Transcript 7.5 Molarity and Dilution

Chapter 7
Solutions
7.5
Molarity and Dilution
Copyright © 2009 by Pearson Education, Inc.
1
Molarity (M)
Molarity (M)
• is a concentration term for solutions.
• gives the moles of solute in 1 L of solution.
• moles of solute
liter of solution
2
Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared
• by weighing out 58.5 g of NaCl
(1.00 mole) and
• adding water to make 1.00 liter
of solution.
Copyright © 2009 by Pearson Education, Inc.
3
Calculation of Molarity
What is the molarity of 0.500 L of NaOH solution if it
contains 6.00 g of NaOH?
STEP 1: Given 6.00 g of NaOH in 0.500 L of solution
Need molarity (mole/L)
STEP 2: Plan g NaOH
mole NaOH
molarity
4
Calculation of Molarity (continued)
STEP 3: Conversion factors 1 mole of NaOH = 40.0 g
of NaOH
1 mole NaOH
and 40.0 g NaOH
40.0 g NaOH
1 mole NaOH
STEP 4: Calculate molarity.
6.00 g NaOH x 1 mole NaOH = 0.150 mole
40.0 g NaOH
0.150 mole = 0.300 mole = 0.300 M NaOH
0.500 L
1L
5
Learning Check
What is the molarity of 325 mL of a solution containing
46.8 g of NaHCO3?
1)
2)
3)
0.557 M
1.44 M
1.71 M
6
Solution
3) 1.71 M
46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3
84.0 g NaHCO3
0.557 mole of NaHCO3 = 1.71 M NaHCO3
0.325 L
7
Learning Check
What is the molarity of 225 mL of a KNO3 solution
containing 34.8 g of KNO3?
1) 0.344 M
2) 1.53 M
3) 15.5 M
8
Solution
2) 1.53 M
34.8 g KNO3 x 1 mole KNO3 = 0.344 mole of KNO3
101.1 g KNO3
M = mole =
0.344 mole KNO3 = 1.53 M
L
0.225 L
In one setup:
34.8 g KNO3 x 1 mole KNO3 x
1
= 1.53 M
101.1 g KNO3 0.225 L
9
Molarity Conversion Factors
The units of molarity are used as conversion factors in
calculations with solutions.
Molarity
3.5 M HCl
Equality
1 L = 3.5 moles of HCl
Written as Conversion Factors
3.5 moles HCl
and
1L
1L
3.5 moles HCl
10
Calculations Using Molarity
How many grams of KCl are needed to prepare 125 mL
of a 0.720 M KCl solution?
STEP 1: Given 125 mL (0.125 L) of 0.720 M KCl
Need g of KCl
STEP 2: Plan
L KCl
moles KCl
g KCl
11
Calculations Using Molarity
STEP 3: Conversion factors
1 mole of KCl = 74.6 g
1 mole KCl
and 74.6 g KCl
74.6 g KCl
1 mole KCl
1 L KCl = 0.720 mole of KCl
1L
and 0.720 mole KCl
0.720 mole KCl
1L
STEP 4: Calculate grams.
0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g of KCl
1L
1 mole KCl
12
Learning Check
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
1)
20.0 g of AlCl3
2)
16.7 g of AlCl3
3)
2.50 g of AlCl3
13
Solution
3)
2.50 g AlCl3
0.125 L x 0.150 mole x 133.5 g = 2.50 g of AlCl3
1L
1 mole
14
Learning Check
How many milliliters of 2.00 M HNO3 contain 24.0 g of
HNO3?
1) 12.0 mL
2) 83.3 mL
3) 190. mL
15
Solution
24.0 g HNO3 x 1 mole HNO3 x
63.0 g HNO3
1000 mL
2.00 mole HNO3
Molarity factor inverted
= 190. mL of HNO3
16
Dilution
In a dilution
• water is added.
• volume increases.
• concentration decreases.
Copyright © 2009 by Pearson Education, Inc.
17
Comparing Initial and Diluted
Solutions
In the initial and diluted solution,
• the moles of solute are the same.
• the concentrations and volumes are related by the
following equations:
For percent concentration:
C1V1 = C2V2
initial
diluted
For molarity:
M1V1 = M2V2
initial
diluted
18
Guide to Calculating Dilution
Quantities
Copyright © 2009 by Pearson Education, Inc.
19
Dilution Calculations with Percent
What volume of a 2.00% (m/v) HCl solution can be
prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?
Prepare a table:
C1= 14.0% (m/v)
V1 = 25.0 mL
C2= 2.00% (m/v)
V2 = ?
Solve dilution equation for unknown and enter values:
C1V1 = C2V2
V2
= V1C1
C2
= (25.0 mL)(14.0%) = 175 mL
2.00%
20
Learning Check
What is the percent (% m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
21
Solution
What is the percent (% m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
Prepare a table:
C1= 9.00 %(m/v)
V1 = 10.0 mL
C 2= ?
V2 = 60.0 mL
Solve dilution equation for unknown and enter values:
C1V1 = C2V2
C2
= C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v)
V2
60.0 mL
22
Dilution Calculations with Molarity
What is the molarity (M) of a solution prepared
by diluting 0.180 L of 0.600 M HNO3 to 0.540 L?
Prepare a table:
M1= 0.600 M
V1 = 0.180 L
M2= ?
V2 = 0.540 L
Solve dilution equation for unknown and enter values:
M1V1 = M2V2
M2
= M1V1
V2
= (0.600 M)(0.180 L) = 0.200 M
0.540 L
23
Learning Check
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
1) 27.0 mL
2) 60.0 mL
3) 90.0 mL
24
Solution
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
Prepare a table:
M1= 1.80 M
V1 = 15.0 mL
M2= 0.300M
V2 = ?
Solve dilution equation for V2 and enter values:
M1V1 = M2V2
V2
= M1V1
M2
= (1.80 M)(15.0 mL) = 90.0 mL
0.300 M
25