(Chapter 13
Andrew Meade
Mitchell Korotkin
Kayla Devin)
Solubility Rules:
-The presence of certain ions may cause a
compound to be soluble or insoluble.
- The solubility rules are a list of ions and
whether they make a compound soluble
or insoluble.
-If there are no solids (all products are
soluble), then there is no reaction
ION
Soluble or
insoluble?
Exceptions?
NO3- (Nitrate)
Soluble
None
C2H3O2- (Acetate)
Soluble
AgCH3COO
ClO3- (Chlorate)
Soluble
None
Cl- (Chloride)
Soluble
AgCl, Hg2Cl2, PbCl2
Br- (Bromide)
Soluble
AgBr, PbBr2, Hg2Br2, HgBr2
I- (Iodide)
Soluble
AgI, Hg2I2, HgI2, PbI2
SO4 2- (Sulfate)
Soluble
BaSO4, PbSO4, Hg2SO4, CaSO4, Ag2SO4, SrSO4
Alkali metal ions, NH4+
Soluble
None
H+
Soluble
None
CO32- (Carbonate)
Insoluble
IA elements, NH4 +
CrO42- (Chromate)
Insoluble
IA elements, NH4 +, CaCrO4, SrCrO4
OH- (Hydroxide)
Insoluble
IA elements, NH4 +, Ba(OH)2, Sr(OH)2, Ca(OH)2
PO4 3- (Phosphate)
Insoluble
IA elements, NH4 +
SO3 2- (Sulfite)
Insoluble
IA elements, NH4 +
S 2- (Sulfide)
Insoluble
IA elements, IIA elements, NH4 +
Example:
Potassium chromate- K2CrO4
1. Look at solubility rules for chromate.
ION
CrO42- (Chromate)
Soluble or
insoluble?
Insoluble
Exceptions?
IA elements, NH4 +, CaCrO4, SrCrO4
Normally, chromates tend to be insoluble. However, one of
the exceptions is the IA elements, and because
potassium(K) is an IA element, Potassium chromate is
soluble.
Answer: SOLUBLE
Example:
Use the solubility rules to predict the
reaction products:
AuCl (aq) + NaNO3 (aq)
AuNO3(aq) + NaCl(aq)
**There is no reaction because both of the
products are aqueous
Complete Ionic Equations
** Notice the Spectator Ions circled- these are
ions that do not take part in a chemical reaction
and are found on both sides of the reaction **
Net Ionic Equations
-Include only the compounds that undergo a
chemical change in a reaction in an aqueous
solution
-To do this simply omit the spectator ions from
the complete ionic equation
Scaffolded Example
Write the complete and net ionic equation for the molecular equation:
HCl(aq) + NaOH (aq)
NaCl(aq) + H2O(l)
1. Split up aqueous reactants and products into ions.
(complete ionic equation)
H+ + Cl- + Na+ + OH-
Na+ +Cl- + H2O
2. Cancel out spectator ions
H+ + Cl- + Na+ + OH-
Na+ +Cl- + H2O
Net ionic equation:
H+ + OH-
H2O
PROBLEM 1
Write the complete and net ionic equation for the molecular equation:
NaCl(aq) + AgNO3 (aq)
AgCl(s) + NaNO3(aq)
Problem 1: ANSWER
Write the complete and net ionic equation for the molecular equation:
NaCl(aq) + AgNO3 (aq)
AgCl(s) + NaNO3(aq)
1. Split up aqueous reactants and products into ions.
(complete ionic equation)
Na+ + Cl- + Ag+ + NO3-
AgCl + Na+ + NO3-
2. Cancel out spectator ions
Na+ + Cl- + Ag+ + NO3NO3-
AgCl(s) + Na+ +
Net ionic equation:
Ag+ + Cl-
AgCl(s)
Problem 2
Write the complete and net ionic equation for the molecular equation:
KI(aq) + AgClO3 (aq)
AgI(s) + KClO3(aq)
Problem 2: ANSWER
Write the complete and net ionic equation for the molecular equation:
KI(aq) + AgClO3 (aq)
AgI(s) + KClO3(aq)
1. Split up aqueous reactants and products into ions.
(complete ionic equation)
K+ + I- + Ag+ + ClO3-
AgI + K+ + ClO3-
2. Cancel out spectator ions
K+ + I- + Ag+ + ClO3-
AgI(s) + K+ + ClO3-
Net ionic equation:
Ag+ + I-
AgI(s)
Colligative Properties
- The physical properties of a solution are different from
those of the pure solvent.
- Some of these differences are due to the mere presence
of solute particles in the solution.
- Colligative properties depend on the number of
particles dissolved in a given mass of
solvent.
- They do not depend on the chemical nature of the
solute or solvent
The 4 Colligative Properties Are:
1. Vapor Pressure Lowering
2. Freezing Point Depression
3. Boiling Point Elevation
4. Osmotic Pressure
Vapor Pressure Lowering
-Vapor pressure occurs because some molecules of a pure
liquid leave the liquid state and enter the gaseous state
(vaporization)
-At the same time, molecules from the gaseous state
return to the liquid state (condensation)
-Equilibrium is established when the rate of vaporization
and condensation becomes equal
-The gas pressure resulting from the vapor molecules over
the liquid is the vapor pressure
-Vapor Pressure of a solvent containing a nonvolatile
solute is lower than the vapor pressure of the pure
solvent
Osmotic Pressure
- Solvents are able to move through a semipermeable
membrane, where they move from a high to a low
concentration.
-Through the movement of the solute, the levels of the
solution becomes uneven.
-The osmosis of the solvent will stop when the pressure
difference becomes large and the levels are very
uneven.
-The pressure difference is called osmotic pressure.
Boiling Point Elevation
-The molal boiling point constant (Kb) is the boiling point
elevation of the solvent in a 1-molal solution of a
nonvolatile, nonelectrolyte solute
-The boiling point elevation is the difference between the
boiling points of the pure solvent and a non electrolyte
solution of that solvent and is directly proportional to
the molal concentration of the solution
-Can be calculated using the formula:
Boiling Point
Elevation
Δ T b = K b (i) (m)
Boiling
Constant
Van't Hoff
Constant
Molality
Boiling Point Elevation
Explanation
Boiling point is essentially the amount of energy it takes to make the vapor
pressure of a solution equal to the external pressure.
100 ºC
With normal water, the amount of energy it
takes for the vapor pressure to equal the
external pressure is reached at 100 ºC.
Boiling
Point
However, when a solute like NaCl is added, the
vapor pressure of the solution is lowered, and it
takes more energy to equalize the vapor and
external pressure, causing a higher boiling point.
Vapor
pressure
Scaffolded Example
1.
What is the boiling point
elevation when 35.0 g of NaCl is
dissolved in
750 g of water?
(Kb for water is .52 ºC/m)
STEP 1:
3:
2: FIND
Plug
Van'tin
MOLALITY
Hoff
values
Factor
ΔTb = Kb (m)(i)
K = .52 ºC/m
m = .799 m
i = 2.00
b
1 mole NaCl = .599 moles NaCl
58.43 g NaCl
b 1 kgb H2O
750g H2O X
Step 1: Find molality
= .750 kg H2O
1000 g H2O
35.0g NaCl X
ΔT = NaCl
K (m)(i)
ΔTb = .52ºC/m(.799m)(2.00)
Na+
molality = moles solute
kg solvent
Step 2: Determine Van't Hoff Factor
Cl-
m= .599 moles NaCl
.750 kg H2O
ΔTb = .831 ºC
molality
.799 m
2 ions == Van't
Hoff Factor of 2
Step 3: Plug in values into the formula
Practice 1
1.
What is the boiling point elevation when 15
g of ammonia (NH3) is dissolved in 250 g of H2O?
(Kb for water is .52 ºC/m)
Δ Tb = Kb(i)(m)
Practice 1: Answer
Δ Tb = Kb (i)(m)
1.
1.
1.
15 g NH3
m=
x
1 mole NH3
17.024 g NH3
.88 moles NH3
.25 kg H20
=
.88 mole NH3
3.52 m
Δ Tb = .52ºC/m(1)(3.52m)
1.
Δ Tb = 1.8 ºC
Practice 2
1.
What is the new boiling point
when 23 g of ammonia (NH3) is
dissolved in 180 g of H2O?
(Kb for water is .52 ºC/m)
Δ Tb = Kb(i)(m)
Practice 2: Answer
Δ Tb = Kb (i)(m)
1.
1.
23.0 g NH3
m=
x
1 mole NH3
17.024 g NH3
1.35 moles NH3
.18 kg H20
=
1.35 mole NH3
7.5 m
1.
Δ Tb = .52ºC/m(1)(7.5m)
1.
Δ Tb = 3.5ºC
1.
100ºC+3.5ºC = 103.5ºC
Freezing Point Depression
-Freezing point depression is the ability of a dissolved
solute to lower the freezing point of its solution.
-When a solute is added to a solvent, more kinetic energy
must be withdrawn from the solution for it to solidify.
The equation for freezing point depression is:
Freezing Point
Depression
ΔTf = Kf (m)(i)
Freezing
Constant
Molalit
y
Van't Hoff
Constant
Freezing Point Depression
Explanation
Free moving molecules
in liquid water.
Water freezing
Water and NaCl
molecules
Water partially
freezing
With pure water, the water molecules move freely
around, and it is easy for the water to freeze into a
lattice structure as solid ice.
However, when a solute like NaCl is added, the additional
movement from the solute particles impairs the freezing
process and makes the freezing process require more
energy, lowering the freezing point.
Scaffolded Example
1.
What is the freezing point
depression when 65.0 g of NaCl is
dissolved in 1250 g of water?
(Kf for water is -1.86 ºC/m)
STEP 1:
3:
2: FIND
Plug
Van'tin
MOLALITY
Hoff
values
Factor
ΔTf = Kf (m)(i)
Kf = -1.86 ºC/m
m = .888 m
i = 2.00
1 mole NaCl = 1.11 moles NaCl
58.43 g NaCl
f 1 kg fH2O
1250g H2O X
Step 1: Find molality
= 1.250 kg H2O
1000 g H2O
65.0g NaCl X
ΔT =NaCl
K (m)(i)
ΔTf = -1.86ºC/m(.888m)(2.00)
Na+
molality = moles solute
kg solvent
Step 2: Determine Van't Hoff Factor
Cl-
m= 1.11 moles NaCl
1.250 kg H2O
ΔTf = -3.31 ºC
2
ions = =Van't
molality
.888Hoff
m Factor of 2
Step
Step 3:
3: Plug
Plug in
in values
values into
into the
the formula
formula
PRACTICE 1
1.
What is the new freezing point when 90.0 g
of oxygen gas is dissolved in 1750 g of water?
(Kf for water is -1.86 ºC/m)
ΔTf = Kf (m)(i)
PRACTICE 1: ANSWERS
1.
1.
1.
90.0 g O2
m=
ΔTf = Kf (m)(i)
x
1 mole O2
32.00 g O2
2.80 moles O2
1.75 kg H20
=
2.80 moles O2
1.60 m
Δ Tf = -1.86ºC/m(1.60m)(1)
Δ Tf = -2.98 ºC
0 - 2.98 = -2.98 ºC
1.
2.
PRACTICE 2
1.
What is the freezing point depression when
48.0 g of oxygen gas is dissolved in 2300 g of water?
(Kf for water is -1.86 ºC/m)
ΔTf = Kf (m)(i)
PRACTICE 2: ANSWERS
ΔTf = Kf (m)(i)
1.
1.
1.
48.0 g O2
m=
x
1 mole O2
32.00 g O2
1.50 moles O2
2.30 kg H20
=
1.50 moles O2
.652 m
Δ Tf = -1.86ºC/m(.652m)(1)
1.
Δ Tf = -1.21ºC