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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”
Homework
Practice pg. 461: #2, 3, 5, 6
Practice pg. 465: #1-7
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?
Solution Dilution Equation
M = mol
LM x L = mol
OR
Solution DilutionM1V1=M2V2
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
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
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
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.
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
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
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
HomeworkPractice pg. 467: #1-3Section 13.2 review, pg. 467: #1-14Concept Review: “Concentration and Molarity”
Next Time:Quiz over this section…
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