1. Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical reactions: precipitation, acid-base and oxidation reactions. Stoichiometry of reactions and balancing the chemical equations.

2. The water molecule is polar. A space-filling model of the water molecule. Water is a dissolving media or solvent Aqueous Solutions

3. Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt assisting in the dissolving process. Cl - Na + Cl - Na + H2OH2O

4. Some Properties of Water 4 Water is “bent” or V-shaped. 4 The O-H bonds are covalent. 4 Water is a polar molecule. 4 Hydration occurs when salts dissolve in water.

5. Figure 4.3: (a) The ethanol molecule contains a polar O—H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O—H bond in ethanol. This is a case of "like dissolving like."

6. Solvent 4 retains its phase (if different from the solute) 4 is present in greater amount (if the same phase as the solute)

7. Solute 4 dissolves in water (or other “solvent”) 4 changes phase (if different from the solvent) 4 is present in lesser amount (if the same phase as the solvent)

8. General Rule for dissolution  Like dissolve likewise  Polar dissolve polar (water dissolves in ethanol)  Non-polar dissolve nonpolar (Alkane H/C dissolves in fat)

9. Figure 4.5: When solid NaCl dissolves, the Na+ and Cl- ions are randomly dispersed in the water. Example

10. Electrolytes Strong - conduct current efficiently NaCl, HNO 3 Weak - conduct only a small current vinegar, tap water Non - no current flows pure water (non-ionic or de-ionized), sugar solution

11. Figure 4.4: Electrical conductivity of aqueous solutions.conductivity

12. Acids Strong acids -dissociate completely to produce H + in solution hydrochloric and sulfuric acid HCl, H 2 SO 4 Weak acids - dissociate to a slight extent to give H + in solution acetic and formic acid CH 3 COOH, CH 2 O

13. Bases Strong bases - react completely with water to give OH  ions. sodium hydroxide Weak bases - react only slightly with water to give OH  ions. ammonia

14. Figure 4.6: HCl (aq) is completely ionized.completely ionized

15. Figure 4.7: An aqueous solution of sodium hydroxide.

16. Figure 4.8: Acetic acid (HC 2 H 3 O 2 ) exists in water mostly as undissociated molecules. Only a small percentage of the molecules are ionized such as 1 in 100 molecules.

17. Molarity Molarity (M) = moles of solute per volume of solution in liters:

18. Calculate the molarity (M) of the solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. Find the concentration of each type of ion in 0.26 M Al(NO 3 ) 3 and 0.15 M CaCl 2 Calculate the number of moles of Cl - ions in 1.75 L of 1.0 x 10 -3 M ZnCl 2.

19. Common Terms of Solution Concentration Stock - routinely used solutions prepared in concentrated form. Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl) Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl): (MV) initial =(MV) Final

20. Figure 4.10: Steps involved in the preparation of a standard aqueous solution.

21. Figure 4.12: Dilution Procedure (a) A measuring pipette is used to transfer 28.7mL of 17.4 M acetic acid solution to a volumetric flask. (b) Water is added to the flask to the calibration mark. (c) The resulting solution is 1.00 M acetic acid. Diluted from 17.4 M to 1.00 M

22. Moles of solute after dilution = moles of solute before dilution Prepare 2.00 L of each of the following solution: a)0.250 M NaOH from solid NaOH b) 0.250 M NaOH from 1.00 M NaOH stock solution M 1 V 1 = M 2 V 2 Before dilution after dilution

23. Types of Solution Reactions 4 Precipitation reactions AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) 4 Acid-base reactions NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l) 4 Oxidation-reduction (redox) reactions Fe 2 O 3 (s) + Al(s)  Fe(l) + Al 2 O 3 (s)

24. Simple Rules for the Solubility of Salts in Water 1.Most nitrate (NO 3  ) salts are soluble. 2.Most alkali (group 1A) salts and NH 4 + are soluble. 3.Most Cl , Br , and I  salts are soluble (NOT Ag +, Pb 2+, Hg 2 2+ ) 4.Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4, CaSO 4 ) 5.Most OH  salts are only slightly soluble (NaOH, KOH are soluble, and Ba(OH) 2, Ca(OH) 2 are marginally soluble) 6.Most S 2 , CO 3 2 , CrO 4 2 , PO 4 3  salts are only slightly soluble.

25. Figure 4.13: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

26. Describing Reactions in Solution Precipitation 1.Molecular equation (reactants and products as compounds) AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) 2.Complete ionic equation (all strong electrolytes shown as ions) Ag + (aq) + NO 3  (aq) + Na + (aq) + Cl  (aq) AgCl(s) + Na + (aq) + NO 3  (aq)

27. Describing Reactions in Solution (continued) 3.Net ionic equation (show only components that actually reacts) Ag + (aq) + Cl  (aq)  AgCl(s) Na + and NO 3  are spectator ions.

28. Stoichiometry of Precipitation Reactions How to calculate quantities of reactants and products involved in a chemical reaction? 1.Convert all quantities to moles. 2.Use the balancing coefficients to relate the moles of reactants & products in a balanced equation. 3.Determine the solid product (precipitate) and spectator ions from the rules of solubility. 4.Calculate the limiting reactant and theoretical yield. 5.Convert moles to mass if required.

29. Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO 3 solution to precipitate all the Ag + ions in the form of AgCl. Hint: Na + and NO 3 - are spectator ions, AgCl is the only product. When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 and 2.00 L of 0.0250 M Na 2 SO 4 are mixed. Hint: Na + and NO 3 - are spectator ions, PbSO 4 is the only product.

30. Acid-Base Reactions An acid produces H + ions in water A base produces OH - ions in water An acid is a proton donor A base is a proton acceptor Arrhenius’s concept Bronsted & Lowrys’ concept

31. Performing Calculations for Acid-Base Reactions 1.List initial species and predict reaction. 2.Write balanced net ionic reaction. 3.Calculate moles of reactants. 4.Determine limiting reactant. 5.Calculate moles of required reactant/product. 6.Convert to grams or volume, as required. Remember: n H+ = n OH- (MV) H+ = (MV) OH-

32. Neutralization Reaction acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H 2 OH + + Cl - + Na + + OH - Na + + Cl - + H 2 O H + + OH - H 2 O 4.3

33. What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of 0.350 M NaOH? In certain experiment, 28.0 ml of 0.0250 M HNO 3 and 53.0 ml of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H + or OH - ions in excess after reaction goes to completion?

34. Neutralization Reaction of Strong Acid and Strong Base M 1 V 1 f 1 = M 2 V 2 f 2 Where: M & V are molarities & volumes of respective acid & base And f = number of H + or OH - per formula unit of acid & base For example in acids: in HCl, f = 1 while in H 2 SO 4, f = 2 and in H 3 PO 4, f = 3 Similarly in bases: In NaOH, f = 1, Ca(OH) 2, f = 2 and so on

35. Key Titration Terms Titrant - solution of known concentration used in titration Analyte - substance being analyzed Equivalence point - enough titrant added to react exactly with the analyte Endpoint - the indicator changes color so you can tell the equivalence point has been reached. movie

36. In one experiment, 1.3009 g sample of potassium hydrogen phthalate (KHC 8 H 4 O 4 often abbreviated KHP) was titrated against NaOH solution. Exactly 41.20 mL of NaOH solution was required for complete titration of 1.3009 g of KHP. Calculate the concentration of the sodium hydroxide solution. KHC 8 H 4 O 4 K + (aq) + HC 8 H 4 O 4 - (aq) NaOH Na + (aq) + OH - (aq) OH - (aq) + HC 8 H 4 O 4 - (aq) H 2 O (l) + HC 8 H 4 O 4 2- (aq)

37. Practice Example How many moles are in 18.2 g of CO 2 ?

38. Practice Example Consider the reaction N 2 + 3H 2 2NH 3 How many moles of H 2 are needed to completely react 56 g of N 2 ?

39. Practice Example How many grams are in 0.0150 mole of caffeine C 8 H 10 N 4 O 2

40. Practice Example A solution containing Ni 2+ is prepared by dissolving 1.485 g of pure nickel in nitric acid and diluting to 1.00 L. A 10.00 mL aliquot is then diluted to 500.0 mL. What is the molarity of the final solution? (Atomic weight: Ni = 58.70).

41. Practice Example Calculate the number of molecules of vitamin A, C 20 H 30 O in 1.5 mg of this compound.

42. Practice Example What is the mass percent of hydrogen in acetic acid HC 2 H 3 O 2

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

44. Reducing agent: Oxidizes it self Oxidizing agent: Reduces it self

45. Redox Reactions Many practical or everyday examples of redox reactions: –Corrosion of iron (rust formation) –Forest fire –Charcoal grill –Natural gas burning –Batteries –Production of Al metal from Al 2 O 3 (alumina) –Metabolic processes combustion

46. Rules for Assigning Oxidation States 1. Oxidation state of an atom in an element = 0 2. Oxidation state of monatomic element = charge 3. Oxygen =  2 in covalent compounds (except in peroxides where it =  1) 4. H = +1 in covalent compounds 5. Fluorine =  1 in compounds 6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

48. Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn is oxidizedZn Zn 2+ + 2e - Cu 2+ is reducedCu 2+ + 2e - 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 2+ + 2e - Ag + + 1e - AgAg + is reducedAg + is the oxidizing agent

49. NaIO 3 Na = +1 O = -2 3x(-2) + 1 + ? = 0 I = +5 IF7IF7 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

50. Balancing by Half-Reaction Method 1.Write separate reduction and oxidation half reactions. 2.For each half-reaction:  Balance other elements first (except H, O)  Balance O using H 2 O  Balance H using H +  Balance charges using electrons

51. Balancing by Half-Reaction Method (continued) 3.If necessary, multiply by integer to equalize number of electrons on both sides. 4.Must cancel electrons on both sides 5.Add both half-reactions 6.Check that number of elements and charges are balanced on both sides. 7.Cancel same number of H + from both sides 8.This will be your balanced redox reaction in acidic medium.

52. Half-Reaction Method - Balancing in Base 1.Start balancing as in acidic medium. 2.Then add OH  ions equal to H + ions present on both sides of the reaction. 3.Form water by combining H + and OH . 4.Cancel same number of H 2 O from both sides 5.Check number of elements and charges are balanced on both sides of the reaction..

53. Balancing Redox Equations Example: Balance the following redox reaction: Cr 2 O 7 2- + Fe 2+ Cr 3+ + Fe 3+ (acidic soln) 1) Break into half reactions: Cr 2 O 7 2- Cr 3+ Fe 2+ Fe 3+

54. Balancing Redox Equations 2) Balance each half reaction: Cr 2 O 7 2- Cr 3+ Cr 2 O 7 2- 2 Cr 3+ Cr 2 O 7 2- 2 Cr 3+ + 7 H 2 O Cr 2 O 7 2- + 14 H + 2 Cr 3+ + 7 H 2 O 6 e - + Cr 2 O 7 2- + 14 H + 2 Cr 3+ + 7 H 2 O

55. Balancing Redox Equations 2) Balance each half reaction (cont) Fe 2+ Fe 3+ Fe 2+ Fe 3+ + 1 e -

56. Balancing Redox Reactions 3) Multiply by integer so e - lost = e - gained 6 e - + Cr 2 O 7 2- + 14 H + 2 Cr 3+ + 7 H 2 O Fe 2+ Fe 3+ + 1 e - x 6

57. Balancing Redox Reactions 3) Multiply by integer so e - lost = e - gained 6 Fe 2+ 6 Fe 3+ + 6 e - 6 e - + Cr 2 O 7 2- + 14 H + 2 Cr 3+ + 7 H 2 O 4) Add both half reactions Cr 2 O 7 2- + 6 Fe 2+ + 14 H + 2 Cr 3+ + 6 Fe 3+ + 7 H 2 O

58. Balancing Redox Reactions Cr 2 O 7 2- + 6 Fe 2+ + 14 H + 2 Cr 3+ + 6 Fe 3+ + 7 H 2 O 5) Check the equation 2 Cr7 O6 Fe14 H +24 + 24

59. Balancing Redox Reactions Procedure for Basic Solutions: –Divide the equation into 2 incomplete half reactions one for oxidation one for reduction

60. Balancing Redox Reactions –Balance each half-reaction: balance elements except H and O balance O atoms by adding H 2 O balance H atoms by adding H + add 1 OH - to both sides for every H + added combine H + and OH - on same side to make H 2 O cancel the same # of H 2 O from each side balance charge by adding e - to side with greater overall + charge different

61. Balancing Redox Equations –Multiply each half reaction by an integer so that # e - lost = # e - gained –Add the half reactions together. Simply where possible by canceling species appearing on both sides of equation –Check the equation # of atoms total charge on each side

62. Balancing Redox Reactions Example: Balance the following redox reaction. NH 3 + ClO - Cl 2 + N 2 H 4 (basic soln) NH 3 N 2 H 4 ClO - Cl 2 1) Break into half reactions:

63. Balancing Redox Reactions NH 3 N 2 H 4 2) Balance each half reaction: 2 NH 3 N 2 H 4 2 NH 3 N 2 H 4 + 2 H + 2 NH 3 + 2 OH - N 2 H 4 + 2 H 2 O + 2 OH - + 2 OH - 2H2O2H2O 2 NH 3 + 2 OH - N 2 H 4 + 2 H 2 O + 2 e -

64. Balancing Redox Reactions 2 ClO - Cl 2 2) Balance each half reaction: 2 ClO - Cl 2 + 2 H 2 O 2 ClO - + 4 H + Cl 2 + 2 H 2 O + 4 OH - + 4 OH - 2 ClO - + 4 H 2 O Cl 2 + 2 H 2 O + 4 OH - 2 ClO - + 2 H 2 O Cl 2 + 4 OH - 2 e - + 2 ClO - + 2 H 2 O Cl 2 + 4 OH - ClO - Cl 2

65. Balancing Redox Reactions 3) Multiply by integer so # e - lost = # e - gained 2 NH 3 + 2 OH - N 2 H 4 + 2 H 2 O + 2 e - 2 e - + 2 ClO - + 2 H 2 O Cl 2 + 4 OH - 4) Add both half reactions 2 NH 3 + 2 OH - + 2ClO - + 2 H 2 O N 2 H 4 + 2 H 2 O + Cl 2 + 4 OH -

66. Balancing Redox Reactions 5) Cancel out common species 2 NH 3 + 2 OH - + 2 ClO - + 2 H 2 O N 2 H 4 + 2 H 2 O + Cl 2 + 4 OH - 2 2 NH 3 + 2 ClO - N 2 H 4 + Cl 2 + 2 OH - 6) Check final equation: 2 N6 H2 Cl2 O-2

69. Practice Example In the following the oxidizing agent is: 5H 2 O 2 + 2MnO 4 - + 6H +  2Mn 2+ + 8H 2 O + 5O 2 a. MnO 4 - b. H 2 O 2 c. H + d. Mn 2+ e. O 2

70. Practice Example Determine the coefficient of Sn in acidic solution Sn + HNO 3  SnO 2 + NO 2 + H 2 O 1

71. Practice Example The sum of the coefficients when they are whole numbers in basic solution: Bi(OH) 3 + SnO 2 2-  Bi + SnO 3 2- 13

72. http://www.chemistrycoach.com/balancing_redox_in_acid.htm#Bal ancingRedoxEquationsinAcidicorBasicMedium

73. http://www.chemistrycoach.com/tutorials-5.htm#Oxidation-Reduction