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1 Reactions in Aqueous Solution Chapter 4. 2 Reaction of lead nitrate with sodium Iodide PbI 2.

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Presentation on theme: "1 Reactions in Aqueous Solution Chapter 4. 2 Reaction of lead nitrate with sodium Iodide PbI 2."— Presentation transcript:

1 1 Reactions in Aqueous Solution Chapter 4

2 2 Reaction of lead nitrate with sodium Iodide PbI 2

3 3 Types of Chemical Reactions 01 Precipitation Reactions: A process in which an insoluble solid (precipitate) drops out of the solution. Most precipitation reactions occur when the anions and cations of two ionic compounds change partners. Pb(NO 3 ) 2 (aq) + 2 NaI(aq)  2 NaNO 3 (aq) + PbI 2 (s)

4 4 Types of Chemical Reactions 02 Acid–Base Neutralization: A process in which an acid reacts with a base to yield water plus an ionic compound called a salt. The driving force of this reaction is the formation of the stable water molecule. HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)

5 5 4.4

6 6 Types of Chemical Reactions 03 Oxidation–Reduction (Redox) Reaction: A process in which one or more electrons are transferred between reaction partners. The driving force of this reaction is the decrease in electrical potential. Mg(s) + I 2 (g)  MgI 2 (s)

7 7 Types of Chemical Reactions 04 Metathesis Reactions: These are reactions where two reactants just exchange parts. AX + BY  AY + BX HNO 3 (aq) + KOH(aq)  KNO 3 (aq) + HOH(l) BaCl 2 (aq) + K 2 SO 4 (aq)  BaSO 4 (s) + 2 KCl(aq)

8 8 Electrolytes in Solution01 Why do ionic compounds conduct electricity when molecular ones generally do not?

9 9 Electrolytes in Solution02 Electrolytes: Dissolve in water to produce ionic solutions. Nonelectrolytes: Do not form ions when they dissolve in water.

10 10 An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte 4.1

11 11 Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Conduct electricity in solution? Cations (+) and Anions (-) 4.1

12 12 Electrolytes in Solution03 Dissociation: The process by which a compound splits up to form ions in the solution.

13 13 Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.   H2OH2O

14 14 Electrolytes in Solution05

15 15 Reaction of lead (II) nitrate with sodium Iodide PbI 2

16 16 Net Ionic Equations in Precipitation Reactions Precipitate – insoluble solid that separates from solution Formula unit equation ionic equation net ionic equation Pb NO Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb I - PbI 2 (s) 4.2

17 17 Writing Net Ionic Equations 1.Write the balanced formula unit equation. 2.Write the ionic equation showing dissociation of strong electrolytes 3.Determine precipitate from solubility rules, leave them un-dissociated. 4.Cancel the spectator ions on both sides of the ionic equation AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO Na + + Cl - AgCl (s) + Na + + NO 3 - Ag + + Cl - AgCl (s) 4.2 Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

18 18 Solubility Rules for Ionic Compounds

19 19 Acids and Bases Blood pH = Regulation of blood pH

20 20 Figure 4.7: Household acids and bases. Photo courtesy of American Color.

21 21 Figure 4.8: Preparation of red cabbage juice as an acid-base indicator. Photo courtesy of James Scherer.

22 22 Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases 4.3

23 23 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water 4.3

24 24 Arrhenius Acid:

25 25 Arrhenius Base Arrhenius Base: A substance that dissociates in, or reacts with, water to form hydroxide ions (OH – ).

26 26 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase 4.3 A Brønsted acid must contain at least one ionizable proton!

27 27 Dissociation of Water: This equilibrium gives us the ion product of water. K w = K c = [H + ][OH – ] = 1.0 x 10 –14 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor

28 28 Acid–Base Concepts05 Lewis Acid: Electron pair acceptor. e.g.: Al 3+, H +, BF 3. Lewis Base: Electron pair donor. e.g.: H 2 O, NH 3, O 2–. Bond formed is called a coordinate bond.

29 29 Lewis Acids and Bases N H H H acidbase F B F F + F F N H H H No protons donated or accepted!

30 30 Strong and Weak Acid-Bases

31 31 Monoprotic acids HCl H + + Cl - HNO 3 H + + NO 3 - CH 3 COOH H + + CH 3 COO - Strong electrolyte, strong acid Weak electrolyte, weak acid Diprotic acids H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Strong electrolyte, strong acid Weak electrolyte, weak acid Triprotic acids H 3 PO 4 H + + H 2 PO 4 - H 2 PO H + + HPO 4 2- HPO H + + PO 4 3- Weak electrolyte, weak acid Strong and Weak Acids

32 32 pH – A Measure of Acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x [H + ] > 1 x [H + ] < 1 x pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ]

33 33 pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x log [H + ] – log [OH - ] = pH + pOH = 14.00

34 34 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = = 1.5 x M The OH - ion concentration of a blood sample is 2.5 x M. What is the pH of the blood? pH + pOH = pOH = -log [OH - ]= -log (2.5 x )= 6.60 pH = – pOH = – 6.60 = 7.40

35 35 Oxidation–Reduction Reactions 02 Oxidation Is Loss (of electrons) Anode Oxidation Reducing Agent

36 36 Oxidation–Reduction Reactions 03 Reduction Is Gain (of electrons) Cathode Reduction Oxidizing Agent

37 37 Oxidation-Reduction Reactions (electron transfer reactions) 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 2Mg + O 2 + 4e - 2Mg O e - 2Mg + O 2 2MgO 4.4

38 38 Types of Oxidation-Reduction Reactions Combination Reaction A + B C S + O 2 SO 2 Decomposition Reaction 2KClO 3 2KCl + 3O 2 C A + B

39 39 NH 3 + H + NH 4 + Zn + 2HCl ZnCl 2 + H 2 Ca + F 2 CaF 2 Precipitation Acid-Base Redox (H 2 Displacement) Redox (Combination) Classify the following reactions. Ca +2 (aq) + CO 3 2- (aq) CaCO3(s)

40 40 4.4

41 41 Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn is oxidizedZn Zn e - Cu 2+ is reducedCu e - Cu Zn is the reducing agent Cu 2+ is the oxidizing agent 4.4 Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) Cu Cu e - Ag + + 1e - AgAg + is reducedAg + is the oxidizing agent

42 42

43 43 RuleApplies toStatement 1ElementsThe oxidation number of an atom in an element is zero. 2Monatomic ions The oxidation number of an atom in a monatomic ion equals the charge of the ion. 3OxygenThe oxidation number of oxygen is –2 in most of its compounds. (An exception is O in H 2 O 2 and other peroxides, where the oxidation number is –1.) Oxidation Number Rules

44 44 Oxidation Number Rules RuleApplies toStatement 4Hydrogen +1 5HalogensFluorine is –1 in all its compounds. Each of the other halogens is –1 in binary compounds unless the other element is oxygen. 6Compounds and ions The sum of the oxidation numbers of the atoms in a compound is zero. The sum in a polyatomic ion equals the charge on the ion.

45 45 HCO 3 - O = -2H = +1 3x(-2) ? = -1 C = +4 Oxidation numbers of all the elements in HCO 3 - ? 4.4

46 46 NaIO 3 Na = +1 O = -2 3x(-2) ? = 0 I = +5 IF 7 F = -1 7x(-1) + ? = 0 I = +7 K 2 Cr 2 O 7 O = -2K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 Oxidation numbers of all the elements in the following ? 4.4

47 47 The Activity Series of the Elements 2Ag 1+ (aq) + Cu(s)2Ag(s) + Cu 2+ (g) Cu 2+ (aq) + 2Ag(s)Cu(s) + 2Ag 1+ (g) Which one of these reactions will occur?

48 48 Activity Series of Elements01

49 49 The Activity Series of the Elements The elements that are higher up in the table are more likely to be oxidized. Thus, any element higher in the activity series will reduce the ion of any element lower in the activity series.

50 50 Activity Series of Elements02 Activity series looks at the relative reactivity of a free metal with an aqueous cation. –Fe(s) + Cu 2+ (aq)  Fe 2+ (aq) + Cu(s) –Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) –Cu(s) + 2 Ag + (aq)  2 Ag(s) + Cu 2+ (aq) –Mg(s) + 2 H + (aq)  Mg 2+ (aq) + H 2 (g) Cu 2+ (aq) + 2 Ag (s)  X  Cu (s) + 2 Ag + (aq)

51 51 The Activity Series of the Elements 2Ag 1+ (aq) + Cu(s)2Ag(s) + Cu 2+ (g) Cu 2+ (aq) + 2Ag(s)Cu(s) + 2Ag 1+ (g) Which one of these reactions will occur?

52 52 Balancing Redox Equations 1.Write the unbalanced equation for the reaction in ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Cr 2 O 7 2- Cr Fe 2+ Fe Balance the atoms other than O and H in each half-reaction. Cr 2 O Cr 3+

53 53 Balancing Redox Equations 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O Cr H 2 O 14H + + Cr 2 O Cr H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe e - 6e H + + Cr 2 O Cr H 2 O 6.If necessary, equalize the number of electrons in the two half- reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe e - 6e H + + Cr 2 O Cr H 2 O Redution Oxidation

54 54 Balancing Redox Equations 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 6e H + + Cr 2 O Cr H 2 O 6Fe 2+ 6Fe e - Oxidation: Reduction: 14H + + Cr 2 O Fe 2+ 6Fe Cr H 2 O 8.Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation.

55 55 MnO 2 (s) + BrO 3 1- (aq)MnO 4 1- (aq) + Br 1- (aq) Balancing Redox Reactions In Basic Media Balance the following net ionic equation in basic solution:

56 56 Write the two unbalanced half-reactions. Br O 3 1- (aq) Br 1- (aq) MnO 2 (s)MnO 4 1- (aq) MnO 2 (s) + BrO 3 1- (aq)MnO 4 1- (aq) + Br 1- (aq)

57 57 Balance both half-reactions for all atoms except O and H. Br O 3 1- (aq) Br 1- (aq) MnO 2 (s)MnO 4 1- (aq)

58 58 Br O 3 1- (aq) + 6H 1+ (aq)3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)4H 1+ (aq) + MnO 4 1- (aq) Balance each half-reaction for O by adding H 2 O, and then balance for H by adding H 1+.

59 59 Br O 3 1- (aq) + 6H 1+ (aq) + 6e - 3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)3e - + 4H 1+ (aq) + MnO 4 1- (aq) Balance each half-reaction for charge by adding electrons to the side with greater positive charge.

60 60 Multiply each half-reaction by a factor to make the electron count the same in both half-reactions. 2 Br O 3 1- (aq) + 6H 1+ (aq) + 6e - 3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)3e - + 4H 1+ (aq) + MnO 4 1- (aq)

61 61 2MnO 2 (s) + H 2 O(l) + BrO 3 1- (aq) 2H 1+ (aq) + 2MnO 4 1- (aq) + Br 1- (aq) Br O 3 1- (aq) + 6H 1+ (aq) + 6e - 3H 2 O(l) + Br 1- (aq) 2MnO 2 (s) + 4H 2 O(l)6e - + 8H 1+ (aq) + 2MnO 4 1- (aq) Add the two balanced half-reactions together and cancel species that appear on both sides of the equation.

62 62 Since the reaction occurs in a basic solution, “neutralize” the excess H 1+ by adding OH 1- and cancel any water (if possible) 2MnO 2 (s) + H 2 O(l) + BrO 3 1- (aq) + 2OH 1- (aq) 2OH 1- (aq) + 2H 1+ (aq) + 2MnO 4 1- (aq) + Br 1- (aq) 2H 2 O 2MnO 2 (s) + BrO 3 1- (aq) + 2OH 1- (aq) H 2 O(l) + 2MnO 4 1- (aq) + Br 1- (aq) Balancing Redox Reactions In Basic Media

63 63

64 64

65 65 Redox Titrations If the unknown concentration is the potassium permanganate solution, MnO 4 1-, it can be slowly added to a known amount of oxalic acid, H 2 C 2 O 4, until a faint purple color persists. Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l)

66 66 Redox Titrations 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) A solution is prepared with g of oxalic acid, H 2 C 2 O mL of an unknown solution of potassium permanganate are needed to titrate the solution. What is the concentration of the potassium permanganate solution? Moles of H 2 C 2 O 4 Mass of H 2 C 2 O 4 Moles of KMnO 4 Molarity of KMnO 4 Mole RatioMolarity of KMnO 4 Molar Mass of H 2 C 2 O 4

67 67 Redox Titrations 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) Moles of H 2 C 2 O 4 available: g 1 mol = mol H 2 C 2 O g H 2 C 2 O 4 Moles of KMnO 4 reacted: 5 mol H 2 C 2 O 4 2 mol KMnO 4 = mol KMnO mol H 2 C 2 O 4 x x

68 68 Redox Titrations 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) Concentration of KMnO 4 solution: = M KMnO mL(1L/1000 ml) mol KMnO 4


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