Chapter 15a
Solutions
Chapter 15
Table of Contents
15.1
15.2
15.3
15.4
15.5
Solubility
Solution Composition: An Introduction
Solution Composition: Mass Percent
Solution Composition: Molarity
Dilution
2
Chapter 15
What is a Solution?
• Solution – homogeneous mixture
 Solvent – substance present in largest
amount
 Solutes – other substances in the solution
 Aqueous solution – solution with water as the
solvent
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3
Section 15.1
Solubility
Various Types of Solutions
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4
Section 15.1
Solubility
Solubility of Ionic Substances
•
Ionic substances breakup into individual
cations and anions.
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5
Section 15.1
Solubility
Solubility of Ionic Substances
•
Polar water molecules interact with the positive
and negative ions of a salt.
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6
Section 15.1
Solubility
Solubility of Polar Substances
•
Ethanol is soluble in water because of the
polar OH bond.
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7
Section 15.1
Solubility
Solubility of Polar Substances
δ+
H
Why is solid sugar soluble in water?
•
•
OδHδ+
Why is isopropyl alcohol
soluble in water?
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8
Section 15.1
Solubility
Solubility of Polar Substances
•
Why is blue food coloring soluble in water?
Blue food dye mixes with alcohol
and water unless a dissolved salt
dominates the water as a solvent.
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9
Section 15.1
Solubility
Substances Insoluble in Water
•
•
Nonpolar oil does not interact with polar water.
Water-water hydrogen bonds keep the water
from mixing with the nonpolar molecules.
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10
Section 15.1
Solubility
How Substances Dissolve
•
•
•
A “hole” must be made in the
water structure for each
solute particle.
The lost water-water
interactions must be replaced
by water-solute interactions.
“like dissolves like”
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11
Section 15.1
Solubility
Concept Check
Which of the following solutes will generally not
dissolve in the specified solvent? Choose the best
answer. (Assume all of the compounds are in the
liquid state.)
a)
b)
c)
d)
CCl4 mixed with water (H2O)
NH3 mixed with water (H2O)
CH3OH mixed with water (H2O)
N2 mixed with methane (CH4)
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12
Section 15.1
Solubility
Factors that affect Solubility
• Molecular structure-like dissolves like
– ionic nature
– polarity
• Temperature
– Increased temperatures increase solubility of solids
dissolved in liquids.
– Increased temperatures decrease solubility of gasses
dissolved in liquids.
• Pressure
– Increased pressures increase solubility of gasses
dissolved in liquids.
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13
Section 15.1
Solubility
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14
Section 15.2
Solution Composition: An Introduction
•
The solubility of a solute is limited.
 Saturated solution – contains as much
solute as will dissolve at that
temperature.
 Unsaturated solution – has not
reached the limit of solute that will
dissolve.
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15
Section 15.2
Solution Composition: An Introduction
•
Supersaturated solution – occurs when a
solution is saturated at an elevated
temperature and then allowed to cool but
all of the solid remains dissolved.
 Contains more dissolved solid than a
saturated solution at that temperature.
 Unstable – adding a crystal causes
precipitation.
http://www.youtube.com/watch?v=HnSg2cl09PI
http://www.youtube.com/watch?v=nvHrXr5Jajg
http://www.youtube.com/watch?v=OjeFliFZQ8A&NR=1
http://axiomsun.com/home/video/supercooled_water.html
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16
Section 15.2
Solution Composition: An Introduction
•
•
Solutions are mixtures.
Amounts of substances can vary in different
solutions.
 Specify the amounts of solvent and
solutes.
 Qualitative measures of concentration
 concentrated – relatively large amount
of solute
 dilute – relatively small amount of solute
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17
Section 15.3
Solution Composition: Mass Percent
mass of solute
Mass percent =
 100%
mass of solution
Mass percent =
grams of solute
 100%
grams of solute + grams of solvent
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18
Section 15.3
Solution Composition: Mass Percent
Exercise
What is the percent-by-mass concentration of
glucose in a solution made my dissolving 5.5 g
of glucose in 78.2 g of water?
6.6%
[5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6% glucose
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19
Section 15.4
Solution Composition: Molarity
•
Molarity (M) = moles of solute per
volume of solution in liters:
moles of solute
M = Molarity =
liters of solution
3 M HCl =
6 moles of HCl
2 liters of solution
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20
Section 15.4
Solution Composition: Molarity
Exercise
You have 1.00 mol of sugar in 125.0 mL of
solution. Calculate the concentration in units
of molarity.
8.00 M
1.00 mol / (125.0 / 1000) = 8.00 M
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21
Section 15.4
Solution Composition: Molarity
Exercise
A 500.0-g sample of potassium phosphate
is dissolved in enough water to make 1.50 L
of solution. What is the molarity of the
solution?
1.57 M
500.0 g is equivalent to 2.355 mol K3PO4 (500.0 g /
212.27 g/mol). The molarity is therefore 1.57 M
(2.355 mol/1.50 L).
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22
Section 15.4
Solution Composition: Molarity
Exercise
You have a 10.0 M sugar solution. What
volume of this solution do you need to have
2.00 mol of sugar?
0.200 L 2.00 mol / 10.0 M = 0.200 L
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23
Section 15.4
Solution Composition: Molarity
Exercise
Consider separate solutions of NaOH and KCl
made by dissolving 100.0 g of each solute in
250.0 mL of solution. Calculate the
concentration of each solution in units of
molarity.
10.01 M NaOH
[100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.01 M NaOH
5.366 M KCl
[100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.366 M KCl
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24
Section 15.4
Solution Composition: Molarity
Concept Check
You have two HCl solutions, labeled Solution A and
Solution B. Solution A has a greater concentration than
Solution B. Which of the following statements are true?
a)
b)
c)
d)
If you have equal volumes of both solutions,
Solution B must contain more moles of HCl.
If you have equal moles of HCl in both solutions,
Solution B must have a greater volume.
To obtain equal concentrations of both solutions,
you must add a certain amount of water to
Solution B.
Adding more moles of HCl to both solutions will make
them less concentrated.
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25
Section 15.4
Solution Composition: Molarity
Concentration of Ions
•
For a 0.25 M CaCl2 solution:
CaCl2 → Ca2+ + 2Cl–
 Ca2+: 1 × 0.25 M = 0.25 M Ca2+
 Cl–: 2 × 0.25 M = 0.50 M Cl–.
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26
Section 15.4
Solution Composition: Molarity
Concept Check
Which of the following solutions contains
the greatest number of ions?
a)
b)
c)
d)
400.0 mL of 0.10 M NaCl.
300.0 mL of 0.10 M CaCl2.
200.0 mL of 0.10 M FeCl3.
800.0 mL of 0.10 M sucrose.
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27
Section 15.4
Solution Composition: Molarity
Let’s Think About It
•
Where are we going?

•
To find the solution that contains the greatest
number of moles of ions.
How do we get there?


Draw molecular level pictures showing each
solution. Think about relative numbers of ions.
How many moles of each ion are in each
solution?
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28
Section 15.4
Solution Composition: Molarity
Notice
•
The solution with the greatest number of
ions is not necessarily the one in which:
 the volume of the solution is the
largest.
 the formula unit has the greatest
number of ions.
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29
Section 15.4
Solution Composition: Molarity
Standard Solution
•
A solution whose concentration is
accurately known.
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30
Section 15.4
Solution Composition: Molarity
To Make a Standard Solution
•
•
•
Weigh out a sample of solute.
Transfer to a volumetric flask.
Add enough solvent to mark on flask.
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31
Section 15.5
Dilution
•
•
•
The process of adding water to a
concentrated or stock solution to achieve
the molarity desired for a particular
solution.
Dilution with water does not alter the
numbers of moles of solute present.
Moles of solute before dilution = moles of
solute after dilution
M1V1 = M2V2
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32
Section 15.5
Dilution
Diluting a Solution
•
•
Transfer a measured amount of original solution
to a flask containing some water.
Add water to the flask to the mark (with swirling)
and mix by inverting the flask.
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33
Section 15.5
Dilution
Concept Check
A 0.50 M solution of sodium chloride in an open
beaker sits on a lab bench. Which of the
following would decrease the concentration of
the salt solution?
a)
b)
c)
d)
Add water to the solution.
Pour some of the solution down the sink drain.
Add more sodium chloride to the solution.
Let the solution sit out in the open air for a
couple of days.
e) At least two of the above would decrease the
concentration of the salt solution.
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34
Section 15.5
Dilution
Exercise
What is the minimum volume of a 2.00 M
NaOH solution needed to make 150.0 mL of
a 0.800 M NaOH solution?
60.0 mL
M1V1 = M2V2
(2.00 M)(V1) = (0.800 M)(150.0 mL)
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35
W Section 15.5
Dilution
Chapter 15b
Solutions
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36
Section 15.5
Dilution
15.6
15.7
15.8
Stoichiometry of Solution Reactions
Neutralization Reactions
Solution Composition: Normality
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37
Section 15.6
Stoichiometry of Solution Reactions
Steps for Solving Stoichiometric Problems Involving Solutions
1. Write the balanced equation for the reaction. For
reactions involving ions, it is best to write the net
ionic equation.
2. Calculate the moles of reactants.
3. Determine which reactant is limiting.
4. Calculate the moles of other reactants or
products, as required.
5. Convert to grams or other units, if required.
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38
Section 15.6
Stoichiometry of Solution Reactions
Concept Check (Part I)
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What precipitate will form?
lead(II) phosphate, Pb3(PO4)2

What mass of precipitate will form?
0.91 g Pb3(PO4)2 see next 2 slides
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39
Section 15.6
Stoichiometry of Solution Reactions
Let’s Think About It
•
Where are we going?

•
To find the mass of solid Pb3(PO4)2 formed.
How do we get there?






What are the ions present in the combined solution?
What is the balanced net ionic equation for the
reaction?
What are the moles of reactants present in the
solution?
Which reactant is limiting?
What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?
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Section 15.6
Stoichiometry of Solution Reactions
2PO4-3 + 3Pb2+
Pb3(PO4)2(s)
0.0100L P(0.30 moles P/L)(1 mole LP/2mole P)
(812g LP/mole) = 1.1g LP
0.0200L Lead(0.20 moles Lead/L)(1 mole LP/3mole
L)(812g LP/mole) = 0.91g LP
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41
Section 15.6
Stoichiometry of Solution Reactions
Concept Check (Part II)
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What is the concentration of nitrate ions
left in solution after the reaction is
complete?
0.27 M see next two slides
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42
Section 15.6
Stoichiometry of Solution Reactions
Let’s Think About It
•
Where are we going?

•
To find the concentration of nitrate ions left in
solution after the reaction is complete.
How do we get there?


What are the moles of nitrate ions present in the
combined solution?
What is the total volume of the combined
solution?
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43
Section 15.6
Stoichiometry of Solution Reactions
2NO3- + Pb2+
Pb(NO3)2(aq)
0.0200L LN(0.20 moles LN/L)(2 mole N/1mole LN)=
0.0080 moles Nitrate ion
0.0080moles N/(0.0100L + 0.0200L) = 0.27 M Nitrate
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Section 15.6
Stoichiometry of Solution Reactions
Concept Check (Part III)
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What is the concentration of phosphate
ions left in solution after the reaction is
complete?
0.02 M see next two slides
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45
Section 15.6
Stoichiometry of Solution Reactions
Let’s Think About It
•
Where are we going?

•
To find the concentration of phosphate ions left in
solution after the reaction is complete.
How do we get there?




What are the moles of phosphate ions present in
the solution at the start of the reaction?
How many moles of phosphate ions were used
up in the reaction to make the solid Pb3(PO4)2?
How many moles of phosphate ions are left over
after the reaction is complete?
What is the total volume of the combined
solution?
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46
Section 15.6
Stoichiometry of Solution Reactions
2PO4-3 + 3Pb2+
Pb3(PO4)2(s)
(1.1g LP-0.91g LP)(1mole LP/812g)(2mole P/1mole LP)/
(0.0100L+0.0200L) = 0.02 M Phosphate Ion
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47
Section 15.7
Neutralization Reactions
•
•
•
An acid-base reaction is called a
neutralization reaction.
Steps to solve these problems are the same
as before.
For a strong acid and base reaction:
H+(aq) + OH–(aq)  H2O(l)
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48
Section 15.7
Neutralization Reactions
Concept Check
For the titration of sulfuric acid (H2SO4) with
sodium hydroxide (NaOH), how many moles
of sodium hydroxide would be required to
react with 1.00 L of 0.500 M sulfuric acid?
1.00 mol NaOH see next 2 slides
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49
Section 15.7
Neutralization Reactions
Let’s Think About It
•
Where are we going?

•
To find the moles of NaOH required for the
reaction.
How do we get there?




What are the ions present in the combined
solution? What is the reaction?
What is the balanced net ionic equation for the
reaction?
What are the moles of H+ present in the solution?
How much OH– is required to react with all of the
H+ present?
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50
Section 15.7
Neutralization Reactions
H2SO4 + 2NaOH
Na2SO4 + 2H2O
1.00L SA(0.500 moles SA/L)(2 mole SH/1mole SA)=
1.00 mol NaOH
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51
Section 15.7
Neutralization Reactions
If I use 8.34 ml of a 0.1562 M solution of NaOH to
titrate or neutralize 10.12 ml of a monoprotic acid
(one H), what is the concentration of the acid?
HA + NaOH
Salt + H2O
0.00834 L SH (0.1562mol SH/L)(1 mole HA/1 mole SH)
/0.01012L HA = 0.129 M HA
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52
Section 15.8
Solution Composition: Normality
Unit of Concentration
•
•
•
One equivalent of acid – amount of acid
that furnishes 1 mol of H+ ions.
One equivalent of base – amount of base
that furnishes 1 mol of OH ions
Equivalent weight – mass in grams of 1
equivalent of acid or base.
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Section 15.8
Solution Composition: Normality
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Section 15.8
Solution Composition: Normality
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Section 15.8
Solution Composition: Normality
number of equivalents
equivalents
equiv
Normality = N =
=
=
1 liter of solution
liter
L
•
To find number of equivalents:
equiv
N  V=
 L = equiv
L
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56
Section 15.8
Solution Composition: Normality
Concept Check
If Ba(OH)2 is used as a base, how many
equivalents of Ba(OH)2 are there in 4 mol
Ba(OH)2?
a)
b)
c)
d)
2
4
8
16
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57
Section 15.8
Colligative
Properties (not Normality
in textbook)
Solution Composition:
•
•
•
A property that does not depend on the identity of a
solute in solution
Vary only with the number of solute particles present in
a specific quantity of solvent
4 colligative properties:
– Osmotic pressure
– Vapor pressure lowering
– Boiling point elevation
– Freezing point depression
11-
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58
Section 15.8
Osmotic
Pressure
Solution
Composition: Normality
•
•
•
Osmosis
– A process in which solvent molecules diffuse through a
barrier that does not allow the passage of solute particles
The barrier is called a semipermeable membrane.
– A membrane that allows the passage of some substances but
not others
Osmotic Pressure
– Pressure that can be exerted on the solution to prevent
osmosis
11-
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59
Section 15.8
Osmotic
Pressure
Solution
Composition: Normality
•
Reverse Osmosis- Ultra Filtration
– A process in which pressure is applied to reverse the osmosis
flow.
– A form of molecular filtration used to remove salt from water.
11-
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60
Section 15.8
Solution Composition: Normality
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p. 479
Section 15.8
Vapor
Pressure
Lowering Normality
Solution
Composition:
Solutes come in 2 forms:
• Volatile
– Solutes that readily form a gas
• Nonvolatile
– Solutes that DO NOT
readily form a gas
•
Generally, the addition of a
solute lowers the vapor
pressure of a solution when
compared to the
pure solvent.
11-
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62
Section 15.8
Vapor
Pressure
Lowering Normality
Solution
Composition:
Solutions under
sealed container
to start with.
Solutions under
sealed container
at later time.
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63
Section 15.8
Phase
Diagram
of Water Normality
Solution
Composition:
11-
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64
Section 15.8
Supercritical
Extraction Normality
Solution Composition:
•Removal of caffeine from
coffee using supercritical CO2
or natural effervescence .
•Removal of oils (fats) from
potato chips.
31.1 °C
72.9 atm
•Dry Cleaning Cloths
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65
Section 15.8
Boiling
Point
Elevation
Solution
Composition:
Normality
•
•
•
The addition of solute affects the boiling point
because it affects the vapor pressure.
The boiling point is raised with the addition of
solute in comparison to the pure solvent.
An equation that gives the increase in boiling
point:
∆Tb = Kbm
Where ∆Tb is the increase in temperature from the
pure solvent’s boiling point, Kb is the boiling point
constant, which is characteristic of a particular
solvent, and m is the molality (moles of solute per
kg of solvent)
11-
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66
Section 15.8
Freezing
DepressionNormality
SolutionPoint
Composition:
•
•
The freezing point is lowered with the
addition of solute in comparison to the
pure solvent.
An equation that gives the decrease in
freezing point:
∆Tf = Kfm
Where ∆Tf is the decrease in temperature
from the pure solvent’s freezing point, Kf is
the freezing point constant, which is
characteristic of a particular solvent, and
m is the molality (moles of solute per kg of
solvent)
11-
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67
Section 15.8
Examples
Colligative Properties
SolutionofComposition:
Normality
Antifreeze lowers the
freezing point of your
radiator fluid and raises the
boiling point.
Salting roads melts the ice.
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68
Section 15.8
Example
of Composition:
Freezing Point Depression
Solution
Normality
Making Ice
Cream
You must add
salt so as to
lower the
freezing point of
the ice water
cold enough to
freeze the ice
cream.
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69
Section 15.8
Calculations
of Freezing Point
Lowering and Boiling Point Elevation
Solution Composition:
Normality
What will be the coldest temperature of an ice water
solution with 5.0 lbs of salt in 3.0 gallon container with
about 5.0 Kg of water? What will be the boiling Point?
5.0 lbs NaCl(454g/lb)(1mole NaCl/58 g) = 39 moles NaCl
39 moles NaCl x 2
(two particles or ions are formed) = 78
ΔTf = Kf mol solute/Kg solvent = 1.858 (78)/5.0= 29oC
0.0oC – 29oC = - 29oC freezing point
ΔTb = Kb mol solute/Kg solvent = 0.52 (78)/5.0= 8.1oC
100.00oC + 8.1oC = 108.1oC boiling point
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70
Section 15.8
Colligative
Properties and Normality
Strong Electrolytes
Solution Composition:
•
•
•
•
Strong electrolytes dissociate most of the time into
their constituent ions.
Therefore, the number of particles (in this case ions)
increases with the number of ions.
Colligative properties are proportional to number of
particles in solution.
Example:
MgCl2(s) Mg2+(aq) + 2 Cl-(aq)
In this case the number of particles increases to 3
particles. Therefore, we would multiply the colligative
property amount by 3.
11-
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71
Section 15.8
Practice
– Strong
vs. Weak Electrolytes
Solution
Composition:
Normality
•
Which of the following aqueous solutions is
expected to have the lowest freezing point?
– 0.5 m CH3CH2OH
Forms 1 particle per molecule or formula
unit: 1 x 0.5 m particles = 0.5 m particles
– 0.5 m Ca(NO3)2
Forms 3 particle per molecule or formula
unit: 3 x 0.5 m particles = 1.5 m particles
– 0.5 m KBr
Forms 2 particle per molecule or formula
unit: 2 x 0.5 m particles = 1.0 m particles
Thus Ca(NO3)2 should the greatest freezing point
lowering or the lowest freezing point.
11-
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72