1 CHAPTER 5 Read Chapter 5 Study examples and exercises.

2 Chapter 5 Outline Electrolytes NIE’s 5 General Chemical Reactions Redox Acid/Bases Molarity Titration

IONIC COMPOUNDS Compounds in Aqueous Solution aqueous solutions.Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO 4 in water K + (aq) + MnO 4 - (aq)

4 Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions conduct electricity! ELECTROLYTES They are called ELECTROLYTES strong electrolytes. HCl, K 2 CrO 4, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. K 2 CrO 4 (aq) ---> 2 K + (aq) + CrO 4 2- (aq)

5 NaCl dissolving in water Cl - Na + Negative O atom Positive H atom

6 Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions conduct electricity! ELECTROLYTES They are called ELECTROLYTES strong electrolytes. HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

7 Figure 5.2 Strong Electrolyte

8 Aqueous Solutions weak electrolyte. Acetic acid ionizes only to a small extent, so it is a weak electrolyte. HC 2 H 3 O 2 (aq)C 2 H 3 O 2 - (aq) + H + (aq)

9 Figure 5.3 Weak Electrolyte

10 Ionized acetic acid H+H+ Acetic acid — Weak Electrolyte

11 Aqueous Solutions weak electrolyte. Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH 3 CO 2 H(aq) - CH 3 CO 2 - (aq) + H + (aq)

12 Aqueous Solutions nonelectrolytes. Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol (in antifreeze)

13 Figure 5.3 b Nonelectrolyte

14 Nonelectrolyte— Ethanol, C 2 H 5 OH

15 WATER SOLUBILITY OF IONIC COMPOUNDS INSOLUBLE. Not all ionic compounds dissolve in water. Some are INSOLUBLE. Figure 5.4 See Figure 5.4 As long as one ion from the list is present in a compound, the compound is water soluble.

16 See your Solubility Table

17 WATER SOLUBILITY OF IONIC COMPOUNDS Common minerals are often formed with anions that lead to insolubility: sulfidefluoride carbonateoxide Azurite, a copper carbonate Iron pyrite, a sulfide Orpiment, arsenic sulfide

18 An acid H + in water ACIDS strong Some strong acids are HClhydrochloric HNO 3 nitric HClO 4 perchloric H 2 SO 4 sulfuric HNO 3

19 An acid H + in water ACIDS HCl(aq) H + (aq) + Cl - (aq)

20 The Nature of Acids HCl H 2 O H 3 O + Cl - hydronium ion

21 Weak Acids WEAK ELECTROLYTES WEAK ACIDS = WEAK ELECTROLYTES HC 2 H 3 O 2 acetic acid H 2 CO 3 carbonic acid H 3 PO 4 phosphoric acid

22 ACIDS Nonmetal oxides can be acids CO 2 (aq) + H 2 O(l) H 2 CO 3 (aq) SO 3 (aq) + H 2 O(l) H 2 SO 4 (aq) and can come from burning coal and oil.

23 Base OH - in water BASES NaOH(aq) Na + (aq) + OH - (aq) NaOH is a strong base.

24 Ammonia, NH 3 An Important Base NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq)

25 Figure 5.11 Ammonia is a weak base

26 BASES Metal oxides are bases CaO(s)+ H 2 O(l) Ca(OH) 2 (aq) CaO in water. Indicator shows solution is basic.

27 Net Ionic Equations Mg(s) + 2 HCl(aq) H 2 (g) + MgCl 2 (aq) We really should write Mg(s) + 2 H + (aq) + 2 Cl - (aq) H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq)

28 Ionic Equations SPECTATOR IONS The two Cl - ions are SPECTATOR IONS — they do not participate. Mg(s) + 2 HCl(aq) H 2 (g) + MgCl 2 (aq) Mg(s) + 2 H + (aq) + 2 Cl - (aq)H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq)

29 Net Ionic Equations Mg(s) + 2 HCl(aq) H 2 (g) + MgCl 2 (aq) Mg(s) + 2 H + (aq) + 2 Cl - (aq) H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) We leave the spectator ions out in writing the NET IONIC EQUATION (NIE) Mg(s) + 2 H + (aq) H 2 (g) + Mg 2+ (aq)

CHEMICAL REACTIONS IN WATER We will look at EXCHANGE REACTIONS The anions exchange places between cations. Pb(NO 3 ) 2 (aq) + 2 KI (aq) ----> PbI 2 (s) + 2 KNO 3 (aq) AX + BY AY X

31 Precipitation Reactions The “driving force” is the formation of an insoluble compound — a precipitate. Pb(NO 3 ) 2 (aq) + 2 KI(aq) 2 KNO 3 (aq) + PbI 2 (s) Net ionic equation Pb 2+ (aq) + 2 I - (aq) PbI 2 (s)

32 Acid-Base Reactions The “ driving force ” is the formation of water.The “ driving force ” is the formation of water. NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) Net ionic equation OH - (aq) + H + (aq) H 2 O(l) OH - (aq) + H + (aq) H 2 O(l) This applies to ALL reactions of STRONG acids and bases.This applies to ALL reactions of STRONG acids and bases.

33 Acid-Base Reactions A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end.A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end. The other product of the A-B reaction is a SALT, MX.The other product of the A-B reaction is a SALT, MX. HX + MOH ---> MX + H 2 O HX + MOH ---> MX + H 2 O M n+ comes from the base & X n- comes from the acid. M n+ comes from the base & X n- comes from the acid. This is one way to make compounds! This is one way to make compounds!

34 Gas-Forming Reactions

35 Gas-Forming Reactions CaCO 3 (s) + H 2 SO 4 (aq) CaSO 4 (s) + H 2 CO 3 (aq) Carbonic acid is unstable and forms CO 2 & H 2 O H 2 CO 3 (aq) CO 2 (g) + water (Antacid tablet has citric acid + NaHCO 3 )

36 Oxidation-Reduction Reactions Fe 2 O 3 (s) + 2 Al(s) 2 Fe(s) + Al 2 O 3 (s) Thermite reaction

37 REDOX REACTIONS EXCHANGEAcid-BaseReactions EXCHANGEGas-FormingReactions EXCHANGE: Precipitation Reactions REACTIONS

38 REDOX REACTIONS Oxidation — ( H, Mg, and Al ) 2 H 2 (g) + O 2 (g) 2 H 2 O(l) Mg(s) + 2 HCl(aq) MgCl 2 (aq) + H 2 (g) All corrosion reactions are oxidations. 2 Al(s) + 3 Cu 2+ (aq) 2 Al 3+ (aq) + 3 Cu(s) Reduction — ( Fe +3 ) Fe 2 O 3 (s) + 2 Al(s) 2 Fe(s) + Al 2 O 3 (s)

39 But notice that in all reactions if something has been oxidized then something has also been reduced. But notice that in all reactions if something has been oxidized then something has also been reduced. Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) REDOX REACTIONS

40 Why Study Redox Reactions Manufacturing metals Fuels Corrosion Batteries

41 Redox reactions are characterized by ELECTRON TRANSFER between an electron donor and electron acceptor. Transfer leads to — 1. Increase in oxidation number of some element = OXIDATION 2. Decrease in oxidation number of some element = REDUCTION REDOX REACTIONS

42 OXIDATION NUMBERS The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules: The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules: 1. Each atom in free element has ox. no. = 0 1. Each atom in free element has ox. no. = 0 Zn O 2 I 2 S 8 2. In simple ions, ox. no. = charge on ion 2. In simple ions, ox. no. = charge on ion -1 for Cl - +2 for Mg for Cl - +2 for Mg 2+

43 OXIDATION NUMBERS 3. O has ox. no. = O has ox. no. = -2 (except in peroxides: in H 2 O 2, O = -1) 4. Ox. no. of H = Ox. no. of H = +1 (except when H is associated with a metal as in NaH where it is -1) 5. Algebraic sum of oxidation numbers 5. Algebraic sum of oxidation numbers = 0 for a compound = 0 for a compound = overall charge for an ion = overall charge for an ion

44 OXIDATION NUMBERS NH 3 N = ClO - Cl = H 3 PO 4 P = MnO 4 - Mn = Cr 2 O 7 2- Cr = C 3 H 8 C = Oxidation number of F in HF?

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

46 Recognizing a Redox Reaction Corrosion of aluminum 2Al (s) + 3Cu 2+ (aq) --> 2Al 3+ (aq) + 3Cu (s) Cu 2+ (aq) + 2 e - --> Cu(s) Ox. no. of Cu decreases as e - are accepted by the ion.Ox. no. of Cu decreases as e - are accepted by the ion. Therefore, Cu is REDUCED and is the OXIDIZING AGENT in this balanced half- reaction.Therefore, Cu is REDUCED and is the OXIDIZING AGENT in this balanced half- reaction.

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

48 Examples of Redox Reactions Metal + acid Mg + HCl Mg = reducing agent H + = oxidizing agent Metal + acid Cu + HNO 3 Cu = reducing agent HNO 3 = oxidizing agent

49 Examples of Redox Reactions Metal + Halogen 2 Al + 3 Br 2 ---> Al 2 Br 6

50 Recognizing a Redox Reaction See Table 5.5 In terms of oxygen gainloss In terms Ox. No. increasedecrease In terms of electronslossgain Reaction Type OxidationReduction

51 Common Oxidizing and Reducing Agents See Table 5.4 Metals (Cu) are reducing agents HNO 3 is an oxidizing agent 2 K + 2 H 2 O --> 2 KOH + H 2 Metals (Na, K, Mg, Fe) are reducing agents Cu + 4 HNO 3 --> Cu(NO 3 ) NO H 2 O

52 Learn to recognize common oxidizing and reducing agents. See Table 5.4.

53 Table 5-4

54 BALANCING REDOX EQUATIONS Section 21.1 The Half-Reaction Method Separate the equation into half-reactions.Separate the equation into half-reactions. Balance the half-reactions.Balance the half-reactions. Combine the half-reactions to form a balanced equation containing no electrons.Combine the half-reactions to form a balanced equation containing no electrons.

55 Balancing Half-Reactions First balance the element changing oxidation state.First balance the element changing oxidation state. Balance the oxygen atoms with water.Balance the oxygen atoms with water. Balance the hydrogen atoms with H +.Balance the hydrogen atoms with H +. Balance the charge with electrons.Balance the charge with electrons. After combining the half-reactions, check for mass and charge balance.

56 Practice Problems Balance the following equations: MnO H 2 SO > Mn +2 + SO 4 -2 Al + NO > Al(OH) NH 3

REACTIONS IN SOLUTION Section 5.8

58 Terminology In solution we need to define the SOLVENTSOLVENT the component whose physical state is preserved when solution forms SOLUTESOLUTE the other solution component

59 Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution

60 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make mL of solution. Calculate molarity.

61 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make mL of solution. Calculate molarity. = mole/L = M 5.00 g mole 5.00 g mole L g L g

62 The Nature of the KMnO 4 Solution KMnO 4 (aq) --> K + (aq) + MnO 4 - (aq) If you make a solution that is 0.30 M in KMnO 4, this means that _ [K + ] = [MnO 4 - ] = 0.30 M

63 The Nature of a Na 2 CO 3 Solution This water-soluble compound is ionic Na 2 CO 3 (aq) --> 2 Na + (aq) + CO 3 2- (aq) If [Na 2 CO 3 ] = M, then [Na + ] = M [CO 3 2- ] = M Na 2 CO 3

64 USING MOLARITY What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a M solution? What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a M solution? Conc (M) = moles/volume = mol/V This means that moles = M V

65 Preparing Solutions From Solids What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a M solution? What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a M solution? L mole 90.0 g L mole = 1.12 g

66 Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent.Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.Dilute a concentrated solution to give one that is less concentrated.

67 Preparing Solutions by Dilution Preparing a 1.64 × M or 1.64 × mol/L solution.

68 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?

69 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? How much water is added? The important point is that moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution

70 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Moles of NaOH in original solution = M V M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Therefore, moles of NaOH in final solution must also = 0.15 mol NaOH (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L 300 mL = volume of final solution. or 300 mL = volume of final solution.

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

72 A shortcut A shortcut M initial V initial = M final V final Preparing Solutions by Dilution

Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? SOLUTION STOICHIOMETRY SOLUTION STOICHIOMETRY Section 5.9

74 Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl 2 (aq) + H 2 (g) Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

75 Step 2: Step 2: Write the given and requested information below the equation. Zn(s) + 2 HCl(aq) --> ZnCl 2 (aq) + H 2 (g) 10.0 g 2.50 M ?mL 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?

76 Step 3: Step 3: Calculate using the information. Zn(s) + 2 HCl(aq) --> ZnCl 2 (aq) + H 2 (g) 10.0 g 2.50 M ?mL = L HCl 10.0 gZn moleZn 2 moleHCl L HCl 65.4 gZn moleZn 2.50 moleHCl Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

77 ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) --->Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) acid base acid base Carry out this reaction using a TITRATION. Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

78 Titration setup Buret contains a solution whose concentration is known exactly. Solution of unknown concentration

79 Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. At equivalence point moles H + = moles OH -

g of H 2 C 2 O 4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

81 Step 1: Write the balanced equation 2 NaOH(aq) + H 2 C 2 O 4 (aq) --> Na 2 C 2 O 4 (aq) + 2 HOH(l) g of H 2 C 2 O 4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

82 Step 2: Step 2: Write the given and requested information below the equation. 2NaOH(aq) + H 2 C 2 O 4 (aq) --> Na 2 C 2 O 4 (aq) +2 HOH(l) mL g ? M g of H 2 C 2 O 4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

83 = M NaOH Step 3: Step 3: Calculate using the information. 2NaOH(aq) + H 2 C 2 O 4 (aq) --> Na 2 C 2 O 4 (aq) +2 HOH(l) mL g ? M 1.065g A mole A 2 mole B L B 90.0 g A mole A = mole B/L B g of H 2 C 2 O 4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

84 LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C 4 H 6 O 5. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) g of apple requires mL of M NaOH for titration. What is mass % of malic acid?

g of apple requires mL of M NaOH for titration. What is weight % of malic acid? Step 1: Write the balanced equation C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(l)

86 Step 2: Step 2: Write the given and requested information below the equation. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(l) ? g mL (? % A) M g of apple requires mL of M NaOH for titration. What is weight % of malic acid?

87 Step 3: Step 3: Calculate using the information. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(l) ? g mL (? % A) M L B.663 mole B mole A g A L B 2 mole B mole A = 1.54 g A 1.54 g A g apple % A = x 100 = 2.00 % A g of apple requires mL of M NaOH for titration. What is weight % of malic acid?

88 Sample Problems 1) What volume of 0.50 M sulfuric acid is needed to react completely with 10.0 mL of 2.0 M potassium hydroxide? H 2 SO KOH --> 2 HOH + K 2 SO M 2.0 M ? mL 10.0 mL.0100 L KOH 2.0 mole KOH mole H 2 SO 4 L H 2 SO 4 L KOH 2 mole KOH.50 mole H 2 SO 4 L KOH 2 mole KOH.50 mole H 2 SO 4 = L H 2 SO 4

89 Sample Problems 2) 16 mL of 2.0 M NaOH neutralizes 25 mL of HCl. What is the molarity of the acid? L NaOH 2.0 mole NaOH mole HCl L HCl L NaOH mole NaOH L HCl L NaOH mole NaOH = 1.3 M HCl HCl + NaOH --> HOH + NaCl HCl + NaOH --> HOH + NaCl ? M 2.0 M ? M 2.0 M 25 mL 16 mL

90 Sample Problems 3) A 0.15 M solution of calcium chloride is added to a solution of ammonium carbonate and g of calcium carbonate is precipitated. What volume of calcium chloride solution was added? CaCl 2 + (NH 4 ) 2 CO 3 --> CaCO NH 4 Cl CaCl 2 + (NH 4 ) 2 CO 3 --> CaCO NH 4 Cl 0.15 M g 0.15 M g ? mL ? mL = L CaCl g CaCO 3 mole CaCO 3 mole CaCl 2 L CaCl gCaCO 3 moleCaCO 3.15 moleCaCl gCaCO 3 moleCaCO 3.15 moleCaCl 2

91 Sample Problems 4) What volume(mL) of M HCl is needed to react completely with 15.7 g of barium hydroxide? 2 HCl + Ba(OH) 2 --> 2 HOH + BaCl 2 2 HCl + Ba(OH) 2 --> 2 HOH + BaCl M 15.7 g M 15.7 g ? mL ? mL = 1.83 L HCl = 1830 mL HCl 15.7 g Ba(OH) 2 mole Ba(OH) 2 2 mole HCl L HCl g Ba(OH) 2 moleBa(OH) 2.100moleHCl g Ba(OH) 2 moleBa(OH) 2.100moleHCl

92 Practice Problems Write the balanced formula, ionic, and net ionic equations for: Mixture of solutions of barium chloride and sodium phosphate. Mixtures of solutions of silver nitrate and sodium carbonate. Mixtures of solutions of nitric acid and barium hydroxide. Mixtures of solutions of ammonia and acetic acid. Silver carbonate solid reacts with nitric acid

93 Practice Problems 1. Write the NIE for each of the following: lead(II) nitrate + potassium iodide perchloric acid + potassium hydroxide sodium sulfite + hydroiodic acid 2. Identify the substance being reduced, the substance being oxidized, the oxidizing agent, and the reducing agent: oxidized, the oxidizing agent, and the reducing agent: 4 Fe + 3 O 2 --> 2 Fe 2 O 3

94 Practice Problems 3. a) Balance the following equation in acidic solution. KMnO 4 + NaCl --> Cl 2 + MnCl 2 KMnO 4 + NaCl --> Cl 2 + MnCl 2 b) Balance the following equation in basic solution. b) Balance the following equation in basic solution. I 2 + NO > IO N 2 O I 2 + NO > IO N 2 O 4. Calculate the molarity (M) of 32.7 g H 3 PO 4 in 250 mL of solution. 5. Explain how to prepare 500. mL of 0.10 M (NH 4 ) 2 CO 3 6. Explain how to prepare 75 mL of M NaOH from 4.0 M NaOH.

95 Practice Problems 7. How many grams of solute is there in 75.0 mL of 0.25 M FeCl 3 ? 8. Explain how to prepare 75.0 mL of M ammonium phosphate mL of a HCl solution is required to react completely with g of CaCO 3. What is the M? 10. How many grams of BaSO 4 can be formed from a barium nitrate solution by adding 42.6 mL of 0.15 M sulfuric acid? 11. How many moles of sulfur dioxide are produced when 25.0 mL of 0.15 M ammonium sulfite reacts with 25.0 mL of 0.25 M chloric acid?

96 Practice Problems Answers 1. Pb 2+ (aq) + 2 I 1- (aq) --> PbI 2 (s) H + (aq) + OH - (aq) --> HOH (l) SO 3 2- (aq) + 2 H + (aq) --> SO 2 (g) + H 2 O (l) 2. Reduced, OA = oxygen Oxidized, RA = iron 3. a) 16 H + (aq) + 2 MnO 4 - (aq) + 10 Cl - (aq) --> 2 Mn 2+ (aq) + 8 H 2 O (l) + 5 Cl 2 (g) b) 2 OH - (aq) + 4 I 2 (aq) + 6 NO 3 - (aq) --> 8 IO 2 - (aq) + 3 N 2 O (g) + H 2 O (l) 8 IO 2 - (aq) + 3 N 2 O (g) + H 2 O (l) M

97 Practice Problems Answers 5. Dissolve 4.8 g ammonium carbonate in some water, then dilute to a total volume of 500. mL. 6. Take 0.47 mL of 4.0 M NaOH and dilute to a total volume of 75 mL g 8. Dissolve 1.17 g ammonium phosphate in some water, then dilute to a total volume of 75.0 mL M g mole End of Chapter 5

98 Sample Problems Write the NIE 1. ferric nitrate + sodium sulfide Fe(NO 3 ) 3 Na 2 S Fe 2 S 3 NaNO 3 +-->+ --> 2 Fe NO Fe(NO 3 ) 3 3 Na 2 S Fe 2 S 3 6 NaNO 3 +-->++ 6 Na + 3 S 2- + Fe 2 S 3 6 NO Na > 2 Fe 3+ 3 S 2- + Fe 2 S 3

99 Sample Problems Write the NIE 2. barium hydroxide + acetic acid Ba(OH) 2 HC 2 H 3 O 2 HC 2 H 3 O 2 Ba(C 2 H 3 O 2 ) 2 HOH HOH+-->+ --> Ba 2+ Ba OH HC 2 H 3 O 2 2 HOH + 2 C 2 H 3 O Ba 2+ Ba(OH) 2 2 HC 2 H 3 O 2 Ba(C 2 H 3 O 2 ) 2 2 HOH +-->+ --> 2 OH - 2 HC 2 H 3 O HOH 2 C 2 H 3 O > OH - OH - HC 2 H 3 O 2 HC 2 H 3 O 2+ HOH HOH C2H3O2- C2H3O2- C2H3O2- C2H3O2-+

100 Sample Problems Write the NIE 3. hydrochloric acid + calcium carbonate HCl HCl CaCO 3 CaCO 3 CaCl 2 CaCl 2 H 2 CO 3 H 2 CO 3+-->+ --> 2 H + 2 H ++ 2 Cl - + CaCO 3 Ca Cl - 2 HCl 2 HCl CaCO 3 CaCO 3 CaCl 2 CaCl 2 H 2 CO 3 H 2 CO 3+-->+ HOH HOH+ CO 2 +--> 2 H + CaCO 3 + CO 2 HOH++ Ca 2+

101 Sample Problems Write the NIE 4. sodium nitrate + potassium chloride NaNO 3 KClNaCl KNO 3 +-->+ --> Na + + NO 3 - NO 3 -+ K+ K+ K+ K+ Cl - + NO 3 - NO 3 -+ K+K+K+K++ Cl - Na + + All spectators, no reaction (N/R)

102 Sample Problems Write the NIE 5. sodium sulfite + sulfuric acid Na 2 SO 3 H 2 SO 4 Na 2 SO 4 H 2 SO 3 +-->+ --> 2 Na + + SO H+H+H+H+ HSO SO H + + HSO > SO SO 2 + H 2 O H2O H2O H2O H2O+ SO 2 + SO 4 2- SO Na + +

103 Sample Problems Write the NIE 6. barium + sodium chloride Ba BaNaCl BaCl 2 Na+-->+ -->Ba + 2 Na + 2 Cl Na + Ba 2+ Ba 2+ 2 Cl > Ba Ba 2 Na + + Ba Na Ba Ba 2 NaCl BaCl 2 2 Na +-->+

104 Practice Problems Balance the following equation: MnO H 2 SO > Mn +2 + SO 4 -2 MnO > Mn +2 H 2 SO > SO Balance atoms: 8 H + + MnO > Mn H 2 O H 2 O + H 2 SO > SO H + 1. Separate into half reactions:

105 Practice Problems 3. Balance charges: 5 e H + + MnO > Mn H 2 O H 2 O + H 2 SO > SO H e - 4. Equal electrons gained and lost: 2(5 e H + + MnO > Mn H 2 O) 5(H 2 O + H 2 SO > SO H e - )

106 Practice Problems 5. Simplify and Add: 2(5 e H + + MnO > Mn H 2 O) 5(H 2 O + H 2 SO > SO H e - ) MnO H 2 SO > 2 Mn SO H H 2 O 4 3

107 Practice Problems Balance the following equation: Al + NO > Al(OH) NH 3 Al ----> Al(OH) 4 - NO > NH 3 2. Balance atoms: 4 H 2 O + Al ----> Al(OH) H + 9 H + + NO > NH H 2 O 9 H + + NO > NH H 2 O 1. Separate into half reactions:

108 Practice Problems 3. Balance charges: 4 H 2 O + Al ----> Al(OH) H e - 4 H 2 O + Al ----> Al(OH) H e - 8 e H + + NO > NH H 2 O 4. Equal electrons gained and lost: 8(4 H 2 O + Al ----> Al(OH) H e - ) 3(8 e H + + NO > NH H 2 O)

109 Practice Problems 5. Simplify and Add: 8(4 H 2 O + Al ----> Al(OH) H e - ) 3(8 e H + + NO > NH H 2 O) H 2 O + 8 Al + 3 NO > 8 Al(OH) H NH

110 Practice Problems 6. Change to basic solution: 23 H 2 O + 8 Al + 3 NO >8 Al(OH) H NH OH OH - 5 H 2 O H 2 O + 5 OH Al + 3 NO >8 Al(OH) NH 3

111 Practice Problems 1. MnO 2 + HBr --> Br 2 + MnBr 2 2e H + + MnO > Mn H 2 O 2 Br > Br 2 + 2e - 2 Br > Br 2 + 2e - 4 H + + MnO Br > Mn H 2 O + Br 2

112 Practice Problems 2. Cl 2 + NaBr --> NaCl + Br 2 2e - + Cl 2 --> 2 Cl - 2e - + Cl 2 --> 2 Cl - 2 Br - --> Br 2 + 2e - 2 Br - --> Br 2 + 2e - Cl Br > 2 Cl - + Br 2 Cl Br > 2 Cl - + Br 2

113 Practice Problems 3. H 2 S + HNO 3 --> S + NO 3(H 2 S ----> S + 2 H e - ) 2(3 e H + + NO > NO + 2 H 2 O) H 2 S + 2 H NO > 3 S + 2 NO + 4 H 2 O

114 Practice Problems 4. PbO 2 + Sb --> PbO + NaSbO 2 (base) 3(2 e- + 2H + + PbO > PbO + H 2 O) 2(2 H 2 O + Sb ----> SbO H e - ) PbO 2 + H 2 O + 2 Sb --->3 PbO + 2 SbO H OH OH OH OH - 2 H 2 O 3PbO OH- + 2Sb --->3PbO + 2SbO H 2 O 1

115 Sample Problems 1) Calculate the molarity(M) of 40.0 g NaOH in 700. mL of solution g L mole 40.0 g = 1.43 M

116 Sample Problems 2) Explain how to prepare 2.0 L of 1.5 M LiBr. 2.0 L 1.5 mole L = 260 g 86.8 g mole mole Dissolve 260 g LiBr in some water, then dilute to a total volume of 2.0 L.

117 Sample Problems 3) Explain how to prepare 100. mL of 0.10 M HCl from 6.0 M HCl M D V D = M C V C mole HCl = mole HCl (0.10M)(100. mL) = (6.0 M)V C V C = 1.7 mL Take 1.7 mL of 6.0 M HCl and dilute to a total volume of 100. mL.

118 Sample Problems 4) How many grams of solute is there in 50.0 mL of 0.15 M NaOH? L 0.15 mole L = 0.30 g 40.0 g mole mole