II

III

I I. The Nature of Mixtures

MixturesMixtures

MixturesMixtures

Heterogeneous Mixture – a non-uniform or unequal blend of two or more pure substances• Suspension – mixture containing

particles that will settle out• Colloid – mixture containing particles

with a size of 1nm – 1000nm, and do not separate – stay suspended

MixturesMixtures

Tyndall Effect – Caused by dilute colloids, which appear to be homogeneous.• Is the scattering of light as it passes

though a dilute colloid

MixturesMixtures

Solutions - Solutions - homogeneous mixtures

Solvent Solvent - present in greater amount

Solute Solute - substance being dissolved

Homogeneous Mixtures

• A substance that dissolves in a solvent is soluble.

• Two liquids that are soluble in each other in any proportion are miscible.

• A substance that does not dissolve in a solvent is insoluble.

• Two liquids that can be mixed but separate shortly after are immiscible.

SolutionsSolutions

Solvation – Solvation – the process of dissolving – sugar dissolves in to water

solute particles are separated and pulled into solutionhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html

solute particles are surrounded by solvent particles

SolvationSolvation

StrongElectrolyte

Non-Electrolyte

solute exists asions only

- +

salt

- +

sugar

solute exists asmolecules

only

- +

acetic acid

WeakElectrolyte

solute exists asions and

molecules

SolvationSolvation

Molecular Molecular SolvationSolvation• molecules

stay intact

C6H12O6(s) C6H12O6(aq)

SolvationSolvation

DissociationDissociation• separation of an ionic solid into

aqueous ions• http://chemmovies.unl.edu/ChemAnime/NACL1D/NACL1D.html

NaCl(s) Na+(aq) + Cl–(aq)

SolvationSolvation

NONPOLAR

NONPOLAR

POLAR

POLAR

““Like Dissolves Like”Like Dissolves Like”““Like Dissolves Like”Like Dissolves Like”

SolvationSolvation

Heat of Solution – energy change that occurs during the formation of the solution.• Exothermic – solvation produces a

warm solution• Endothermic – solvation produces a

cold solution

Factors Affecting Solvation (solubility)

Factors Affecting Solvation (solubility)

Agitation – stirring or shaking – allows for more collisions (mixing) between solute and solvent

Surface Area – smaller pieces – more collisions

Temperature – in general solids dissolve faster in higher temps. – more collisions• Opposite for gases – colder the better

SolubilitySolubility

SolubilitySolubility• maximum grams of solute that will

dissolve in 100 g of solvent at a given temperature

• varies with temp• based on a saturated soln• Book Reference: p 969 R-3 and p 974

R-8• http://chemmovies.unl.edu/ChemAnime/SLBIL1D/SLBIL1D.html

Solubility RulesSolubility Rules

In general:

• Group I ions and Ammonium are soluble

• Acetates and Nitrates are soluble

• Cl, Br, I are soluble, except with Pb, Ag, Hg2+2

• Sulfates are soluble, except with Ba, Sr, Pb, Ca, Ag, Hg2

+2

• Carbonates, Hydroxides, oxides, sulfides, phosphates are INsoluble

SolubilitySolubility

SATURATED SOLUTION

no more solute dissolves

UNSATURATED SOLUTIONmore solute dissolves

SUPERSATURATED SOLUTION

becomes unstable, crystals form

concentration

SolubilitySolubility

Solubility CurveSolubility Curve• shows the

dependence of solubility on temperature

SolubilitySolubility

Solids are more soluble at...Solids are more soluble at...• high temperatures.

Gases are more soluble at...Gases are more soluble at...

• low temperatures &

• high pressures (Henry’s Law).

Types of Reactions in Aqueous Solutions

• When two solutions that contain ions as solutes are combined, the ions might react.

• If they react, it is always a double replacement reaction.

• Three products can form: precipitates, water, or gases.

SolubilitySolubility

Replacement Reactions • Double Replacement Reactions occur

when ions exchange between two compounds.

• This figure shows a generic double replacement equation.

SolubilitySolubility

SolubilitySolubility

• Aqueous solutions of sodium hydroxide and copper(II) chloride react to form the precipitate copper(II) hydroxide.

2NaOH(aq) + CuCl2(aq) → 2NaCl(aq) + Cu(OH)2(s)

• Ionic equations that show all of the particles in a solution as they actually exist are called complete ionic equations.

2Na+(aq) + 2OH–(aq) + Cu2+ (aq)+ 2Cl–(aq) → 2Na+

(aq) + 2Cl–(aq) + Cu(OH)2(s)

Types of Reactions in Aqueous Solutions (cont.)

• Ions that do not participate in a reaction are called spectator ions and are not usually written in ionic equations.

• Formulas that include only the particles that participate in reactions are called net ionic equations.

2OH–(aq) + Cu2+(aq) → Cu(OH)2(s)

A. Concentration is...

a measure of the amount of solute dissolved in a given quantity of solvent

A concentrated solution has a large amount of solute

A dilute solution has a small amount of solute• thus, only qualitative descriptions

A. ConcentrationA. Concentration

Describing Concentration Quantitatively

• % by mass

• % by volume

• Molarity

• Molality

• Mole Fraction

B. PercentB. Percent

100xsolution of mass

solute of massMassby Percent

100xsolution ofvolume

solute ofvolume Volumeby Percent

B. PercentB. Percent

What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with water?

17% 4.8 g of NaCl are dissolved in 82.0 mL of

solution. What is the percent of the solution? 5.53% How many grams of salt are there in 52 mL of

a 6.3 % solution? 3.3g

C. MolarityC. Molarity

solvent of L

solute of moles(M)molarity

Volume of solvent only

1 kg water = 1 L water

L 1

mol0.25 0.25M

C. MolarityC. Molarity

Find the molarity of a solution containing 75 g of MgCl2 in 250 mL of water.

75 g MgCl2 1 mol MgCl2

95.21 g MgCl2

= 3.2 M MgCl2

0.25 L water

L

molM

C. MolarityC. Molarity

How many grams of NaCl are req’d to make a 1.54 M solution using 0.500 L of water?

0.500 L water 1.54 mol NaCl

1 L water

= 45.0 g NaCl

58.44 g NaCl

1 mol NaCl

L 1

mol1.54 1.54M

D. MolalityD. Molality

1 L water = 1 Kg water so … Molality (m) = moles of solute / Kg of solvent

Basically the same calculations, but different units

solvent ofKg

solute of moles(m)molality

2211 VMVM

E. DilutionE. Dilution

Preparation of a desired solution by adding water to a concentrate.

Moles of solute remain the same.

E. DilutionE. Dilution

What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:

M1 = 15.8M

V1 = ?

M2 = 6.0M

V2 = 250 mL

WORK:

M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

F. Preparing Solutions F. Preparing Solutions

500 mL of 1.54M NaCl

500 mLwater

45.0 gNaCl

• mass 45.0 g of NaCl• add water until total

volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water

500 mLmark

500 mLvolumetric

flask

1.54m NaCl in 0.500 kg of water

F. Preparing SolutionsF. Preparing Solutions

250 mL of 6.0M HNO3 by dilution

• measure 95 mL of 15.8M HNO3

95 mL of15.8M HNO3

water for

safety

250 mL mark

• combine with water until total volume is 250 mL

• Safety: “Do as you oughtta, add the acid to the watta!”

Solution Preparation LabSolution Preparation Lab Turn in one paper per team. Complete the following steps:

A) Show the necessary calculations.

B) Write out directions for preparing the solution.

C) Prepare the solution. For each of the following solutions:

1) 100.0 mL of 0.500M NaCl

2) 0.250m NaCl in 100.0 mL of water

3) 100.0 mL of 1.00M Red Solution from 12.1M concentrate.

II

III

I III. Colligative Properties

SolutionsSolutions

A. DefinitionA. Definition

Colligative PropertyColligative Property

• property that depends on the

concentration of solute particles, not

their identity

A. ElectrolytesA. Electrolytes

Dissolved Solute particles disrupt the normal Intermolecular Forces of the Solvent

Molecules – Count as one dissolved solute particle• Example: CH3OH = 1 solute particle

Ionic Compounds – number of solute particles is equal to the total number of ions in the neutral formula• Example: AlCl3 = 4 solute particles • 1 Al and 3 Cl ions

B. TypesB. Types

Vapor Pressure Lowering• Solutions have a lower vapor pressure

than the original pure solvent• Because the solute particles are

attracted to solvent particles cause more IMF

• So more IMF = less solvent particles becoming vapors

B. TypesB. Types

Freezing Point DepressionFreezing Point Depression (tf)

• f.p. of a solution is lower than f.p. of the pure solvent

Boiling Point ElevationBoiling Point Elevation (tb)

• b.p. of a solution is higher than b.p. of the pure solvent

B. TypesB. Types

Freezing Point Depressionhttp://chemmovies.unl.edu/ChemAnime/SOLND/SOLND.html

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html

B. TypesB. Types

Solute particles increase IMF of the solvent.

Boiling Point Elevation

C. Phase DiagramC. Phase Diagram

D. ApplicationsD. Applications

salting icy roadsmaking ice creamAntifreeze - cars (-64°C to 136°C)

E. CalculationsE. Calculations

t: change in temperature (°C)

k: constant based on the solvent (°C·kg/mol)

m: molality (m)

n or i: # of particles

t = k · m · n

E. CalculationsE. Calculations

# of Particles# of Particles

• Nonelectrolytes (covalent)• remain intact when dissolved • 1 particle

• Electrolytes (ionic)• dissociate into ions when dissolved• 2 or more particles

E. CalculationsE. Calculations

At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?

m = 3.2mn = 1tb = kb · m · n

WORK:

m = 0.73mol ÷ 0.225kg

GIVEN:b.p. = ?tb = ?

kb = 3.56°C·kg/moltb = (3.56°C·kg/mol)(3.2m)(1)

tb = 11°C

b.p. = 181.8°C + 11°C

b.p. = 193°C

E. CalculationsE. Calculations

Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.

m = 4.8m

n = 2

tf = kf · m · n

WORK:

m = 0.48mol ÷ 0.100kg

GIVEN:

f.p. = ?

tf = ?

kf = 1.86°C·kg/mol

tf = (1.86°C·kg/mol)(4.8m)(2)

tf = 18°C

f.p. = 0.00°C - 18°C

f.p. = -18°C