# Tuesday, Feb. 25 th : “A” Day Wednesday, Feb. 26 th : “B” Day (1:05 out) Agenda  Homework Questions/Collect  Finish Section 13.2: “Concentration and.

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<ul><li>Slide 1</li></ul> <p> Slide 2 Slide 3 Tuesday, Feb. 25 th : A Day Wednesday, Feb. 26 th : 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 Slide 4 Homework Practice pg. 461: #2, 3, 5, 6 Practice pg. 465: #1-7 Slide 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? Slide 6 Solution Dilution Equation M = mol L M x L = mol OR Solution Dilution M 1 V 1 =M 2 V 2 Slide 7 Solution Dilution Demo The concentration of the CuSO 4 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: M 1 V 1 = M 2 V 2 M 1 : the original concentration I made (0.50 M) V 1 : what volume of that solution will I need (?) M 2 : the new concentration Im trying to make (0.05 M) V 2 : the volume of the new concentration Im making: 250 mL (0.50 M) V 1 = (0.05 M) (250 mL) V 1 = 25 mL Slide 8 How many liters of 0.155 M Ni(NO 3 ) 2 can be made from 75.0 mL of 12.0 M Ni(NO 3 ) 2 ? Use the solution dilution equation: M 1 V 1 = M 2 V 2 M 1 : 12.0 M V 1 : 75.0 mL M 2 : 0.155 M V 2 : ? (12.0 M) (75.0 mL) = (0.155 M) (V 2 ) V 2 = 5,806 mL 5.81 L Dilution Example #1 Slide 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: M 1 V 1 = M 2 V 2 M 1 : 19 M V 1 : ? M 2 : 0.15 M V 2 : 1.0 L (19 M) V 1 = (0.15 M) (1.0 L) V 1 =.00789 L 7.9 mL Slide 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. Slide 11 Sample Problem C, Pg. 466 What volume (in mL) of a 0.500 M solution of copper (II) sulfate, CuSO 4, is needed to react with an excess of aluminum to provide 11.0 g of copper? 3 CuSO 4 (aq) + 2 Al (s) 3 Cu (s) + Al 2 (SO 4 ) 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 CuSO 4 : 0.173 mole Cu X 3 mol CuSO 4 = 0.173 mole CuSO 4 3 mole Cu Last, use molarity to find volume of solution and convert to mL: 0.173 mole CuSO 4 X 1 L CuSO 4 X 1,000 mL = 346 mL 0.500 mole CuSO 4 1 L CuSO 4 Slide 12 Additional Example A zinc bar is placed in 435 mL of a 0.770 M solution of CuCl 2. What mass of zinc would be replaced by copper if all of the copper ions were used up? Zn + CuCl 2 Cu + ZnCl 2 First, convert to L and then use molarity to find moles of CuCl 2 : 435 mL CuCl 2 X 1 L X 0.770 mol CuCl 2 = 0.335 1,000 mL 1 L CuCl 2 mol CuCl 2 Next, use mole ratio to change mol CuCl 2 mol Zn: 0.335 mol CuCl 2 X 1 mol Zn = 0.335 mol Zn 1 mol CuCl 2 Last, use molar mass mol Zn gram Zn: 0.335 mol Zn X 65.4 g Zn = 21.9 g Zn 1 mol Zn Slide 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 Na 2 CO 3 to produce water, CO 2, and NaCl? Na 2 CO 3 + 2 HCl H 2 O + CO 2 + 2 NaCl First, change grams Na 2 CO 3 mole Na 2 CO 3 : 28.4 g Na 2 CO 3 X 1 mol Na 2 CO 3 = 0.268 mol Na 2 CO 3 106 g Na 2 CO 3 Next, use mole ratio to change mol Na 2 CO 3 mol HCl: 0.268 mol Na2CO3 X 2 mol HCl = 0.536 mol HCl 1 mol Na 2 CO 3 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 mL HCl 1.50 mol HCl 1 L Slide 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 </p>