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Chapter 7 Solutions and Colloids
7.5
Solution Preparation
1
Molarity (M)
Molarity (M) is
• a concentration term for solutions.
• gives the moles of solute in 1 L solution.
• moles of solute
liter of solution
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Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared
• by weighing out 58.5 g NaCl (1.00 mole) and
• adding water to make 1.00 liter of solution.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Calculation of Molarity
What is the molarity of 0.500 L NaOH solution if it
contains 6.00 g NaOH?
STEP 1 Given 6.00 g NaOH in 0.500 L solution
Need molarity (mole/L)
STEP 2 Plan g NaOH
mole NaOH
molarity
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Calculation of Molarity (cont.)
STEP 3 Conversion factors 1 mole NaOH = 40.0 g
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
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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
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Solution
3) 1.71 M
46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3
84.0 g NaHCO3
0.557 mole NaHCO3 = 1.71 M NaHCO3
0.325 L
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Learning Check
What is the molarity of 225 mL of a KNO3 solution
containing 34.8 g KNO3?
1) 0.344 M
2) 1.53 M
3) 15.5 M
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Solution
2) 1.53 M
34.8 g KNO3 x 1 mole KNO3
101.1 g KNO3
= 0.344 mole 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
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Molarity Conversion Factors
The units of molarity are used as conversion factors in
calculations with solutions.
TABLE 7.8
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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 Grams of KCl
STEP 2 Plan
L KCl
moles KCl
g KCl
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Calculations Using Molarity
STEP 3 Conversion factors
1 mole 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 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
1L
1 mole KCl
= 6.71 g KCl
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Learning Check
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
1)
20.0 g AlCl3
2)
16.7g AlCl3
3)
2.50 g AlCl3
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Solution
3)
2.50 g AlCl3
0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3
1L
1 mole
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Learning Check
How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3?
1) 12.0 mL
2) 83.3 mL
3) 190. mL
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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 HNO3
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Dilution
In a dilution
• water is added.
• volume increases.
• concentration decreases.
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1 container
orange juice
concentrate
3 containers
of H2O
+
Mix
The amount of
solute in the
concentrated
solution =
amount of
solute in the
diluted solution
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
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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%
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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?
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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
C2= ?
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
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Dilution Calculations with Molarity
What is the molarity (M) of a solution prepared
by diluting 0.180L of 0.600 M HNO3 to 0.540 L?
Prepare a table:
M1= 0.600 M
V1 = 0.180 L
M 2= ?
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
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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
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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
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