Ch. 13 & 14 - Solutions
I
I. The Nature of Solutions
II
III
(p. 401 - 410, 425 - 433)
C. Johannesson
A. Definitions
 Solution
- homogeneous mixture
Solute - substance
being dissolved
Solvent - present in
greater amount
C. Johannesson
A. Definitions
Solute - KMnO4
C. Johannesson
Solvent - H2O
C. Johannesson
Terms
Suspensions – a mixture when particles in a solvent are so large that
they settle out unless constantly stirred or agitated. (Ex: muddy water)
Colloids – when particles that are intermediate in size between those in
solutions and suspensions form mixtures
C. Johannesson
B. Solvation
 Solvation
– the process of dissolving
solute particles are surrounded by
solvent particles
solute particles are separated and
pulled into solution
C. Johannesson
B. Solvation
-
-
+
sugar
-
+
+
salt
acetic acid
NonElectrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules
only
solute exists as
ions and
molecules
solute exists as
ions only
View animation online.
C. Johannesson
DISSOCIATION
IONIZATION
B. Solvation
 Dissociation
• separation of an
ionic solid into
aqueous ions
NaCl(s)  Na+(aq) + Cl–(aq)
C. Johannesson
B. Solvation
 Ionization
• breaking apart
of some polar
molecules into
aqueous ions
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
C. Johannesson
B. Solvation
 Molecular
Solvation
• molecules
stay intact
C6H12O6(s)  C6H12O6(aq)
C. Johannesson
B. Solvation
“Like Dissolves Like”
NONPOLAR
POLAR
NONPOLAR
POLAR
C. Johannesson
B. Solvation
 Soap/Detergent
• polar “head” with long nonpolar “tail”
• dissolves nonpolar grease in polar water
C. Johannesson
C. Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
concentration
C. Johannesson
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
C. Solubility
 Solubility
• maximum grams of solute that will
dissolve in 100 g of solvent at a given
temperature
• varies with temp
• based on a saturated soln
C. Johannesson
C. Solubility
 Solubility
Curve
• shows the
dependence of
solubility on
temperature
C. Johannesson
C. Solubility
 Solids
are more soluble at...
• high temperatures.

Gases are more soluble at...
• low temperatures &
• high pressures
(Henry’s Law).
• EX: nitrogen narcosis,
the “bends,” soda
C. Johannesson
Ch. 13 & 14 - Solutions
I
II. Concentration
II
III
(p. 412 - 418)
C. Johannesson
A. Concentration
 The
amount of solute in a solution.
 Describing
Concentration
• % by mass - medicated creams
• % by volume - rubbing alcohol
• ppm, ppb - water contaminants
• molarity - used by chemists
• molality - used by chemists
C. Johannesson
A. Concentration
SAWS Water Quality
Report - June 2000
C. Johannesson
B. Molality
moles of solute
molality (m) 
kg of solvent
0.25 mol
0.25m 
1 kg
mass of solvent only
1 kg water = 1 L water
C. Johannesson
B. Molality
 Find
the molality of a solution containing
75 g of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2 0.25 kg water
mol
m
kg
= 3.2m MgCl2
C. Johannesson
B. Molality
 How
many grams of NaCl are req’d to make
a 1.54m solution using 0.500 kg of water?
0.500 kg water 1.54 mol NaCl
1 kg water
1.5 mol
1.5m 
1 kg
58.44 g NaCl
1 mol NaCl
= 45.0 g NaCl
C. Johannesson
B. Molarity
moles of solute
molarity (M) 
L of solution
0.25 mol
0.25M 
1L
L of solution
C. Johannesson
Practice
 You
have 0.8 L of a 0.5 M HCl solution.
How many moles of HCl does this
solution contain?
C. Johannesson
Practice
 What
volume of 3.00 M NaCl is needed
for a reaction that requires 146.3g of
NaCl?
C. Johannesson
C. Dilution
 Preparation
of a desired solution by
adding water to a concentrate.
 Moles of solute remain the same.
M1V1  M 2V2
C. Johannesson
C. Dilution
 What
volume of 15.8M HNO3 is required
to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
C. Johannesson
D. Preparing Solutions

1.54m NaCl in
0.500 kg of water
• mass 45.0 g of NaCl
• add 0.500 kg of water
 500
mL of 1.54M NaCl
• mass 45.0 g of NaCl
• add water until total
volume is 500 mL
500 mL
water
500 mL
mark
45.0 g
NaCl
C. Johannesson
500 mL
volumetric
flask
D. Preparing Solutions
Copyright © 1995-1996 NT Curriculum Project, UW-Madison
(above: “Filling the volumetric flask”)
C. Johannesson
D. Preparing Solutions
Copyright © 1995-1996 NT Curriculum Project, UW-Madison
(above: “Using your hand as a stopper”)
C. Johannesson
D. Preparing Solutions

250 mL of 6.0M HNO3
by dilution
95 mL of
15.8M HNO3
• measure 95 mL
of 15.8M HNO3
• combine with water until
total volume is 250 mL
250 mL
mark
• Safety: “Do as you
oughtta, add the acid to
the watta!”
water
for
safety
C. Johannesson
Solution Preparation Lab

Turn in one paper per team.

Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution.

For each of the following solutions:
1) 1 L of 0.21M NaCl
2) 0.65m NaCl in 200.0 mL of water
3) 100.0 mL of 3.0M HCl from 12.1M concentrate.
C. Johannesson
Ch. 13 & 14 - Solutions
I
III. Colligative Properties
II
III
(p. 436 - 446)
C. Johannesson
A. Definition
 Colligative
Property
• property that depends on the
concentration of solute particles, not
their identity
C. Johannesson
B. Types
 Freezing
Point Depression (tf)
• f.p. of a solution is lower than f.p. of
the pure solvent
 Boiling
Point Elevation (tb)
• b.p. of a solution is higher than b.p. of
the pure solvent
C. Johannesson
B. Types
Freezing Point Depression
View Flash animation.
C. Johannesson
B. Types
Boiling Point Elevation
Solute particles weaken IMF in the solvent.
C. Johannesson
B. Types
 Applications
• salting icy roads
• making ice cream
• antifreeze
• cars (-64°C to 136°C)
• fish & insects
C. Johannesson
C. Calculations
t = k · m · n
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n: # of particles
C. Johannesson
C. Calculations

# of Particles
• Nonelectrolytes (covalent)
• remain intact when dissolved
• 1 particle
• Electrolytes (ionic)
• dissociate into ions when dissolved
• 2 or more particles
C. Johannesson
C. Calculations

At what temperature will a solution that is
composed of 0.73 moles of glucose in 225 g of
phenol boil?
GIVEN:
WORK:
b.p. = ?
m = 0.73mol ÷ 0.225kg
tb = ?
tb = (3.60°C·kg/mol)(3.2m)(1)
kb = 3.60°C·kg/mol
tb = 12°C
m = 3.2m
b.p. = 181.8°C + 12°C
n=1
b.p. = 194°C
tb = kb · m · n
C. Johannesson
C. Calculations

Find the freezing point of a saturated solution of
NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C·kg/mol)(4.8m)(2)
m = 4.8m
n=2
tf = kf · m · n
f.p. = 0.00°C - 18°C
tf = 18°C
f.p. = -18°C
C. Johannesson
Ch. 15 & 16 - Acids &
Bases
I. Introduction to
Acids & Bases
I
II
(p. 453 - 473)
III
C. Johannesson
A. Properties

electrolytes





sour taste
turn litmus red
react with metals
to form H2 gas
vinegar, milk, soda,
apples, citrus fruits
C. Johannesson
 electrolytes

bitter taste
turn litmus blue

slippery feel

ammonia, lye,
antacid, baking soda
ChemASAP
B. Definitions
 Arrhenius
- In aqueous solution…
• Acids form hydronium ions (H3O+)
HCl + H2O 
+
H3O
H
H
Cl
acid
C. Johannesson
O
H
H
–
+
O
H
+
Cl
H
–
Cl
B. Definitions
 Arrhenius
- In aqueous solution…
• Bases form hydroxide ions (OH-)
NH3 + H2O 
+
NH4
H
H
H
N
H
base
C. Johannesson
O
H
H
–
+
O
N
H
+
OH
H
H
H
B. Definitions
 Brønsted-Lowry
• Acids are proton (H+) donors.
• Bases are proton (H+) acceptors.
HCl + H2O 
acid
–
Cl
+
+
H3O
base
conjugate base
C. Johannesson
conjugate acid
B. Definitions
H2O + HNO3  H3O+ + NO3–
B
C. Johannesson
A
CA
CB
B. Definitions
NH3 + H2O 
B
A
 Amphoteric
C. Johannesson
+
NH4
CA
+
OH
CB
- can be an acid or a base.
B. Definitions

Give the conjugate base for each of the following:
HF
F
H3PO4
H2PO4
+
H3O
H2O
 Polyprotic
C. Johannesson
-
- an acid with more than one H+
B. Definitions

Give the conjugate acid for each of the following:
Br
C. Johannesson
-
HBr
HSO4
H2SO4
2CO3
HCO3
B. Definitions
 Lewis
• Acids are electron pair acceptors.
• Bases are electron pair donors.
Lewis
base
C. Johannesson
Lewis
acid
C. Strength
 Strong
Acid/Base
• 100% ionized in water
• strong electrolyte
HCl
HNO3
H2SO4
HBr
HI
HClO4
C. Johannesson
-
+
NaOH
KOH
Ca(OH)2
Ba(OH)2
C. Strength
 Weak
- +
Acid/Base
• does not ionize completely
• weak electrolyte
HF
CH3COOH
H3PO4
H2CO3
HCN
C. Johannesson
NH3
Ch. 15 & 16 - Acids &
Bases
I
II. pH
II
III
(p. 481 - 491)
C. Johannesson
A. Ionization of Water
H 2O + H 2 O
Kw =
H3
+
[H3O ][OH ]
C. Johannesson
+
O
+
= 1.0 
OH
-14
10
A. Ionization of Water
 Find
the hydroxide ion concentration of
3.0  10-2 M HCl.
[H3O+][OH-] = 1.0  10-14
[3.0  10-2][OH-] = 1.0  10-14
[OH-] = 3.3  10-13 M
Acidic orAcidic
basic?
C. Johannesson
B. pH Scale
14
0
7
INCREASING
ACIDITY
pH =
NEUTRAL
+
-log[H3O ]
pouvoir hydrogène (Fr.)
“hydrogen power”
C. Johannesson
INCREASING
BASICITY
B. pH Scale
pH of Common Substances
C. Johannesson
B. pH Scale
pH =
+
-log[H3O ]
pOH =
-log[OH ]
pH + pOH = 14
C. Johannesson
B. pH Scale
 What
is the pH of 0.050 M nitric acid
(HNO3)?
pH = -log[H3O+]
pH = -log[0.050]
pH = 1.3
Acidic orAcidic
basic?
C. Johannesson
B. pH Scale
 What
is the molarity of hydrobromic acid
(HBr) in a solution that has a pOH of 9.6?
pH + pOH = 14
pH = -log[H3O+]
pH + 9.6 = 14
4.4 = -log[H3O+]
pH = 4.4
10-4.4 = [H3O+]
Acidic
[H3O+] = 4.0  10-5 M HBr
C. Johannesson
Ch. 15 & 16 - Acids &
Bases
I
III. Titration
II
III
(p. 493 - 503)
C. Johannesson
A. Neutralization
 Chemical
reaction between an acid and
a base.
 Products are a salt (ionic compound)
and water.
C. Johannesson
A. Neutralization
ACID + BASE  SALT + WATER
HCl + NaOH  NaCl + H2O
strong
strong
neutral
HC2H3O2 + NaOH  NaC2H3O2 + H2O
weak
strong
basic
• Salts can be neutral, acidic, or basic.
• Neutralization does not mean pH = 7.
C. Johannesson
B. Titration
standard solution
 Titration
• Analytical method
in which a standard
solution is used to
determine the
concentration of an
unknown solution.
unknown solution
C. Johannesson
B. Titration
 Equivalence
point (endpoint)
• Point at which equal
amounts of H3O+ and OHhave been added.
• Determined by…
• indicator color change
• dramatic change in pH
C. Johannesson
B. Titration
+
O
moles H3 = moles
MVn = MVn
OH
M: Molarity
V: volume
n: # of H+ ions in the acid
or OH- ions in the base
C. Johannesson
B. Titration
 42.5
mL of 1.3M KOH are required to
neutralize 50.0 mL of H2SO4. Find the
molarity of H2SO4.
H3O+
OH-
M=?
M = 1.3M
V = 50.0 mL
n=2
V = 42.5 mL
n=1
C. Johannesson
MV# = MV#
M(50.0mL)(2)
=(1.3M)(42.5mL)(1)
M = 0.55M H2SO4