1. Chapter 4 Types of Chemical Reactions and Solution Chemistry

2. Topics  Water as a solvent  Electrolytes and nonelectrolytes  Calculations involving molarity of solutions  Precipitation reactions  Acid base reactions  Oxidation reduction reactions

3. The Water Molecule, Polarity δ-δ- δ+δ+ δ+δ+ H H O δ - means a partial negative charge δ+ means a partial positive charge Thus, water has a partial negative end (0xygen) and a partial positive end (Hydrogen) – and it is called “polar” because of the unequal charge distribution 4.1 Water, the common solvent bond angle of water = 105 o

4.  Ions have charges and attract the opposite charges on the water molecules.  The process of breaking the ions of salts apart is called hydration Dissolving ionic salts in water and Hydration and Hydration NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) H 2 O (l) Designates hydration of ions

5. How Ionic solids dissolve in water H H O H H O H H O H H O H H O H H O H H O H H O H H O These ions have been surrounded by water, and are now dissolved! These ions have been pulled away from the main crystal structure by water’s polarity. Hydration

6. Solubility in water and Aqueous Solutions  Water dissolves ionic compounds (NaCl) and polar covalent molecules (ethanol C 2 H 5 OH)  The rule is: “like dissolves like”  Polar dissolves polar.  Nonpolar dissolves nonpolar.  Oil is nonpolar. –Oil and water don’t mix.  Many Salts are ionic- sea water

7. The Solution Process  Called “solvation”.  Water breaks the + and - charged pieces apart and surrounds them.  Solubility in water depends on the relative attractions of ions for each other and attraction of ions for water molecules  In some ionic compounds, the attraction between ions is greater than the attraction exerted by water (slightly soluble slats) –Barium sulfate and calcium carbonate

8.  Solids will dissolve if the attractive force of the water molecules is stronger than the attractive force of the crystal.  If not, the solids are insoluble.  Water doesn’t dissolve nonpolar molecules (like oil) because the water molecules can’t hold onto them.

9. How does ethanol dissolve in water? Ethanol Molecule Contains a Polar O-H Bond Similar to Those in the Water Molecule

10. The polar water molecule interacts strongly with the polar-O-H bond in ethanol

11. homogenous mixture A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount SolutionSolventSolute(s) Soft drink (l) Air (g) Soft Solder (s) H2OH2O N2N2 Pb Sugar, CO 2 O 2, Ar, CH 4 Sn 4.2 The nature of aqueous solutions: strong and weak electrolytes

12. Electrolytes and Nonelectrolytes  Electrolytes- compounds that conduct an electric current in aqueous solution, or in the molten state –all ionic compounds are electrolytes because they dissociate into ions (they are also called “salts”)  barium sulfate- will conduct when molten, but is insoluble in water!

13.  Nonelectrolytes- Do not conduct an electric current –Most are molecular materials, because they do not have ions  Not all electrolytes conduct to the same degree –there are strong electrolytes, and weak electrolytes – Conductivity depends on: degree of dissociation or ionization

14. ______________ Nonelectrolyte Weak electrolyte Strong electrolyte Ethanol and table sugar Acetic acid ammonia Sodium chloride Hydrochloric acid

15. Dissociation of acids and bases: Strong and weak acids and bases  Acids- form H + ions when dissolved in water (According to Arrhenius) (According to Arrhenius)  Strong acids dissociate completely into H + and anions  Strong acids- H 2 SO 4 HNO 3 HCl HBr HI HClO 4  Bases - form OH - ions when dissolved in water  Strong bases- KOH, NaOH

16.  Weak acids- dissociate partially  Acetic acid: HC 2 H 3 O 2 has 1% dissociation in aqueous solutions  The most common weak base is ammonia, NH 3

17. HCl H + + Cl - HNO 3 H + + NO 3 - HC 2 H 3 O 2 (aq) H + + C 2 H 3 O 2 - Strong electrolyte, strong acid Weak electrolyte, weak acid H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Strong electrolyte, strong acid Weak electrolyte, weak acid H 3 PO 4 H + + H 2 PO 4 - H 2 PO 4 - H + + HPO 4 2- HPO 4 2- H + + PO 4 3- Weak electrolyte, weak acid H2OH2O H2OH2O

18.  NaOH(s) Na + (aq) + OH - (aq)  NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) H2OH2O

19. 4.3 The composition of solutions Molarity (M) Molarity: A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute liters solution liters solution

20. Molarity Calculation If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

21. 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH 40.0 g NaOH 40.0 g NaOH 500. mL x 1 L _ = 0.500 L 1000 mL 1000 mL 0.10 mole NaOH = 0.20 mole NaOH 0.500 L 1 L = 0.20 M NaOH = 0.20 M NaOH Calculating Molarity

22. 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L 1 L An acid solution is a 0.10 M HCl. How many moles of HCl are in 1500 mL of this acid solution?

23. How many grams of KCl are present in 2.5 L of 0.50 M KCl? 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1 L 1 mole KCl 1 L 1 mole KCl

24. How many milliliters of an acid solution, which is 0.10 M HCl, contain 0.15 mole HCl? 0.15 mole HCl x 1 L soln x 1000 mL 0.15 mole HCl x 1 L soln x 1000 mL 0.10 mole HCl 1 L 0.10 mole HCl 1 L = 1500 mL HCl = 1500 mL HCl

25. How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 400. mL x 1 L = 0.400 L 1000 mL 1000 mL 0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1 L 1 mole NaOH 0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1 L 1 mole NaOH = 48 g NaOH

26. 4.5

27. 1.0mg NaCl = 1.2X10 -4 L 1L 0.14mol NaCl x 1 mol NaCl 58.5g NaCl x 1 g 1000 mg x A sample of 0.14 M NaCl. What volume of sample contains 1.0 mg NaCl? # mol = M X V (L) 

28. Dilution  Adding more solvent to a known solution.  The moles of solute stay the same.  #moles = M x volume (L)  # moles before dilution (1) = # moles after dilution  M 1 V 1 = M 2 V 2  Stock solution is a solution of known concentration used to make more dilute solutions

29. Preparing a less concentrated solution from a more concentrated solution by dilution Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf =

30. M 1 V 1 = M 2 V 2 M 1 = 4.00 M 2 = 0.200V 2 = 60.0 m L V 1 = ?mL V 1 = M2V2M2V2 M1M1 = = How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? Example 0.200 M X 60.0 mL 4.00 M 3.00 mL

31. Dilution process pipits Wash bottle Funnel Volumetric flask

32. 4.4 Types of Chemical Reactions  Precipitation reactions  Acid-base reactions  Oxidation-reduction reactions

33. 4.5 Precipitation Reactions  When aqueous solutions of ionic compounds are mixed together a solid forms.  A solid that forms from mixed solutions is called precipitate  If the substance is not part of the solution, it is a precipitate

35. Precipitation Reactions ________________________ Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Spectator Na + and NO 3 - are Spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb 2+ + 2I - PbI 2 (s) Anions and cations switch partners Anions and cations switch partners

36. Solubility rules for common ionic compounds in water at 25 0 C Soluble Compounds Exceptions Compounds containing alkali metal ions and NH 4 + NO 3 -, HCO 3 -, ClO 3 - Cl -, Br -, I - Halides of Ag +, Hg 2 2+, Pb 2+ SO 4 2- Sulfates of Ag +, Ca 2+, Sr 2+, Ba 2+, Hg 2+, Pb 2+ Slightly soluble Compounds Exceptions CO 3 2-, PO 4 3-, CrO 4 2-, S 2- Compounds containing alkali metal ions and NH 4 + OH - Compounds containing alkali metal ions and Ba 2+, Ca 2+, Mg 2+ are marginally soluble

37. Solubility Rules Predicting reaction’s product  All nitrates are soluble  Alkali metals ions and NH 4 + ions are soluble  Halides are soluble except Ag +, Pb +2, and Hg 2 +2  Most sulfates are soluble, except Pb +2, Ba +2, Hg +2,and Ca +2

38. Solubility Rules  Most hydroxides are slightly soluble (insoluble) except NaOH and KOH  Sulfides, carbonates, chromates, and phosphates are insoluble  Lower number rules supersede so Na 2 S is soluble

39. 4.6 Describing reactions in solution  Types of equations used to represent chemical reactions: –Formula equation –Complete ionic equation –Net ionic equation

40. Writing Net Ionic Equations 1.Write the balanced formula equation. 2.Write the net ionic equation showing the strong electrolytes 3.Determine precipitate from solubility rules 4.Cancel the spectator ions on both sides of the ionic equation AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

41.  What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide?  Ba(Cl) 2 +2NaOH Ba(OH) 2 (s) + 2NaCl  Ba 2+ + 2Cl - + 2Na + + 2OH - Ba(OH) 2 (s) + 2Na + +2Cl - 4.7 Stoichiometry of precipitation reactions

42. Example  Calculate mass of solid NaCl that must be to 1.50 L 0f 0.100M AgNO 3 solution to precipitate all of the Ag + ions in the form of AgCl Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) AgCl(s) #moles Ag + = #moles AgNO 3 = 1.50L X = 0.15 mol Ag + = 1.50L X = 0.15 mol Ag + 1 L 0.100 mol Ag + #mole Cl - required =#mole NaCl = #moles Ag + (1:1 mole ratio) 0.15 mol NaCl X = 0.15 mol NaCl X 1mol NaCl 58.45 g NaCl = 8.77 g NaCl

43. 4.8 Acid-Base Reactions  For the purpose of this chapter: An acid is a proton donor An acid is a proton donor  a base is a proton acceptor usually (donates OH - ions to the solution) acidbaseacidbase

44. Describing acid-base reactions HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O H + + Cl - + Na + + OH - Na + + Cl - + H 2 O H + (aq) + OH - (aq) H 2 O (l) HC 2 H 3 O 2 (aq)+NaOH (aq) NaC 2 H 3 O 2 (aq) +H 2 O HC 2 H 3 O 2 (aq) +Na + + OH - C 2 H 3 O 2 - +Na + +H 2 O HC 2 H 3 O 2 (aq) + OH - C 2 H 3 O 2 - +H 2 O Weak acid Strong acid

45. When acid and bases with equal amounts of hydrogen ion H + and hydroxide ions OH - are mixed, the resulting solution is neutral. When acid and bases with equal amounts of hydrogen ion H + and hydroxide ions OH - are mixed, the resulting solution is neutral. This reaction is called This reaction is called Neutralization reaction HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) H 2 SO 4 (aq) + 2KOH (aq)  K 2 SO 4 (aq) + 2H 2 O(l) Notice salt (NaCl) and water are the products Salt =Salt = ionic compound whose cation comes from a base and anion from an acid. Neutralization between acid and metal hydroxide produces water and a salt. Neutralization Reactions and Salts

46. Example How many mL of 2.00 M H 2 SO 4 are required to neutralize 50.0 mL of 1.00 M KOH? H 2 SO 4 + 2KOH K 2 SO 4 + 2H 2 O H 2 SO 4 + 2KOH K 2 SO 4 + 2H 2 O 0.0500 L x 1.00 mole KOH x 1 mole H 2 SO 4 x 0.0500 L x 1.00 mole KOH x 1 mole H 2 SO 4 x 1 L 2 mole KOH 1 L 2 mole KOH 1 L x 1000 mL = 12.5 mL 1 L x 1000 mL = 12.5 mL 2 mole H 2 SO 4 1 L 2 mole H 2 SO 4 1 L

47. Another solution  H 2 SO 4 + 2KOH K 2 SO 4 + 2H 2 O #mole KOH = (M) KOH X (V) KOH #mole KOH = 0.05 L X1M = 0.05 mol KOH #mole H 2 SO 4 = # mole KOH X 2mol KOH 1 mol H 2 SO 4 = 0.025 mol H 2 SO 4 #mole H 2 SO 4 = M H 2 SO 4 X vol (L) H 2 SO 4 0.025 mol H 2 SO 4 = 2.00 M X V V = 0.0125 L = 12.5 mL

48. Acid - Base Titrations  Often called a neutralization titration Because the acid neutralizes the base  Often called volumetric analysis since titration is made to determine concentrations  It involves delivery of a measured volume of solution of known concentration (titrant)  Titrant is added to the unknown (analyte)  until the equivalence (stoichiometric) point is reached where enough titrant has been added to neutralize it.

49. Titration A solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point– the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) Endpoint –the point at which the color of indicator changes

50. Titration  Equivalence point is marked by using an indicator  Where the indicator changes color is the endpoint  End point does not match always the equivalence point.  A successful titration requires: –A rapid known exact reaction –Endpoint is very close to the equivalence point –Accurate Measurement of volume of titrant Accurate determination of a solution concnetration is called

51. Titration  A 50.00 mL sample of aqueous Ca(OH) 2 requires 34.66 mL of 0.0980 M Nitric acid for neutralization. What is [Ca(OH) 2 ]?  # of H + x M A x V A = # of OH - x M B x V B

52. Example  In the titration of KHP (molar mass = 204.22g/mol), 34.67 mL of NaOH solution was required to react with 0.1082 g KHP. Calculate the molarity of NaOH solution.

53. Solution NaOH(aq) + KHP(aq) → NaKP(aq) + H 2 O(l) There are 5.298X10 -4 mol of sodium hydroxide in 34.67 mL of solution. #mole NaOH = M NaOH X V NaOH = 1.528 M NaOH 0.1082 g KHP × # mol NaOH = = 5.298X10 -4 mol NaOH = M NaOH =

54. Each sodium atom loses one electron: Each chlorine atom gains one electron: 4.9 Oxidation - Reduction (Redox) Reactions

55. Definitions Loss of Electrons = Oxidation Sodium is oxidized Gain of Electrons = Reduction Chlorine is reduced Increase of the oxidation state Decrease of the oxidation state

56. Definitions - The substance that loses electrons is (Electron donor) called reducing agent (Electron donor) - The substance that gains electrons is called oxidizing agent (Electron acceptor) Mg (s) + S (s) → MgS (s) Mg is oxidized – loses e - S is reduced – gains e - Mg is the reducing agent S is the oxidizing agent 00 +2 -2

57. Not All Reactions are Redox Reactions - Reactions in which there has been no change in oxidation number are not redox reactions. Examples:

58. Assigning Oxidation States An “oxidation state” (oxidation number) is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction. Generally, a bonded atom’s oxidation number is the charge it would have if the electrons in the bond were assigned to the atom of the more electronegative element

59. Rules for Assigning Oxidation States State 1)The oxidation State of any uncombined element is zero. State 2)The oxidation State of a monatomic ion equals its charge.

60. Rules for Assigning Oxidation States state 3)The oxidation state of oxygen in compounds is -2, except in peroxides, such as H 2 O 2 where it is -1. state 4)The oxidation state of hydrogen in compounds is +1, except in metal hydrides, like NaH, where it is -1.

61. Rules for Assigning Oxidation states 5)The sum of the oxidation states of the atoms in the compound must equal 0. 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H

62. Rules for Assigning Oxidation States State 6)The sum of the oxidation States in the formula of a polyatomic ion is equal to its ionic charge. X + 3(-2) = -1 N O  X = +5  X = +6 X + 4(-2) = -2 S O

63. NaIO 3 Na =? O = ? 3x-2 + 1 + ? = 0 I = IF 7 F = ? 7x-1 + ? = 0 I = K 2 Cr 2 O 7 O = ?K = ? 7x-2 + 2x+1 + 2x? = 0 Cr =?? What is the oxidation states of all atoms in the following ?

64. 4.10 Balancing oxidation-reduction Equations Two systematic methods for balancing redox equations are available, and are based on the fact that the total electrons gained in reduction equals the total lost in oxidation. The two methods: 1)Use oxidation state changes 2)Use half-reactions (the method to be used her)

65. 1: write separate half-reaction equations for oxidation and reduction 2. balance the atoms in the half reactions 3. add enough electrons to one side of each half-reaction to balance the charges 4. multiply each half-reaction by a number to make the electrons equal in both 5. add the balanced half-reactions to show an overall equation and cancel identical species 6. Check that elements and charges are balanced Balancing redox equations using half-reactions

66. Example 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

67. Balancing redox equations in an acidic solution For reactions in acidic solution an 8 step procedure. For reactions in acidic solution an 8 step procedure.  Write separate half reactions  For each half reaction balance all reactants except H and O  Balance O using H 2 O

68.  Balance H using H +  Balance charge using e -  Multiply equations to make electrons equal  Add equations and cancel identical species  Check that charges and elements are balanced.

69. Example: Balance the following equation I - + OCI -  I 3 - + Cl - (acidic solution) Oxidation : I -  I 3 - 3I -  I 3 - 3I -  I 3 - + 2e - Reduction : OC1 -  C1 - OC1 -  C1 - + H 2 O

70. 2H + + OC1 -  C1 - + H 2 O 1+ -1 2H + + OC1 - + 2e -  C1 - + H 2 O (2) 3I -  I 3 - + 2e - (1) (1) + (2) 2H + + OC1 - + 3I -  C1 - + I 3 - + H 2 O

71. Do everything you would with acid, but add one more step. Add enough OH - to both sides to neutralize the H + For each H atom that is lacking, add one molecule of H 2 O to the side that requires it. At the same time add one unit of OH - in the opposite side of the half-reaction Balancing equations in basic solution

72. Example First half reaction: MnO 4 -  MnO 2 (a)MnO 4 -  MnO 2 + 2H 2 O (b)4 H 2 O + MnO 4 -  MnO 2 + 2H 2 O + 4 OH - (c)4 H 2 O + MnO 4 - + 3 e -  MnO 2 + 2H 2 O + 4 OH - (d)2 H 2 O + MnO 4 - + 3 e-  MnO 2 + 4 OH - (1) Balance the following equation in basic solution MnO 4 - + C 2 O 4 2-  MnO 2 + CO 3 2-

73. Second half reaction : C 2 O 4 2-  CO 3 2- (a)C 2 O 4 2-  2 CO 3 2- (b)2 H 2 O + C 2 O 4 2-  2 CO 3 2- (c)4 OH - + 2 H 2 O + C 2 O 4 2-  2 CO 3 2- + 4 H 2 O (d)4 OH - + 2 H 2 O + C 2 O 4 2-  2 CO 3 2- + 4 H 2 O + 2 e - (e)4 OH - + C 2 O 4 2-  2 CO 3 2- + 2 H 2 O + 2 e - (2) 2 H 2 O + MnO 4 - + 3 e -  MnO 2 + 4 OH - (1) We eliminate electrons by multiplying (1) by 2 and (2) by 3.

74. ( 2) X 3: 12 OH - + 3 C 2 O 4 2-  6 CO 3 2- + 6 H 2 O + 6 e - (3) 4 H 2 O + 2 MnO 4 - + 6 e -  2 MnO 2 + 8 OH - (4) (1) X 2: (3) + (4) to eliminate the electrons and common species. 4 OH - + 3 C 2 O 4 2- + 2 MnO 4 -  6 CO 3 2- + 2 H 2 O + 2 MnO 2