Transcript Introductory Chemistry, 2nd Edition Nivaldo Tro

Chapter 13
Solutions
1
Solutions
• Recall: Solutions are homogeneous
mixtures.
Appears to be one substance, though really
contains multiple materials.
• Most homogeneous materials we encounter
are actually solutions.
E.g., air and lake water.
2
Solutions, Continued
• Solutions = Solute + Solvent
• Solute is the dissolved substance.
Seems to “disappear.”
“Takes on the state” of the solvent.
• Solvent is the substance solute dissolves in.
Does not appear to change state.
E.g. Salt dissolved in water.
3
Solutions, Continued
• When both solute and solvent have the same
state, the solvent is the component present in the
highest percentage.
• E.g. mixture of 2 mL of ethanol and 10 mL of
water.
Ethanol = solute
Water = solvent
4
Types of Solutions
• Saturated solutions have the maximum
amount of solute that will dissolve in that
solvent at that temperature.
• Unsaturated solutions can dissolve more
solute.
• Supersaturated solutions are holding more
solute than they should be able to.
Unstable.
5
Solubility
• Solubility: The amount of solute that dissolves in 100 g
of solvent at a specific temperature.
In solids: Higher temperature = Higher solubility.
In gases: Higher temperature = Lower solubility.
Above line = Supersaturated
On line = saturated
Below line = Unsaturated
6
Solubility and Pressure
• The solubility of gases in water depends on the
pressure of the gas.
• Pressure does not affect solubility of solids
• Higher pressure = higher solubility.
7
Amount of Solute in a solution
• Mostly determined by:
Mass percent
Concentration (Molarity)
8
Mass Percent
Mass of Solute, g
Mass Percent
100%
Mass of Solution,g
Mass of Solute  Mass of Solvent  Mass of Solution
If a solution is 0.9% by mass, then there are 0.9
grams of solute in every 100 grams of solution.
9
Example 13.1—Calculate the Mass Percent of a
Solution Containing 27.5 g of Ethanol in 175 mL H2O.
Given: 27.5 g ethanol, 202.5
175 mL
g solution
H2 O
Find: % by mass
Solution Map: mL
g EtOH,
H2O g H2O g H2O
g sol’n
%
1.00 g H 2O
g solute
g solute  g solvent  g solution % by Mass
 100%
1 mL H 2O
g solution
Relationships: 1 mL H2O = 1.00 g
Solve:
1.0027.5
g H 2gOethanol
175
H 2 O 
 175 g H2100
O %
% mL
by Mass
1 20
mL2H
.52gOsolution
27.5
g ethanol 175 g H2O  202.5 g solution
 13.6%
10
Practice—Calculate the Mass Percent of a Solution that
Has 10.0 g of I2 Dissolved in 150.0 g of Ethanol.
6.25 %
11
Using Mass Percent as
Conversion Factors
Example—How Would You Prepare 250.0 g of
5.00% by Mass Glucose Solution ?
12
Example—How Would You Prepare 250.0 g of
5.00% by Mass Glucose Solution ?
Given:
Find:
Equivalence:
Solution Map:
250.0 g solution
g glucose
5.00 g glucose  100 g solution
g solution
g glucose
5.00 g Glucose
100 g solution
Apply Solution Map:
5.00 g glucose
250.0 g solution 
 12.5 g glucose
100 g solution
Answer: Dissolve 12.5 g of glucose in enough water to
total 250.0 g.
13
Practice—Milk Is 4.5% by Mass Lactose.
Determine the Mass of Lactose in 175 g of Milk.
7.9 g
14
Practice—How Would You Prepare 450.0 g of
15.0% by Mass Aqueous Ethanol?
67.5 g in 382.5 g water
15
Solution Concentrations
also known as molarity
• Concentration = Moles of solute in liters of
solution.
moles of solute
Molarity =
liters of solution
• If a sugar solution concentration is 2.0 M ,
1 liter of solution contains 2.0 moles of sugar
2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
16
Example 13.3—Calculate the Molarity of a Solution
Made by Dissolving 15.5 g of NaCl in 1.50 L of Solution
Given: 15.5 g NaCl, 1.50 L solution
Find: M
Solution Map:
g NaCl
mol NaCl
L solution
mol
M
L
Relationships: M = mol/L, 1 mol NaCl = 58.44 g
M
Solve:
1 mol NaCl
15.5 g NaCl 
 0.2652 mol NaCl
58.44 g NaCl
0.2652 mol NaCl
M
1.50L
M  0.177M
17
Practice—What Is the Molarity of a Solution
Containing 3.4 g of NH3 (MM 17.03) in 200.0 mL
of Solution?
1.0 M
18
Practice—Determine the Mass of CaCl2
(MM = 110.98) in 1.75 L of 1.50 M Solution
291 g
19
Example—How Would You Prepare 250 mL
of 0.20 M NaCl?
Given:
250 mL solution
Find:
g NaCl
Equivalence: 0.20 moles NaCl  1 L solution; 0.001 L = 1 mL;
58.44 g = 1 mol NaCl
Solution Map:
0.20 mol NaCl
mL
solution 0.001 L
1 mL
L
solution
1L
moles
NaCl
58.44 g
1 mol NaCl
g
NaCl
Apply Solution Map:
0.001 L 0.20 mol NaCl
58.44 g
250 mL 


 2.9 g NaCl
1 mL
1L
1 mol NaCl
Answer:
Dissolve 2.9 g of NaCl in enough water to total 250 mL.
20
Practice—How Would You Prepare 100.0 mL
of 0.100 M K2SO4 (MM = 174.26)?
1.74 g of K2SO4 in enough water to total 100.0 mL.
21
Molarity and Dissociation
• Strong electrolytes dissociate completely in water.
NaCl(aq)  Na+(aq) + Cl-(aq)
1 mole NaCl = 1 mole Na+
1 mole NaCl = 1 mole ClCaCl2(aq)  Ca2+(aq) + 2 Cl-(aq)
1 mole CaCl2 = 1 mole Ca2+
1 mole CaCl2 = 2 moles Cl-(aq)
22
Example 13.5—Determine the Molarity of the
Ions in a 0.150 M Na3PO4(aq) Solution.
Given: 0.150 M Na3PO4(aq)
Find: concentration of Na+ and PO43−, M
Relationships: Na3PO4(aq)  3 Na+(aq) + PO43−(aq)
Solve:
1 mol PO43
0.150 M Na 3PO4 
 0.150 M PO43
1 mol Na 3PO4
3 mol Na 
0.150 M Na 3PO4 
 0.450 M Na 
1 mol Na 3PO4
23
Practice—Find the Molarity of All Ions in the
Given Solutions of Strong Electrolytes.
• 0.25 M MgBr2(aq).
• 0.33 M Na2CO3(aq).
• 0.0750 M Fe2(SO4)3(aq).
24
Dilution
• Dilution is adding extra solvent to decrease the
concentration of a solution.
• The amount of solute stays the same, but the
concentration decreases.
• Dilution Formula:
Concstart solnx Volstart soln = Concfinal solnx Volfinal sol
• M1V1 = M2V2
25
*Example*—What Volume of 12.0 M KCl Is
Needed to Make 5.00 L of 1.50 M KCl Solution?
26
Example—Dilution Problems
• What is the concentration of a solution made by
diluting 15 mL of 5.0% sugar to 135 mL?
M1 = 5.0 %
M2 = ? %
V1 = 15 mL
V2 = 135 mL
(5.0%)(15 mL) = M2 x (135 mL)
M2 = 0.55%
• How would you prepare 200 mL of 0.25 M
NaCl solution from a 2.0 M solution?
M1 = 2.0 M
V1 = ? mL
M2 = 0.25 M
V2 = 200 mL
(2.0 M) x V1 = (0.25 M)(200 mL)
V1 = 25 mL
Dilute 25 mL of 2.0 M NaCl solution to 200 mL.
27
Practice—Determine the Concentration of
the Following Solutions.
• Made by diluting 125 mL of 0.80 M HCl to 500
mL.
0.2 M
28
Practice Question 1—How Would You Prepare
400 mL of a 4.0% Solution From a 12%
Solution?
Practice Question 2—How Would You Prepare 250
mL of a 3.0% Solution From a 7.5% Solution?
Q 1: Dilute 133 mL of 12% solution to 400 mL
Q2: Dilute 100 mL of 7.5% solution to 250 mL..
29
Molality, m
• Moles of solute per 1 kilogram of solvent.
Defined in terms of amount of solvent, not
solution.
Does not vary with temperature.
Because based on masses, not volumes.
moleof solute
molality
kg of solvent
Mass of solution = volume of solution x density.
Mass of solution = mass of solute + mass of solvent.
30
Example 13.8—What Is the Molality of a Solution
Prepared by Mixing 17.2 g of C2H6O2 with
0.500 kg of H2O?
Given: 17.2 g C2H6O2, 0.500 kg H2O
Find: m
Concept Plan: g C2H6O2
mol C2H6O2 m  mol
kg
kg H2O
m
Relationships: m = mol/kg, 1 mol C2H6O2 = 62.07 g
1 molC2 H 6O2
Solve:
17.2 g C2 H 6O2 
62.07g C2 H 6O 2
 0.2771 mol
0.2771 molC 2 H 6 O 2
m
0.500kg H 2 O
m  0.554m
31
Practice—What Is the Molality of a Solution that Is
Made by Dissolving 3.4 g of NH3 (MM 17.03) in
1500 mL of H2O (d =0.90 g/mL).
0.15 m
32
Colligative Properties
• Colligative properties-- properties whose value
depends only on the number of solute particles
and not on the type.
• Examples:
Vapor pressure
Freezing point
Boiling point
Osmotic Pressure
33
33
Freezing Point
• The freezing point of a solution is lower than the freezing
point of the pure solvent.
• Freezing point depression, DTf :-- the difference between the
freezing point of the solution and freezing point of the pure
solvent
DTf  (FPpure solvent – FPsolution)
• DTf is proportional to the molality of the solution
DTf = m∙Kf
• Kf is the Freezing Point Depression constant
34
34
Example 13.9—What Is the Freezing Point of a 1.7 m
Aqueous Ethylene Glycol Solution, C2H6O2?
Given: 1.7 m C2H6O2(aq)
Find: Tf, °C
Solution Map:
m
DTf
DT f  m  K f
FP
FPsolv − FPsol’n = DT
Relationships: DTf = m ∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C
Solve: DT f  m  K f ,H 2O
FPH O  FPsol'n  DT f
(
 (1.7 m  1.86
DT f  3.2 C
C
m

2
0.00C  FPsol'n  3.2C
FPsol'n  3.2C
35
Practice—What Is the Freezing Point of a Solution
that Has 0.20 moles of Sulfur Dissolved in 0.10 kg
of Cyclohexane?
(FPcyclohexane = 6.5 C, Kf = 20.0 C/m)
-33.5 oC
36
Boiling Point
• The boiling point of a solution is higher than the
boiling point of the pure solvent.
• Boiling Point Elevation--difference between the boiling
point of the solution and boiling point of the pure
solvent.
DTb  BPsolution – BPpure solvent = m∙Kb
• DTb is proportional to molality, m.
DTb = m∙Kb
• Kb is the Boiling Point Elevation Constant
37
37
Example 13.10—What Is the Boiling Point of a 1.7-m
Aqueous Ethylene Glycol Solution, C2H6O2?
Given: 1.7 m C2H6O2(aq)
Find: Tb, °C
Solution Map:
m
DTb
DTb  m  Kb
BP
BPsol’n − BPsolv = DT
Relationships: DTb = m ∙Kb, Kb H2O = 0.512 °C/m, BPH2O = 100.00 °C
Solve:
BPsolution BPsolvent  DTb
DT  m  K
b
(
b ,H 2O
 (1.7 m  0.512 mC
DTb  0.87C

BPsolution 100.00C  0.87C
BPsol'n  100.87C
38
Practice—What Is the Boiling Point of a Solution
that Has 0.20 moles of Sulfur Dissolved in 0.10 kg
of Cyclohexane?
(BPcyclohexane = 80.7 C, Kb= 2.79 C/m)
86.3 oC
39
Osmosis
• Osmosis--flow of solvent from a solution of
low concentration into a solution of high
concentration.
• The solutions may be separated by a
semipermeable membrane.
• Semipermeable membrane-- allows solvent
to flow through it, but not solute.
40
40
Osmotic Pressure
Solvent flows through a semipermeable membrane to make the
solution concentration equal on both sides of the membrane.
The pressure required to stop this process is osmotic pressure.
41
Types of Osmotic Solutions
• Isotonic solution: Same solute
concentration relative to another solution
• Hypotonic solution: Low solute
concentration relative to another solution
• Hypertonic solution: High solute
concentration relative to another solution
42
Hemolysis and Crenation
Normal red blood
cell in an isotonic
Solution.
Red blood cell in Red blood cell in
a hypotonic
hypertonic
solution.
solution. Water
Water flows into
flows out
the cell, eventually
of the cell,
causing
eventually causing
the cell to burst.
the cell to distort
and shrink. 43
Drinking Seawater
Because seawater has
a higher salt concentration
than your cells, water flows
out of your cells into the
seawater to try to decrease
its salt concentration.
The net result is that, instead
of quenching your thirst,
you become dehydrated.
44
Recommended Study Problems: Chapter 13
NB: Study problems are used to check the student’s understanding
of the lecture material. Students are EXPECTED TO BE ABLE
TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS. If
you encounter any problems, please talk to your professor or
seek help at the HACC-Gettysburg learning center.
Questions from text book Chapter 13
5, 6, 7, 11, 13, 17, 21, 22, 23, 24, 25, 27, 27, 33, 37,
43, 45, 49, 51, 65, 67, 69, 73, 75, 79, 83, 85, 87, 91,
93, 99, 103, 105, 107
ANSWERS
-The answers to the odd-numbered study problems are found at
the back of your textbook (see Appendix III)
45