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SolutionsSolutionsCh. 15 (p. 501-529)Ch. 15 (p. 501-529)

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Sugar in waterAir

Dental fillingsSaline

Sugar in waterAir

Dental fillingsSaline

SolutionsSolutions

Solution:Solution: A homogeneous mixture of twoor more components in a single physical state

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Characteristics of a SolutionCharacteristics of a Solution

•The solute can’t be filtered out.•The solute always stays mixed.•Particles are always in motion.•Volumes may not be additive.•A solution will have different

properties than the solvent.

A solution consists of two components:

solventsolvent - component in the greater extent

solutesolute - component in the lesser extent(There may be more than

one.)

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Physical states of solutionsPhysical states of solutions

Solutions can be made that exist in any of the three states. (See Fig. 15-3, p. 503)

Solid solutionsSolid solutionsdental fillings, 14K gold, sterling silver

Liquid solutionsLiquid solutionssaline, liquor, antifreeze, vinegar

Gas solutionsGas solutionsair, anesthesia gases

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Solution DefinitionsSolution Definitions

Alloy: Alloy: solution of two or more metals

Soluble: Soluble: able to dissolve in another substance

Insoluble: Insoluble: not able to dissolve in another substance

Miscible: Miscible: pairs of liquids that mix in any amount (Ex. ethanol and water)

Immiscible: Immiscible: pairs of liquids that do not mix in any amount (Ex. oil and water)

Aqueous: Aqueous: solutions with water as the solvent

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Amount of Solute in SolutionAmount of Solute in Solution

ConcentrationConcentration: the amount of solute per quantity of solvent or solution

There are many systems - we will cover four.

• Molarity (M)• Mass percent• Mole fraction (X)• Molality (m)

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Example: MolarityExample: Molarity

Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH.

MNaOH = 10 mol / 2.0 L

= 5.0 M

Molarity (M) = moles solute molliters of solution L=

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Mass PercentMass Percent

Mass % = Mass SoluteTotal Mass

x 100

If a ham contained 5 grams of fat in 200 gof ham, what is the mass percent?

Use the same units for bothUse the same units for both

5 g / 200g * 100 = 2.5 %

On the label, it would say 97.5 % fat free.

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MolalityMolality

Example: Calculate the molality of a 10.0 g of KCl in 80.0 grams of water.

Molality (m) = moles solute molkilograms of solvent kg

=

m 1.68kg 0.080

g 74.6mole 1

g 10 molality

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Mole fractionMole fraction

Mole fractionMole fractionThe moles of solute, expressed as a fraction of the total number of moles in the solution.

XA =

Because the units in the numerator and denominator are the same, mole fraction is a unitless quantity.

The sum of all components must equal one.

nA

nA + nB + nC + . . . .

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Example: Mole FractionExample: Mole Fraction

What is the mole fraction of sulfur dioxide (SO2) in an industrial exhaust gas containing 128.0 g of sulfur dioxide dissolved in 1500. g of carbon dioxide.

g 44.01mole

g 1500g 64.1

mole 1g 128

g 64.1mole 1

g 128

moles totalSO of moles

X 2SO2

0.05539moles 34.09 moles 1.999

moles 1.999

XCO2 = 1 – XSO2 = 1 – 0.05539 = .9446

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SaturationSaturation

When a solution contains as much solute as it can at a given temperature.

UnsaturatedUnsaturated Can still dissolve more.

SaturatedSaturated Have dissolved all you can.

SupersaturatedSupersaturated Temporarily have dissolved

too much.

PrecipitatePrecipitate Excess solute that falls out of solution.

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Properties of Aqueous SolutionsProperties of Aqueous Solutions

There are two general classes of solutes.

ElectrolytesElectrolytes•ionic compounds in polar solvents•dissociate in solution to make ions•conduct electricity•may be strong (100% dissociation) or

weak (less than 100%)

NonelectrolytesNonelectrolytes•do not conduct electricity•solute is dispersed but does not

dissociate

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Colligative PropertiesColligative Properties

“Bulk” properties that change when you add a solute to make a solution.

•Based on how much you add but not what the solute is.•Effect of electrolytes is based on

number of ions produced.

Colligative propertiesColligative properties•vapor pressure lowering•freezing point depression•boiling point elevation•osmotic pressure

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Vapor Pressure LoweringVapor Pressure Lowering

The introduction of a nonvolatile solute will reduce the vapor pressure of the solvent in the resulting solution.

• The vapor pressure of a nonvolatile component is essentially zero.

• It does not contribute to the vapor pressure of the solution.

• However, the solution’s vapor pressure is dependent on the solute mole fraction.

Volatile: has a measurable vapor pressure

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Demo: Vapor Pressure LoweringDemo: Vapor Pressure Lowering

Water will end up in the ‘salt’ solution because it’svapor pressure is lower than the pure water.

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Vapor Pressure (A Dynamic Process)Vapor Pressure (A Dynamic Process)

liquid

vapor

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Boiling Point ElevationBoiling Point Elevation

When you add a nonvolatile solute to a solvent, the boiling point goes up. This is because the vapor pressure has been lowered.

Tb = Kb m

The boiling point will continue to be elevated as you add more solute until you reach saturation.

ExamplesExamples Cooking pasta in salt water Antifreeze

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Example: Boiling point elevationExample: Boiling point elevation

ExampleExampleDetermine the boiling point for a 0.222 m aqueous solution of sucrose.

Kb = 0.512 oC/m for water.

Tb = 0.512 oC/m (0.222 m)

= 0.114 oC

Tb = 100.00 oC + 0.114 oC = 100.11 oC

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Phase Diagram (Boiling Point Phase Diagram (Boiling Point Elevation)Elevation)

Diagram: www.sparknotes.com

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Freezing Point DepressionFreezing Point Depression

When you add a solute to a solvent, the freezing point goes down.

Tf = Kf m

The more you add, the lower it gets.

This will only work until you reach saturation.

ExamplesExamples “Salting” roads in winter Making ice cream

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Example: Freezing Point Example: Freezing Point DepressionDepression

ExampleExampleDetermine the freezing point for a 0.222 m aqueous solution of sucrose.

Kf = 1.86 oC/m for water.

Tf = -1.86 oC/m (0.222 m)

= -0.413 oC

Tf = 0.00 oC - 0.413 oC = -0.41 oC

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Example Constants (p. 522/526) Example Constants (p. 522/526)

Normal Kb Normal Kf

Solvent Tb, oCoC/m Tf, oC oC/m

Water 100.0+0.512 0.0 -1.86

Benzene 80.1+2.53 5.5-5.12

Camphor 207.4+5.61 178.8 -39.7

Ethanol 78.3 +1.22 -117.3 -1.99

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Phase Diagram (Freezing Point Depression)Phase Diagram (Freezing Point Depression)

Diagram: www.sparknotes.com

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Electrolytes and Colligative PropertiesElectrolytes and Colligative Properties

Ionic substances have a greater effect per mole than covalent.

• 1 mol/kg of water for glucose = 1 molal

• 1 mol/kg of water for NaCl = 2 molal ions

• 1 mol/kg of water for CaCl2 = 3 molal ions

Effects are based on the number of particles!Effects are based on the number of particles!

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Van’t Hoff FactorVan’t Hoff Factor

Van’t Hoff factor (i): number of ions created.

• Glucose (nonelectrolyte)—i = 1

• NaCl (Na+ + Cl-)—i = 2

• CaCl2(Ca2+ + 2 Cl-)—i = 3

Tb = iKbm

Tf = iKfm

Equations become:

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Example: ElectrolytesExample: ElectrolytesExampleExample

Determine the boiling point for a 1.35 m aqueous solution of MgCl2.

i = 3 (Mg2+ + 2 Cl-)

Kb = 0.512 oC/m for water.

Tb = (3)0.512 oC/m (1.35 m)

= 2.07oC

Tb = 100.00 oC + 2.1 oC = 102.1 oC

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Determining Molar MassDetermining Molar Mass

EXAMPLE: A 10.0 gram sample of an unknown nonvolatile, nonelectrolyte compound is dissolved in 100 g of water. The boiling point of the solution is elevated to 0.433 ◦C. What is the molar mass of the unknown sample? (Kb for water is 0.52 ◦C/m)

Given: munk=10.0 gmsolv=100 g = .100 kgTb = 0.433 ◦Ci=1Kb=0.52 ◦C/m

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Determining Molar Mass (page 2)Determining Molar Mass (page 2)

Given: munk=10.0 gmsolv=100 g = .100 kgTb = 0.433 ◦CKb = 0.52 ◦C/m

Equation: Tb = Kbm

m= Tb/Kb = 0.433/0.52 = 0.83 m (Step 1)

By Definition: m = mol solute/kg solvent

Rearranging: mol solute = m x kg solvent (Step 2)= 0.83 m * .100 kg = 0.083 mol

Conclusion: 10 g/0.083 mol = 120 g/mole (Step 3)

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How Solutions FormHow Solutions Form

When an ionic solid is placed in a polar solvent, the outer ions are exposed to the polar molecules.

Water will pull the ions from the solid and surround them - solvate solvate them.

Solvation of ions is an exothermic process which helps overcome the lattice energy that holds the crystal together. Hydration is this same process—with water as the solvent. (Solvation is entropically favorable, as well.)

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Solution of solidsSolution of solids

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Factors Affecting SolubilityFactors Affecting Solubility

1) Nature of Solvent and Solute

2) Temperature

3) Pressure

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Nature of Solvent & SoluteNature of Solvent & Solute

““Like dissolves like.”Like dissolves like.”Materials with similar polarity are soluble in

each other. Dissimilar ones are not

Ionic substances are not soluble in nonpolar solvents like hexane.

• A large amount of energy is need to separate the ions.

• A nonpolar solvent can’t solvate ions so there is no solvation energy to offset the lattice energy.

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Pressure and solubility of Pressure and solubility of gasesgases

Henry’s LawHenry’s LawAt constant temperature, the solubility of a gas is directly proportional to the pressure of the gas above the solution.

cg = kpgas

This law is accurate to within 1-3% for slightlysoluble gases and pressures up to oneatmosphere.

0 1 2 Pressure (atm)

Solu

bili

ty(g

/10

0g

wate

r)

0.010

0.005

0.000

O2

N2

He

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Solubilities of solidsSolubilities of solids

Ionic substances are not soluble in nonpolar solvents like hexane.

• A large amount of energy is need to separate the ions.

• A nonpolar solvent can’t solvate ions so there is no solvation energy to offset the lattice energy.

Predicting the solubility of ionic solids in water is difficult because a number of competing factors are involved.

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Solution of solidsSolution of solids

While covalent compounds do not dissociate, they are solvated in solution.

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Saturated SolutionsSaturated Solutions

At saturation, the solute is in dynamic equilibrium. The concentration is constant.

Solute species areconstantly inmotion, movingin and out of solution.

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Why do some Why do some curves curves decrease with decrease with increasing increasing temperature temperature and some and some increase?increase?