calculating Molarity Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na 2 SO 4 ) in enough water to form 125 mL of solution.

76. Aqueous Reactions Solution Analyze: We are given the number of grams of solute (23.4 g), its chemical formula (Na 2 SO 4 ), and the volume of the solution (125 ml). We are asked to calculate the molarity of the solution. Plan: We can calculate molarity using Equation 4.33. To do so, we must convert the number of grams of solute to moles and the volume of the solution from milliliters to liters. Solve: The number of moles of Na 2 SO 4 is obtained by using its molar mass: Converting the volume of the solution to liters: Thus, the molarity is

77. Aqueous Reactions Sample Exercise 4.11 Calculating Molarity Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C 6 H 12 O 6 ) in sufficient water to form exactly 100 mL of solution. Practice Exercise

78. Aqueous Reactions Sample Exercise 4.12 Calculating Molar Concentrations of Ions What are the molar concentrations of each of the ions present in a 0.025 M aqueous solution of calcium nitrate? What is the molar concentration of K + ions in a 0.015 M solution of potassium carbonate? Practice Exercise

79. Aqueous Reactions Sample Exercise 4.13 Using Molarity to Calculate Grams of Solute How many grams of Na 2 SO 4 are required to make 0.350 L of 0.500 M Na 2 SO 4 ?

80. Aqueous Reactions Solution Analyze: We are given the volume of the solution (0.350 L), its concentration (0.500 M), and the identity of the solute Na 2 SO 4 and asked to calculate the number of grams of the solute in the solution. Plan: We can use the definition of molarity (Equation 4.33) to determine the number of moles of solute, and then convert moles to grams using the molar mass of the solute. Solve: Calculating the moles of Na 2 SO 4 using the molarity and volume of solution gives Because each mole of Na 2 SO 4 weighs 142 g, the required number of grams of Na 2 SO 4 is Check: The magnitude of the answer, the units, and the number of significant figures are all appropriate.

81. Aqueous Reactions Sample Exercise 4.13 Using Molarity to Calculate Grams of Solute (a) How many grams of Na 2 SO 4 are there in 15 mL of 0.50 M Na 2 SO 4 ? (b) How many milliliters of 0.50 M Na 2 SO 4 solution are needed to provide 0.038 mol of this salt? Practice Exercise

82. Aqueous Reactions Sample Exercise 4.14 Preparing A solution by Dilution How many milliliters of 3.0 M H 2 SO 4 are needed to make 450 mL of 0.10 M H 2 SO 4 ?

83. Aqueous Reactions Sample Exercise 4.14 Preparing A solution by Dilution (a) What volume of 2.50 M lead(II) nitrate solution contains 0.0500 mol of Pb 2+ ? (b) How many milliliters of 5.0 M K 2 Cr 2 O 7 solution must be diluted to prepare 250 mL of 0.10 M solution? (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution? Practice Exercise

84. Aqueous Reactions Sample Exercise 4.15 Using Mass Relations In a Neutralization Reaction How many grams of Ca(OH) 2 are needed to neutralize 25.0 mL of 0.100 M HNO 3 ?

85. Aqueous Reactions Sample Exercise 4.15 Using Mass Relations In a Neutralization Reaction (a) How many grams of NaOH are needed to neutralize 20.0 mL of 0.150 M H 2 SO 4 solution? (b) How many liters of 0.500 M HCl(aq) are needed to react completely with 0.100 mol of Pb(NO 3 ) 2 (aq), forming a precipitate of PbCl 2 (s)? Practice Exercise

86. Aqueous Reactions Sample Exercise 4.16 Determining the Quality of Solute by Titration The quantity of Cl – in a municipal water supply is determined by titrating the sample with Ag +. The reaction taking place during the titration is The end point in this type of titration is marked by a change in color of a special type of indicator. (a) How many grams of chloride ion are in a sample of the water if 20.2 mL of 0.100 M Ag + is needed to react with all the chloride in the sample? (b) If the sample has a mass of 10.0 g, what percent Cl – does it contain?

87. Aqueous Reactions Sample Exercise 4.16 Determining the Quality of Solute by Titration Solution (continued) (b) Plan: To calculate the percentage of Cl – in the sample, we compare the number of grams of Cl – in the sample, 7.17  10 –2 g, with the original mass of the sample, 10.0 g. Comment: Chloride ion is one of the most common ions in water and sewage. Ocean water contains 1.92% Cl –. Whether water containing Cl – tastes salty depends on the other ions present. If the only accompanying ions are Na +, a salty taste may be detected with as little as 0.03% Cl –. A sample of an iron ore is dissolved in acid, and the iron is converted to Fe 2+. The sample is then titrated with 47.20 mL of 0.02240 M MnO 4 – solution. The oxidation-reduction reaction that occurs during titration is as follows: (a) How many moles of MnO 4 – were added to the solution? (b) How many moles of Fe 2+ were in the sample? (c) How many grams of iron were in the sample? (d) If the sample had a mass of 0.8890 g, what is the percentage of iron in the sample? Practice Exercise

88. Aqueous Reactions Answers: (a) 1.057  10–3 mol MnO4– (b) 5.286  10–3 mol Fe2+, (c) 0.2952 g, (d) 33.21%

89. Aqueous Reactions Sample Exercise 4.17 Determining Solution Concentration Via an Acid-Base Titration One commercial method used to peel potatoes is to soak them in a solution of NaOH for a short time, remove them from the NaOH, and spray off the peel. The concentration of NaOH is normally in the range of 3 to 6 M. The NaOH is analyzed periodically. In one such analysis, 45.7 mL of 0.500 M H 2 SO 4 is required to neutralize a 20.0-mL sample of NaOH solution. What is the concentration of the NaOH solution?

90. Aqueous Reactions Sample Exercise 4.17 Determining Solution Concentration Via an Acid-Base Titration What is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 M H 2 SO 4 ? Practice Exercise Solution (continued) Knowing the number of moles of NaOH present in 20.0 mL of solution allows us to calculate the molarity of this solution:

91. Aqueous Reactions Sample Integrative Exercise A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) Calculate the theoretical yield, in grams, of the precipitate that forms.

92. Aqueous Reactions Solution (a) Potassium phosphate and silver nitrate are both ionic compounds. Potassium phosphate contains K + and PO 4 3– ions, so its chemical formula is K 3 PO 4. Silver nitrate contains Ag + and NO 3 – ions, so its chemical formula is AgNO 3. Because both reactants are strong electrolytes, the solution contains K +, PO 4 3–, Ag +, and NO 3 – ions before the reaction occurs. According to the solubility guidelines in Table 4.1, and PO 4 3– form an insoluble compound, so Ag 3 PO 4 will precipitate from the solution. In contrast, K + and NO 3 – will remain in solution because KNO 3 is water soluble. Thus, the balanced molecular equation for the reaction is K 3 PO 4 (aq) + 3 AgNO 3 (aq) → Ag 3 PO4(s) + 3 KNO 3 (aq) (b) To determine the limiting reactant, we must examine the number of moles of each reactant. (Section 3.7) The number of moles of K 3 PO 4 is calculated from the mass of the sample using the molar mass as a conversion factor. (Section 3.4) The molar mass of K 3 PO 4 is 3(39.1) + 31.0 + 4(16.0) = 212.3 g/mol. Converting milligrams to grams and then to moles, we have

93. Aqueous Reactions Sample Integrative Exercise Solution (continued) We determine the number of moles of AgNO 3 from the volume and molarity of the solution. (Section 4.5) Converting milliliters to liters and then to moles, we have Comparing the amounts of the two reactants, we find that there are (7.5  10 –4 )/(3.32  10 –4 ) = 2.3 times as many moles of AgNO 3 as there are moles of K 3 PO 4. According to the balanced equation, however, 1 mol K 3 PO 4 requires 3 mol of AgNO 3. Thus, there is insufficient AgNO 3 to consume the K 3 PO 4, and AgNO 3 is the limiting reactant. (c) The precipitate is Ag 3 PO 4, whose molar mass is 3(107.9) + 31.0 + 4(16.0) = 418.7 g/mol. To calculate the number of grams of Ag 3 PO 4 that could be produced in this reaction (the theoretical yield), we use the number of moles of the limiting reactant, converting mol. We use the coefficients in the balanced equation to convert moles of AgNO 3 to moles Ag 3 PO 4, and we use the molar mass of Ag 3 PO 4 to convert the number of moles of this substance to grams. The answer has only two significant figures because the quantity of AgNO 3 is given to only two significant figures.

94. Aqueous Reactions Sample Integrative Exercise A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) Calculate the theoretical yield, in grams, of the precipitate that forms.

95. Aqueous Reactions Solution (a) Potassium phosphate and silver nitrate are both ionic compounds. Potassium phosphate contains K + and PO 4 3– ions, so its chemical formula is K 3 PO 4. Silver nitrate contains Ag + and NO 3 – ions, so its chemical formula is AgNO 3. Because both reactants are strong electrolytes, the solution contains K +, PO 4 3–, Ag +, and NO 3 – ions before the reaction occurs. According to the solubility guidelines in Table 4.1, and PO 4 3– form an insoluble compound, so Ag 3 PO 4 will precipitate from the solution. In contrast, K + and NO 3 – will remain in solution because KNO 3 is water soluble. Thus, the balanced molecular equation for the reaction is K 3 PO 4 (aq) + 3 AgNO 3 (aq) → Ag 3 PO4(s) + 3 KNO 3 (aq) (b) To determine the limiting reactant, we must examine the number of moles of each reactant. (Section 3.7) The number of moles of K 3 PO 4 is calculated from the mass of the sample using the molar mass as a conversion factor. (Section 3.4) The molar mass of K 3 PO 4 is 3(39.1) + 31.0 + 4(16.0) = 212.3 g/mol. Converting milligrams to grams and then to moles, we have

96. Aqueous Reactions Sample Integrative Exercise Solution (continued) We determine the number of moles of AgNO 3 from the volume and molarity of the solution. (Section 4.5) Converting milliliters to liters and then to moles, we have Comparing the amounts of the two reactants, we find that there are (7.5  10 –4 )/(3.32  10 –4 ) = 2.3 times as many moles of AgNO 3 as there are moles of K 3 PO 4. According to the balanced equation, however, 1 mol K 3 PO 4 requires 3 mol of AgNO 3. Thus, there is insufficient AgNO 3 to consume the K 3 PO 4, and AgNO 3 is the limiting reactant. (c) The precipitate is Ag 3 PO 4, whose molar mass is 3(107.9) + 31.0 + 4(16.0) = 418.7 g/mol. To calculate the number of grams of Ag 3 PO 4 that could be produced in this reaction (the theoretical yield), we use the number of moles of the limiting reactant, converting mol. We use the coefficients in the balanced equation to convert moles of AgNO 3 to moles Ag 3 PO 4, and we use the molar mass of Ag 3 PO 4 to convert the number of moles of this substance to grams. The answer has only two significant figures because the quantity of AgNO 3 is given to only two significant figures.