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Chapter 12 Solutions. 12.5 Molarity and Dilution. 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. Preparing a 6.0 M Solution. A 6.00 M NaOH solution is prepared - PowerPoint PPT Presentation
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Basic Chemistry Copyright © 2011 Pearson Education, Inc.1
12.5
Molarity and Dilution
Chapter 12 Solutions
Basic Chemistry Copyright © 2011 Pearson Education, Inc.2
Molarity (M)Molarity (M) is
• a concentration term for solutions
• the moles of solute in 1 L of solution
• moles of soluteliter of solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.3
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
Basic Chemistry Copyright © 2011 Pearson Education, Inc.4
Calculating Molarity
Basic Chemistry Copyright © 2011 Pearson Education, Inc.5
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 moles of NaOH molarity
Example of Calculating Molarity
Basic Chemistry Copyright © 2011 Pearson Education, Inc.6
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 NaOHSTEP 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 mol0.500 L solution 1 L
= 0.300 M NaOH solution
Example of Calculating of Molarity (continued)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.7
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
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.8
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 moles of NaHCO3 molarity
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.9
STEP 3 Write equalities and conversion factors needed.
1 mol of NaHCO3 = 84.01 g of NaHCO3
1 mol NaHCO3 and 40.01 g NaHCO3
40.01 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
Solution (continued)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.10
Molarity Conversion Factors
The units of molarity are used as conversion factors in calculations with solutions.
Basic Chemistry Copyright © 2011 Pearson Education, Inc.11
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 moles of KCl grams of KCl
Basic Chemistry Copyright © 2011 Pearson Education, Inc.12
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
Basic Chemistry Copyright © 2011 Pearson Education, Inc.13
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 1 L 1 mol KCl
Basic Chemistry Copyright © 2011 Pearson Education, Inc.14
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
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.15
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 liters of AlCl3 solution
moles of AlCl3 grams of AlCl3
Basic Chemistry Copyright © 2011 Pearson Education, Inc.16
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
Basic Chemistry Copyright © 2011 Pearson Education, Inc.17
Solution (continued)
STEP 4 Set up problem to calculate mass or volume.
37.8 mL x 1 L x 0.150 mol x 133.33 g 1000 mL 1 L 1 mol
= 0.756 g of AlCl3 (B)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.18
How many milliliters of a 2.00 M HNO3 solutioncontain 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
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.19
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 moles of HNO3 mL of HNO3
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.20
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
Solution (continued)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.21
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)
Solution (continued)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.22
DilutionIn a dilution,
• water is added
• volume increases
• concentration decreases
Basic Chemistry Copyright © 2011 Pearson Education, Inc.23
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 = M2V2
initial diluted
Basic Chemistry Copyright © 2011 Pearson Education, Inc.24
Calculating Dilution Quantities
Basic Chemistry Copyright © 2011 Pearson Education, Inc.25
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
M1 = 0.600 M M2 = ?
V1 = 0.180 L V2 = 0.540 L
Basic Chemistry Copyright © 2011 Pearson Education, Inc.26
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 MV2 0.540 L
Basic Chemistry Copyright © 2011 Pearson Education, Inc.27
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
Basic Chemistry Copyright © 2011 Pearson Education, Inc.28
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= 0.300 M V2 = ?
Basic Chemistry Copyright © 2011 Pearson Education, Inc.29
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 )