CHAPTER 9
Water and
Solutions
9.2 Concentration
and Solubility
Concentration
concentration: the amount of each solute compared to
the total solution.
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9.2 Concentration and Solubility
Concentration
More solute
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Less solute
9.2 Concentration and Solubility
Concentration
More solute
Less solute
How can we express concentration quantitatively
(with numbers)?
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9.2 Concentration and Solubility
Concentration
In a healthy person, potassium is dissolved in
blood at a concentration of 140 to 200 mg/L.
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9.2 Concentration and Solubility
Concentration
In a healthy person, potassium is dissolved in
blood at a concentration of 140 to 200 mg/L.
The concentration is expressed as
the mass of potassium per volume unit of blood
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9.2 Concentration and Solubility
Concentration
In a healthy person, potassium is dissolved in
blood at a concentration of 140 to 200 mg/L.
If the concentration is less than 130 mg/L:
 muscle weakness and heart rhythm
instability (hypokalemia)
The concentration is expressed as
the mass of potassium per volume unit of blood
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9.2 Concentration and Solubility
Concentration
In a healthy person, potassium is dissolved in
blood at a concentration of 140 to 200 mg/L.
If the concentration is less than 130 mg/L:
 muscle weakness and heart rhythm
instability (hypokalemia)
If the concentration is higher than 215 mg/L:
 heart instability (hyperkalemia)
The concentration is expressed as
the mass of potassium per volume unit of blood
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9.2 Concentration and Solubility
Concentration
There are several ways to express concentration
mass of solute (g )
concentration (g / L ) 
volume of solution (L )
concentration (%) 
mass of solute (g )
 100
mass of solution (g )
moles of solute (mole )
concentration (molarity , M ) 
volume of solution (L )
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9.2 Concentration and Solubility
Concentration
There are several ways to express concentration
mass of solute (g )
concentration (g / L ) 
volume of solution (L )
concentration (%) 
mass of solute (g )
 100
mass of solution (g )
moles of solute (mole )
concentration (molarity , M ) 
volume of solution (L )
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9.2 Concentration and Solubility
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?
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9.2 Concentration and Solubility
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?
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Asked:
The mass percent concentration
Given:
10 g of solute (sugar) and 90 g of solvent (water)
Relationships:
concentration 
mass of solute
 100%
total mass of solution
9.2 Concentration and Solubility
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?
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Asked:
The mass percent concentration
Given:
10 g of solute (sugar) and 90 g of solvent (water)
Relationships:
concentration 
mass of solute
 100%
total mass of solution
Solve:
concentration 
10 g sugar
 100%  10% sugar
10  90 g of solution
9.2 Concentration and Solubility
Concentration
There are several ways to express concentration
mass of solute (g )
concentration (g / L ) 
volume of solution (L )
concentration (%) 
mass of solute (g )
 100
mass of solution (g )
moles of solute (mole )
concentration (molarity , M ) 
volume of solution (L )
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9.2 Concentration and Solubility
Concentration
Calculate the molarity of a salt solution made by adding 6.0 g of NaCl to
100 mL of distilled water.
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9.2 Concentration and Solubility
Concentration
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
moles
M
L
Relationships:
Formula mass of NaCl  22.99  35.45  58.44 g / mole
1,000 mL  1.0 L, therefore 100 mL  0.10 L
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9.2 Concentration and Solubility
Concentration
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
moles
M
L
Relationships:
Formula mass of NaCl  22.99  35.45  58.44 g / mole
1,000 mL  1.0 L, therefore 100 mL  0.10 L
Solve:
moles NaCl  6.0 g NaCl 
M
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1 mole NaCl
 0.103 moles NaCl
58.44 g NaCl
0.103 moles
 1.03 M
0.100 L
9.2 Concentration and Solubility
Concentration
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
moles
M
L
Relationships:
Formula mass of NaCl  22.99  35.45  58.44 g / mole
1,000 mL  1.0 L, therefore 100 mL  0.10 L
Solve:
moles NaCl  6.0 g NaCl 
M
Answer:
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1 mole NaCl
 0.103 moles NaCl
58.44 g NaCl
0.103 moles
 1.03 M
0.100 L
1.03 M solution of NaCl
9.2 Concentration and Solubility
Solubility
What happens when you add 10 g of sugar to 100 mL of water?
10 g
sugar
100 mL
H2O
Conc. (%) = 10 g/110 g
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9.2 Concentration and Solubility
Solubility
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
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9.2 Concentration and Solubility
Solubility
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…
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9.2 Concentration and Solubility
Solubility
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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9.2 Concentration and Solubility
Solubility
What happens when you add 10 g of sugar to 100 mL of water?
Equilibrium
This is an aqueous equilibrium!
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9.2 Concentration and Solubility
Solubility
Concentration
low
Equilibrium
“undissolving”
dissolving
10 g
sugar
100 mL
H 2O
20oC
Conc. (%) = 10 g/110 g
high
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9.2 Concentration and Solubility
Solubility
Concentration
low
Equilibrium
dissolving
“undissolving”
high
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9.2 Concentration and Solubility
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.
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9.2 Concentration and Solubility
Solubility
204 g
sugar
100 mL
H 2O
20oC
Conc. = 204 g/100 mL
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.
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9.2 Concentration and Solubility
Solubility
low
Concentration
250 g
sugar
100 mL
H2O
20oC
Undissolved
sugar
Conc. = 250 g/100 mL
Equilibrium
high
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dissolving
“undissolving”
9.2 Concentration and Solubility
Temperature and solubility
20oC
30oC
210 g
sugar
100 mL
H2O
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Undissolved
sugar
210 g
sugar
100 mL
H2O
All the sugar
is dissolved
9.2 Concentration and Solubility
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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9.2 Concentration and Solubility
solubility: the amount of a solute that will dissolve in a
particular solvent at a particular temperature and pressure.
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9.2 Concentration and Solubility
Temperature and solubility
You can dissolve (a lot) more
sugar at higher temperatures
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9.2 Concentration and Solubility
Temperature and solubility
Sugar becomes “undissolved” (solid)
as the temperature goes down
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9.2 Concentration and Solubility
Temperature and solubility
Temperature does not have the same effect on
the solubility of all solutes
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9.2 Concentration and Solubility
Temperature and solubility
For some solutes, solubility changes a lot with temperature
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9.2 Concentration and Solubility
Temperature and solubility
For other solutes, solubility changes very little with temperature
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9.2 Concentration and Solubility
Temperature affects:
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- the solubility of solutes
how much
- the rate of solubility
how fast
9.2 Concentration and Solubility
Dissolving rate
Dissolving is a collision process
Slow (cold) molecules are not as effective as fast (hot) molecules
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9.2 Concentration and Solubility
Dissolving rate
Dissolving is a collision process
Slow (cold) molecules are not as effective as fast (hot) molecules
Salt dissolves faster in hot water
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9.2 Concentration and Solubility
Dissolving rate
Substances are often ground up into
powder to make them dissolve faster
A 1 cm cube has a
surface area of 6 cm2
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The same volume has a
surface area of 9 cm2 when
divided up into smaller cubes
9.2 Concentration and Solubility
The rate of solubility increases:
- with an increase in temperature
- with an increase in surface area of the solute
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9.2 Concentration and Solubility
Gases are soluble in liquids
Dissolved O2 and CO2 allow
animal and plant life to exist
under water
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9.2 Concentration and Solubility
At higher temperatures:
- solid solutes (like salt and
sugar) are more soluble
- gases are less soluble
Solubility of common gases in water at 25oC
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9.2 Concentration and Solubility
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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9.2 Concentration and Solubility
Preparing a solution
How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution.
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9.2 Concentration and Solubility
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
 40.078   2  35.43 
 110.98 g / mole
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9.2 Concentration and Solubility
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 
2. 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
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9.2 Concentration and Solubility
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 
2. 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
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9.2 Concentration and Solubility
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.
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9.2 Concentration and Solubility
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.
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9.2 Concentration and Solubility
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
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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.
9.2 Concentration and Solubility
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.
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9.2 Concentration and Solubility
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.
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9.2 Concentration and Solubility
Ways to express concentration:
concentration (g / L ) 
concentration (%) 
concentration (molarity , M ) 
‹#›
mass of solute (g )
volume of solution (L )
mass of solute (g )
 100
mass of solution (g )
moles of solute (mole )
volume of solution (L )
9.2 Concentration and Solubility
Ways to express concentration:
concentration (g / L ) 
concentration (%) 
concentration (molarity , M ) 
mass of solute (g )
volume of solution (L )
mass of solute (g )
 100
mass of solution (g )
moles of solute (mole )
volume of solution (L )
A higher temperature causes higher:
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- solubility of solutes
how much
- rates of solubility
how fast
9.2 Concentration and Solubility