Mark S. Cracolice
Edward I. Peters
www.cengage.com/chemistry/cracolice
Chapter 16
Solutions
Mark S. Cracolice • The University of Montana
Solution
Solution is a homogenous mixture.
A solution can exist in any of the three states:
Solid solution: steel, brass, bronze
Liquid solution: alcohol in water, sugar water.
Gaseous solution: air, any mixture of gases
.
Solution Terminology
Solute
The substance present in a relatively small amount in a
solution; the solid or gas when a substance in that
state is dissolved in a liquid to make a solution.
Solvent
The substance present in a relatively large amount in a
solution; the liquid when a solid or gas is dissolved
to make a solution.
Solution Terminology
Concentrated Solution
has a relatively large quantity of a specific solute
per unit amount of solution.
Dilute Solution
has a relatively small amount of a specific solute
per unit amount of solution.
Solution Terminology
Saturated solution
A solution that can exist in equilibrium with
undissolved solute is a saturated solution.
Solubility
The concentration of the saturated solution is called
the solubility of the solute.
Unsaturated
A solution is unsaturated if its concentration is
smaller than the solubility.
Solution Terminology
Supersaturated
A solution is said to be supersaturated if its
concentration is greater than the solubility.
Supersaturated solution is not stable,
a slight physical disturbance can start crystallization
The Formation of a Solution
The water molecule is polar.
A polar molecule is one with an asymmetrical distribution of
charge, resulting in positive and negative poles.
The Formation of a NaCl Solution
The Formation of a NaCl Solution
The formation of a sodium chloride solution from
sodium chloride crystals is made possible by the
interaction of water molecules with sodium and
chloride ions.
NaCl (s) ↔ Na+ (aq) + Cl- (aq)
The sodium and chloride ions in solution are
surrounded by polar water molecules, and are said to
be hydrated.
The Formation of a Solution
The rate of dissolving depends on
1 The surface area. A finely divided solid dissolves
more rapidly.
2 The diffusion of the solute from the surface. Stirring
increases the rate.
3 Temperature. The higher the temperature, the
higher the rate.
The Formation of a Solution
Crystallization rate
The rate per unit of surface area increases as the
solution concentration at the surface increases.
Equilibrium in a Saturated Solution
When the rates of dissolving and crystallization are the
same the solution is saturated.
The Formation of a Saturated Solution
When crystallization rate is equal to the dissolving rate
the solution is saturated.
Factors That Determine Solubility
Solubility depends on three factors:
Intermolecular forces
Gas pressure (for a gas)
Temperature
Factors That Determine Solubility
Intermolecular Forces
Generally, if the forces between molecules A are about
the same as the forces between molecules B, A and
B will probably dissolve in each other.
The rule of solubility is like dissolves like.
A nonpolar solvent can dissolve a nonpolar solute.
A polar solvent can dissolve a polar or ionic solute
Factors That Determine Solubility
Ethylene glycol HO-CH2–CH2–OH, which is polar and
can form hydrogen bond, is soluble in water. Motor oil
is not soluble in water because hydrocarbon
molecules are nonpolar.
Factors That Determine Solubility
Partial Pressure of Solute Gas Over Liquid Solution
• Pressure has little effect on the solubility of solids or
liquids but has a pronounced effect on the solubility
of gases.
• For an ideal solution, the solubility of a gas in a liquid
is directly proportional to the partial pressure of the
gas over the surface of the liquid.
Factors That Determine Solubility
Solubility of carbon dioxide decreases when partial
pressure of carbon dioxide drops
Factors That Determine Solubility
Temperature
The solubility of most solids increases with rising
temperature (but there are notable exceptions).
The solubilities of gases in liquids are generally
lower at higher temperatures.
Factors That Determine Solubility
Factors That Determine Solubility
Percentage by Mass
g solute
% by mass =
 100
g solution
g solute
% by mass =
 100
g solute + g solvent
Percentage by mass can be used as conversion factor between
mass of solution and mass of solute.
Percentage by Mass
• When 125 grams of a solution is evaporated to
dryness, 42.3 grams of solute was recovered. What
was the percentage of the solute?
g solute
% by mass =
 100
g solution
42.3g
=
 100 = 33.8% NaCl
125g
Molarity
Molarity, M
Moles of solute per liter of solution:
moles solute
mol
M º
=
liter solution
L
Molarity can be used as conversion factor between liters of
solution and moles of solute.
Volume of solution x molarity = number of moles
Molarity
Example:
Calculate the molarity of a solution made by dissolving 15 g of
NaOH in water and diluting to 1.00 x 102 mL
Solution:
First calculate the number of moles NaOH
1 mol NaOH
15 g NaOH x
 0.38mol NaOH
40.00g NaOH
Molarity
Next calculate the volume of solution in liters
1.00 x 102
x
1L
 0.100 L
1000 mL
Calculate the answer using the definition
0.38 mol NaOH
M
 3.8 M NaOH
0.100 L solution
Molarity
Example:
• How many grams of silver nitrate must be dissolved to prepare
5.00 x102mL of 0.150 M AgNO3
• 5.00 x102mL x ( 1 L/ 1000 mL)
x(0.150 mol AgNO3/L)
x(169.9 g AgNO3/mol AgNO3)
= 12.7 g AgNO3
Molarity
Example
Find the volume of a 1.40 M solution that contains 0.287 mole of
ammonia.
Solution
0.287 mole NH3 x ( 1L/ 1.40 mole NH3) = 0.205 L
Molarity
To prepare a solution of a specified molarity:
1. Weigh the appropriate amount of solute.
2. Add less than the total volume of solvent.
3. Mix to completely dissolve the solute.
4. Add additional solvent until the total solution volume is
appropriate.
Preparation of 250 mL Solution KMnO4 0.0100M
250 mL x (1L/1000 mL) x (0.00100 mol KMnO4 /1L) x
(158.04 g KMnO4/ 1 mol KMnO4) = 0.395 g KMnO4
Molality
Molality, m
The number of moles of solute dissolved
in one kilogram of solvent:
mol solute
m º
kg solvent
Molarity (mol/L) is temperature dependent;
molality is temperature independent.
Normality: Equivalent
Equivalent, eq
One equivalent of acid is the quantity that yields
one mole of hydrogen ions in a chemical reaction.
Once equivalent of base is the quantity that reacts
with one mole of hydrogen ions.
Normality: Equivalent
eq /mol
H3PO4 + NaOH → NaH2PO4 + H2O
1
H3PO4 + 2 NaOH → Na2HPO4 + 2 H2O
2
H3PO4 + 3 NaOH → Na3PO4 + 3 H2O
3
Number of equivalents of acid = Number of equivalents of base
Normality: Equivalent Mass
Equivalent mass: The number of grams per equivalent.
H3PO4 + NaOH → NaH2PO4 + H2O
97.99 g/eq
H3PO4 + 2 NaOH → Na2HPO4 + 2 H2O
97.99 g/2 eq
H3PO4 + 3 NaOH → Na3PO4 + 3 H2O
97.99 g/3 eq
Equivalent mass of H3PO4 for the three reactions are
respectively: 97.99 g/eq, 49.00 g/eq, and 32.66 g/eq .
Normality
Normality, N
The number of equivalents, eq, per liter of solution:
N º
equivalents solute
eq
=
liter solution
L
Normality
Example:
What is the normality of a solution made by dissolving 2.50
grams of sodium hydroxide in water to make 5.00 × 102 mL of
solution?
Solution:
By definition,
equivalents solute
eq
N º
=
liter solution
L
Therefore, we need to find equivalents of solute and liters of
solution.
Normality
What is the normality of a solution made by dissolving 2.50
grams of sodium hydroxide in water to make 5.00 × 102 mL of
solution?
Equivalents of solute:
GIVEN: 2.5 g NaOH
WANTED: eq NaOH
1 eq NaOH
2.5 g NaOH
= 0.0625eq NaOH
40.00g NaOH
Normality
What is the normality of a solution made by dissolving 2.50
grams of sodium hydroxide in water to make 5.00 × 102 mL of
solution?
GIVEN: 5.00 × 102 mL
WANTED: L
1L
5.00  10 mL 
= 0.500 L
1000 mL
2
Normality
What is the normality of a solution made by dissolving 2.50
grams of sodium hydroxide in water to make 5.00 × 102 mL of
solution?
We now have both parts of the normality fraction,
0.0625 eq NaOH and 0.500 L solution:
eq 0.0625 eq NaOH
N
=
= 0.125 N NaOH
L
0.500 L
Solution Concentration: Summary
Dilution of Concentrated Solutions
Concentrated solutions are diluted by adding more solvent.
The amount of solute remains the
same before and after a dilution.
Molarit y Volume = Number of moles
mol
MV =
 L = mol = n
L
Mc x Vc = number of moles = Md x Vd
Mc x Vc = Md x Vd
Dilution of Concentrated Solutions
Example: How many milliliters of concentrated hydrochloric acid,
which is 11.6 M, should be used to prepare 5.50 liters of
0.500 M HCl ?
Vc x Mc =
Vc x 11.6 M =
Vc =
Vd x Md
5.50 L x 0.500 M
5.50 L x 0.500 M /11.6 M = 0.237 L = 237 mL
Dilution of Concentrated Solutions
If the volume of solution is increased 10 times, the
concentration of diluted solution is 10 times smaller.
Solution Stoichiometry
How many grams of lead(II) iodide will precipitate when
excess potassium iodide solution is added to 50.0 mL of
0.811 M lead(II) nitrate.
Pb(NO3)2 (aq) + 2 KI(aq)
 PbI2(s) + 2 KNO3(aq)
Given: 50.0 mL of 0.811 M lead(II) nitrate
Wanted: gram of lead(II) iodide.
Need to convert volume of solution to moles of solute
Solution Stoichiometry
How many grams of lead(II) iodide will precipitate when
excess potassium iodide solution is added to 50.0 mL of 0.811
M lead(II) nitrate.
Pb(NO3)2 (aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq)
mL of Pb(NO3)2 liters of Pb(NO3)2moles of Pb(NO3)2
moles of PbI2gram of PbI2
50.0 mL of Pb(NO3)2 x (L of Pb(NO3)2/ 1000 mLof Pb(NO3)2)
x (0.811 mol of Pb(NO3)2/L Pb(NO3)2)
x (1 mol PbI2 /1 mol Pb(NO3)2)
x (461.0 g PbI2 /mol PbI2)
= 18.7 g PbI2
Titration Using Molarity
Titration: the very careful addition of one solution to
another by a burette.
Titration Using Molarity
Standardize
Determination of the concentration of a solution to be used in a
titration by titrating it against a primary standard.
Primary Standard
A soluble solid of reasonable cost that is very stable and pure,
preferably with a high molar mass, that can be weighed
accurately for use in a titration.
Titration Using Molarity
1.18 g oxalic acid, H2C2O4. 2H2O (126.07 g/mol), are dissolved
in water and the solution is titrated with a solution of NaOH of
unknown concentration. 28.3 ml NaOH are required to
neutralize the acid. Calculate the molarily of the NaOH
solution.
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
Given 1.18 g H2C2O4. 2H2O, 28.3 mL (0.0283L) NaOH
Wanted mol/L NaOH
g H2C2O4. 2H2O mol H2C2O4. 2H2Omol H2C2O4
 mol NaOH  molarity NaOH
Titration Using Molarity
1.18 g oxalic acid, H2C2O4. 2H2O (126.07 g/mol), are dissolved
in water and the solution is titrated with a solution of NaOH of
unknown concentration. 28.3 ml NaOH are required to
neutralize the acid. Calculate the molarily of the NaOH
solution.
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
1.18 g H2C2O4. 2H2O
x (1 mol H2C2O4. 2H2O/ 126.07g H2C2O4. 2H2O )
x (1 mol H2C2O4. / 1 mol H2C2O4. 2H2O)
x ( 2 mol NaOH/1 mol H2C2O4.) x ( 1/ 0.0283L NaOH)
= 0.661 mol/L
Titration Using Normality
The number of equivalents of all species
in a reaction is the same.
For an acid–base reaction,
equivalents of acid = equivalents of base
Titration Using Normality
A 25.0 mL sample of an electroplating solution is
analyzed for its sulfuric acid concentration. It takes
46.8 mL of the 0.661 N NaOH to neutralize the
sample, find the normality of the acid.
Na x Va = Nb x Vb
Na x 25.0 mL = 46.8 mL x 0.661 N
Na = (46.8 mL / 25.0 mL ) x 0.661 N = 1.24 N H2SO4
Colligative Properties of Solutions
A pure solvent has distinct physical properties.
Introducing a solute into the solvent affects these properties.
In dilute solutions of certain solutes, the change in
some of these properties is proportional
to the molal concentration of the solute particles.
Colligative Property
Solution property that is determined only by the number of solute
particles dissolved in a fixed quantity of solvent and not by the
identity of the solute particles.
Colligative Properties of Solutions
Boiling Point Elevation
The boiling point of a solution is higher
than the boiling point of the pure solvent.
Example:
The normal boiling point of water is 100.0°C.
The normal boiling point of a 1 m
solution of sugar water is 100.5°C.
Colligative Properties of Solutions
Freezing Point Depression
The freezing point of a solution is lower
than the freezing point of the pure solvent.
Example:
The normal freezing point of water is 0.0°C.
The normal boiling point of a 1 m
solution of sugar water is –1.9°C.
Colligative Properties of Solutions
Colligative Properties of Solutions
Boiling point elevation and freezing point depression
are colligative properties:
They depend on the number of solute
particles but not their identity.
∆Tb = change in boiling temperature
∆Tf = change in freezing temperature
Colligative Properties of Solutions
∆Tb = Kb × m
∆Tf = Kf × m
Kb = molal boiling-point elevation constant
Kf = molal freezing-point depression constant
Colligative Properties of Solutions
Molal Boiling Point Elevation Constant Values
Substance
Boiling Point (°C)
Benzene
80
Carbon disulfide
46
Carbon tetrachloride
77
Water
100
Kb (°C/m)
2.5
2.4
5.0
0.52
Molal Freezing Point Depression Constant Values
Substance
Freezing Point (°C)
Kf (°C/m)
Benzene
6
5.1
Carbon disulfide
–112
3.8
Carbon tetrachloride
– 23
30
Water
0
1.86
Colligative Properties of Solutions
What is the freezing point of a solution made by adding
12.0 g of urea, CO(NH2)2 , to 2.50 × 102 grams of water?
1 molCO(NH2 ) 2
12.0g CO(NH2 ) 2 
= 0.200molCO(NH2 ) 2
60.06g CO(NH2 ) 2
molsolute 0.200molCO(NH2 ) 2
m
=
kg solvent
2.50 102 g H 2O
1000g H 2O

= 0.800m CO(NH2 ) 2
kg H 2O
Colligative Properties of Solutions
What is the freezing point of a solution made by adding
12.0 g of urea, CO(NH2)2 , to 2.50 × 102 grams of water?
∆Tf = Kf x m = 1.86 °C/m x 0.800 m = 1.49 0C
Tf = 0°C – 1.49°C = –1.49 °C
Determination of molar mass
Calculate the Molar Mass of a Solute from
Freezing Point Depression or Boiling Point Elevation Data
1. Calculate molality from m = ∆Tf/Kf or m = ∆Tb/Kb. Express as
mol solute/kg solvent.
2. Using molality as a conversion factor between moles of solute
and kilograms of solvent, find the number of moles of solute.
3. Use the defining equation for molar mass, MM = g/mol, to
calculate the molar mass of the solute.
Determination of molar mass from ∆Tb
Example. Kb of benzene is 2.5 0C/m. A solution of 15.2 g
of unknown solute in 91.0 g benzene boils at a
temperature 2.1 C higher than the boiling point of pure
benzene. Calculate the molar mass of the solute.
Given: ΔTb = 2.1 0C, Kb = 2.5 0C/m
Need: molar mass
First, calculate molality m : ∆Tb = Kb × m
Determination of molar mass from ∆Tb
Example. Kb of benzene is 2.5 0C/m. A solution of 15.2 g of
unknown solute in 91.0 g benzene boils at a temperature
2.1 C higher than the boiling point of pure benzene.
Calculate the molar mass of the solute.
∆Tb = Kb × m
m = ΔTb / Kb = 2.1 0C/(2.5 0C/m) = 0.84 mol/kg
Then calculate moles of solute and molar mass
Moles of solute in 91.0 g of benzene = 0.84 mol/kg x
( 1kg/1000g) x 91.0 g = 0.077 mol
Molar mass = g/ mol = 15.2 g/ 0.077 mol
= 197.40 g/mol = 200 g/mol
Homework
• Homework: 33, 43, 45, 49, 61, 71, 85, 89, 93, 103, 111, 121,
125, 127, 132.