Transcript 12.5_Molarity_and_Dilutions

Chapter 12 Solutions
12.5
Molarity and Dilution
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Molarity (M)
Molarity (M) is
• a concentration term for solutions
• the moles of solute in 1 L of solution
• moles of solute
liter of solution
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Preparing a 6.0 M Solution
A 6.00 M NaOH solution is
prepared
• by weighing out 60.0 g of
NaOH (1.50 mol) and
• adding water to make 0.250 L
of a 6.00 MNaOH solution
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Preparation of Solutions
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Preparation of Solutions

A solution of a desired concentration can be prepared by
diluting a small volume of a more-concentrated solution,
a stock solution, with additional solvent.
– Calculate the number of moles of solute desired in the final
volume of the more-dilute solution and then calculate the
volume of the stock solution that contains the amount of solute.
– Diluting a given quantity of stock solution with solvent does not
change the number of moles of solute present.
– The relationship between the volume and concentration of the
stock solution and the volume and concentration of the desired
diluted solution is
(Vs) (M s) = moles of solute = (Vd) (M d).
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Preparation of Solutions
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Calculating Molarity
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Example of Calculating Molarity
What is the molarity of 0.500 L of a NaOH solution if
it contains 6.00 g of NaOH?
STEP 1 State the given and needed quantities.
Given 6.00 g of NaOH in 0.500 L of solution
Need molarity (M)
STEP 2 Write a plan to calculate molarity.
molarity (M) = moles of solute
liters of solution
grams of NaOH
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moles of NaOH
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molarity
Example of Calculating of
Molarity (continued)
STEP 3 Write equalities and conversion factors
needed.
1 mol of NaOH = 40.01 g of NaOH
1 mol NaOH
and 40.01 g NaOH
40.01 g NaOH
1 mol NaOH
STEP 4 Set up problem to calculate molarity.
6.00 g NaOH x 1 mol NaOH = 0.150 mol of NaOH
40.01 g NaOH
0.150 mol NaOH = 0.300 mol
0.500 L solution
1L
= 0.300 M NaOH solution
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Learning Check
What is the molarity of 325 mL of a solution
containing 46.8 g of NaHCO3?
A.
0.557 M NaHCO3 solution
B.
1.44 M NaHCO3 solution
C.
1.71 M NaHCO3 solution
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Solution
STEP 1 State the given and needed quantities.
Given 46.8 g of NaHCO3 in 0.325 L of solution
Need molarity (M)
STEP 2 Write a plan to calculate molarity.
molarity (M) = moles of solute
liters of solution
grams of NaHCO3
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moles of NaHCO3
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molarity
Solution (continued)
STEP 3 Write equalities and conversion factors
needed.
1 mol of NaHCO3 = 84.01 g of NaHCO3
1 mol NaHCO3 and 84.01 g NaHCO3
84.01 g NaOH
1 mol NaHCO3
STEP 4 Set up problem to calculate molarity.
46.8 g NaHCO3 x 1 mol NaHCO3 = 0.557 mol of NaHCO3
84.01 g NaHCO3
0.557 mol NaHCO3 = 1.71 M NaHCO3 solution
0.325 L
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Molarity Conversion Factors
The units of molarity are used as conversion
factors in calculations with solutions.
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Example of Using Molarity in
Calculations
How many grams of KCl are needed to prepare
0.125 L of a 0.720 M KCl solution?
STEP 1 State the given and needed quantities.
Given 0.125 L of a 0.720 M KCl solution
Need grams of KCl
STEP 2 Write a plan to calculate mass or
volume.
liters of KCl solution
grams of KCl
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moles of KCl
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Example of Using Molarity in
Calculations (continued)
STEP 3 Write equalities and conversion factors
needed.
1 mol of KCl = 74.55 g of KCl
1 mol KCl
and 74.55 g KCl
74.55 g KCl
1 mol KCl
1 L of KCl solution = 0.720 mol of KCl
1 L KCl solution and 0.720 mol KCl
0.720 mol KCl
1 L KCl solution
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Example of Using Molarity in
Calculations (continued)
STEP 4 Set up problem to calculate mass or volume.
0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g of KCl
1L
1 mol KCl
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Learning Check
How many grams of AlCl3 are needed to prepare
37.8 mL of a 0.150 M AlCl3 solution?
A. 0.00567 g of AlCl3
B. 0.756 g of AlCl3
C. 5.04 g of AlCl3
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Solution
STEP 1 State the given and needed quantities.
Given 37.8 mL of a 0.150 M AlCl3 solution
Need grams of AlCl3
STEP 2 Write a plan to calculate mass or volume.
milliliters of AlCl3 solution
moles of AlCl3
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liters of AlCl3 solution
grams of AlCl3
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Solution (continued)
STEP 3 Write equalities and conversion factors
needed.
1 mol of AlCl3 = 133.33 g of AlCl3
1 mol AlCl3
and 133.33 g AlCl3
133.33 g AlCl3
1 mol AlCl3
1000 mL of AlCl3 solution = 1 L of AlCl3 solution
1000 mL AlCl3 solution and 1 L AlCl3 solution
1 L AlCl3 solution
1000 mL AlCl3 solution
1 L of AlCl3 solution = 0.150 mol of AlCl3
1 L AlCl3 solution and 0.150 mol AlCl3
0.150 mol AlCl3
1 L AlCl3 solution
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Solution (continued)
STEP 4 Set up problem to calculate mass or
volume.
37.8 mL x
1L
x 0.150 mol x 133.33 g
1000 mL
1L
1 mol
= 0.756 g of AlCl3 (B)
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Learning Check
How many milliliters of a 2.00 M HNO3 solution
contain 24.0 g of HNO3?
A. 12.0 mL of HNO3 solution
B. 83.3 mL of HNO3 solution
C. 190. mL of HNO3 solution
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Solution
STEP 1 State the given and needed quantities.
Given 24.0 g of HNO3; 2.00 M HNO3 solution
Need milliliters of HNO3 solution
STEP 2 Write a plan to calculate mass or volume.
g of solution
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moles of HNO3
mL of HNO3
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Solution (continued)
STEP 3 Write equalities and conversion factors
needed.
1 mol of HNO3 = 63.02 g of HNO3
1 mol HNO3
and 63.02 g HNO3
63.02 g HNO3
1 mol HNO3
1000 mL of HNO3 = 2.00 mol of HNO3
1000 mL HNO3 and 2.00 mol HNO3
2.00 mol HNO3
1000 mL HNO3
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Solution (continued)
STEP 4 Set up problem to calculate mass or
volume.
24.0 g HNO3 x 1 mol HNO3 x
1000 mL
63.02 g HNO3 2.00 mol HNO3
= 190. mL of HNO3 solution (C)
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Dilution
In a dilution,
• water is added
• volume increases
• concentration decreases
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NO3Na+
Moles = 1.0
Volume = 1.0 L
Molarity = 1.0 M
NaNO3 solution
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NO3Na+
Moles = 1.0
Volume = 2.0 L
Molarity = 0.50 M
• Solution volume is doubled
• Moles of solute remain the same
• Solution concentration is halved
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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 equation
M1V1
initial
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= M2V2
diluted
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Calculating Dilution Quantities
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Example of Dilution
Calculations
What is the final molarity of the solution when
0.180 L of 0.600 M KOH is diluted to 0.540 L?
STEP 1 Prepare a table of the initial and diluted
volumes and concentrations.
Initial Solution
Diluted Solution
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M1 = 0.600 M
M2 = ?
V1 = 0.180 L
V2 = 0.540 L
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Example of Dilution
Calculations (continued)
STEP 2 Solve the dilution expression for the
unknown quantity.
M1V1 = M2V2
M1V1 = M2V2
V2
V2
M2
= M1V1
V2
STEP 3 Set up the problem by placing known
quantities in the dilution expression.
M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M
V2
0.540 L
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Learning Check
What is the final volume, in milliliters, if 15.0 mL
of a 1.80 M KOH solution is diluted to give a
0.300 M KOH solution?
A. 27.0 mL of 0.300 M KOH solution
B. 60.0 mL of 0.300 M KOH solution
C. 90.0 mL of 0.300 M KOH solution
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Solution
STEP 1 Prepare a table of the initial and diluted
volumes and concentrations.
Initial Solution Diluted Solution
M1= 1.80 M
V1 = 15.0 mL
M2= 0.300M
V2 = ?
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Solution (continued)
STEP 2 Solve the dilution expression for the
unknown quantity.
M1V1 = M2V2
M1V1 = M2V2
M2
M2
V2 = M1V1
M2
STEP 3 Set up the problem by placing known
quantities in the dilution expression.
V2 = M1V1 = (1.80 M)(15.0 mL)
M2
0.300 M
= 90.0 mL (C )
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