1 © 2009 brooks/cole - cengage solutions chapter 14 why does a raw egg swell or shrink when placed...

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© 2009 Brooks/Cole - Cengage

SolutionsSolutionsChapter 14Chapter 14

Why does a raw egg swell or shrink when Why does a raw egg swell or shrink when placed in different solutions?placed in different solutions?

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Some DefinitionsSome Definitions

A solution is a A solution is a HOMOGENEOUSHOMOGENEOUS mixture mixture of 2 or more substances of 2 or more substances in a single phase. in a single phase.

One constituent is usually One constituent is usually regarded as the regarded as the SOLVENTSOLVENT and the others as and the others as SOLUTESSOLUTES..

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Solutions can be Solutions can be classified as classified as saturatedsaturated or or

ununsaturatedsaturated..

A saturated solution A saturated solution contains the maximum contains the maximum quantity of solute that quantity of solute that dissolves at that dissolves at that temperature.temperature.

DefinitionsDefinitions

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Solutions can be classified Solutions can be classified as as unsaturatedunsaturated or or saturatedsaturated..

A saturated solution A saturated solution contains the maximum contains the maximum quantity of solute that quantity of solute that dissolves at that dissolves at that temperature.temperature.

SUPERSATURATED SUPERSATURATED SOLUTIONSSOLUTIONS contain contain more than is possible and more than is possible and are unstable.are unstable.

DefinitionsDefinitions

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Dissolving An Ionic Dissolving An Ionic SolidSolid

See Active Figure 14.9

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An An IDEAL SOLUTIONIDEAL SOLUTION is is one where the properties one where the properties depend only on the depend only on the concentration of solute.concentration of solute.

Need concentration units to Need concentration units to tell us the number of solute tell us the number of solute particles per solvent particles per solvent particle.particle.

The unit molarity (review) The unit molarity (review) does not do this!does not do this!

Concentration UnitsConcentration Units

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Concentration UnitsConcentration Units

MOLE FRACTION, XMOLE FRACTION, X

For a mixture of A, B, and CFor a mixture of A, B, and C

MOLE FRACTION, XMOLE FRACTION, X

For a mixture of A, B, and CFor a mixture of A, B, and C

XA mol fraction A =

mol A

mol A + mol B + mol C XA mol fraction A =

mol A

mol A + mol B + mol C

molality of solute (m) =

mol solute

kilograms solvent molality of solute (m) =

mol solute

kilograms solvent

WEIGHT %WEIGHT % = grams solute per 100 g solution = grams solute per 100 g solution

MOLALITY, mMOLALITY, m

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Calculating Calculating ConcentrationsConcentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate mol fraction, molality, and weight O. Calculate mol fraction, molality, and weight

% of glycol.% of glycol.

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250. g H250. g H22O = 13.9 molO = 13.9 mol

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

Xglycol= 1.00 mol glycol

1.00 mol glycol + 13.9 mol H2O

Xglycol= 1.00 mol glycol

1.00 mol glycol + 13.9 mol H2O

X X glycolglycol = 0.0672 = 0.0672

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Calculate molalityCalculate molality

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

conc (molality) = 1.00 mol glycol

0.250 kg H2O 4.00 molal

conc (molality) = 1.00 mol glycol

0.250 kg H2O 4.00 molal

% glycol =

62.1 g

62.1 g + 250. g x 100% = 19.9%

% glycol =

62.1 g

62.1 g + 250. g x 100% = 19.9%

Calculate weight %Calculate weight %

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Energetics Energetics of the Solution of the Solution

ProcessProcess

See Energetics of Dissolution of KF simulation

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Energetics Energetics of the Solution of the Solution

ProcessProcessIf the enthalpy of If the enthalpy of

formation of the formation of the solution is more solution is more negative than that negative than that of the solvent and of the solvent and solute, the enthalpy solute, the enthalpy of solution is of solution is negative. negative.

The solution process The solution process is is exothermicexothermic!!PLAY MOVIE

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SupersaturatedSupersaturatedSodium Sodium

AcetateAcetate

• One application of a One application of a supersaturated supersaturated solution is the solution is the sodium acetate sodium acetate “heat pack.”“heat pack.”

• Sodium acetate has Sodium acetate has an an ENDOthermicENDOthermic heat of solution. heat of solution.

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SupersaturatedSupersaturated Sodium AcetateSodium Acetate

Sodium acetate has an Sodium acetate has an ENDOthermicENDOthermic heat of heat of solution. solution.

NaCHNaCH33COCO22 (s) + (s) + heatheat Na Na++(aq) + CH(aq) + CH33COCO22

--(aq)(aq)

Therefore, formation of solid sodium acetate Therefore, formation of solid sodium acetate from its ions is from its ions is EXOTHERMICEXOTHERMIC..

NaNa++(aq) + CH(aq) + CH33COCO22--(aq) (aq)

NaCH NaCH33COCO22 (s) + (s) + heatheat

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Dissolving Gases & Dissolving Gases & Henry’s LawHenry’s Law

Gas solubility (mol/L) = kGas solubility (mol/L) = kHH ·· P Pgasgas

kkHH for O for O22 = 1.66 x 10 = 1.66 x 10-6-6 M/mmHg (1.3 x 10 M/mmHg (1.3 x 10-3 -3 mol/kg * bar)mol/kg * bar)

When When PPgasgas drops, solubility drops. drops, solubility drops.

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PPsolventsolvent = X = Xsolventsolvent · P · Poosolventsolvent


Raoult’s LawRaoult’s LawAn ideal solution obeys this law. Because solvent An ideal solution obeys this law. Because solvent

vapor pressure & the relative # of solvent vapor pressure & the relative # of solvent molecules are proportional:molecules are proportional:

Because mole fraction of solvent, XBecause mole fraction of solvent, XAA, is always , is always

less than 1, then Pless than 1, then PAA is always less than P is always less than PooAA..

The vapor pressure of solvent over a solution is The vapor pressure of solvent over a solution is

always always LOWEREDLOWERED!!

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Raoult’s LawRaoult’s LawAssume the solution containing 62.1 g of glycol in Assume the solution containing 62.1 g of glycol in

250. g of water is ideal. What is the vapor pressure 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 of water over the solution at 30 ooC? C? (The VP of pure (The VP of pure HH22O is 31.8 mm Hg; see App. E.)O is 31.8 mm Hg; see App. E.)

SolutionSolution

XXglycolglycol = 0.0672 = 0.0672 and so Xand so Xwaterwater = ? = ?

Because XBecause Xglycolglycol + X + Xwaterwater = 1 = 1

XXwaterwater = 1.000 - 0.0672 = 0.9328 = 1.000 - 0.0672 = 0.9328

PPwaterwater = X = Xwaterwater ·· P Poowaterwater = (0.9328)(31.8 mm Hg) = (0.9328)(31.8 mm Hg)

PPwaterwater = 29.7 mm Hg = 29.7 mm Hg

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Colligative PropertiesColligative PropertiesOn adding a solute to a solvent, the properties On adding a solute to a solvent, the properties

of the solvent are modified.of the solvent are modified.

• Vapor pressure Vapor pressure decreasesdecreases

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)

These changes are called These changes are called COLLIGATIVE COLLIGATIVE PROPERTIESPROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of solute of solute particles relative to solvent particles, not on particles relative to solvent particles, not on the the KINDKIND of solute particles. of solute particles.

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Understanding Understanding Colligative PropertiesColligative Properties

To understand colligative properties, study To understand colligative properties, study the the LIQUID-VAPOR EQUILIBRIUMLIQUID-VAPOR EQUILIBRIUM for a for a solution.solution.

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To understand To understand colligative colligative properties, properties, study the study the LIQUID-VAPOR LIQUID-VAPOR EQUILIBRIUMEQUILIBRIUM for a solution.for a solution.

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VP of HVP of H22O over a solution depends on the O over a solution depends on the

number of Hnumber of H22O molecules per solute O molecules per solute

molecule.molecule.

PPsolventsolvent proportional toproportional to X Xsolventsolvent

VP of solvent over solution VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)= (Mol frac solvent)•(VP pure solvent)

RAOULT’S LAWRAOULT’S LAW

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Changes in Freezing and Changes in Freezing and Boiling Points of SolventBoiling Points of Solvent

See Figure 14.13See Figure 14.13

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

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Vapor Vapor PressurPressur

e e LowerinLowerin

gg

See Figure 14.13

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The boiling point of a The boiling point of a solution is higher than solution is higher than

that of the pure that of the pure solvent.solvent.

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Elevation of Boiling Point Elevation of Boiling Point Elevation in BP = ∆TElevation in BP = ∆TBPBP = K = KBPBP·m·m

(where K(where KBPBP is characteristic of solvent from table 14.3 p.633) is characteristic of solvent from table 14.3 p.633)

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

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Change in Boiling Point Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g Dissolve 62.1 g of glycol (1.00 mol) in 250. g

of water. What is the BP of the solution?of water. What is the BP of the solution?

KKBPBP = +0.512 = +0.512 ooC/molal for water (see Table C/molal for water (see Table 14.3).14.3).

SolutionSolution1.1. Calculate solution molality = 4.00 mCalculate solution molality = 4.00 m

2.2. ∆T∆TBPBP = K = KBPBP ·· m m

∆∆TTBPBP = +0.512 = +0.512 ooC/molal (4.00 molal)C/molal (4.00 molal)

∆∆TTBPBP = +2.05 = +2.05 ooCC BP = 102.05 BP = 102.05 ooCC

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Change in Freezing Change in Freezing Point Point

The freezing point of a solution is The freezing point of a solution is LOWERLOWER than that of the pure solvent.than that of the pure solvent.

FP depression = ∆TFP depression = ∆TFPFP = K = KFPFP·m·m

Pure waterPure water Ethylene glycol/water Ethylene glycol/water solutionsolution

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Lowering the Freezing Lowering the Freezing PointPoint

Water with and without antifreeze When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

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Calculate the FP of a 4.00 molal glycol/water Calculate the FP of a 4.00 molal glycol/water solution.solution.

KKFPFP = -1.86 = -1.86 ooC/molal (Table 14.3)C/molal (Table 14.3)

SolutionSolution

∆∆TTFPFP = K = KFPFP ·· m m

= (-1.86 = (-1.86 ooC/molal)(4.00 m)C/molal)(4.00 m)

∆∆TTFP FP = -7.44 = -7.44 ooCC

Recall that ∆TRecall that ∆TBPBP = +2.05 ˚C for this solution. = +2.05 ˚C for this solution.

Freezing Point Freezing Point DepressionDepression

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How much NaCl must be dissolved in 4.00 How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 kg of water to lower FP to -10.00 ooC?C?

SolutionSolution

Calc. required molalityCalc. required molality

∆∆TTFPFP = K = KFPFP ·· m m

-10.00 -10.00 ooC = (-1.86 C = (-1.86 ooC/molal)(Conc)C/molal)(Conc)

Conc = 5.38 molal Conc = 5.38 molal


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How much NaCl must be dissolved in 4.00 kg of water to lower How much NaCl must be dissolved in 4.00 kg of water to lower

FP to -10.00 FP to -10.00 ooC?C?..

SolutionSolution

Conc req’d = 5.38 molalConc req’d = 5.38 molal

This means we need 5.38 mol of dissolved This means we need 5.38 mol of dissolved particles per kg of solvent. particles per kg of solvent.

Recognize that m represents the Recognize that m represents the total concentration of all total concentration of all dissolved particles.dissolved particles.

Recall that 1 mol NaCl(aq) Recall that 1 mol NaCl(aq) 1 mol Na 1 mol Na++(aq) + 1 mol (aq) + 1 mol

ClCl--(aq)(aq)


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How much NaCl must be dissolved in 4.00 kg of How much NaCl must be dissolved in 4.00 kg of

water to lower FP to -10.00 water to lower FP to -10.00 ooC?C?..SolutionSolution

Conc req’d = 5.38 molalConc req’d = 5.38 molal

We need 5.38 mol of dissolved particles per kg of We need 5.38 mol of dissolved particles per kg of solvent. solvent.

NaCl(aq) NaCl(aq) Na Na++(aq) + Cl(aq) + Cl--(aq)(aq)

To get 5.38 mol/kg of particles we needTo get 5.38 mol/kg of particles we need

5.38 mol / 2 = 2.69 mol NaCl / kg5.38 mol / 2 = 2.69 mol NaCl / kg

2.69 mol NaCl / kg 2.69 mol NaCl / kg 157 g NaCl / kg 157 g NaCl / kg

(157 g NaCl / kg)(4.00 kg) = (157 g NaCl / kg)(4.00 kg) = 629 g NaCl629 g NaCl


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Boiling Point Elevation Boiling Point Elevation and Freezing Point and Freezing Point

DepressionDepression ∆∆T = K·m·iT = K·m·iA generally useful equation A generally useful equation

i = van’t Hoff factor = number of particles i = van’t Hoff factor = number of particles produced per formula unit.produced per formula unit.

CompoundCompound Theoretical Value of iTheoretical Value of i

glycolglycol 11

NaClNaCl 22

CaClCaCl22 33

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OsmosisOsmosis

Dissolving the shell in vinegar

Egg in corn syrupEgg in pure water

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OsmosisOsmosis

The semipermeable membrane allows only the The semipermeable membrane allows only the movement of solvent molecules.movement of solvent molecules.

Solvent Solution

Semipermeable membrane

Solvent Solution

Semipermeable membrane

Solvent molecules move from pure Solvent molecules move from pure solvent to solution in an attempt solvent to solution in an attempt

to make both have the same to make both have the same concentration of solute.concentration of solute.

Driving force is entropyDriving force is entropy

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Process of OsmosisProcess of Osmosis

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Osmotic Pressure, ∏Osmotic Pressure, ∏Equilibrium is reached when Equilibrium is reached when

pressure — the pressure — the OSMOTIC OSMOTIC PRESSUREPRESSURE, , ∏∏ — — produced by extra solution produced by extra solution counterbalances pressure counterbalances pressure of solvent molecules of solvent molecules moving thru the moving thru the membrane.membrane.

∏ ∏ = cRT= cRT

((cc is conc. in mol/L) is conc. in mol/L)

(R = 0.082057 L*atm/K*mol)(R = 0.082057 L*atm/K*mol)

Osmotic pressure

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OsmosisOsmosis

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Osmosis Osmosis at the Particulate Levelat the Particulate Level

See Figure 14.17

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OsmosiOsmosiss

• Osmosis of solvent Osmosis of solvent from one solution to from one solution to another can continue another can continue until the solutions are until the solutions are ISOTONICISOTONIC — they — they have the same have the same concentration.concentration.

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Osmosis and Living Osmosis and Living CellsCells

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Reverse OsmosisReverse OsmosisWater DesalinationWater Desalination

Water desalination plant in Tampa

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Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Dissolve 35.0 g of hemoglobin in enough water Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. to make 1.00 L of solution. ∏∏ measured to measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.hemoglobin.

SolutionSolution(a)(a) Calc. Calc. ∏∏ in atmospheres in atmospheres ∏∏ = (10.0 mmHg)(1 atm / 760 mmHg)= (10.0 mmHg)(1 atm / 760 mmHg)

= 0.0132 atm= 0.0132 atm(b)(b) Calc. concentrationCalc. concentration

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Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Conc = 5.39 x 10Conc = 5.39 x 10-4-4 mol/L mol/L

(c)(c) Calc. molar massCalc. molar mass

Molar mass = 35.0 g / 5.39 x 10Molar mass = 35.0 g / 5.39 x 10-4-4 mol/L mol/L

Molar mass = 65,100 g/molMolar mass = 65,100 g/mol

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. of solution. ∏ ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.molar mass of hemoglobin.

SolutionSolution

(b)(b)Calc. concentration from Calc. concentration from ∏∏ = cRT = cRT

Conc. =0.0132 atm / (0.0821) (298 K)Conc. =0.0132 atm / (0.0821) (298 K)

L*atm/K*molL*atm/K*mol

1 © 2009 brooks/cole - cengage solutions chapter 14 why does a raw egg swell or shrink when placed...

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