CHAPTER 12SOLUTIONS

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 gasA 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

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.

12.2 - THE SOLUTION PROCESS

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

gas + solvent solution

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

gas + solvent solution

PRESSURE AND SOLUBILITY CONT.

Increasing the pressure…causes gas particles to collide with the liquid

surface and forces more gas into the 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)

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.

12.3 - CONCENTRATION OF SOLUTIONS

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 soluteliters 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 NaCl90.0 g NaCl 1.54 mol NaCl

58.44 g NaCl

1.54 mol NaCl

3.50 L of 0.44

solu0 M

til

onNaC

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

1.0 L of solut0.4 mol

ion HCl

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

2 42 4

2 4

0.120 mol K CrO6.0 M K CrO

L K CrO solutionx

EXAMPLE PROBLEM C SOLUTION

Find Molar Mass of Solute:

Find Moles of Solute:

Find Volume of Solution:

2 4 2 41 mol K CrO 194.2 g K CrO

2 42 4 2 4

2 4

1 mol K CrO23.4 g K CrO 0.120 mol K CrO

194.2 g K CrO

2 40.020 L K CrO solutionx

MOLALITY

Molality(m): moles of solute per kilogram of solvent

Molality is used when studying properties of solutions related to vapor pressure and temperature changes, because molality does not change with temperature.

Molality (m) = mol of solute kg of solvent

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:

Find Molality:

12 22 1112 22 11 12 22 11

12 22 11

1 mol C H O17.1 g C H O 0.0500 mol C H O

342.34 g C H O

22

125 g H O0.125 kg H O

1000 g/kg

12 22 112 22 11

1

2

0.0500 mol C H O

0.125 0.400 C

kg H OH Om

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:

Find moles:

Convert to grams:

44

100.0 g CCl0.100 kg CCl

1000 g/kg

x

m x22

2

mol I0.480 0.0480 mol I

0.1 kg H O

2 22

2

253.8 g I0.480 mol I

mol12.2 I

I g

CHAPTER 13IONS 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.

s aq + aq2H O –NaCl( ) Na ( ) Cl ( )

s aq + aq2H O 2 –2CaCl ( ) Ca ( ) 2Cl ( )

1 mol 1 mol 1 mol

1 mol 1 mol 2 mol

DISSOCIATION

Dissociation: separation of ions that occurs when an ionic compound dissolves

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:

How many moles of each ion?

What is the total number of ions present?

s aq + aq2H O 3 2–2 4 3 4Al (SO ) ( ) 2Al ( ) 3SO ( )

+ 3 2–2 4 3 41 mol Al (SO ) 2 mol Al 3 mol. SOa

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

aq + aq + aq + aq

s + aq + aq

2 3– 4 2–

3– 4

Cd ( ) 2NO ( ) 2NH ( ) S ( )

CdS( ) 2NO ( ) 2NH ( )

aq + aq s 2 2–Cd ( ) S ( ) CdS( )

TOTAL IONIC EQUATION

NET IONIC EQUATION

REACTION

Cd(NO3)2(aq) + (NH4)2S(aq) CdS(s) + 2NH4(NO3)(aq)

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 double-displacement reaction. Then write the formula equation, overall ionic equation, and net ionic equation for the reaction.

EXAMPLE PROBLEM 2 SOLUTION

1. Write the reaction:

2. Identify precipitate:

aq + aq ? + ?3 2 4 2 4 3Zn(NO ) ( ) (NH ) S( ) ZnS( ) 2NH NO ( )

aq + aq s + aq3 2 4 2 4 3Zn(NO ) ( ) (NH ) S( ) ZnS( ) 2NH NO ( )

3. Write the ionic equation:

4. Write the net ionic equation:

aq + aq + aq + aq

s + aq + aq

2 – 2–3 4

–4 3

Zn ( ) NO ( ) NH ( ) S ( )

ZnS( ) NH ( ) NO ( )

aq + aq s 2 2–Zn ( ) S ( ) ZnS( )

aq + aq2H O –HCl H ( ) Cl ( )

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

DISSOCIATION AND IONIZATION VIDEO

aq + aq2H O –3HCl H O ( ) Cl ( )

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+

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?

[HF] >> [H+] and [F–]

aq + aq–3HF H O ( ) F ( )

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?

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 LoweringFreezing Point DepressionBoiling Point ElevationOsmotic Pressure

FREEZING POINT DEPRESSION

Freezing Point Depression (∆tf )Difference between the freezing points of the

pure solvent and a solution

Molal 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 depressionb. freezing point of the solution

EXAMPLE PROBLEM 3 SOLUTION

Calculate Moles:

Calculate Molality:

m

12 22 11

12 22 11

0.0500 mol C H O 1000 g water

200. g water 1 kg water

0.250 mol C H O.250

kg water

12 22 11 12 22 1112 22 11

1 mol solute17.1 g C H O 0.0500 mol C H O

342.34 g C H O

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 Point Elevation (∆tb)Difference between the boiling points of the pure

solvent and a solution

Molal 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:

Calculate Molality:

Calculate Boiling Point Elevation:

1 mol solute20.1 g of solute 0.324 mol of solute

62.0 g of solute

m

0.324 mol of solute 1000 g water

400.0 g water 1 kg water

mol solute0.810 0.810

kg water

∆tb = 0.51°C/m × 0.810 m = 0.41°C

s aq + aq2H O –NaCl( ) Na ( ) Cl ( )

s aq + aq2H O 2 –2CaCl ( ) Ca ( ) 2Cl ( )

aq2H O12 22 11 12 22 11C H O C H O ( )

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:

Moles of Solute?

___

___

___

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 FWhat 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 SolutionGiven: 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)

molmolality of solute

kg

Sample Problem F Solution, continuedSolution:

f

K

2

mol mol ionsmolality of solute molality conversion

kg mol

C kg H Oexpected freezing - point depression ( C)

mol ions

3 2 3 2 3 2

2 3 2 2

62.5 g Ba(NO ) mol Ba(NO ) 0.239 mol Ba(NO )

1.00 kg H O 261.35 g Ba(NO ) kg H O

3 2 2

2 3 2

0.239 mol Ba(NO ) -1.86 C kg H O3 mol ions

kg H O mol Ba(NO ) mol

-1.33

i ns

C

o

Sample Problem F Solution, continuedSolution:

s aq + aq2H O 2 –3 2 3Ba(NO ) ( ) Ba ( ) 2NO ( )

Each formula unit of barium nitrate yields three ions in solution.

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

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.