Colligative
Properties &
Solubility Rules
Ariel Autrey, Katie Elicker, and
Alyssa Truong
Colligative Properties
•
• Properties that depend on the number of
particles dissolved in a given mass of
solvent are called colligative properties
• They do not depend on the chemical
nature of the solute or the solvent
The Four Colligative Properties
1. Vapor Pressure Reduction
2. Osmotic Pressure
3. Freezing Point Depression
4. Boiling Point Elevation
Vapor Pressure Reduction
• Nonvolatile solute: a solute that can not
escape to the vapor phase.
• The solvent will go to the vapor phase,
•
and the solution's volume will increase
because the solvent will go into the vapor
phase more easily than the solution will.
The solution's volume increases in an
attempt to decrease the vapor pressure.
Vapor Pressure Reduction
Raoult's Law: A nonvolatile solute lowers the
vapor pressure of a solvent because solute
particles become surrounded by solvent
particles when dissolved. This causes the
particles of solvent to evaporate less.
Therefore, the vapor pressure of a solution
is less than that of the pure solvent it was
created with.
Osmotic Pressure
• Osmotic pressure is the pressure required
•
to prevent solvent from passing through a
semipermeable membrane from the
higher concentration to the lower
concentration of solute.
The result is a net flow of solvent
molecules from the less concentrated
solution to the more concentrated
solution.
Osmotic Pressure
• As the solvent moves back and forth
•
•
through the membrane, the levels of
solution become uneven, and the process
continues until the difference of pressures
is large.
Two solutions with identical osmotic
pressures are called isotonic.
Two solutions with a different osmotic
pressures are said to be hypotonic.
Freezing Point Depression
•
•
•
•
Ability of a dissolved solute to lower the
freezing point of its solution
Freezing point: the temperature at which the
vapor pressures of the solid and liquid
phases are the same
solute + solvent
solute molecules disrupt
the orderly pattern of the solid
More kinetic energy must be withdrawn from
the solution than from the pure solvent for it
to solidify
Freezing Point Depression Problem
1.60 g of naphthalene (C10H8) is dissolved in
20.0 g of benzene. The freezing point of pure
benzene is 5.5 oC, and the freezing point of
the mixture is 2.8 oC. What is the molal
freezing point depression constant, Kf of
benzene?
Step 1
Calculate the freezing point depression of
benzene.
Tf = (Freezing point of pure solvent) (Freezing point of solution)
(5.5 oC) - (2.8 oC) = 2.7oC
Step 2
• Calculate the molal concentration of the
solution
-molality = moles of solute / kg of solvent
-moles of naphthalene = (1.60 g) (1 mol / 128 g)
= 0.0125 mol
naphthalene
-molality of solution = (0.0125 mol) / (0.0200 kg)
= 0.625m
Answer
• Calculate K of the solution.
f
Tf = (Kf) (m)
(2.7 oC) = (Kf) (0.625 m)
Kf = 4.3 oC/m
http://chemed.chem.purdue.edu/genchem/probsolv/colligative/kf1.3.html
Practice Problems
1.What is the freezing point depression when
85.3 g of oxygen gas is dissolved in 1500 g of
water? Kf (ºC/molal) is -1.86.
2. Ethylene glycol (C2H6O2) is the principal
ingredient in antifreeze. How many grams of
ethylene glycol will be needed to lower the
freezing point of 2100 g of water by 20ºC?
Kf (ºC/molal) is -1.86.
Answers to Practice Problems
1. 3.31ºC
2. 1.4 kg
http://home.comcast.net/~cochranjim/PDFS3/COLIGWS1A.pdf
Boiling Point Elevation
• The difference in temperature between
•
the boiling points of a solution and a pure
solvent
Boiling point of a substance is the
temperature at which the vapor pressure
of a liquid is equal to the external
pressure on its surface. (one example of
external pressure is atmospheric
pressure)
Boiling Point Elevation
• Tb= imKb
• Tb= change in boiling point
• i =Van't Hoff factor
• m = molality (moles solute/ kg solvent)
• Kb = boiling point elevation constant
(water Kb = +0.52oC/m)
• normal boiling point of water = 100.oC
Boiling Point Elevation
• Adding a nonvolatile solute reduces the
vapor pressure of the solution
• A higher temperature is now necessary to
•
get the vapor pressure of the solution up
to atmospheric pressure so that the
solution boils
The amount by which the boiling temp. is
raised is the boiling point elevation
Problem (boiling point
elevation)
31.65 g of sodium chloride is added to 220.0
mL of water at 34 °C. How will this affect
the boiling of the water? Assume the
sodium chloride completely dissociates in
the water.
Given: density of water at 35 °C = 0.994
g/mL
Kf water = 1.86 °C kg/mol
Step 1
• žCalculate the molality of the NaCl
molality (m) of NaCl = moles of NaCl/kg
water
mNaCl = moles of NaCl/kg water
mNaCl = 0.542 mol/0.219 kg
mNaCl = 2.477 mol/kg
Step 2
• Determine the Van't Hoff Factor
• The Van't Hoff Factor, i, is a constant
associated with the amount of dissociation of
the solute in the solvent. For substances that
do not dissociate in water, such as sugar, i =
1. For solutes that completely dissociate into
two ions, i = 2. NaCl completely dissociates
into the two ions, Na+ and Cl-. Therefore, i =
2.
Step 3
• Find ΔT
ΔT = imKf
ΔT = 2 x 0.51 °C kg/mol x 2.477 mol/kg
ΔT = 2.53 °C
Answer
Adding 31.65 g of NaCl to 220.0 mL of water
will lower the boiling point 2.53°C
This means the new boiling point is 36.53°C
o Because the original boiling point was 34°C,
you have to add 2.53°C to the original boiling
point to get the new boiling point.
(http://chemistry.about.com/od/workedchemistryproblems/a/Boiling-PointElevation-Example-Problem.htm
Practice Problems
1. What is the boiling point elevation when 11.4
g of ammonia (NH3) is dissolved in 200. g of
water? Kb for water is 0.52 °C/m.
2. Determine the molecular mass of a nonionizing, non-volatile solute if 0.546g of it are
dissolved in 15.0g of benzene ( Kf=5.12 ºC/
F.P. = 5.5ºC) and the freezing point
depression was 0.240ºC.
Answer to Practice Problem 1
1. Δt = 1.74 °C
A) Determine molality of 11.4 g of ammonia in 200. g of
water:
11.4 g / 17.031 g/mol = 0.6693676 mol
0.6693676 mol / 0.200 kg = 3.3468 m
B) Determine bp elevation:
Δt = i Kb m
Δt = (1) (0.52 °C/m) (3.3468 m)
Δt = 1.74 °C
Answer to Practice Problem 2
2. 776 g/mol
m= T/kf
m=0.240C/5.12C/m
m=0.0469 mol/kg
0.0469 mol/kgx0.0150kg=7.03x10^-4mol
0.546g/0.000703mol=776g/mol
Solubility Rules
• The solubility of a substance varies from
one substance to the next.
• The solubility of a substance can be
determined on the solubility rules table.
• If a substance is mainly soluble in water, it
is aqueous.
• If a substance is mainly insoluble in water,
it is a solid.
Solubility Rules
• An example of a soluble substance is KCl
•
because according to the solubility rules
table, all chlorates are soluble.
An example of an insoluble substance is
MgSO3 because according to the
solubility rules table, all sulfites are
insoluble except those of the IA elements
and NH4+
o Because Mg+2 is not part of the IA elements,
the substance is insoluble.
Solubility Problems
Use the solubility rules table to find out if the
substance is soluble or insoluble.
1. HCl
2. Ba(OH)2
3. (NH4)2CrO4
4. CaCrO4
5. AgI
Complete Ionic Equations
• Complete Ionic Equations (CIE) are used
•
to describe the complete reaction and
show all of the reactants and products as
ions.
Only split the aqueous substances on
both sides of the reaction.
How to Write a CIE
1. Start with a balanced equation.
2. Break up all of the aqueous compounds
into ions on both sides of the equation.
3. Write all of the ions together to form a
CIE.
4. Reminders:
a. Make sure you count for the coefficients in
front of the compounds.
b. Leave all solid, liquid, and gaseous
compounds alone.
c. Be sure to keep all of the (s), (l), (g), (aq)
Example: Writing a CIE
Mixing together the aqueous solutions
Sodium Phosphate and Calcium Chloride
makes an insoluble white solid, Calcium
Phosphate.
The balanced equation looks like this:
2 Na3PO4 (aq)+ 3 CaCl2 (aq)
Ca3(PO4)2 (s)
6 NaCl (aq)+
Example: Writing a CIE
Break up the aqueous compounds on the
reactants side of the equation:
• 2 Na3PO4 (aq) has 6 sodium ions and 2
•
phosphate anions. This looks like 6
Na+1(aq) + 2 PO4-3(aq)
3 CaCl2 (aq) has 3 calcium ions and 6
chloride anions. This looks like 3 Ca+2(aq)
+ 6 Cl-1(aq)
Example: Writing a CIE
Break up the aqueous compounds on the
products side of the equation:
• 6 NaCl(aq) has 6 sodium ions and 6
•
chloride anions. This looks like 6 Na+1(aq)
+ 6 Cl-1(aq)
Because Ca3(PO4)2 (s) is a solid, the
compound does not change; it stays
together.
Example: Writing a CIE
Now, put all of the information together.
6 Na+1(aq) + 6 Cl-1(aq) + 3 Ca+2(aq) + 2 PO4-3(aq)
6 Na+1(aq) + 6 Cl-1(aq) + Ca3(PO4)2 (s)
This is your complete ionic equation.
http://www.occc.edu/kmbailey/chem1115tutorials/Net_Ionic_Eqns.htm
Net Ionic Equations
• A net ionic equation (NIE) is a little
different than a CIE.
• In a NIE, the ions that are not used in the
•
equation to make a solid, called spectator
ions, are not written.
The spectator ions are present during the
reaction, but were not actively
participating in the reaction.
How to write a NIE
1. Start with a balanced equation.
2. Write a complete ionic equation.
3. Remove all the spectator ions from both
sides of the reaction.
4. All of the leftover ions make the NIE.
a. Reminder: Don't forget the (aq), (s), (l), (g)
labels!
b. Hint: the spectator ions always look exactly
the same on both sides of the reaction. They
will be in the same state and have the same
formula, charge, and number of ions.
Example: Writing a NIE
Using the same equation we used to write a
CIE, write a NIE.
Since we have already written the CIE within
the last example, let's remind ourselves
what it looks like:
6 Na+1(aq) + 6 Cl-1(aq) + 3 Ca+2(aq) + 2 PO4-3(aq)
6 Na+1(aq) + 6 Cl-1(aq) + Ca3PO4 (s)
Example: Writing a NIE
Identify the spectator ions, and cross them
out.
6 Na+1(aq) + 6 Cl-1(aq) + 3 Ca+2(aq) + 2 PO4-3(aq)
6 Na+1(aq) + 6 Cl-1(aq) + Ca3(PO4)2 (s)
When the spectators are crossed out:
6 Na+1(aq) + 6 Cl-1(aq) + 3 Ca+2(aq) + 2 PO4-3(aq)
6 Na+1(aq) + 6 Cl-1(aq) + Ca3(PO4)2 (s)
Example: Writing a NIE
This will leave you with the net ionic
equation:
3 Ca+2(aq) + 2 PO4-3(aq)
Ca3(PO4)2 (s)
http://www.occc.edu/kmbailey/chem1115tutorials/Net_Ionic_Eqns.htm
Practice: Writing CEIs and NEIs
Balance the equations, if needed, and write a
complete ionic equation and a net ionic
equation.
•
•
NaOH(aq) + H2SO4 (aq)
(NH4)2CO3 (aq) + Al(NO3)3 (aq)
(s)
Na2SO4 (aq) + H2O(l)
NH4NO3 (aq) + Al2(CO3)3
Answers to #1
Complete Ionic Equation:
2 Na+(aq)+ 2 OH-(aq) + 2 H+(aq) + SO42-(aq)
2 Na+(aq) + SO42-(aq) + 2 H2O(l)
Net Ionic Equation:
2 OH-(aq) + 2 H+(aq)
2 H2O(l)
Which Simplifies To:
OH-(aq) + H+(aq)
H2O(l)
Wrong Answer?
• Make sure these spectator ions are the
ones you crossed out:
2 Na+(aq)+ 2 OH-(aq) + 2 H+(aq) + SO42-(aq)
2 Na+(aq) + SO42-(aq) + 2 H2O(l)
• That should leave you with this net ionic
equation:
•
2 OH-(aq) + 2 H+(aq)
2 H2O(l)
Because of the same coefficients in front,
the NIE becomes: OH-(aq) + H+(aq)
H2O(l)
Answers to #2
Complete Ionic Equation:
6 NH4+(aq) + 3 CO32-(aq) + 2 Al3+(aq) + 6 NO3-(aq)
6 NH4+(aq) + 6 NO3-(aq) + Al2(CO3)3 (s)
Net Ionic Equation:
2 Al3+(aq) + 3 CO32-(aq)
Al2(CO3)3 (s)
http://www.occc.edu/kmbailey/chem1115tutorials/Net_Ionic_Eqns_Answers.ht
m
Wrong Answer?
• Make sure these spectator ions are the
ones you crossed out:
6 NH4+(aq) + 3 CO32-(aq) + 2 Al3+(aq) + 6 NO3-(aq)
6 NH4+(aq) + 6 NO3-(aq) + Al2(CO3)3 (s)
• That should leave you with this net ionic
equation:
2 Al3+(aq) + 3 CO32-(aq)
Al2(CO3)3 (s)