NAME: _______________________________
UNIT #9:
1.
MOLARITY
DILUTIONS
SOLUBILITY CURVES
COLLIGATIVE PROPERTIES OF SOLUTIONS
MOLARITY
a) Molarity is a measurement of the concentration of a solution in Chemistry.
b) When making solutions, the solute is the substance being dissolved in solution.
c) The solvent is the substance doing the dissolving, usually water.
Ex. When making Kool-Aid, water is the solvent and the Kool-Aid is the solute.
d) Molarity (M) = moles solute
Liter of solution
e) 1 Liter = 1,000 mLs.
To convert mLs. to Liters: divide by 1,000 (move decimal point 3 places to left)
To convert Liters to mLs.: multiply by 1,000 (move decimal point 3 places to right)
f)
Example Problems:
1.
What is the concentration of a solution prepared by dissolving 5 moles of CaCl2 in
400 mLs. of water?
400 mLs. = 0.40 Liters
M = 5 moles CaCl2 = 12.5M CaCl2
0.4 L
2.
What is the molarity of a solution prepared by dissolving 116 grams of NaCl in 800
mLs. of water?
116g NaCl x
1 mole = 2.0 moles NaCl
58g. NaCl
800 mLs. = 0.80 Liters
M = 2.0 moles NaCl = 2.5M NaCl
0.80 L
3.
Which solution contains the greatest number of moles of solute?
a. 0.5L of 0.5M solution
0.5M = moles solute moles solute = 0.25
0.5 L
b. 0.5L of 2M solution
2M = moles solute moles solute = 1.0
0.5 L
c. 2L of 0.5M solution
0.5M = moles solute moles solute = 1.0
2L
d. 2L of 2M solution
2M =
moles solute moles solute = 4.0
2L
2.
DILUTIONS
a)
Solutions of different molarities (concentrations) are often created in the lab by
diluting a more concentrated solution known as a stock solution with water.
b) Dilutions can be determined by using the equation:
M1V1 = M2V2 where M1 = molarity of stock solution (higher concentration)
V1 = volume of stock solution
M2 = molarity of diluted solution (lower concentration)
V2 = final volume of diluted solution
c)
The difference between V2 (final volume of diluted solution) and V1 (volume of
concentrated stock solution used) is equal to the amount of water added to make the
dilution.
V2 - V1 = volume of water added
d) Example Problems:
1. How would you prepare 500 mLs. of a 3M HCl solution using a 6M solution of HCl?
M1 = 6M HCl (stock solution)
V1 = ?
M2 = 3M HCl (diluted solution)
V2 = 500 mLs.
M1V1 = M2V2
(6M HCl) V1 = (3M HCl)(500 mLs.)
V1 = (3M HCl)(500 mLs.) = 250 mLs.
6M HCl
V2 - V1 = mLs. water added
500 mLs. - 250 mLs. = 250 mLs. water added
Solution is made by adding 250 mLs. water to 250 mLs. of 6M HCl solution.
2.
What is the concentration of a solution created by diluting 10 mLs. of a 2.5M stock
solution with 60 mLs. of water?
M1V1 = M2V2
M1
V1
M2
V2
= 2.5M (stock solution)
= 10 mLs.
= ?
= 10 mLs. stock solution + 60 mLs. water = 70 mLs.
(2.5M)(10 mLs.) = M2(70 mLs.)
(2.5M)(10 mLs.) = M2
70 mLs.
0.36M = M2
3.
SOLUBILITY CURVES
a)
Solubility curves show the amounts of different solutes that will dissolve in 100 grams
(100 mLs) of water at varying temperatures.
b) The solubility of a most solid solutes increases as the water temperature
increases.
c) The solubility of gases in water decreases as the water temperature increases.
d) A solution is considered saturated when it contains the maximum amount of solute
that will dissolve in a solvent at a given temperature. If more solute is added to a
saturated solution, it will not dissolve and instead, will fall to the bottom of the
container.
Any point on the line of a solubility curve represents the exact saturation point for that
compound at a given temperature.
e) A supersaturated solution is one that contains more solute than the maximum that
will dissolve in a solvent at a given temperature.
The area above the line of a solubility curve represents the amounts of solute in a
supersaturated solution at a given temperature.
f) An unsaturated solution is one that contains less solute than the maximum that will
dissolve in a solvent at a given temperature. More solute can be added to an
unsaturated solution and will dissolve, but only up to its saturation point.
g) Example Problems: referencing the chart on page 4
1. What is the maximum amount of KCl that will dissolve in 100 grams of water at
10°C?
On the solubility chart, find 10°C on the x-axis and move up to the curve for KCl.
Read across to the y-axis to find the grams of solute that will exactly saturate the
solution.
ANSWER: 30 grams KCl will exactly saturate 100 grams of water at 10°C.
2. Define whether the following solutions are unsaturated, saturated or
supersaturated.
a. 20 grams NaCl/100 grams water @ 50°C. ANSWER: Unsaturated
b. 70 grams NH4Cl/100 grams water @80°C. ANSWER: Supersaturated
c. 110 grams KNO3/100 grams water @60°C ANSWER: Saturated
For each, you need to find the given temperature on the x-axis and move up to
the curve for that substance. If the grams of solute/100 grams water falls above
the curve, the solution is supersaturated; if it falls exactly on the curve, the
solution is saturated; if it falls below the curve, the solution is unsaturated.
4.
COLLIGATIVE PROPERTIES OF SOLUTIONS
a)
Colligative properties of solutions pertains to the effect of solute particles in a solution
on the freezing and boiling point of the solution.
b) Solutes lower the freezing point of a solution.
1. Liquids freeze when the kinetic energy of the particles is no longer great enough
to overcome the intermolecular attraction of the particles. Freezing occurs at the
temperature in which solvent particles attract and the solution solidifies.
2. Solute particles interfere with the solvent particles and thus, more energy must be
withdrawn from the solution before the solvent particles are able to attract to one
another. This results in a lower freezing temperature.
c) Solutes raise the boiling point of a solution.
1. Liquids boil when the kinetic energy of the particles is high enough to overcome
the intermolecular attraction of the particles and they emerge into a gaseous
vapor.
2. Vapor pressure is the pressure exerted upwards of liquid particles that have
escaped into a gaseous form.
3. Liquids boil when vapor pressure exceeds atmospheric pressure.
4. Solute particles interfere with the number of solvent particles exposed at the
liquid’s surface and thus, more energy is required to vaporize enough solvent
particles to exceed atmospheric pressure. This results in a higher boiling
temperature.
d) The NUMBER of particles in solution (not the size) determines the effect.
1. Dissociation factors refer to the number of particles that dissociate into solution
for a given compound.
2. Ionic solutes have a greater effect on freezing point depression and boiling point
elevation because they dissociate into 2 or more particles when they are dissolved
in water. Remember, ionic compounds contain a metal and a non-metal.
Examples:
NaCl(s) → Na+(aq) + Cl-(aq) dissociation factor is 2 since 2 particles dissociate
MgCl2(s) → Mg2+(aq) + 2Cl- dissociation factor is 3 since 3 particles dissociate
Al2S3(s) → 2Al3+(aq) + 3S2-(aq) dissociation factor is 5 since 5 particles dissociate
Hint: Dissociation factors for ionic compounds can usually be determined by
adding up the subscripts of the compound’s formula.
3. Covalent solutes (molecules) have less of an effect on freezing point depression
and boiling point elevation because they do not dissociate into multiple particles
in solution. Thus, covalent solutes always have a dissociation factor of 1.
Remember, covalent compounds contain all non-metals.
Example:
C12H22O11 (s) sucrose → C12H22O11(aq)
e)
The concentration of solute particles has an effect on freezing point depression
and boiling point elevation. The more concentrated a solution, the more freezing
is depressed and the more boiling point is increased.
1. Molality (m) is the measurement of concentration in colligative properties.
2. Molality (m) = moles of solute
kilogram of solvent
One Liter of water has a mass of 1 kilogram
1,000 mLs. x 1 gram
= 1,000 g. = 1 kg.
1 mL. H2O
Therefore, for water based solutions Molality (m) = moles of solute
Liter of water
f)
The number of degrees by which freezing point is depressed is calculated as:
∆T = (m)(df)(kF) where m = molality
df = dissociation factor
kF = freezing point depression constant
kF for water is 1.86°C/m
Water is the most common solvent and pure water freezes at 0°C; therefore, for water
based solutions: Freezing Point of Solution = 0°C - ∆T
g)
The number of degrees by which boiling point is increased is calculated as:
∆T = (m)(df)(kB) where m = molality
df = dissociation factor
kB = boiling point elevation constant
kB for water is 0.52°C/m
Water is the most common solvent and pure water boils at 100°C; therefore, for water
based solutions: Boiling Point of Solution = 100°C + ∆T
h) Example Problems:
a. What is the freezing point of a solution prepared by dissolving 1.2 moles of CaF2 in
3.0 Liters of water?
m = 1.2 moles CaF2 = 0.4m
3.0 kg. water
df = 3 because CaF2(s) → Ca2+(aq) + 2F-(aq) 3 ions in solution
kF = 1.86°C/m
∆T = (0.4m)(3)(1.86°C/m)
∆T = 2.23°C
Freezing point of solution = 0°C - 2.23°C = -2.23°C
b.
i)
What is the boiling point of a solution containing 5 moles of C6H12O6 dissolved in
3.5 Liters of water?
m = 5 moles C6H12O6 = 1.43m
3.5 kg. water
df = 1 since C6H12O6 is a covalent compound
kB = 0.52°C/m
∆T = (1.43m)(1)(0.52°C/m)
∆T = 0.74°C
Boiling point of solution = 100°C + 0.74°C = 100.74°C
Practical applications of colligative properties
a. Anti-freeze is usually mixed 50/50 with water in a car’s radiator. It contains
alcohols and other ingredients such as propylene glycol. These solutes lower the
freezing point and raise the boiling point of the water in the car radiator to
prevent it from freezing in the winter and boiling over in the summer or during
operation.
b. Salts are put on roads in the winter to melt ice and to mix with water on the roads
to create a solution with a much lower freezing point than pure water.
Consequently, the water on roads does not freeze as air temperatures fall well
below freezing.
c. Brine chills are used in the meat packing industry to quick chill meat coming out of
smokehouses. The brine is a mixture of NaCl and water and can be cooled to a
very low temperature without freezing. This super-cooled brine is sprayed on the
meat to chill it.
d. Ice skating rinks have super-cooled brine that circulates beneath the ice to keep
the ice frozen while maintaining a comfortable air temperature in the rink. The
brine continues to flow as a liquid at very low temperatures because the salt in
the water significantly lowers the freezing point of the water.
Unit 9 Note Quiz Questions
Unit 9.1: Solutions and solubility curves
1. A
5.5 a
2. a
3. a
6. a
4. a
7. a
8. a
9. a
10. a
Unit 9.2: Units of Concentration & Dilutions
1. A
2. A
3. A
4. A
5. a
6. a
7. a
8. a
9.
10. a