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

2. 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.....

3. 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)

4. 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!

5. 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)

6. 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 !

7. 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.

8. 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”

9. 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?

10. 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 :

11. Mass of solute Volume of soltn Molar mass, solute Conversion to L

12. 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

13. Molarity Molar Mass conversion

14. 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

15. 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)

16. 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

18. 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

19. 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 13.67 g Fe 2 O 3 = ? mL soltn

20. Molar Mass Balanced Equation Molarity

21. 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  mol HCl  mol Fe 2 O 3  g Fe 2 O 3 25.0 mL soltn = ? g Fe 2 O 3

22. 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

23. Limiting Reagent, Solutions Suppose you mixed 20.00 mL of.250 M Pb(NO 3 ) 2 solution with 30.00 mL of.150 KI solution. How many g of PbI 2 precipitate might you theoretically obtain? Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M 461.0 g/mol 20.00 mL 30.00 mL ? g 1Pb = 1 X 207.2 = 207.2 2 I = 2 X 126.9 = 253.8 461.0 g/mol

24. Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M 461.0 g/mol 20.00 mL 30.00 mL ? g Pathway: mL  mol “A”  mol “C”  g “C” mL  mol “B”  mol “C”  g “C” “A”“B”“C” Low number wins!

25. Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M 461.0 g/mol 20.00 mL 30.00 mL ? g 20.00 ml soltn A.250 mol A 1 mol C 461.0 g C = 2.3050 g C 1000 mL 1 mol A 1 mol C = 2.30 g C 30.00 mL soltn B.150 mol B 1 mol C 461.0 g C = 1.037 g C 1000 mL 2 mol B 1 mol C = 1.04 g C

26. How many mL of solution A,.250 M Pb(NO 3 ) 2, would be required to react with 30.00 mL of solution B,.150 M KI? Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M ? mL 30.00 mL Excess reagent: Pathway: mL “B” soltn  mol “B”  mol “A”  mL “A” soltn 30.00 mL B soltn = ? mL A soltn

27. Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M ? mL 30.00 mL 30.00 mL soltn.150 mol B 1 mol A 1000 mL soltn 1000 mL soltn 2 mol B.250 mol A = 9.00 mL soltn A Molarity equation Molarity

28. Proof: 9.00 mL soltn A.250 Mol A 2 mol B =.00450 mol B 1000 mL soltn 1 mol A Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M 9.00 mL 30.00 mL 30.00 mL soltn B.150 mol B =.00450 mol B 1000 mL soltn

29. Group Work: How many mL of.150 M KI solution would be required to react with 20.00 mL of.250 M Pb(NO 3 ) 2 ? Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M 20.00 mL ? mL

30. Answer: Pb(NO 3 ) 2 (aq) + 2 KI (aq) ---> PbI 2 (s) + 2 KNO 3 (aq).250 M.150 M 20.00 mL ? mL 20.00 mL soltn A.250 mol A 2 mol B 1000 mL soltn B 1000 mL soltn A 1 mol A.150 mol B = 66.67 mL soltn B

31. Summary 1. Molarity, M: useful description of a solution; gives number of moles of solute in 1 L of solution. Useful to convert mL of solution to moles of solute in equation situations. 2. Molarity of solution can also yield moles per liter of the ions produced when the solute ionizes in water, using the formula of the solute as conversion factor 3. Adds another way to calculate mass or moles from volume (compare to density, #grams of substance or solution in 1 mL or 1 cm 3 of solution or substance).