Tuesday, Feb. 25 th : “A” Day Wednesday, Feb. 26 th : “B” Day (1:05 out) Agenda

3. Tuesday, Feb. 25th: “A” DayWednesday, Feb. 26th: “B” Day (1:05 out)Agenda • Homework Questions/Collect • Finish Section 13.2: “Concentration and Molarity” • Solution dilution equation, using molarity in stoichiometric calculations  • Homework: • Practice Pg. 467: #1-3 • Section 13.2 review, pg. 467: #1-14 • Concept Review: “Concentration and Molarity”

4. Homework • Practice pg. 461: #2, 3, 5, 6 • Practice pg. 465: #1-7

5. Solution Dilution • Often, solutions such as acids are sold in very concentrated form, such as 12 M HCl. • However, in the lab, we rarely use such concentrated acids. • How do scientists dilute these very concentrated solutions to get the concentration or molarity that is needed for their experiment?

6. Solution Dilution Equation OR M x L = mol Solution Dilution M1V1=M2V2 M = mol L

7. Solution Dilution Demo The concentration of the CuSO4 solution I made last time was supposed to be 0.05 M, NOT 0.50 M. Do I have to throw away the solution I made last time, or can I somehow fix it? Use the solution dilution equation: M1V1 = M2V2 M1: the original concentration I made (0.50 M) V1: what volume of that solution will I need (?) M2: the new concentration I’m trying to make (0.05 M) V2: the volume of the new concentration I’m making: 250 mL (0.50 M) V1 = (0.05 M) (250 mL) V1 = 25 mL

8. Dilution Example #1 • How many liters of 0.155 M Ni(NO3)2 can be made from 75.0 mL of 12.0 M Ni(NO3)2 ? Use the solution dilution equation: M1V1 = M2V2 M1: 12.0 M V1: 75.0 mL M2: 0.155 M V2: ? (12.0 M) (75.0 mL) = (0.155 M) (V2) V2 = 5,806 mL → 5.81 L

9. Dilution Example #2 • What volume of 19 M NaOH must be used to prepare 1.0 L of a 0.15 M NaOH solution? Use the solution dilution equation: M1V1 = M2V2 M1: 19 M V1: ? M2: 0.15 M V2: 1.0 L (19 M) V1 = (0.15 M) (1.0 L) V1 = .00789 L → 7.9 mL

10. Using Molarity in Stoichiometric Calculations  • There are many instances in which solutions of known molarity are used in chemical reactions in the laboratory. • Instead of starting with a known mass of reactant or with a desired mass of product, the process involves a solution of known molarity. • The substances are measured out by volume, instead of being weighed on a balance.

11. Sample Problem C, Pg. 466 What volume (in mL) of a 0.500 M solution of copper (II) sulfate, CuSO4, is needed to react with an excess of aluminum to provide 11.0 g of copper? 3 CuSO4(aq) + 2 Al(s) 3 Cu(s) + Al2(SO4)3 (aq) First, use molar mass to change gram Cu → moles Cu: 11.0 g Cu X 1 mole Cu = 0.173 mole Cu 63.5 g Cu Next, use mole ratio to change mole Cu → mole CuSO4: 0.173 mole Cu X 3 mol CuSO4 = 0.173 mole CuSO4 3 mole Cu Last, use molarity to find volume of solution and convert to mL: 0.173 mole CuSO4 X 1 L CuSO4 X 1,000 mL = 346 mL 0.500 mole CuSO4 1 L CuSO4

12. Additional Example A zinc bar is placed in 435 mL of a 0.770 M solution of CuCl2. What mass of zinc would be replaced by copper if all of the copper ions were used up? Zn + CuCl2→ Cu + ZnCl2 First, convert to L and then use molarity to find moles of CuCl2: 435 mL CuCl2 X 1 L X 0.770 mol CuCl2 = 0.335 1,000 mL 1 L CuCl2 mol CuCl2 Next, use mole ratio to change mol CuCl2 → mol Zn: 0.335 mol CuCl2 X 1 mol Zn = 0.335 mol Zn 1 mol CuCl2 Last, use molar mass mol Zn→ gram Zn: 0.335 mol Zn X 65.4 g Zn= 21.9 g Zn 1 mol Zn

13. Another Example What volume, in mL, of a 1.50 M HCl solution would be needed to react completely with 28.4 g of Na2CO3 to produce water, CO2, and NaCl? Na2CO3 + 2 HCl H2O + CO2 + 2 NaCl First, change grams Na2CO3 → mole Na2CO3: 28.4 g Na2CO3 X 1 mol Na2CO3 = 0.268 mol Na2CO3 106 g Na2CO3 Next, use mole ratio to change mol Na2CO3→mol HCl: 0.268 mol Na2CO3 X 2 mol HCl= 0.536 mol HCl 1 mol Na2CO3 Last, use molarity to find volume of solution and convert to mL: 0.536 mol HCl X 1 L HCl X 1,000 mL = 357 mLHCl 1.50 mol HCl 1 L

14. Homework • Practice pg. 467: #1-3 • Section 13.2 review, pg. 467: #1-14 • Concept Review: “Concentration and Molarity” Next Time: Quiz over this section…