1. Types of Chemical Reactions & Solution Stoichiometry Chapter 4

2. Aqueous Solutions Water is the dissolving medium, or solvent.

3. Some Properties of Water Water is able to dissolve so many substances because: -Water is “bent” or V-shaped. -The O-H bonds are covalent. -Water is a polar molecule. -Hydration occurs when salts dissolve in water.

4. Water is a polar molecule because it is a bent molecule. The hydrogen end is  + while the oxygen end is  -, Delta (  ) is a partial charge--less than 1.

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

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

7. Polar water molecules interact with the positive and negative ions of a salt, assisting in the dissolving process. This process is called hydration.

9. Solubility The general rule for solubility is: “Like dissolves like.” Polar water molecules can dissolve other polar molecules such as alcohol and, also, ionic substances such as NaCl. Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.

10. Miscibility Miscible -- two substances that will mix together in any proportion to make a solution. Alcohol and water are miscible because they are both polar and form hydrogen bonds. Immiscible -- two substances that will not dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and vinegar is polar.

11. Solubility How does the rule “Like dissolves like.” apply to cleaning paint brushes used for latex paint as opposed to those used with oil-based paint?

12. Simple Rules for Solubility 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, 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.

13. Electrolytes & Nonelectrolytes An electrolyte is a material that dissolves in water to give a solution that conducts an electric current. A nonelectrolyte is a substance which, when dissolved in water, gives a nonconducting solution.

14. Electrolytes Strong - conduct current efficiently and are soluble salts, strong acids, and strong bases. NaCl, KNO 3, HNO 3, NaOH NaCl, KNO 3, HNO 3, NaOH Weak - conduct only a small current and are weak acids and weak bases. HC 2 H 3 O 2, aq. NH 3, tap H 2 O Non - no current flows and are molecular substances pure H 2 O, sugar solution, glycerol

15. Electrical conductivity of aqueous solutions. a) strong electrolyte b) weak electrolyte c) nonelectrolyte in solution. Svante Arrhenius first identified these electrical properties.

16. When BaCl 2 dissolves, the Ba 2+ and Cl - ions are randomly dispersed in the water. BaCl 2 is a strong electrolyte.

17. Acids Strong acids -dissociate completely (~100 %) to produce H + in solution HCl, H 2 SO 4, HNO 3, HBr, HI, & HClO 4 Weak acids - dissociate to a slight extent (~ 1 %) to give H + in solution HC 2 H 3 O 2, HCOOH, HNO 2, & H 2 SO 3

18. HCl is completely ionized and is a strong electrolyte.

19. Bases Strong bases - react completely with water to give OH  ions. sodium hydroxide NaOH (s) ---> Na + (aq) + OH - (aq) Weak bases - react only slightly with water to give OH  ions. ammonia NH 3(aq) + HOH (l) NH 4 + (aq) + OH - (aq)

20. An aqueous solution of sodium hydroxide which is a strong bases dissociating almost 100 %.

21. Acetic acid(CH 3 COOH) exists in water mostly as undissociated molecules. Only a small percent of the molecules are ionized.

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

23. Molarity Calculations Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. (11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH) = 0.192M NaOH

24. Molarity Calculations Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution. (1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl) (1000mL/1L) = 1.60M HCl

25. Molarity Calculations How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO 3 ) 2 solution? (25.00mL)(1L/1000mL)(0.75mol Co(NO 3 ) 2 /1L) (2 mol NO 3 - /1 mol Co(NO 3 ) 2 ) = 3.8 x 10 -2 mol NO 3 -

26. Molarity Calculations Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0 mg of NaCl? (1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) = 1.2 x 10 -4 L blood serum

27. Standard Solution A standard solution is a solution whose concentration is accurately known. Standard solutions are made using a volumetric flask as follows: mass the solute accurately and add it to the volumetric flaskmass the solute accurately and add it to the volumetric flask add a small quantity of distilled HOHadd a small quantity of distilled HOH dissolve the solute by gently swirling the flaskdissolve the solute by gently swirling the flask add more distilled HOH until the level of the solution reaches the mark on the neckadd more distilled HOH until the level of the solution reaches the mark on the neck invert the capped volumetric 25X to thoroughly mix the solution.invert the capped volumetric 25X to thoroughly mix the solution.

28. Steps involved in making a standard solution.

29. 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)

30. Dilution of Stock Solutions When diluting stock solutions, the moles of solute after dilution must equal the moles of solute before dilution. Stock solutions are diluted using either a measuring or a delivery pipet and a volumetric flask.

31. Steps to dilute a stock solution.

32. Dilution Calculations What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H 2 SO 4 solution? (0.10mol H 2 SO 4 /1L)(1.5L)(1L/16mol)(1000mL/1L) = 9.4 mL conc H 2 SO 4

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

34. Describing Reactions in Solution 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)

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

36. Simple Rules for Solubility 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, 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.

37. Solubility Using the solubility rules, predict what will happen when the following pairs of solutions are mixed. a) KNO 3(aq) & BaCl 2(aq) b) Na 2 SO 4(aq) & Pb(NO 3 ) 2(aq) c) KOH (aq) & Fe(NO 3 ) 3(aq) No precipitate forms PbSO 4(s) forms Fe(OH) 3(s) forms

39. Precipitation Calculations When aqueous solutions of Na 2 SO 4 & 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 & 2.00 L of 0.0250 M Na 2 SO 4 are mixed. 1. Species present are Na +, SO 4 2-, Pb 2+, & NO 3 -. 2. Pb 2+ (aq) + SO 4 2- (aq) ----> PbSO 4(s)

40. Precipitation Calculations Continued 3. (1.25L)(0.500mol Pb 2+ /L) = 0.0625 mol Pb 2+ (2.00L)(0.0250mol SO 4 2- /L) = 0.0500 mol SO 4 2- 4.(0.0625mol Pb 2+ )(1mol SO 4 2- /1mol Pb 2+ ) = 0.0625 mol SO 4 2- SO 4 2- is the limiting reactant. 5.(0.0500mol SO 4 2- )(1mol PbSO 4 /1mol SO 4 2- ) (303.3g/1mol) = 15.2 g PbSO 4

41. 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.

42. Acid-Base Calculations What volume of a 0.100M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? 1. H +, Cl -, Na +, & OH - 2. H + (aq) + OH - (aq) ----> HOH (l) (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solution

43. 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.

44. Titration Calculations If 41.20 mL of NaOH is required to neutralize 1.3009 g of KHP (KHC 8 H 4 O 6 ), what is the concentration of the NaOH solution? Assume there is one acidic hydrogen in KHP. HC 8 H 4 O 6 - (aq) + OH - (aq) ----> HOH (l) + C 8 H 4 O 6 2- (aq) (1.3009g KHP)(1mol/204.22g)(1mol NaOH/1mol KHP)(1/41.20mL)(1000mL/1L) = 0.1546 M NaOH

45. Redox Reactions Redox reactions are reactions in which electrons are transferred. Decomposition and synthesis reactions may be redox. Single replacement reactions are always redox. Double replacement reactions are never redox. Combustion reactions are always redox.

46. OIL RIG Oxidation Is Loss. Reduction Is Gain.

47. Redox Oxidizing agent is the electron acceptor-- usually a nonmetal. Reducing agent is the electron donor--usually a metal. CH 4(g) + 2O 2(g) ----> CO 2(g) + 2HOH (g) Carbon is oxidized. Oxygen is reduced. CH 4 is the reducing agent. O 2 is the oxidizing agent.

48. 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

49. Determining Oxidation States SF 6 [NO 3 ] - +6 -6 = 0 +5 -6 = -1 (-1 for each F)(-2 for each O)

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

51. Balancing by Half-Reaction Method (continued) 3.If necessary, multiply by integer to equalize electron count. 4.Add half-reactions. 5.Check that both elements and charges are balanced.

52. Balancing By Half-Reaction Acidic Solution H + (aq) + Cr 2 O 7 2- (aq) + C 2 H 5 OH (l) ---> Cr 3+ (aq) + CO 2(g) + HOH (l) Red Cr 2 O 7 2- (aq) ---> Cr 3+ (aq) OxC 2 H 5 OH (l) ---> CO 2(g) Red 2(6e - + 14H + (aq) + Cr 2 O 7 2- (aq) ---> 2Cr 3+ (aq) + 7HOH (l) ) Ox C 2 H 5 OH (l) + 3HOH (l) ---> 2CO 2(g) + 12H + (aq) + 12e - 16H + (aq) + 2Cr 2 O 7 2- (aq) + C 2 H 5 OH (l) ---> 4Cr 3+ (aq) + 11HOH (l) + 2CO 2(g) 12+ = 12+

53. Half-Reaction Method - Balancing in Base 1.Balance as in acid. 2.Add OH  that equals H + ions (both sides!) 3.Form water by combining H +, OH . 4.Check elements and charges for balance.

54. Balancing By Half-Reaction Basic Solution Ag (s) + CN - (aq) + O 2(g) ---> Ag(CN) 2 - (aq) (Basic) Ox CN - (aq) + Ag (s) ---> Ag(CN) 2 - (aq) Red O 2(g) ---> Ox 4(2CN - (aq) + Ag (s) ---> Ag(CN) 2 - (aq) + e - ) Red O 2(g) + 4H + (aq) + 4e - ---> 2HOH (l) 8CN - (aq) + 4Ag (s) + O 2(g) + 4H + (aq) ---> 4Ag(CN) 2 - (aq) + 2HOH (l)

55. Balancing By Half-Reaction Basic Solution(Continued) 8CN - (aq) + 4Ag (s) + O 2(g) + 4H + (aq) + 4OH - (aq) ---> 4Ag(CN) 2 - (aq) + 2HOH (l) + 4OH - (aq) 8CN - (aq) + 4Ag (s) + O 2(g) + 4HOH (l) ---> 4Ag(CN) 2 - (aq) + 2HOH (l) + 4OH - (aq) 8CN - (aq) + 4Ag (s) + O 2(g) + 2HOH (l) ---> 4Ag(CN) 2 - (aq) + 4OH - (aq) 8 - = 8 -