CHAPTER 12
SOLUTIONS
CHAPTER 12
Section 1 – Types of Mixtures
 Section 2 – The Solution Process
 Section 3 – Concentration of Solutions

12.1 - TYPES OF MIXTURES
Distinguish between electrolytes and
nonelectrolytes.
List three different solute-solvent combinations.
Compare the properties of suspensions, colloids,
and solutions.
Distinguish between electrolytes and
nonelectrolytes.
DEFINITIONS

Soluble: capable of being dissolved
What is something that is soluble?

Solution: homogeneous mixture of two or more
substances in a single phase
Review: What is a homogeneous mixture?
HOMOGENEOUS VS HETEROGENEOUS
DEFINITIONS CONT.
 Solvent:
substance doing the dissolving
 Solute: substance dissolved in solution
 Can
be any combination of solid, liquid, or
gas

A solid solution is called an alloy

Example: brass is zinc and copper
EXAMPLES
In these examples, the substance
with the greater volume is the solvent
PARTICLE MODELS FOR GOLD AND
GOLD ALLOY
SUSPENSIONS

Suspension: particles in a solvent are so large
they settle out unless the mixture is constantly
stirred
What is an example of a suspension?
SUSPENSIONS
COLLOIDS

Colloid: particles that are intermediate in size
between those in solutions and suspensions
the particles are small enough to be suspended
throughout the solvent by the constant movement
of the surrounding molecules

Colloidal particles make up the dispersed phase,
and water is the dispersing medium.

Example: mayonnaise
TYNDALL EFFECT

Tydnall Effect: when light is scattered by colloidal
particles dispersed in a transparent medium


This is used to distinguish between a solution and a
colloid
Remember the Types of Solutions Lab from the
beginning of the year?
COLLOIDS
EMULSIONS
a mixture of two or more immiscible (un-blendable) liquids
SOLUTIONS, COLLOIDS, AND SUSPENSIONS
ELECTROLYTES

Electrolyte: a substance that dissolves in water to
give a solution that conducts electric current

What would make a good electrolyte – ionic or
covalent compounds? Why?

Ionic – the positive and negative ions separate from
each other in solution and are free to move making it
possible for an electric current to pass through the
solution
NONELECTROLYTE

Nonelectrolyte: a substance that dissolves in
water to give a solution that does not conduct
electricity

Neutral solute molecules do not contain mobile
charged particles so they do not conduct electric
current

Example: sugar
ELECTRICAL CONDUCTIVITY OF SOLUTIONS
12.2 - THE SOLUTION PROCESS
List and explain three factors that affect the rate at
which a solid solute dissolves in a liquid solvent.
Explain solution equilibrium, and distinguish among
saturated, unsaturated, and supersaturated solutions.
Explain the meaning of “like dissolves like” in terms of
polar and nonpolar substances.
List the three interactions that contribute to the
enthalpy of a solution, and explain how they combine to
cause dissolution to be exothermic or endothermic.
Compare the effects of temperature and pressure on
solubility.
DISSOLVING PROCESS VIDEO
FACTORS AFFECTING THE
RATE OF DISSOLUTION
1. Increase the surface area
-- because dissolution occurs at the surface
2. Stirring or shaking
-- increases contact between solvent and solute
3. Increase the temperature
-- increases collisions between solute and solvent
and are of higher energy
FACTORS AFFECTING THE RATE OF DISSOLUTION
SOLUBILITY

If you add spoonful after spoonful of sugar to tea,
eventually no more sugar will dissolve.

There is a limit to the amount of solute that can
dissolve in a solvent.

The limit depends on:
Solute
 Solvent
 Temperature

PARTICLES OF SOLUBLE SUBSTANCES
PARTICLES OF INSOLUBLE SUBSTANCES
SOLUBILITY CONT.
1.
When a solute is added, the molecules leave the
solid surface and move about at random.
2.
As more solute is added, more collisions occur
between dissolved solute particles. Some of the
solute molecules return to the crystal.
3.
When maximum solubility is reached molecules
are returning to the solid form at the same rate
they are going into solution (much like the vapor
pressure of a liquid)

This is called solution equilibrium
SOLUTION EQUILIBRIUM VIDEO
SOLUTIONS

Saturated Solution: contains the maximum
amount of dissolved solute

Supersaturated Solution: contains more dissolved
solute than a saturated solution

Unsaturated Solution: contains less solute than a
saturated solution under the same condition
MASS OF SOLUTE ADDED VERSUS MASS OF SOLUTE
DISSOLVED
SOLUTIONS CONT.

When a saturated solution is cooled, the excess
solute usually comes out of solution, leaving the
solution saturated at the lower temperature.

A supersaturated solution will form crystals of solute
if disturbed or more solute is added.
SOLUBILITY VALUES

Solubility: the amount of substance required to
form a saturated solution with a specific amount of
solvent at a specified temperature


Example: solubility of sugar is 204 g per 100 g of water
at 20OC
Solubilities vary widely and must be determined
experimentally
SOLUBILITY OF COMMON COMPOUNDS
SOLUBILITY OF COMMON COMPOUNDS
SATURATED SOLUTION AND TEMPERATURE
SOLUTE-SOLVENT INTERACTIONS

Solubility depends on the compounds involved.

“LIKE DISSOLVES LIKE” is used to predict
whether one substance will dissolve in another

What makes substances similar depends on:
Type of bond
 Polar/Nonpolar molecules
 Intermolecular forces

LIKE DISSOLVES LIKE
DISSOLVING IONIC COMPOUNDS IN
AQUEOUS SOLUTIONS

What type of forces are present in water?

These forces attract the ions in ionic compounds
and surround them, separating them from the
crystal surface and drawing them into solution.

Hydration: when water is the solvent

Ions are said to be hydrated
HYDRATION OF LITHIUM CHLORIDE
NONPOLAR SOLVENTS

Ionic compounds are generally not soluble in
nonpolar solvents

Example: carbon tetrachloride, CCl4
toluene, C6H5CH3

The nonpolar solvent does not attract the ions to
break apart the forces holding the crystal together.

There are differences in:
__________ , ___________ , __________
LIQUID SOLUTES AND SOLVENTS

Oil and water do not mix because oil is nonpolar and
water is polar. The hydrogen bonding squeezes out
whatever oil molecules may come between them.

Two polar substances or two nonpolar substances
easily form solutions.
Immiscible: liquids that are not soluble in each other
 Miscible: liquids that dissolve freely in one another in
proportion

COMPARING MISCIBLE AND
IMMISCIBLE LIQUIDS VIDEO
PRESSURE AND SOLUBILITY

Increases in pressure, increase gas solubilities in
liquids

An equilibrium is established between a gas above
a liquid solvent and the gas dissolved in a liquid

 solution
gas + solvent 

PRESSURE AND SOLUBILITY CONT.

Increasing the pressure…
causes gas particles to collide with the liquid
surface and forces more gas into the solution
 gas + solvent 

 solution

Decreasing the pressure…
allows more dissolved gas to escape from solution


 gas + solvent 
 solution
SODA CARBONATION VIDEO
HENRY’S LAW

Henry’s Law: solubility of a gas in a liquid is directly
proportional to the partial pressure of that gas on the
surface of the liquid
 In carbonated beverages, the solubility of carbon dioxide is
increased by increasing the pressure. The sealed containers
contain CO2 at high pressure, which keeps the CO2 dissolved
in the beverage, above the liquid.
 When the beverage container is opened, the pressure above
the solution is reduced, and CO2 begins to escape from the
solution.

Effervescence: the rapid escape of a gas from a liquid
HENRY’S LAW VIDEO
EFFERVESCENCE
TEMPERATURE AND SOLUBILITY OF GAS

Increasing the temperature…
Increases the average kinetic energy and allows more
solute to escape the attraction of the solvent and
escape to the gas phase

At higher temperatures, equilibrium is reached with
fewer gas molecules in solution
TEMPERATURE AND SOLUBILITY OF
LIQUIDS AND SOLIDS

Increasing the temperature…
Increases in temperature usually increases the solubility of
solids in liquids

The solubility depends on the solid and properties
of the solid.

A few solid solutes are actually less soluble at
higher temperatures.
SOLUBILITY VS. TEMPERATURE
ENTHALPIES OF SOLUTION

The formation of a solution is accompanied by an
energy change.
 If you dissolve some potassium iodide, KI, in
water, you will find that the outside of the
container feels cold to the touch.
 But if you dissolve some sodium hydroxide,
NaOH, in the same way, the outside of the
container feels hot.
 The formation of a solid-liquid solution can
either absorb energy (KI in water) or release
energy as heat (NaOH in water)
ENTHALPIES OF SOLUTION CONT
Before dissolving begins, solute particles are held
together by intermolecular forces. Solvent particles
are also held together by intermolecular forces.
Energy changes occur during solution formation
because energy is required to separate solute
molecules and solvent molecules from their
neighbors.
A solute particle that is surrounded by solvent
molecules is solvated.
ENTHALPY CHANGES DURING THE
FORMATION OF A SOLUTION
ENTHALPIES OF SOLUTION CONT.

Enthalpy of Solution: net amount of energy
absorbed as heat by the solution when a specific
amount of solute dissolves in a solvent

Negative – energy is released


(Steps 1/2 attractions is less than Step 3)
Positive – energy is absorbed

(Steps 1/2 attractions is greater than Step3)
12.3 - CONCENTRATION OF SOLUTIONS
Given the mass of solute and volume of solvent,
calculate the concentration of solution.
Given the concentration of a solution, determine the
amount of solute in a given amount of solution.
Given the concentration of a solution, determine the
amount of solution that contains a given amount of
solute.
CONCENTRATION

Concentration: measure of the amount of solute in
a given amount of solvent or solution

Concentration is a ratio – any amount of a given
solution has the same concentration

Dilute: the opposite of concentrated

These terms DO NOT relate to saturation!
CONCENTRATION UNITS
CONCENTRATION
MOLARITY

Molarity(M): moles of solute in one liter of solution
M =
moles of solute
liters of solution
PREPARING A SOLUTION

NOTE: 1 M solution is NOT made by adding 1 mol of
solute to 1 L of solvent
EXAMPLE PROBLEM A

You have 3.50 L of solution that contains 90.0 g of
sodium chloride. What is the molarity of that solution?
Given: solute mass = 90.0 g sodium chloride
solution volume = 3.50 L
 Unknown: molarity of NaCl solution

1 mol NaCl
90.0 g NaCl 
 1.54 mol NaCl
58.44 g NaCl
1.54 mol NaCl
 0.440 M NaCl
3.50 L of solution
EXAMPLE PROBLEM B
You
have 0.8 L of a 0.5 M HCl solution. How many
moles of HCl does this solution contain?
Given: volume of solution = 0.8 L
concentration of solution = 0.5 M HCl
 Unknown: moles of HCl in a given volume

0.5 mol HCl
 0.8 L of solution  0.4 mol HCl
1.0 L of solution
EXAMPLE PROBLEM C
To
produce 40.0 g of silver chromate, you will need at
least 23.4 g of potassium chromate in solution as a
reactant.
If you have 5 L of a 6.0 M K2CrO4 solution. What
volume of the solution is needed to give you the
23.4 g K2CrO4 needed for the reaction?

Given: volume of solution = 5 L
concentration of solution = 6.0 M K2CrO4
mass of solute = 23.4 K2CrO4
mass of product = 40.0 g Ag2CrO4

Unknown: volume of K2CrO4 solution in L
EXAMPLE PROBLEM C SOLUTION

Find Molar Mass of Solute:
1 mol K 2CrO4  194.2 g K 2CrO4

Find Moles of Solute:
1 mol K 2CrO4
23.4 g K 2CrO4 
 0.120 mol K 2CrO4
194.2 g K 2CrO4

Find Volume of Solution:
0.120 mol K 2CrO4
6.0 M K 2CrO4 
x L K 2CrO4 solution
x  0.020 L K 2CrO4 solution
MOLALITY

Molality(m): moles of solute per kilogram of solvent
Molality (m) = mol of solute
kg of solvent

Molality is used when studying properties of solutions related
to vapor pressure and temperature changes,
because molality does not change with temperature.
MAKING A MOLAL SOLUTION
EXAMPLE PROBLEM A
A
solution was prepared by dissolving 17.1 g of
sucrose (table sugar, C12H22O11) in 125 g of water.
Find the molal concentration of this solution.
Given: solute mass = 17.1 C12H22O11
solvent mass = 125 g H2O
 Unknown: molal concentration

SAMPLE PROBLEM A SOLUTION

Convert the Given:
1 mol C12H22O11
17.1 g C12H22O11 
 0.0500 mol C12H22O11
342.34 g C12H22O11
125 g H2O
 0.125 kg H2O
1000 g/ kg

Find Molality:
0.0500 mol C12H22O11
 0.400 m C12H22O11
0.125 kg H2O
EXAMPLE PROBLEM B

A solution of iodine, I2, in carbon tetrachloride, CCl4,
is used when iodine is needed for certain chemical
tests. How much iodine must be added to prepare a
0.480 m solution of iodine in CCl4 if 100.0 g of CCl4
is used?
Given: molality of solution = 0.480 m I2
mass of solvent = 100.0 g CCl4
 Unknown: mass of solute

MOLALITY, CONTINUED

Convert Solvent:
100.0 g CCl4
 0.100 kg CCl4
1000 g/ kg

Find moles:
x mol I2
0.480 m 
0.1 kg H2O

x  0.0480 mol I2
Convert to grams:
253.8 g I2
0.480 mol I2 
 12.2 g I2
mol I2
CHAPTER 13
IONS IN SOLUTION AND
COLLIGATIVE PROPERTIES
CHAPTER 13

Section 1 – Compounds in Aqueous Solutions

Section 2 – Colligative Properties
13.1 COMPOUNDS IN AQ SOLUTIONS
Write equations for the dissolution of soluble ionic
compounds in water.
 Predict whether a precipitate will form when
solutions of soluble ionic compounds are combined,
and write net ionic equations for precipitation
reactions.
 Compare dissociation of ionic compounds with
ionization of molecular compounds.
 Draw the structure of the hydronium ion, and
explain why it is used to represent the hydrogen ion
in solution.
 Distinguish between strong electrolytes and weak
electrolytes.

DISSOCIATION

Dissociation: separation of ions that occurs when
an ionic compound dissolves
H2O
NaCl(s ) 
 Na  (aq ) + Cl– (aq )
1 mol
1 mol
1 mol
H2O
CaCl2 (s ) 
 Ca2 (aq ) + 2Cl– (aq )
1 mol
1 mol
2 mol
DISSOCIATION OF NaCl
EXAMPLE PROBLEM 1

Write the equation for the dissolution of aluminum
sulfate, Al2(SO4)3 , in water.
How many moles of aluminum ions and sulfate ions are
produced by dissolving 1 mol of aluminum sulfate?
 What is the total number of moles of ions produced by
dissolving 1 mol of aluminum sulfate?

EXAMPLE PROBLEM 1 SOLUTION

Equation:
3
Al2 (SO4 )3 (s) 
 2Al (aq ) + 3SO (aq )
H2O

2–
4
How many moles of each ion?
3
a. 1 mol Al2 (SO4 )3  2 mol Al

+ 3 mol SO
What is the total number of ions present?
2–
4
PRECIPITATION REACTIONS

No ionic compound is completely insoluble, but
ones with very low solubility can be considered to
be insoluble and to form a precipitate.
REMEMBER
THE SOLUBILITY
RULES!!!
GENERAL SOLUBILITY GUIDELINES
WHICH IONIC COMPOUNDS ARE INSOLUBLE?
NiCl2
KMnO4 CuSO4 Pb(NO3)2
AgCl
CdS
PARTICLE MODEL FOR THE
FORMATION OF A PRECIPITATE
PRECIPITATION REACTIONS VIDEO
NET IONIC EQUATIONS

Total Ionic Equation: contains all compounds and
ions including any substances that are not involved
in the reaction (spectator ions)

Spectator ions: do not take part in a chemical
reaction and are found in solution before and after
the reaction

Net Ionic Equation: includes only those
compounds and ions that undergo a chemical
change in a reaction in an aqueous solution
REACTION
Cd(NO3)2(aq) + (NH4)2S(aq)  CdS(s) + 2NH4(NO3)(aq)
TOTAL IONIC EQUATION
Cd2 (aq ) + 2NO3– (aq ) + 2NH4 (aq ) + S2– (aq ) 
CdS(s ) + 2NO3– (aq ) + 2NH4 (aq )
NET IONIC EQUATION
2
Cd (aq ) + S (aq )  CdS(s )
2–
EXAMPLE REACTION
Starting with two
solutions of
Potassium Sulfate
and Barium Nitrate
At the end of the
reaction, there is
one solution and
one precipitate
WRITING NET IONIC EQUAITONS
The ions crossed out in blue are the
spectator ions and do not take part
in the reaction.
NET IONIC EQUATION VIDEO
EXAMPLE PROBLEM 2

Identify the precipitate that forms when aqueous
solutions of zinc nitrate and ammonium sulfide are
combined. Write the equation for the possible doubledisplacement reaction. Then write the formula
equation, overall ionic equation, and net ionic
equation for the reaction.
EXAMPLE PROBLEM 2 SOLUTION
1. Write the reaction:
Zn(NO3 )2 (aq ) + (NH4 )2 S(aq )  ZnS(? ) + 2NH4NO3 (? )
2. Identify precipitate:
Zn(NO3 )2 (aq ) + (NH4 )2 S(aq )  ZnS(s ) + 2NH4NO3 (aq )
3. Write the ionic equation:
Zn2 (aq ) + NO3– (aq ) + NH4 (aq ) + S2– (aq )
 ZnS(s ) + NH4 (aq ) + NO3– (aq )
4. Write the net ionic equation:
Zn2 (aq ) + S2– (aq )  ZnS(s )
IONIZATION

Ionization: formation of ions from solute molecules
by the action of the solvent

When a molecular compound dissolves and ionizes
in a polar solvent, ions are formed where none
existed in the undissolved compound

Example: HCl is molecular but ionizes in water
H2O
HCl 
 H (aq ) + Cl– (aq )
DISSOCIATION AND IONIZATION VIDEO
HYDRONIUM ION

When the H+ ion is formed, it attracts other
molecules or ions so it does not exist alone.

It forms the hydronium ion: H3O+
H2O
HCl 
 H3O (aq ) + Cl– (aq )
STRONG AND WEAK ELECTROLYTES

What is an electrolyte?

The strength of an electrolyte depends on the ability
of a substance to form ions, or the degree of
ionization or dissociation.
MODELS FOR STRONG AND WEAK
ELECTROLYTES AND NONELECTROLYTES
STRONG ELECTROLYTES

Strong Electrolyte: conducts electricity well due to
presence of all or almost all of the dissolved
compound as ions

What is an example of a strong electrolyte?
WEAK ELECTROLYTES

Weak Electrolyte: conducts electricity poorly due to
presence of a small amount of the dissolved
compound as ions

What is an example of a weak electrolyte?

–


HF 
H
O
(
aq
)
+
F
(aq )
 3
[HF] >> [H+] and [F–]
13.2 COLLIGATIVE PROPERTIES

List four colligative properties, and explain why
they are classified as colligative properties.

Calculate freezing-point depression, boiling-point
elevation, and solution molality of nonelectrolyte
solutions.

Calculate the expected changes in freezing point
and boiling point of an electrolyte solution.

Discuss causes of the differences between
expected and experimentally observed colligative
properties of electrolyte solutions.
COLLIGATIVE PROPERTIES
 Colligative
Properties: properties that depend
on the concentration of solute particles but not
on their identity
Vapor Pressure Lowering
 Freezing Point Depression
 Boiling Point Elevation
 Osmotic Pressure

FREEZING POINT DEPRESSION
 Freezing

Difference between the freezing points of the
pure solvent and a solution
 Molal

Point Depression (∆tf )
freezing point constant (Kf)
Freezing point depression of a solvent in 1 molal
solution solution (nonvolatile, nonelectrolyte)
∆tf = Kfm
MOLAL FREEZING POINT AND
BOILING POINT CONSTANTS
FREEZING-POINT DEPRESSION
EXAMPLE PROBLEM 3

What is the freezing-point depression of water in a
solution of 17.1 g of sucrose, C12H22O11, in 200. g
of water? What is the actual freezing point of the
solution?
Given: 17.1 g C12H22O11 of solute
200. g H2O of solvent
 Unknown: a. freezing-point depression
b. freezing point of the solution

EXAMPLE PROBLEM 3 SOLUTION

Calculate Moles:
17.1 g C12H22O11 

1 mol solute
 0.0500 mol C12H22O11
342.34 g C12H22O11
Calculate Molality:
0.0500 mol C12H22O11
1000 g water

200. g water
1 kg water

0.250 mol C12H22O11
kg water
 .250 m
EXAMPLE PROBLEM 3 CONT.

Calculate Freezing Point Depression:
∆tf = Kfm
∆tf = 0.250 m × (−1.86°C/m) = −0.465°C

Calculate New Freezing Point:
Freezing Pointsolution= Freezing Pointsolvent + ∆tf
FPsolution= 0.000°C + (−0.465°C) = −0.465°C
BOILING POINT ELEVATION
 Boiling

Difference between the boiling points of the pure
solvent and a solution
 Molal

Point Elevation (∆tb)
boiling point constant (Kb)
Boiling point elevation of a solvent in 1 molal solution
solution (nonvolatile, nonelectrolyte)
∆tb = Kbm
BOILING-POINT ELEVATION AND THE
PRESENCE OF SOLUTES
EXAMPLE PROBLEM 4

What is the boiling-point elevation of a solution made
from 20.1 g of a nonelectrolyte solute and 400.0 g of
water? The molar mass of the solute is 62.0 g.
Given: solute mass = 20.1 g
solute molar mass = 62.0 g/mol
solvent mass and identity = 400.0 g H2O
 Unknown: boiling-point elevation

EXAMPLE PROBLEM 4 SOLUTION

Calculate Moles:
1 mol solute
20.1 g of solute 
 0.324 mol of solute
62.0 g of solute

Calculate Molality:
0.324 mol of solute
1000 g water

400.0 g water
1 kg water
mol solute
 0.810
 0.810 m
kg water

Calculate Boiling Point Elevation:
∆tb = 0.51°C/m × 0.810 m = 0.41°C
ELECTROLYTES AND COLLIGATIVE PROPERTIES
Electrolytes depress the freezing point and elevate
the boiling point of a solvent MORE than expected.
 Electrolytes produce MORE THAN 1 mol of solute
particles for each mole of a compound:

C12H22O11
Moles of Solute?
HO
___

 C12H22O11(aq )
2
H2O
CaCl2 (s ) 
 Ca2 (aq ) + 2Cl– (aq )
___
H2O
NaCl(s ) 
 Na  (aq ) + Cl– (aq )
___
CALCULATIONS FOR ELECTROLYTES

Colligative properties depend on the total
concentration of solute particles.

The changes in colligative properties cause by
electrolytes is proportional to the TOTAL molality of
all dissolved particles, not to formula units.

For the same molal concentrations of sucrose and
sodium chloride…

What would you expect the effect on the colligative
property to be? By what proportion is there a difference?
Sample
Problem F
What is the expected change in the freezing point
of water in a solution of 62.5 g of barium nitrate,
Ba(NO3)2, in 1.00 kg of water?
Sample
Problem F Solution
Given:
solute mass and formula = 62.5 g Ba(NO3)2

solvent mass and identity = 1.00 kg water

∆tf = Kfm
Unknown: expected freezing-point depression
Solution:
mass of solute (g)
1 mol solute


mass of solvent (kg)
molar mass solute (g)
 mol 
molality of solute 

 kg 
Sample
Problem F Solution, continued
Solution:
 mol 
 mol ions 
molality of solute 
  molality conversion 

kg
mol




 C  kg H2O 
 Kf 
 exp ected freezing - point depression (C)

 mol ions 
62.5 g Ba(NO3 )2
mol Ba(NO3 )2
0.239 mol Ba(NO3 )2


1.00 kg H2O
261.35 g Ba(NO3 )2
kg H2O
Sample
Problem F Solution, continued
Solution:
H2O
Ba(NO3 )2 (s ) 
 Ba2 (aq ) + 2NO3– (aq )
Each formula unit of barium nitrate yields three
ions in solution.
0.239 mol Ba(NO3 )2
-1.86C kg H2O
3 mol ions


kg H2O
mol Ba(NO3 )2
mol ions
 -1.33C
ACTUAL VALUES FOR ELECTROLYTES
The actual values of the colligative properties for all strong
electrolytes are almost what would be expected based
on the number of particles they produce in solution.
ACTUAL VALUES FOR ELECTROLYTES

The differences between the expected and
calculated values are caused by the attractive
forces that exist between dissociated ions in
aqueous solution.

According to Debye and Hückel a cluster of
hydrated ions can act as a single unit rather than as
individual ions, causing the effective total
concentration to be less than expected.

Ions of higher charge have lower effective
concentrations than ions with smaller charge.