Solutions
Chapter 14
Hein and Arena
Version 2.0
12th Edition
Eugene Passer
Chemistry Department
Bronx Community
1 College
© John Wiley and Sons, Inc
Chapter Outline
14.1 General Properties of
Solutions
14.2 Solubility
14.3 Factors Related to
Solubility
14.4 Rate of Dissolving Solids
14.5 Solutions: A Reaction
Medium
14.6 Concentration of
Solutions
14.7 Colligative Properties
of Solutions
14.8 Osmosis and Osmotic
Pressure
2
14.1
General Properties of
Solutions
3
• A solution is a system in which one or
more substances are homogeneously
mixed or dissolved in another substance.
• The solute is the component that is
dissolved or is the least abundant
component of the solution.
• The solvent is the dissolving agent or the
most abundant component in the solution.
4
5
Properties of True Solutions
6
• A mixture of two or more components–
solute and solvent –is homogeneous and
has a variable composition.
– This means that the ratio of solvent to solute
can be varied.
• The dissolved solute is molecular or ionic
in size.
• It is either colored or colorless and is
usually transparent.
7
• The solute remains uniformly distributed
throughout the solution and will not
settle out with time.
• The solute can generally be separated
from the solvent by purely physical
means such as evaporation.
• The solute particles of a true solution
are molecular or ionic in size.
– The particle size range is 0.1 nm to 1 nm
(10-8 cm to10-7cm).
8
Formation of a Solution
solute
solvent
solution
9
14.2
Solubility
10
Solubility describes the amount of solute that
will dissolve in a specified amount of solvent.
11
Solubilities of substances vary widely.
12
• Terms that describe the extent of
solubility of a solute in a solvent:
– very soluble
– soluble
– moderately soluble
– slightly soluble
– insoluble
13
• Terms that describe the solubility of
liquids:
– miscible: liquids that are capable of mixing
and forming homogeneous solutions.
methyl alcohol and water
– immiscible: liquids that are insoluble in
each other.
oil and water
14
Solubility of Various Common Ions in Cold Water
15
14.2
14.3
Factors Related to
Solubility
16
• Factors that affect solubility are:
– ion size
– interactions between solute and solvent
– temperature
17
The Nature of the
Solute and Solvent
18
• The general rule for predicting solubility
is “like dissolves like”.
19
• Polar compounds tend to be more
soluble in polar solvents than
nonpolar solvents.
Solvent
Polarity
– NaCl (sodium chloride) is
• soluble in water
• slightly soluble in ethyl alcohol
• insoluble in ether and benzene
20
Dissolution of sodium chloride in water.
The hydrated ions slowly
diffuse away from the crystal
to
dissolved
in
Asbecome
the attraction
between
Polar
water molecules are
solution.
the ions weakens,
the ions
+
attracted to Na and Clmove apart and become
ions in the salt or crystal,
surrounded by water
weakening the attraction
dipoles.
between the ions.
21
14.3
• Nonpolar compounds tend to be more
soluble in nonpolar solvents than in
polar solvents.
Solvent
Polarity
– benzene is
• insoluble in water
• soluble in ether
22
The Effect of
Temperature on Solubility
23
Most solutes have a limited solubility in
a specific solvent at a fixed temperature.
24
Effect of Temperature on the
Solubility of a Solid in a Liquid
• For most solids dissolved in a liquid,
an increase in temperature results in
increased solubility.
– Some solids increase in solubility only
slightly with increasing temperature.
– Some solids decrease in solubility with
increasing in temperature.
– Any point on any solubility curve
represents a saturated solution of that
solute
25
large increase in solubility
with temperature
decrease in solubility with
increasing temperature
slight increase in
solubility with
temperature
26
27
Effect of Temperature on the
Solubility of a Gas in Water
• The solubility of a gas in water usually
decreases with temperature.
– Kinetic molecular theory accounts for
this decreased solubility.
– For the gas to dissolve it must form bonds
of some sort with the molecules of the
liquid.
– At higher temperatures the kinetic energy
of the gas molecules is sufficient to break
28
the gas water bond.
large decrease in solubility
with increasing temperature
29
The Effect of
Pressure on Solubility
30
• Small pressure changes have little
effect on the solubility of
– solids in liquids
– liquids in liquids
• Small pressure changes have a great
effect on the solubility of gases in
liquids.
– The solubility of a gas in a liquid is
directly proportional to the pressure of
that gas above the liquid.
31
Effect of Pressure on the Solubility of a Gas
number of gas
particles doubles
solubility of
gas doubles
liquid
32
Saturated, Unsaturated and
Supersaturated Solutions
33
• At a specific temperature there is a
limit to the amount of solvent that will
dissolve in a given amount of solute.
• When this condition occurs the solution
is said to be saturated.
34
• In a saturated solution two processes are
occurring simultaneously
– The solute is crystallizing out of solution.
– The solid is dissolving into the solution.
solute (dissolved)
solute (undissolved)
35
Dissolving a Solid in a Liquid
A saturated solution may
Solution is
is
be Solution
either dilute or
unsaturated.
saturated. depending
concentrated,
Solute dissolves
KCl of the
on No
themore
solubility
can dissolve.
solute.
35
34
KCl
10
15
20
25
30
5 ggKCl
100 g H2O
20oC
36
Dissolving a Solid in a Liquid
A solution that is saturated at one temperature
may not be saturated at another temperature.
Raise temperature
35 g KCl
It may increase or
decrease
depending on the
Solubility
of a solute
solute-solvent
in a solvent
changes
system.
with temperature.
Solute dissolves
The solution is unsaturated
100 g H2O
A saturated solution
may be either dilute
or concentrated.
50oC
20
C
37
Dissolving a Solid in a Liquid
A stress to the system will cause the solute in
Whenever a solution contains solute in excess of
solution
excessA supersaturated
of the saturation
limit istounstable.
come out of
its solubility limit the solution is supersaturated.
solution.
Cool Solution to 20oC
35 g KCl
At 20oC the solubility of KCl
in water is 34 g/100 g H2O.
100 g H2O
20oC
50
The
Nosolution
KCl precipitates.
contains 1
gram of KCl in excess
of the solubility limit.
38
14.4
Rate of
Dissolving Solids
39
• Particle Size
– A solid can dissolve only at the surface
that is in contact with the solvent.
– Smaller crystals have a larger surface to
volume ratio than large crystals.
– Smaller crystals dissolve faster than
larger crystals.
40
41
14.5
• Temperature
– In most cases, the rate of dissolving of a
solid increases with temperature.
– This occurs because solvent molecules
strike the surface of the solid more
frequently, causing the solid to dissolve
more rapidly.
42
• Concentration of the Solution
Δc
Δt
Δc
Δt
As solution
concentration
increases, the rate of
dissolving decreases.
The rate of dissolving
is at a maximum when
solute and solvent are
first mixed.
43
• Agitation or stirring.
– When a solid is first put into water, it
comes in contact only with water. The
rate of dissolving is a maximum.
44
• Agitation or stirring.
– When a solid is first put into water, it
comes in contact only with water. The
rate of dissolving is a maximum.
– As the solid dissolves, the amount of
dissolved solute around the solid
increases and the rate of dissolving
decreases.
45
• Agitation or stirring.
– When a solid is first put into water, it
comes in contact only with water. The
rate of dissolving is then a maximum.
– Stirring
As the solid
distributes
dissolves,
thethedissolved
amount ofsolute
dissolved solute
throughout
the water;
around more
the solid
water is in
increaseswith
contact
and the
theratesolid
of dissolving
causing it to
decreases.
dissolve
more rapidly.
46
14.5
Solutions:
A Reaction Medium
47
Many solids must be put into
solution to undergo appreciable
chemical reactions.
48
sodium chloride reacts
with silver nitrate
49
Ag+
NO-3
Na+
Cl-
NaCl(s) + AgNO3(s) → no reaction
If the reactants are in the solid phase, no
50
reaction occurs.
Ag+
NO-3
Na+
Cl-
NaCl(s) + AgNO3(s) → no reaction
NaCl(s) and AgNO3(s) do not react because their
51
ions are securely locked in their crystal lattices.
NO-3
Na+
Cl-
Ag+
If the reactants are dissolved in water, an
52
immediate reaction occurs.
NO-3
Na+
ClAg+
NaCl(s) + AgNO3(s) → AgCl(s) +NaNO3(aq)
Mobile Ag+ and Cl- ions come into contact and
53
form insoluble AgCl which precipitates.
NO-3
Na+
ClAg+
NaCl(s) + AgNO3(s) → AgCl(s) +NaNO3(aq)
Na+
3
and NO remain in solution.
54
14.6
Concentration
of Solutions
55
The concentration of a solution expresses
the amount of solute dissolved in a given
quantity of solvent or solution.
56
Dilute and
Concentrated Solutions
57
• The terms dilute and concentrated are
qualitative expressions of the amount
of solute present in a solution.
• A dilute solution contains a relatively
small amount of dissolved solute.
• A concentrated solution contains a
relatively large amount of solute.
58
• Concentrated HCl contains 12 mol of
HCl per liter of solution.
• In some laboratories a dilute solution
might be 6 mol of HCl per liter of
solution.
• In another laboratory dilute HCl might
be 4 mol of HCl per liter of solution.
59
Mass Percent Solution
60
Mass percent expresses the concentration
of solution as the percent of solute in a
given mass of solution.
g solute
g solute
mass percent =
x 100 =
x 100
g solute + g solvent
g solution
61
What is the mass percent of sodium hydroxide in a
solution that is made by dissolving 8.00 g NaOH in
50.0 g H2O?
grams of solute (NaOH) = 8.0 g
grams of solvent (H2O) = 50.0 g
g solute


mass percent = 
x 100

 g solute + g solvent 


8.00 g NaOH
 8.00 g NaOH + 50.0 g H O  x 100 = 13.8% NaOH solution

2 
62
What masses of potassium chloride and water are
needed to form 250. g of 5.00% solution?
The percent expresses the mass of the solute.
Dissolving 12.5 g
250.g = total mass of solution
KCl
in
238
g
H
O
2
 g solute 
mass percent = 
x 100 gives a 5.00% KCl

 g solution 
solution.
 mass percent x g solution  = g solute


100


 5.00 x 250. g  = 12.5 KCl solute


100


250.g – 12.5 g = 238 g H2O
63
A 34.0% sulfuric-acid solution has a density of 1.25
g/mL. How many grams of H2SO4 are contained in
1.00 L of this solution?
Step 1. Determine
grams of solution.
Mass
Density =
Volume
Grams of solution are
determined from the
solution density.
Density x Volume = Mass
1.00 L = 1.00 x 103 mL
 1.25 g  (1.00 x 103 mL) = 1250 g (mass of solution)


 1 mL 
64
A 34.0% sulfuric-acid solution has a density of 1.25
g/mL. How many grams of H2SO4 are contained in
1.00 L of this solution?
Solve the mass percent equation for grams of solute.
 g solute 
mass percent = 
x 100

 g solution 
mass percent x g solution
g solute =
100
mass of solution = 1250 g
 34.0 g H 2SO4 1250 g  =
g H 2SO 4 =
425 g H 2SO4
100 g
65
Mass/Volume Percent (m/v)
66
Mass /volume percent expresses the
concentration as grams of solute per
100 ml solution.
g solute
mass/volume percent =
x 100
mL solution
67
A 3.0% H2O2 solution is commonly used as a topical
antiseptic to prevent infection. What volume of this
solution will contain 10. g of H2O2?
Solve the mass/volume equation for grams of solute.
g solute
mass/volume percent =
x 100
mL solution
 g solute 
mL solution = 
x 100

 m/v percent 
mL solution =
10 g H 2O2 
 3.0 m/v percent 
x 100 = 330 mL
68
Volume Percent
69
Solutions that are formulated from
The
liquids
volumearepercent
oftenis the
expressed
volume of
as a
liquid
volume
in 100
percent
ml of solution.
with respect to the
solute.
volume of liquid in question
volume percent =
x 100
total volume of solution
70
Volumes are not necessarily additive.
• A bottle of rubbing alcohol reads 70%
by volume.
• The alcohol solution could be prepared
by mixing 70 mL of alcohol with water
to make a total volume of 100 mL of
solution.
• 30 mL of water could not be added to
70 mL of alcohol because the volumes
are not necessarily additive.
71
Molarity
72
Molarity of a solution is the number of
moles of solute per liter of solution.
number of moles of solute
moles
molarity = M =
=
liter of solution
liter
73
Molarity can be expressed in grams of
solute rather than moles of solute.
number of moles of solute
moles
molarity =
=
liter of solution
liter
g solute
moles solute =
molar mass
 g solute 


moles
molar mass 

molarity =
=
liter of solution
liter
g solute
moles
molarity =
=
molar mass x liter of solution
liter74
Preparation of a 1 molar solution
75
14.7
What is the molarity of a solution containing 1.4 mol
of acetic acid (HC2H3O2) in 250. ml of solution?
It is necessary to convert 250. mL to L since
molarity = mol/L.
mol mol
The conversion is:

=M
mL
L
 1.4 mol   1000 mL  = 5.6 mol = 5.6 M



L
L
 250. mL  

76
How many grams of potassium hydroxide are required
to prepare 600. mL of 0.450 M KOH solution?
Convert: mL  L  mol  g
The data are:
volume = 600 mL
0.450 mol
M=
L
56.11 g
molar mass KOH =
mol
The calculation is:
1 L   0.450 mol   56.11 g KOH 

 600 mL  


 = 15.1 g KOH
L
mol

 1000 mL  

77
Calculate the number of moles of nitric acid in 325 mL
of 16 M HNO3.
Use the equation: moles = liters x M
Convert: mL  L
1L 

(325 mL) 
 = 0.325 L
 1000 mL 
Substitute the data given in the problem and solve:
16 mol HNO3 

moles =  0.325 L  
 = 5.2 mol HNO3
1L


78
What volume of 0.250 M solution can be prepared
from 16.0 g of potassium carbonate?
Convert: g K 2CO3  mol K 2CO3  L solution
The data are:
mass K 2CO3 = 16.0 g
0.250 mol
M=
L
138.2 g
molar mass K 2CO3 =
mol
The conversion is:
 1 mol K 2CO3  

1L
= 0.463 L
16.0 g K2CO3  



 138.2 g K 2CO3   0.250 mol K 2CO3 
79
How many mL of 2.00 M HCL will react with 28.0 g
NaOH?
Step 1 Write and balance the equation for the
reaction.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(aq)
Step 2 Find the number of moles of NaOH in 28.0 g
NaOH.
Convert: g NaOH  mol NaOH
 1 mol 
= 0.700 mol NaOH
 28.0 g NaOH  

 40.0 g 
80
How many mL of 2.00 M HCL will react with 28.0 g
NaOH?
Step 3 Solve for moles and volume of HCl.
Convert: mol NaOH  mol HCl  L HCl  mL HCl
 1 mol HCl   1 L HCl  =
 0.700 mol NaOH  

 0.350 L HCl
 1 mol NaOH   2.00 mol HCl 
 1000 mL  =
0.350 L HCl 
 350 mL HCl
 1L 
81
82
Dilution Problems
83
• If a solution is diluted by adding pure
solvent:
– the volume of the solution increases.
– the number of moles of solute remain the
same.
84
NO-3
Na+
NaNO3 solution
85
NO-3
Na+
Solution volume is doubled.
Solution concentration is halved.
86
Moles of solute remain the same.
Calculate the molarity of a sodium hydroxide solution
that is prepared by mixing 100. mL of 0.20 M NaOH
with 150. mL of water. Assume volumes are additive.
Step 1 Calculate the moles of NaOH in the original
solution.
mol
M=
L
mol = L  M
0.20 mol NaOH 

 0.100 L  
 = 0.020 mol NaOH
1L


87
Calculate the molarity of a sodium hydroxide solution
that is prepared by mixing 100. mL of 0.20 M NaOH
with 150. mL of water. Assume volumes are additive.
Step 2 Solve for the new molarity.
solution volume = 100. mL + 150. mL = 250. mL
0.020 M NaOH
M=
= 0.080 M NaOH
0.250 L
88
Alternative Solution
When the moles of a solute in a solution
before and after dilution are the same, the
moles before and after dilution may be set
equal to each other.
mol1 = mol2
before
dilution
after
dilution
mol1 = L1 × M1
mol 2 = L2 × M 2
L1 × M1 = L2 × M 2
V1 and V2 must
have the same
volume units.
V1 × M1 = V2 × M 2
89
Calculate the molarity of a sodium hydroxide solution
that is prepared by mixing 100. mL of 0.20 M NaOH
with 150. mL of water. Assume volumes are additive.
V1 = 100. mL
V2 = 250. mL
M1 = 0.20 M
M2 = unknown
V1 × M1 = V2 × M 2
(100. mL)(0.20 M) = (250. mL)M2
(100. mL)(0.20 M)
M2 =
= 0.080 M NaOH
250 mL
90
How many grams of silver chloride will be
precipitated by adding sufficient silver nitrate to react
with 1500. mL of 0.400 M barium chloride solution?
2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)
Step 1 Determine the number of moles of BaCl2 in
1500 mL of 0.400 M solution.
mol
M=
L
mol = L  M
1500 mL = 1.500 L
0.400 mol BaCl 2 

1.500 L  
 = 0.600 mol BaCl 2
L


91
How many grams of silver chloride will be
precipitated by adding sufficient silver nitrate to react
with 1500. mL of 0.400 M barium chloride solution?
2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)
Step 2 Use the mole-ratio method to calculate the
moles and grams of AgCl.
Convert: mol BaCl 2  mol AgCl  g AgCl
 2 mol AgCl   143.4 g AgCl 
= 172 g AgCl
 0.600 mol BaCl2 



 1 mol BaCl 2   mol AgCl 
92
14.7
Colligative Properties
of Solutions
93
When a nonvolatile solute is added to a
solvent, three physical properties of the
solvent will change:
Boiling point elevation.
Colligative
Freezing point depression.
Properties
Vapor pressure.
94
Colligative properties of a solution
depend only on the number of solute
particles in a solution and not on the
nature of those particles.
Boiling point elevation.
Colligative
Freezing point depression.
Properties
Vapor pressure.
95
Each solvent shows a characteristic:
freezing point depression constant
boiling point elevation constant
96
The vapor pressure of a liquid depends
on the ease with which its molecules can
escape from the liquid’s surface.
H2O(l) → H2O(g)
97
If 10% of the molecules in a solution are nonvolatile
solute molecules, the vapor pressure of the solution is
10% lower than that of the pure solvent.
H2O(l) → H2O(g)
98
A liquid
solution
boils
willwhen
haveitsa vapor
lower pressure
vapor pressure,
equals
atmospheric
and
consequently
pressure.
a higher boiling point.
Vapor Pressure Curve of
Pure Water and Water
Solution: Boiling Point
Elevation
99
A solution
liquid freezes
will have
when
a lower
its vapor pressure,
pressure
equals
and
consequently
the vapor pressure
a lower freezing
of its solid.
point.
Vapor Pressure Curve of
Pure Water and Water
Solution: Freezing Point
Depression
water vapor
pressure curve
solution vapor
pressure curve
ice vapor
pressure curve
100
Molality
101
The freezing point depression and the
mol solute
boiling point
elevation
are
directly
molality =
solution
proportional to the kg
number
of moles of
solute per kilogram of solvent.
102
•
•
•
•
•
Symbols used in the calculation of
colligative properties
m = molality
Δtf = freezing point depression: oC
Δtb = boiling point elevation: oC
Kf = freezing point depression constant
Kb = boiling point elevation constant
103
What is the molality (m) of a solution prepared by
dissolving 2.70 g CH3OH in 25.0 g H2O?
The conversion is:
2.70 g CH 3OH
mol CH 3OH
mol CH 3OH


25.0 g H 2O
25.0 g H 2O
1 kg H 2O
The molar mass of CH3OH 32.04 g/mol
 2.70 g CH 3OH   1 mol CH 3OH
 25.0 g H O  

  32.04 g CH 3OH
2
  1000 g H 2O 
  1 kg H O 

2

3.37 mol CH 3OH
=
1 kg H 2O
104
A solution is made by dissolving 100. g of ethylene
glycol (C2H6O2) in 200. g H2O. What is the freezing
point of the solution?
Calculate moles of C2H6O2:
 1 mol C2 H 6O2 
= 1.61 mol C2 H 6O2
100 g C2 H 6O2  

 62.07 g C2 H 6O2 
Calculate kilograms of H2O
 1 kg 
200. g H 2O 
= 0.200 kg H 2O

 1000 g 
105
A solution is made by dissolving 100. g of ethylene
glycol (C2H6O2) in 200. g H2O. What is the freezing
point of the solution?
Calculate the freezing point depression.
 1.61 mol C2 H 6O2  1.86 C kg H 2O 
o
=
15.0
C
Δt f = 



 0.200 kg H 2O  1 mol C2 H 2O 2 
o
freezing point solution = freezing point solvent – Δtf.
= 0oC – 15oC = – 15oC
106
A solution made by dissolving 4.71 g of a compound
of unknown molar mass in 100.0 g of H2O has a
freezing point of –1.46oC. What is the molar mass of
the compound?
Δtf = + 1.46oC
Δt f = mK f
Δt f
m=
Kf
1.46oC × mol solute 0.785 mol solute
m=
=
o
1.86 C × kg H 2O
kg H 2O
107
Convert 4.71 g solute/100 g H2O to g/mol.
 4.71 g solute   1000 g H 2O  
1 kg H 2O

 100 g H O   1 kg H O   0.785 mol solute 




2
2
= 60.0 g mol
molar mass of
the compound
108
14.8
Osmosis and
Osmotic Pressure
109
Osmosis is the diffusion of water, either from a
dilute solution or from pure water, through a
semipermeable membrane into a solution of
higher concentration.
110
112
14.9
• A 0.90% solution of sodium chloride in
water is known as physiological saline
solution. It is isotonic with blood
plasma.
• Red blood cells do not shrink or swell
in an isotonic solution.
113
• Red blood cells swell in a hypotonic
solution.
• Red blood cells shrink in a hypertonic
solution.
114
Osmtoic Pressure
• Osmotic pressure is dependent only on the
concentration of the solute particles in a
solution and is independent of their nature.
• The osmotic pressure of a solution can be
measured by applying enough pressure to
stop the flow of water due to osmosis.
• The difference between the applied pressure
and the atmospheric pressure is the osmotic
pressure.
Osmotic pressure is a colligative property.
118
In osmosis,the net transfer of
solvent is always from the more
concentrated to the less
concentrated solution.
water passes
through the
cellophane
119
14.9
120