Tuesday, Feb. 25th: “A” Day
Wednesday, 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”
Homework
Practice pg. 461: #2, 3, 5, 6
Practice pg. 465: #1-7
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?
Solution Dilution Equation
mol
M =
L
OR
M x L = mol
Solution Dilution
M1V1=M2V2
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:
M 1V 1 = M 2V 2
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
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
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
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. 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 =
0.500 mole CuSO4 1 L
346 mL
CuSO4
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
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 =
1.50 mol HCl
1L
357 mL HCl
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…