1. LECTURE Thirteen CHM 151 ©slg Topics: 1. Solution Stoichiometry: Molarity 2. Titration Problems

2. GROUP WORK, BALANCE C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O (sugar) C 5 H 11 OH + O 2 --> CO 2 + H 2 O (alcohol) a) Balance C b) Balance H c) Balance O d) Check

3. C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O C 6 H 12 O 6 + O 2 --> 6 CO 2 + H 2 O C 6 H 12 O 6 + O 2 --> 6 CO 2 + 6 H 2 O C 6 H 12 O 6 + 6 O 2 --> 6 CO 2 + 6 H 2 O 6C 12H 6 O + 12 O ---> 6 C 12 O + 12 H 6 O Solution, Sugar 12 O + 6 O = 18 O

4. C 5 H 11 OH + O 2 --> CO 2 + H 2 O C 5 H 11 OH + O 2 --> 5 CO 2 + H 2 O C 5 H 11 OH + O 2 --> 5 CO 2 + 6 H 2 O C 5 H 11 OH + 7.5 or 15/2 O 2 --> 5 CO 2 + 6 H 2 O 1 O 15 O 10 O 6 O #1 #2 #3

5. C 5 H 11 OH + 7.5 or 15/2 O 2 --> 5 CO 2 + 6 H 2 O 10 O 6 O X 2 2 C 5 H 11 OH + 15 O 2 --> 10 CO 2 + 12 H 2 O [10C 24H 2 O] + 30 O --> [10C 20 O] + [24H 12 O]

6. Group work 2: H 2 SO 4 (aq) + Al(OH) 3 (s) -----> ? H 3 PO 4 (aq) + KOH (aq) ----> HCH 3 CO 2 (aq) + Mg(OH) 2 (s) ----->? 1. Form H 2 O and use leftover ions to form salt 2. Write balanced equation 3. Do total ionic (watch out for weak acids!) 4. Do net ionic equation

7. Net Ionic Equations: What to “Break Up”: A) all soluble salts “(aq)” B) All soluble bases “(aq)” C) All strong acids: HCl HBr HI; HNO 3 H 2 SO 4 HClO 4 What not to “break up”: A) all insoluble salts or bases “(s)” B) H 2 O (and all molecules) C) All weak acids: H 3 PO 4, HCH 3 CO 2, H 2 CO 3.....

8. H 2 SO 4 (aq) + Al(OH) 3 (s) -----> ? ----> H 2 O + salt Step One: Formula of salt: Al 3+ + SO 4 2- -----> Al 2 (SO 4 ) 3 acid base Always!

9. Step Two: Complete Equation: H 2 SO 4 (aq) + Al(OH) 3 (s) -----> H 2 O (l) + Al 2 (SO 4 ) 3 (aq) Balance: 3 H 2 SO 4 (aq) + 2 Al(OH) 3 (s) -----> 6 H 2 O + Al 2 (SO 4 ) 3 (aq)

10. Step 3: Total Ionic Equation: 6 H + (aq) + 3 SO 4 2- (aq) + 2 Al(OH) 3 (s) -----> 6 H 2 O(l) + 2 Al 3+ (aq) + 3 SO 4 2- (aq) Strong acid, Break up! Insoluble base, Don’t break up Molecule, Don’t... Soluble, Do...

11. Step 3: Total Ionic Equation: 6 H + (aq) + 3 SO 4 2- (aq) + 2 Al(OH) 3 (s) -----> 6 H 2 O(l) + 2 Al 3+ (aq) + 3 SO 4 2- (aq) Step Four: Net Ionic Equation 6 H + (aq) + 2 Al(OH) 3 (s) -----> 6 H 2 O (l) + 2 Al 3+ (aq)

12. H 3 PO 4 (aq) + KOH (aq) ----> ? -----> H 2 O + salt Step One: salt formula K + + PO 4 3- -----> K 3 PO 4 (aq) Step Two: Write Equation; Balance H 3 PO 4 (aq) + KOH (aq) ----> H 2 O (l) + K 3 PO 4 (aq) H 3 PO 4 (aq) + 3 KOH (aq) ---->3 H 2 O (l) + K 3 PO 4 (aq)

13. Step Three: Total Ionic: H 3 PO 4 (aq) + 3 K + (aq) + 3 OH - (aq) -----> 3 H 2 O (l) + 3 K + (aq) + PO 4 3- (aq) Step Four: Net Ionic: H 3 PO 4 (aq) + 3 OH - (aq) -----> 3 H 2 O (l) + PO 4 3- (aq) Weak acid, no! Yes!No!Yes!

14. HCH 3 CO 2 (aq) + Mg(OH) 2 (s) ----->? -----> H 2 O + salt Step One: Salt Formula: Mg 2+ + CH 3 CO 2 - -----> Mg(CH 3 CO 2 ) 2 (aq) Step Two: Write Equation, Balance: HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> H 2 O + Mg(CH 3 CO 2 ) 2 (aq) 2 HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> 2 H 2 O + Mg(CH 3 CO 2 ) 2 (aq)

15. Step Three: Total Ionic Equation: 2 HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> 2 H 2 O(l) + Mg 2+ (aq) +2 CH 3 CO 2 - (aq) Step Four: Net Ionic Equation SAME! NO, weak acid! NO! Yes !

16. Concentrations of Compounds in Aqueous Solutions (Chapter 5, Section 5.8, p. 213) Usually the reactions we run are done in aqueous solution, and therefore we need to add to our study of stoichiometry the concentration of compounds in aqueous solution. We will utilize a very useful quantity known as “molarity,” the number of moles of solute per liter of solution.

17. Concentration (molarity) = # moles solute L solution If we placed 1.00 mol NaCl (58.4 g) in a 1 L volumetric flask, dissolved it in water, swirled to dissolve and diluted the solution to the 1.00 liter mark you would have a solution that contains 1.00 mol NaCl per liter of solution. This may be represented several ways: Concentration (molarity) = 1.00 mol NaCl / L soln = 1.00 M NaCl = [1.00] NaCl Chemists call this a “1.00 molar solution”

18. Calculating Molar Amounts TYPICAL PROBLEMS: What is the molarity of a solution made by dissolving 25.0 g of BaCl 2 in sufficient water to make up a solution of 500.0 mL? How many g of BaCl 2 would be contained in 20.0 mL of this solution? How many mL of this solution would deliver 1.25 g of BaCl 2 ? How many mol of Cl - ions are contained in 10.00 ml of this solution?

19. What is the molarity of a solution made by dissolving 25.0 g of BaCl 2 in sufficient water to make up a solution of 500.0 mL? 25.0 g BaCl 2 = ? mol/ L BaCl 2 (= ? M BaCl 2 ) 500.0 mL soln 1Ba = 1 X 137.33g = 137.33 2Cl = 2 X 35.45g = 70.90 208.23 g/mol Molar mass, BaCl 2 :

21. How many g of BaCl 2 would be contained in 20.0 mL of this solution? (.240 M BaCl 2 ) Question: 20.0 mL soln = ? g BaCl 2 Relationships: 1000 mL = 1 L 1 L soln =.240 mol BaCl 2 1 mol BaCl 2 = 208.23 g BaCl 2 20.0 mL soln = ? g BaCl 2 mL  L  mol  g

22. Molarity Molar Mass

23. How many mL of this solution would deliver 1.25 g of BaCl 2 ? 1.25 g BaCl 2 = ? mL soln.240 mol BaCl 2 = 1 L soln 208.23 g BaCl 2 = 1 mol BaCl 2 Molar Mass Molarity

24. How many mol of Cl - ions are contained in 10.0 ml of this solution? 10.0 mL soln = ? Mol Cl -.240 mol BaCl 2 = 1000 mL soln 1 mol BaCl 2 = 2 mol Cl - Note: BaCl 2(aq) ---> Ba 2+ (aq) + 2 Cl - (aq)

25. GROUP WORK: If 35.00 g CuSO 4 is dissolved in sufficient water to makeup 750. mL of an aqueous solution, a) what is the molarity of the solution? b) how many mL of the solution will deliver 10.0 g of CuSO 4 ? c) How many moles of sulfate ion (SO 4 2- ) will be delivered in 10.0 mL of the solution? 1 Cu = 1 X 63.55 = 63.55 1 S = 1 X 32.07 = 32.07 4 O = 4 X 16.00 = 64.00 159.62 g/mol

27. STOICHIOMETRY OF REACTIONS IN AQUEOUS SOLUTION: Chapter 5, Section 5.9 Let’s use our favorite reaction to add another dimension to calculating from balanced equations: How many ml of 3.00 M HCl solution would be required to react with 13.67 g of Fe 2 O 3 according to the following balanced equation: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 3.00 M 13.67 g ? mL

28. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 3.00 M 13.67 g ? mL Pathway: g Fe 2 O 3 ---> mol Fe 2 O 3 ---> mol HCl --- > mL soln 13.67 g Fe 2 O 3 = ? mL soln 1000 mL soln = 3.00 mol HCl 6 mol HCl = 1 mol Fe 2 O 3 159.70 g Fe 2 O 3 = 1 mol Fe 2 O 3

29. Molar Mass Balanced Equation Molarity

30. Group Work How many g of Fe 2 O 3 would react with 25.0 mL of 3.00 M HCl? Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 3.00 M ? g 25.0 mL mL  L  mol HCl  mol Fe 2 O 3  g Fe 2 O 3

31. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 3.00 M ? g 25.0 mL 25.0 mL soln 3.00 mol HCl 1 mol Fe 2 O 3 159.70 g Fe 2 O 3 1000 mL soln 6 mol HCl 1 mol Fe 2 O 3 = 1.996 g Fe 2 O 3 = 2.00 g Fe 2 O 3 Molarity Equation Molar mass

32. Constant mass, Bromine, variable masses Fe Ch 4, #60

33. (a) What mass of Br 2 is used when the reaction consumes 2.0 g Fe? Product  10.8 g; 10.8 - 2.0 = 8.8 g Br (b) What is the mole ratio of Fe to Br in this reaction? 2.0 g Fe 1 mol Fe =.036 mol Fe.036 = 1 55.85 g Fe.036 8.8 g Br 1 mol Br =.11 mol Br.11 = 3 79.9 g Br.036

34. (c) What is the empirical formula of the product? FeBr 3 (d) Balanced chemical equation: Fe + 3 Br 2 -----> 2 FeBr 3 (e) Name? Iron (III) Bromide

35. Which best describes graph: 1. When 1.00 g Fe is added, Fe is LR Yes 2. When 3.50 g of Fe is added, there is an excess of Br 2 3. When 2.50 g of Fe is added to the Br 2, both reactants are used up completely 4. When 2.00 g of Fe is added to the Br 2, 10.0 g of product is formed. The percent yield must be 20%

36. TITRATION CALCULATIONS (Titrations and makeup of standard solutions seen on CD ROM) Titration: Procedure in which measured increments of one reactant are added to a known amount of a second reactant until some indicator signals that the reaction is complete. This point is called “the equivalence point.” Indicators include many acid/base dyes, potentiometers, color change in one reagent...

37. Titrations are run with buret and Erlenmeyer flasks as seen on the CD ROM or demo just observed... Two types of problems are encountered: Standardizing an acid or a base solution (determining the exact molar concentration of the acid or base solution) Determining amount of acidic or basic material in a sample To be continued.....Happy Spring Break!