Natural Approach to Chemistry
Chapter 9
Water & Solutions
9.1 Solutes, Solvents, and Water
9.2 Concentration and Solubility
9.3 Properties of Solutions
1
9.1 Assignments
• 290/1-11 (complete sentences, please),36,4042
2
Special properties of water
•
•
•
•
Cohesive nature
Ability to moderate temperature
Expands upon freezing
Versatile solvent
3
Cohesive nature
There is a strong attraction among water
molecules due to hydrogen bonding
Substances are generally denser
in the solid phase than in the
liquid phase.
Water is different
In ice, hydrogen bonds force water
molecules to align in a crystal
structure where molecules are
farther apart than they are in a
liquid.
4
Moderates temperature
•If ice did not float, ponds would freeze from
the bottom up killing everything inside.
•Water boils at 100oC because hydrogen
bonding keeps the molecules together and they
cannot separate easily
•Called the universal solvent because it
dissolves both ionic and covalent compounds.
5
Water as a solvent
Not chemically bonded
hydration: the process of molecules with any charge separation
to collect water molecules around them.
6
Tap water contains dissolved salts and minerals.
(Tap water will conduct electricity)
Distilled water and deionized water have been
processed to remove dissolved salts and minerals.
Deionization is a
Distillation boils water
specific filtration
to steam which is then
process to remove
condensed back to
all ions.
liquid water
(Won’t conduct
(Won’t conduct
electricity)
electricity)
7
Phases and Chemical Reactions
• Solids – chem rxn occur, but very slowly
• Gases – occur and VERY rapidly
– Low density and high mobility of molecules
– Fire needs oxygen to support burning
8
Chemical reactions in Liquids – occur easily
because of high density and mobility.
9
Reactions in liquids
Life involves many complex chemical reactions
that only occur in aqueous solutions!
A step in the Krebs cycle – this is how energy is
extracted from glucose
10
Not everything dissolves in water. Why not?
In general,
“like” dissolves
“like”
Polar solvents dissolve polar solutes
Nonpolar solvents dissolve nonpolar solutes
11
9.2 Assignments
• 290/76-80
12
9.2 Concentration & Solubility
concentration: the amount of each solute compared to the
total solution.
13
There are several ways to express concentration
mass of solute (g )
concentration (g / L ) 
volume of solution (L )
mass of solute (g )
concentration (%) 
 100
mass of solution (g )
moles of solute (mole)
concentration (molarity , M ) 
volume of solution (L )
molality, m = moles solute
kg solvent
14
Suppose you dissolve 10.0 g of sugar in 90.0 g of
water. What is the mass percent concentration of
sugar in the solution?
Asked:
Given:
The mass percent concentration
10 g of solute (sugar) and 90 g of
solvent (water)
Relationships:
mass of solute
concentration 
 100%
total mass of solution
Solve:
10 g sugar
concentration 
 100%  10% sugar
10  90 g of solution
15
Calculate the molarity of a salt solution made by
adding 6.0 g of NaCl to 100 mL of distilled water.
Asked:
Molarity of solution
Given:
Volume of solution = 100.0 mL,
mass of solute (NaCl) = 6.0 g
Relationships: M = moles
L
Formula mass NaCl = 22.99 + 35.45 = 58.44 g/mole
1,000 mL = 1.0 L, therefore 100 ml = 0.10 L
Moles NaCl = 6.0g NaCl x 1 mole NaCl = 0.103 moles NaCl
58.44 g NaCl
Answer:
M = 0.103 moles = 1.03 M solution of NaCl
0.100 L
16
Calculate the molality, m of a solution containing 350.9
g of NaCl in 750.0 g of water. (Density of water is 1g/L.)
Find the molar mass of NaCl:
22.99 + 35.45 = 58.44 g/mol
Mass to mole conversion:
350.9 g NaCl 1 mole = 6.004 mole NaCl
58.44g
m = moles solute = 6.004 mole NaCl = 8.005 m
kg solvent
0.7500kg
17
What happens when you add 10 g of sugar to
100 mL of water?
10 g
sugar
Water molecules
dissolve sugar
molecules
100 mL
H2O
Conc. (%) = 10 g/110 g
18
What happens when you add 10 g of sugar to
100 mL of water?
But when two sugar molecules find each other,
they will become “undissolved” (solid) again…
… then, they become redissolved in water again.
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What happens when you add 10 g of sugar to 100 mL of water?
Equilibrium
This is an aqueous equilibrium!
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Equilibrium
dissolving
“undissolving”
saturation: situation that occurs when
the amount of dissolved solute in a
solution gets high enough that the rate
of “undissolving” matches the rate of
dissolving.
21
Temperature and solubility
20oC
30oC
210 g
sugar
100 mL
H2O
Undissolved
sugar
210 g
sugar
100 mL
H2O
All the
sugar is
dissolved
Temperature has an effect on solubility
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solubility: the amount of a solute that
will dissolve in a particular solvent at a
particular temperature and pressure.
23
Temperature and solubility
Temperature does not have the same
effect on the solubility of all solutes
24
Temperature affects:
- the solubility of solutes
how much
- the rate of solubility
how fast
25
Dissolving is a collision process
Slow (cold) molecules are not as effective as fast
(hot) molecules
Salt dissolves faster in hot water
26
The rate of solubility increases:
- with an increase in temperature
- with an increase in surface area of the solute
At higher temperatures:
- solid solutes (like salt and sugar) are more
soluble
- gases are less soluble
27
Seltzer water is a
supersaturated solution
of CO2 in water
This solution is unstable, and
the gas “undissolves” rapidly
(bubbles escaping)
supersaturation: term used to describe
when a solution contains more
dissolved solute than it can hold.
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Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
M, molarity = moles solute / liter solution
Molar mass of CaCl2
 40.078   2  35.43 
1.Determine the formula mass
of the solute.
 110.98 g / mole
29
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
1. Determine the formula mass of the solute.
Molar mass of CaCl2:
110.98 g/mole
1.0 M 
1.0 mole 0.5 mole

1.0 L
0.5 L
2. Use the formula mass of the solute
to determine the grams of solute
needed.
We need 0.5 moles CaCl2
30
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
Molar mass of CaCl2:1.
2.
110.98 g/mole
1.0 M 
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the
grams of solute needed.
1.0 mole 0.5 mole

1.0 L
0.5 L
We need 0.5 moles CaCl2
0.5 moles 
110.98 g
1 mole
 55.49 g CaCl 2
We need 55.49 g CaCl2
31
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
1. Determine the formula mass of the solute.
2. Use the formula mass of the solute to determine the
grams of solute needed.
3. Weigh the grams of solute on the balance.
32
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
1. Determine the formula mass of the solute.
2. Use the formula mass of the solute to determine the
grams of solute needed.
3. Weigh the grams of solute on the balance.
4. Add the solute to a volumetric flask or graduated
cylinder.
33
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
1. Determine the formula mass of the solute.
2. Use the formula mass of the solute to determine the
grams of solute needed.
3. Weigh the grams of solute on the balance.
500.0 mL
mark
Do not fill all
the way up
4. Add the solute to a volumetric flask or graduated
cylinder.
5. Fill the flask about two thirds of the way up with
distilled water.
34
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
1. Determine the formula mass of the solute.
2. Use the formula mass of the solute to determine the
grams of solute needed.
3. Weigh the grams of solute on the balance.
4. Add the solute to a volumetric flask or graduated
cylinder.
5. Fill the flask about two thirds of the way up with
distilled water.
6. Mix the solution until the solid dissolves completely.
35
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
1. Determine the formula mass of the solute.
2. Use the formula mass of the solute to determine the
grams of solute needed.
3. Weigh the grams of solute on the balance.
4. Add the solute to a volumetric flask or graduated
cylinder.
5. Fill the flask about two thirds of the way up with
distilled water.
6. Mix the solution until the solid dissolves completely.
7. Fill the volumetric flask or graduated cylinder up to the
correct volume marker.
36
Ways to express concentration:
mass of solute (g )
concentration (g / L ) 
volume of solution (L )
mass of solute (g )
concentration (%) 
 100
mass of solution (g )
moles of solute (mole )
concentration (molarity , M ) 
volume of solution (L )
Molality, m = moles solute
kg solvent
A higher temperature causes higher:
- solubility of solutes
how much
- rates of solubility
how fast
37
9.3 Assignments
• 290/84-87.
38
9.3 Properties of Solutions
Reaction rate is generally dependent upon
concentration – greater concentration means
reaction occurs faster
Heat of solution – energy absorbed or released
when a solute dissolves in a particular solvent
exothermic, loss of energy (gives off energy)
or negative heat of solution (feels hot)
endothermic, energy absorbed (feels cold) or
positive heat of solution
39
Exothermic – energy lost
Endothermic – energy gained
40
Heat loss must equal heat gained: net change is zero
The energy
inside the
system is
constant
41
What changes is the enthalpy
NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = +25.7 kJ/mole
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
∆H = –56 kJ/mole
enthalpy: the energy potential of a chemical
reaction measured in joule per mole (J/mole)
or kilojoules per mole (kJ/mole).
42
Enthalpy
“∆” means
“change”
Endothermic reaction
NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = +25.7 kJ/mole
Positive value
Exothermic reaction
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
∆H = –56 kJ/mole
Negative value
43
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
∆H = –56 kJ/mole
Heat released by
the reaction
=
Heat gained by
the solution
The energy inside the
system is constant
44
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
∆H = –56 kJ/mole
Heat released by
the reaction
∆Hreaction
= –56 kJ/mole
=
Heat gained by
the solution
∆Hsolution
= +56 kJ/mole
Opposite signs!
45
When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0
M HCl in a coffee cup calorimeter, the final temperature of the
mixture rises from 22.0oC to 27oC. Calculate the enthalpy change
for the reaction of hydrochloric acid (HCl) with sodium hydroxide
(NaOH). Assume that the coffee cup calorimeter loses negligible
heat, that the density of the solution is that of pure water (1.0
g/mL), and that the specific heat of the solution is the same as that
of pure water.
Break down the problem!
- Experimental setup Given: 40.0 mL of NaOH (1.0 M)
+ 40.0 mL of HCl (1.0 M)
NaOH + HCl  NaCl + H2O
Tinitial = 22.0oC and Tfinal = 27oC
46
When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0
M HCl in a coffee cup calorimeter, the final temperature of the
mixture rises from 22.0oC to 27oC. Calculate the enthalpy change
for the reaction of hydrochloric acid (HCl) with sodium hydroxide
(NaOH). Assume that the coffee cup calorimeter loses negligible
heat, that the density of the solution is that of pure water (1.0
g/mL), and that the specific heat of the solution is the same as that
of pure water.
Break down the problem!
- Experimental setup
- What is asked
Asked:
Amount of heat change (DH) for
NaOH and HCl reaction
47
When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0
M HCl in a coffee cup calorimeter, the final temperature of the
mixture rises from 22.0oC to 27oC. Calculate the enthalpy change
for the reaction of hydrochloric acid (HCl) with sodium hydroxide
(NaOH). Assume that the coffee cup calorimeter loses negligible
heat, that the density of the solution is that of pure water (1.0
g/mL), and that the specific heat of the solution is the same as that
of pure water.
Break down the problem! Given:
- Experimental setup
- What is asked
- Assumptions
Isolated system:
∆Hreaction = ∆Hsolution
Density (H2O) = 1.0 g/mL
Cp  solution   Cp water   4.18
J
g  oC
48
Relationships: qsolution  msolution  Cp solution  DT
Solve:
First note that the temperature increased, so the
reaction released energy to the solution. This means
the reaction is exothermic and will have a negative
DH.
Total volume of solution is 40.0 mL + 40.0 mL = 80.0
mL
Total mass of solution is 80.0 g using the densitywater
(1.0 g/mL).
qsolution  80.0 g  4.18 J  g  oC    27.0 oC  22.0 oC 
qsolution   1,672 J   1.67 kJ
49
Relationships:
Solve:
First note that the temperature increased, so the
reaction released energy to the solution. This means
the reaction is exothermic and will have a negative DH.
Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mL
Total mass of solution is 80.0 g using the densitywater
(1.0 g/mL).
qsolution  msolution  Cp solution  DT
qsolution  80.0 g  4.18 J  g  oC    27.0 oC  22.0 oC 
qsolution   1,672 J   1.67 kJ
The positive sign indicates heat is absorbed.
We reverse the sign as heat gained by the solution is
lost by the reaction. Therefore qrxn = –1.67 kJ.
50
Solve:
qrxn = –1.67 kJ
To find heat on a per mole basis, we use the
molarity times the volume in liters to calculate
moles; 1.0 M x 0.040 L = 0.040 moles of both
reactants (where NaOH and HCl are in
equimolar amounts).
51
Solve:
qrxn = –1.67 kJ
To find heat on a per mole basis, we use the molarity
times the volume in liters to calculate moles; 1.0 M x
0.040 L = 0.040 moles of both reactants (where NaOH
and HCl are in equimolar amounts).
Lastly,
Since the temperature increased, heat was released
form the reaction, making DH negative.
1.67 kJ
  41.8 kJ mole  DH
0.040 moles
Answer:
DH = –41.8 kJ/mole
52
What we have seen so far…
Reaction rates increase with:
increasing concentrations
increasing temperatures
53
Solution vs. pure solvent
Salt dissociates into ions,
which fit in between water
molecules
20 g salt
80 mL
water
87 mL
solution!
Volumes of solute and
solvent do not add up
to the volume of solution
54
Why does ice melt
when salt is
sprinkled on it?
55
Freezing point depression
Why does ice
melt when salt is
sprinkled on it?
Pure water freezes
at 0oC, but a water
and salt solution
freezes at a lower
temperature.
56
Freezing point depression
Entropy
Order
more less
less more
Pure solvent
Solid formation is not hindered
Solution
Solute particles “get in
the way” of solid
formation
The freezing point is lowered in the presence of salt
57
more
less
Pure solvent
Entropy
Order
Solid formation is not hindered
Solution
Freezing point
depression is a
colligative property
Solute particles “get in the way” of
solid formation
less
more
colligative property: physical property of a solution that
depends only on the number of dissolved solute particles not
on the type (or nature) of the particle itself.
58
To calculate the freezing point of a solution:
Freezing point
depression constant
D Tf = Kf x m
Change in freezing
Point, oC
molality
Do not get
confused with
molarity, M
(moles solute / L of
solution)
59
Calculate the freezing point of a 1.8 m aqueous solution
of antifreeze that contains ethylene glycol (C2H6O2) as
the solute.
Asked:
Given:
The freezing point of a 1.8 m solution of ethylene glycol
molality, m = 1.8 m; Kf = 1.86oC/m (Kf , freezing point
depression for water, the solvent)
Relationships:
DTf  Kf  m
Solve:
DTf  Kf  m  1.86o C m  1.8 m  3.35 oC
Freezing point of antifreeze solution  0 oC  3.35 oC   3.35 oC
Answer:
The freezing point is lowered by 3.35oC.
60
Electrolyte solutions
Aqueous solutions containing dissolved ions are able to conduct
electricity
1 mole of solute → 2 moles of ions
1 mole of solute → 3 moles of ions
electrolyte: solute capable of conducting
electricity when dissolved in an aqueous
solution.
61
Electrolyte solutions
Aqueous solutions containing dissolved ions are able to conduct electricity
1 mole of solute → 2 moles of ions
1 mole of solute → 3 moles of ions
The greater the number of particles in solution, the greater the effects.
colligative property: physical property of a
solution that depends only on the number of
dissolved solute particles not on the type (or
nature) of the particle itself.
62
Reaction rates increase with:
increasing concentrations
increasing temperatures
Chemical reactions are accompanied by changes in
enthalpy, ΔH q
q
m
C
 DT
reaction
DHreaction 
solution
solution
p, solution
qreaction
# of moles
Solution vs. pure solvent
density (solution) > density (pure solvent)
colligative properties: freezing point depression is an
example
63