Mass-related concentration units
Example: 5.00 g of NaCl are dissolved in 50.0 g of water.
What is the concentration of this solution in mass
percent?
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
mass percent =
mass percent -- mass of solute per mass of solution
(expressed as a percentage)
mass percent =
( % m/m )
mass of solute
total mass of solution
g of solute
g solution
x 100
mass of solute = 5.00 g
mass of solution = mass of solute + mass of solvent
x 100
= 5.00 g + 50.0 g
= 55.0 g
mass percent =
5.00 g
55.0 g
x 100 = 9.09% (m/m)
Mass-related concentration units
Mass-related concentration units
mass percent -- unit of solute per 100 units of solution
mass percent -- unit of solute per 100 units of solution
mass percent =
( % m/m )
mass of solute
total mass of solution
x 100
Alternative concentration units for very dilute solutions:
mass of solute
total mass of solution
mass of solute
total mass of solution
x 100
Alternative concentration units for very dilute solutions:
parts per million -- unit of solute per 106 units of solution
parts per million =
( ppm )
mass percent =
( % m/m )
x 106
parts per billion -- unit of solute per 109 units of solution
parts per billion =
( ppb )
mass of solute
total mass of solution
x 109
Example: A sample of groundwater with a mass of 2.5 g was
found to contain 5.4 µg of Zn2+. What is the concentration of
Zn2+ in the water in ppm?
mass of solute
parts per million =
( ppm )
total mass of solution
5.4 x 10-6 g
=
=
2.5 g
ml solution
volume percent -- mililiters of solute per
mililiters of solution, expressed as a
percentage
x 106
volume percent =
( % v/v )
2.2 ppm
ml of solute
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
x 106
Example: 12.0 ml of ethanol are dissolved in 100.0 g of
water. What is the concentration of this solution in
volume percent?
volume percent =
Concentration of solutions
ml of solute
ml solution
x 100
Concentration of solutions
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
x 100
mass / volume percent -- grams of
solute per mililiters of solution,
expressed as a percentage
volume of solute = 12.0 ml
volume of solution = volume of solute + volume of solvent
= 12.0 ml + 100.0 ml
= 112.0 ml
mass percent =
12.0 ml
112.0 ml
x 100 = 10.7% (v/v)
mass / volume
percent
( % m/v )
=
g of solute
ml solution
x 100
Example: 3.5 g of KOH are dissolved in water to make
80.0 ml of solution. What is the concentration of this
solution in mass/volume percent?
mass/volume percent =
g of solute
ml solution
x 100
Concentration units based on
the number of moles of solute
molarity -- number of moles of solute per liter of solution
mass of solute = 3.5 g
moles of solute
Molarity (M ) =
liters solution
volume of solution = 80.0 ml
mass/volume percent =
3.5 g
80.0 ml
x 100 = 4.4% (m/v)
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
a) What is the molarity of ascorbic acid in the solution?
molar mass C6H8O6 = 6 (12.01 g/mol) + 8 (1.008 g/mol) + 6 (16.00 g/mol)
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
a) What is the molarity of ascorbic acid in the solution?
Calculate the volume of solution
= 176.12 g / mol
mass of solution = 80.5 g + 215 g = 296 g
Calculate the number of moles of ascorbic acid in solution
80.5 g C6H8O6
1 mol C6H8O6
176.12 g C6H8O6
= 0.457 mol C6H8O6
296 g
1 ml
1.22 g
=
243 ml
=
0.243 L
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
a) What is the molarity of ascorbic acid in the solution?
Concentration units based on
the number of moles of solute
mole fraction -- the portion of the total moles of all components
of a solution represented by the solute, expressed as a fraction
Calculate the molarity of ascorbic acid
Molarity =
=
moles of solute
liters of solution
Mole fraction (X ) =
moles of solute
total moles of all components
0.457 mol
0.243 L
= 1.88 M
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
b) What is the mole fraction of ascorbic acid in the solution?
Calculate the number of moles of ascorbic acid in solution
80.5 g C6H8O6
1 mol C6H8O6
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
b) What is the mole fraction of ascorbic acid in the solution?
Calculate the mole fraction of ascorbic acid
= 0.457 mol C6H8O6
Mole fraction (X ) =
176.12 g C6H8O6
moles of solute
total moles of all components
Calculate the number of moles of water in solution
molar mass H2O = 2 (1.008 g/mol) + (16.00 g/mol) = 18.02 g / mol
215 g H2O
1 mol H2O
18.02 g H2O
= 11.9 mol H2O
=
0.457 mol
(11.9 mol + 0.457 mol)
= 0.0370
Concentration units based on
the number of moles of solute
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
c) What is the molality of ascorbic acid in the solution?
molality -- number of moles of solute per kilogram of solvent
Calculate the number of moles of ascorbic acid in solution
Molality (m ) =
moles of solute
kilograms of solvent
80.5 g C6H8O6
1 mol C6H8O6
= 0.457 mol C6H8O6
176.12 g C6H8O6
Don!t confuse molality with molarity
Example: A solution was prepared by dissolving 80.5 g of
ascorbic acid (C6H8O6) in 215 g of water. The density of the
solution is 1.22 g/ml.
c) What is the molality of ascorbic acid in the solution?
General solubility rule:
“Like dissolves like”
Polar compounds tend to be more soluble in
polar solvents than in nonpolar solvents
Calculate the molality of ascorbic acid
Molality =
=
moles of solute
Example: NaCl (polar compound)
kg of solvent
0.457 mol
0.215 kg
= 2.12 m
• soluble in water
Solvent
Polarity
• slightly soluble in ethyl alcohol
• insoluble in ether and benzene
Formation of solutions
General solubility rule:
“Like dissolves like”
The ability of substances to form solutions depends on
two general factors:
Nonpolar compounds tend to be more soluble in
nonpolar solvents than in polar solvents
1. Intermolecular forces affecting solute and solvent particles
2. The natural tendency of substances to spread into
larger volumes when not restrained in some way
Example: benzene (nonpolar compound)
Types of intermolecular forces associated with solutions:
• dispersion forces
• dipole-dipole (ion-dipole)
• hydrogen bonding
• insoluble in water
Solvent
Polarity
• soluble in ether
Solutions tend to form when the attractive forces between
solute and solvent particles are comparable to or stronger than
the solute– solute attractions or solvent –solvent attractions
Example: Dissolution of NaCl in water
Example: Dissolution of NaCl in water
Attractions between water molecules and Na+ and Cl – ions are strong
enough to overcome solute-solute and solvent-solvent attractions
Once separated from the crystal, Na+ and Cl –
ions are surrounded by water molecules
H
!+
!+
Na+
!"
O
H
!+
Cl –
O
!+
Cl –
H
H
!+
!+
Na+
Cl –
H
H
!+
H
O
!+
H
Na+
O
!"
!+ H
Cl –
Na+
H
H
!"
O H
!+
!"
Na+
!"
!"
!"
H
!+
!+
!+
H
!"
H
O
!"
O
!+ H O !"
!+
H
!"
H
!+
Na+
!"
!+
O
H
!+
O H
H
!+
!+
O H !+
H
!+
O
H
!+
!+
Cl –
H !+
H
!+ O
!+ H
!+
!"
H
!+ O
O H !+
H
!"
!+ !+
!"
H
H
O
!"
This process is referred to as solvation
• specifically called hydration when water is the solvent
Example: Dissolution of NaCl in water
+
Example: Dissolution of NaCl in water
+
–
–
–
Polar water molecules
are attracted to Na+ and
Cl– ions in the crystal,
weakening the attraction
between the ions
As the attraction between
the ions weakens, the ions
move apart and become
surrounded by water dipoles
+
+
–
+ – + –
!"
–
–
+
+
–
+ – + –
!"
–
O
+
–
–
O
H
H
+
!+
!+
–
+
Example: Dissolution of NaCl in water
The hydrated ions diffuse
away from the crystal to
become dissolved in solution
+
H
H
+
!+
!+
–
Qualitative terms used to describe
solubility
–
–
–
+
+
–
+ – + –
!"
–
insoluble
O
+
H
H
+
!+
!+
–
slightly
soluble
moderately
soluble
very
soluble
Qualitative terms used to describe
the solubility of liquids
miscible -- liquids that are capable of mixing and forming a solution
immiscible -- liquids that do not form solutions (i.e., they are insoluble
in each other)
Solubility: Exact definition
solubility -- the amount of a solute that will dissolve in a
specified amount of a solvent under stated conditions
(i.e., temperature, pressure)
Example: 36.0 g of sodium chloride will dissolve in 100 g of
water at 20 °C
H 2O
methanol
miscible
H 2O
oil
immiscible
Solubilities of different substances
vary widely
If more than 36.0 g
of NaCl is added, it
will not dissolve
36.0 g
NaCl
100 g
H 2O
20 °C
Saturation
At a specific temperature, there is a limit to the amount
of solute that will dissolve in a given amount of solvent
Example: The solubility of KCl at 20 °C is 34 g / 100 g H2O
When this condition occurs, the solution is saturated
Saturation
Saturation
In a saturated solution two processes are occurring
simultaneously:
• the solute is dissolving into the solution
An undersaturated solution contains less solute than the
solubility limit at a given temperature
-- i.e., more solute can be dissolved in the solution
• the solute is crystallizing out of solution
solute (undissolved)
solute (dissolved)
In some circumstances, solutions can contain more solute
than the solubility limit at a given temperature
-- such a solution is called supersaturated
In a saturated solution, these
two opposing processes occur
at the same rate
-- supersaturated solutions are unstable (excess solute
will precipitate quickly upon disturbance)
• dynamic equilibrium
Dissolving a Solid in a Liquid
Dissolving a Solid in a Liquid
A solution that is saturated at one temperature
may not be saturated at another temperature
Solution
Solution is
is
saturated
unsaturated
34
35
20
10
30 gg KCl
KCl
-- solute dissolves
No more KCl
can dissolve
100 g H2O
20 °C
Raise temperature
35 g KCl
It may increase or
decrease depending
on the soluteSolubility
a solute
solventofsystem
in a solvent changes
with temperature
Solute dissolves
• The solution is now
unsaturated
100 g H2O
20
50 °C
°C
Dissolving a Solid in a Liquid
Whenever
solution will
contains
in excess
of
A stress
to thea system
causesolute
the solute
in excess
A supersaturated
solution
is unstable
its
solubility
limit
the
solution
is
supersaturated
of the saturation limit to come out of solution
Problem: Will a solution made by adding 62.0 g of NaNO3
to 100 g of H2O be saturated or undersaturated at 20°C?
Determine the solubility of NaNO3 in water at 20°C (you can
look this up in a solubility table or diagram)
Cool Solution to 20 °C
35 g KCl
Solubility of NaNO3 (20°C) = 88.0 g / 100 g H2O
At 20 °C the solubility of KCl
in water is 34 g/100 g H2O
62.0 g < 88.0 g
undersaturated
The solution contains 1
No of
KClKCl
precipitates
gram
in excess
of the solubility limit
100 g H2O
50 °C
20
Problem: Will a solution made by adding 3.6 g of NaCl
to 10 g of H2O be saturated or undersaturated at 20°C?
Step 1: Determine the solubility of NaCl in water at 20°C
(you can look this up in a solubility table or diagram)
Solubility of NaCl (20°C) = 36.0 g / 100 g H2O
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
BUT…
Step 2: Adjust the solubility value to match your given
amount of solvent
(10 g H2O)
36.0 g NaCl
100 g H2O
3.60 g NaCl = x
=
x
10 g H2O
(10 g H2O)
3.60 g = 3.60 g
saturated
– some solids increase in solubility only slightly with
increasing temperature
– some solids decrease in solubility with increasing in
temperature
Effect of Temperature on the
Solubility of a Solid in a Liquid
+
Solubility of ionic compounds in water
–
–
+
+
–
+ – + –
–
At higher temperatures,
solvent molecules have
higher kinetic energy
+
–
Effect of Temperature on the
Solubility of a Gas in a Liquid
The solubility of a gas in water usually decreases with
increases in temperature
– at higher temperatures, gas molecules have higher
kinetic energy
– gas molecules gain sufficient energy to overcome
the interactions with solvent molecules that keep the
gas in solution
Effect of Temperature on the
Solubility of a Gas in a Liquid
Solubility of gases in water
Effect of pressure on solubility
Small pressure changes have little effect on the solubility of:
– solids in liquids
– liquids in liquids
Small pressure changes have a large 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
higher pressure = higher solubility
lower pressure = lower solubility
Effect of Pressure on the Solubility of a Gas
pressure
doubles
number of gas
particles doubles
Henry’s law:
The solubility of a gas
in a liquid is directly
proportional to the
pressure of that gas
above the liquid.
solubility of
gas doubles
liquid
Effect of pressure on gas solubility