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Transcript of Tuesday, Feb. 25 th : “A” Day Wednesday, Feb. 26 th : “B” Day (1:05 out) Agenda Homework...
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Tuesday, Feb. 25th: “A” DayWednesday, Feb. 26th: “B” Day (1:05 out)
AgendaHomework Questions/CollectFinish Section 13.2: “Concentration and Molarity”
Solution dilution equation, using molarity in stoichiometric calculations
Homework:Practice Pg. 467: #1-3Section 13.2 review, pg. 467: #1-14Concept Review: “Concentration and Molarity”
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Homework
Practice pg. 461: #2, 3, 5, 6
Practice pg. 465: #1-7
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Solution DilutionOften, 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?
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Solution Dilution Equation
M = mol
LM x L = mol
OR
Solution DilutionM1V1=M2V2
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Solution Dilution DemoThe 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
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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 MV1: 75.0 mLM2: 0.155 MV2: ?
(12.0 M) (75.0 mL) = (0.155 M) (V2)
V2 = 5,806 mL → 5.81 L
Dilution Example #1
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Dilution Example #2What 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
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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.
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Sample Problem C, Pg. 466What 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 CuNext, use mole ratio to change mole Cu → mole CuSO4:
0.173 mole Cu X 3 mol CuSO4 = 0.173 mole CuSO4
3 mole CuLast, 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
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Additional ExampleA 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
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Another ExampleWhat 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 NaClFirst, 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 mL HCl 1.50 mol HCl 1 L
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HomeworkPractice pg. 467: #1-3Section 13.2 review, pg. 467: #1-14Concept Review: “Concentration and Molarity”
Next Time:Quiz over this section…