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CHAPTER 5 Reactions in Aqueous Solution

CHAPTER 5 Reactions in Aqueous Solution. 5.1 Properties of Compounds in Aqueous Solution 5.2 Precipitation reactions 5.3 Acids and Bases 5.4 Reactions of Acids and Bases 5.5 Gas-forming reactions 5.6 Classifying Reactions in Aqueous Solution 5.7 Oxidation-Reduction Reactions

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CHAPTER 5 Reactions in Aqueous Solution

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  1. CHAPTER 5Reactions in Aqueous Solution • 5.1 Properties of Compounds in Aqueous Solution • 5.2 Precipitation reactions • 5.3 Acids and Bases • 5.4 Reactions of Acids and Bases • 5.5 Gas-forming reactions • 5.6 Classifying Reactions in Aqueous Solution • 5.7 Oxidation-Reduction Reactions • 5.8 Measuring Concentrations of Compounds in solution • 5.9 pH, a Concentration scale for acids and bases • 5.10 Stoichiometry of Reactions in Aqueous Solutions Kull Chem 105 Chapter 2

  2. 5.1 Properties of Compounds in Aqueous Solution • Solute • Solvent • Solution • Electrodes (conductor of electricity) • Electrolytes (ionic compound soluble in water, conducts electricity) • Ions in Aqueous Solution: electrolytes, complete/nearly complete dissociation • Electrolytes: All ionic compounds soluble in water • Strong – nearly complete dissociation, good conductor • Weak – only partial dissociation, weak conductor • Figure 5.3, page 179 (be able to use) • Non-electrolyte: dissolve in water, but do not conduct

  3. Water Solubility of Ionic Compounds If one ion from the “Soluble Compd.” list is present in a compound, the compound is water soluble.

  4. 5.2 Precipitation reactions Writing Equations for Aqueous Ionic Reactions • Produces a water-insoluble product: precipitate • Writing the equation: use state symbols, (s) denotes the precipitate • Three types of equations are used to represent aqueous ionic reactions: molecular, total ionic, and net ionic equations. • molecular equation:shows all reactants and products as if they were intact, un-dissociated • total ionic equation: shows all the soluble ionic substances dissociated into ions. Charges must balance • net ionic equation: it eliminates the spectator ions and shows the actual chemical change taking place. • Spectator ions not involved in chemical change.

  5. Precipitation Reactions The “driving force” is the formation of an insoluble compound — a precipitate. Molecular equation Pb(NO3)2(aq) + 2 KI(aq) -----> 2 KNO3(aq) + PbI2(s) Net ionic equation Pb2+(aq) + 2 I-(aq) ---> PbI2(s)

  6. Precipitation Reactions: Predicting Whether a Precipitate Will Form • Predict whether a reaction occurs, and write balanced total and net ionic equations. (a) iron(III) chloride(aq) + cesium phosphate(aq) → (b) sodium hydroxide(aq) + cadmium nitrate(aq)→ (c) magnesium bromide(aq) + potassium acetate(aq)→ (d) silver sulfate(aq) + barium chloride(aq) →

  7. Problem 5.2.1 • molecular equation Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2 KNO3(aq) • Write the Total Ionic and Net Ionic equations • Total • Net

  8. Problem 5.2.2 • Write the Total Ionic and Net Ionic equations for CaCl2(aq) + Na2CO3(aq) → CaCO3(s) + 2 NaCl(aq) • Total • Net

  9. Problem 5.2.3 This problem ties in concepts from at least 3 chapters: 3, 4, & 5. • Solutions of iron (III) chloride and potassium hydroxide are combined. • Naming, balancing formulas, balancing reactions, solubility • Write the: • Molecular equation • Total ionic equation • Net ionic equation

  10. Review: Combining skills Problem 5.2.4 • Combine calcium chloride and potassium phosphate • Write the molecular equation

  11. Review: Combining skills Problem 5.2.4 3 CaCl2(aq) + 2 K3PO4(aq) → Ca3PO4(s) + 6 KCl(aq) • Write the Total Ionic and Net Ionic equations • Combining 5 grams CaCl2 with 3.5 grams of K3PO4 produced only a 67% yield of KCl. • What is the limiting reactant? • What mass of product did you make?

  12. Road Map • Where we were • Total & Net ionic equations • Precipitation reactions • Where we are going • Acids and Bases • Classifying reactions • Oxidation reactions • Measuring concentrations of compounds in solution • pH • Stoichiometry of reactions in aqueous solution

  13. 5.3 – Acids and Bases(know table 5.2, pg 187) • Acid: increases the H+ concentration • Base: increases the OH- concentration • Strong acid or base: completely dissociates/ionizes • Weak acid or base: partial ionization • Oxides of nonmetals and metals • Acidic oxides (C,N,S) give H+ • Basic oxides (CaO) give OH-

  14. Know the strong acids & bases!

  15. 5.4 Acid-Base Reactions • The “driving force” is the formation of water. NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(liq) • Net ionic equation OH-(aq) + H+(aq) ---> H2O(liq) • This applies to ALL reactions of STRONG acids and bases.

  16. 5.4 - Reactions of Acids and Bases • An Acid reacting with a Base produces a salt and water Acid(aq) + Base(aq)  salt (s) + H2O (l) HX + MOH ---> MX + H2O This is one way to make compounds! Neutralization reaction: a strong acid with a strong base

  17. 5.4 Acid-Base Reactions: Neutralization • What volume of 0.1292 M Ba(OH)2 would neutralize 50.00 mL of 0.325 M HCl solution?

  18. 5.5 Gas-forming reactions This is primarily the chemistry of metal carbonates. CO2 and water ---> H2CO3 H2CO3(aq) + Ca2+(aq) ---> 2 H+(aq) + CaCO3(s) (limestone) Adding acid reverses this reaction. MCO3 + acid ---> CO2 + salt

  19. 5.5 Gas-Forming Reactions CaCO3(s) + H2SO4(aq) ---> 2 CaSO4(s) + H2CO3(aq) Carbonic acid is unstable and forms CO2 & H2O H2CO3(aq) ---> CO2 (g) + water (l) (Antacid tablet has citric acid + NaHCO3)

  20. CHAPTER 5Reactions in Aqueous Solution 5.1 Properties of Compounds in Aqueous Solution 5.2 Precipitation reactions Be familiar with and know how to use the solubility rules 5.3 Acids and Bases 5.4 Reactions of Acids and Bases 5.5 Gas-forming reactions 5.6 Classifying Reactions in Aqueous Solution Know the strong acids & bases, table 5.2 5.7 Oxidation-Reduction Reactions 5.8 Measuring Concentrations of Compounds in solution 5.9 pH, a Concentration scale for acids and bases 5.10 Stoichiometry of Reactions in Aqueous Solutions Kull Chem 105 Chapter 2

  21. 5.6 Classifying reactions in aqueous solution • Balance the following reactions and then classify each as precipitation (s), acid-base, or gas-forming reaction (g). a) Ba(OH)2 (aq) + 2 HCl (aq)  BaCl2 (aq) + 2 H2O (l) b) 2 HNO3 (aq) + CoCO3 (s)  Co(NO3)2 (aq) + H2O (l) + CO2 (g) c) 2 Na3PO4 (aq) + 3 Cu(NO3)2 (aq)  Cu3(PO4)2 (s) + 6 NaNO3 (aq)

  22. 5.7 Oxidation-Reduction Reactions • Oxidation: any process in which oxygen is added to another substance; gains oxygen, • OIL: Oxidation Is Loss of electrons • RIG: Reduction Is Gain of electrons • Reducing agent gets oxidized • Oxidizing agent gets reduced

  23. 5.7 Oxidation-Reduction Reactions Oxidation Numbers: ReDox Reactions • 5-36 (pg 225) Oxidation number • PF6- H2AsO4- UO2- N2O5 POCl5XeO42- • 5-37 (a) • Zn (s)  Zn2+ (aq) 0  2+ (ox) • N5+  N4+ (aq) 5+  4+ (red) • O2-  O2- (aq) no change • H+  H+ (aq) no change • Oxidation (loss of e-) • Reduction (gain of e-)

  24. 5.7 Recognizing a Redox Reaction Corrosion of aluminum 2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s) Al(s) --> Al3+(aq) + 3 e- • Ox. no. of Al increases as e- are donated by the metal. • Therefore, Al is OXIDIZED • Al is the REDUCING AGENT in this balanced half-reaction.

  25. 5.7 Recognizing a Redox Reaction Notice that the 2 half-reactions add up to give the overall reaction if we use 2 mol of Al and 3 mol of Cu2+. 2 Al(s) --> 2 Al3+(aq) + 6 e- 3 Cu2+(aq) + 6 e- --> 3 Cu(s) ---------------------------------------------------------------------------------------------------------------------------------- 2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s) Final eqn. is balanced for mass and charge.

  26. Metals (Cu) are reducing agents Metals (Na, K, Mg, Fe) are reducing agents HNO3 is an oxidizing agent 5.7 Common Oxidizing and Reducing AgentsSee Table 5.4 Cu + HNO3 --> Cu2+ + NO2 2 K + 2 H2O --> 2 KOH + H2

  27. 5.7 Oxidation-Reduction Reactions Thermite reaction Fe2O3(s) + 2 Al(s) ----> 2 Fe(s) + Al2O3(s) X Reactant Product Fe +3 + 3 e- 0 O -2 -2 Al 0 +3 + 3e-

  28. 5.8 Measuring Concentrations • How many moles of each ion are in each solution? (a) 2 mol of potassium perchlorate dissolved in water (b) 354 g of magnesium acetate dissolved in water (c) 1.88 x 1024 formula units of ammonium chromate dissolved in water (d) 1.32 L of 0.55 M sodium bisulfate

  29. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

  30. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?

  31. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M • V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL

  32. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

  33. 5.8 Measuring Concentrations of Compounds in Solution • Concentration (Molarity) = moles/L = [ ] • How many moles of H+(aq) are present in 451 mL of 3.20 M hydrobromic acid? • 5-46 What volume 2.06 M KMnO4, in liters, contains 322 g of solute? • 0.989 L

  34. 5.8 Preparing solutions of Known Concentrations • 5-50 What mass of oxalic acid, H2C2O4, is required to prepare 250. mL of a solution that has a concentration or 0.15 M H2C2O4? • 3.4 g H2C2O4

  35. 5.9 pH, a Concentration scale for acids and bases pH: a way to express acidity -- the concentration of H+ in solution. Low pH: high [H+] High pH: low [H+] Acidic solution pH < 7 Neutral pH = 7 Basic solution pH > 7

  36. 5.9 pH, a Concentration scale for acids and bases • pH = -log [H+] • pOH = - log [OH-] • Kw = [H+][OH-] = 1.00 x 10-14 • pH + pOH = 14 • 5-56 A saturated solution of milk of magnesia, Mg(OH)2, has a pH of 10.5. What is the hydrogen ion concentration of the solution? Is the solution acidic or basic? • 3 x 10-11 • basic

  37. [H+] and pH If the [H+] of soda is 1.6 x 10-3 M, the pH is ____? Because pH = - log [H+] then pH= - log (1.6 x 10-3) pH = 2.80

  38. 5.10 Stoichiometry of Reactions in Aqueous Solutions • 5-64 Hydrazine, N2H4, a base like ammonia, can react with an acid such as sulfuric acid. What mass of hydrazine reacts with 250. mL of 0.146 M H2SO4? 2 N2H4 (aq) + H2SO4 (aq)  2 N2H5+ (aq) + SO42- (aq) • 2.34 g N2H4

  39. 5.10 SOLUTION STOICHIOMETRY Zinc reacts with acids to produce H2 gas. • We have 10.0 g of Zn • What volume of 2.50 M HCl is needed to convert the Zn completely?

  40. 5.10 SOLUTION STOICHIOMETRY Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g) Step 2: Calculate moles of Zn

  41. 5.10: Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 3: Use the stoichiometric factor to get Moles of HCl Step 4: Calculate volume of HCl req’d

  42. 5.10: Standardize a solution of NaOH: i.e., accurately determine its concentration. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

  43. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H2C2O4 Step 2: Calculate amount of NaOH req’d

  44. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amountof H2C2O4 = 0.0118 mol acid Step 2: Calculate amount of NaOH req’d = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH [NaOH] = 0.663 M

  45. 5.10: Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C4H6O5. C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(liq) 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

  46. 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1:Calculate moles of NaOH used. C • V = (0.663 M)(0.03456 L) = 0.0229 mol NaOH Step 2:Calculate amount of acid titrated. Step 3:Calculate mass of acid titrated. Step 4:Calculate % malic acid. = 0.0115 mol acid

  47. Challenge • 5- 72 • 5- 75 • 5-76 • First 3 teams (max 4 members) to correctly solve the problem receive 5 bonus points

  48. Next Lesson • Quiz Chapter 5 • Chapter 6

  49. Chapter 6 Principles of Reactivity: Energy and Chemical Reactivity • Energy transfer • Calorie burning, gravitational, chemical, electrostatic • Heat- mostly seen in chemical processes • Thermodynamics – transfer of heat between objects; science of heat and work • Energy: Some basic principles – capacity to do work • Kinetic – energy of motion • Potential – energy of position • Conservation of energy (aka First law of thermodynamics) • Total energy of universe is constant

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