Their Concentrations and Colligative Properties

Home , Colligative properties, Ideal solution, Lattice energy

5/31/2018

Chapter 11: Properties of Solutions - Their Concentrations and Colligative Properties

Chapter Outline

. 11.1 Energy Changes when Substances Dissolve . 11.2 Vapor Pressure . 11.3 Mixtures of Volatile Substances . 11.4 Colligative Properties of Solutions . 11.5 Osmosis and Osmotic Pressure . 11.6 Using Colligative Properties to Determine Molar Mass

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Enthalpy of Solution

. Dissolution of Ionic Solids:

• Enthalpy of solution (∆Hsoln) depends on: » Energies holding solute ions in crystal lattice » Attractive force holding solvent molecules together » Interactions between solute ions and solvent molecules

• ∆Hsoln = ∆Hion-ion + ∆Hdipole-dipole + ∆H ion-dipole • When solvent is water: = » ∆Hsoln ∆Hion-ion + ∆Hhydration

Enthalpy of Solution for NaCl

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Calculating Lattice Energies Using the Born-Haber Cycle

Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. It is always endothermic.

2+ - e.g. MgF2(s)  Mg (g) + 2F (g)

푄 푥푄 Lattice energy (E) increases as Q 퐸 = 2.31 푥 10−19퐽 ∙ 푛푚 1 2 푒푙 푑 increases and/or as r decreases.

Lattice Q1 is the charge on the cation Compound Energy (U) Q2 is the charge on the anion

d is the distance between the ions MgF2 2957 Q= +2,-1

MgO 3938 Q= +2,-2

LiF 1036 radius F < LiCl 853 radius Cl

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Born-Haber Cycles – Calculating Ulattice

Hess’ Law is used to calculate Ulattice The calculation is broken down into a series of steps (cycles)

Steps:

1. Sublimate the metal = ΔHsub

2. if necessary, breaking bond of a diatomic = ΔHBE

3. ionization of the metal = IE1 + IE2 etc

4. electron affinity of the nonmetal = EA1 + EA2 etc

5. formation of 1 mole of the salt from ions = ΔHlast = -Ulattice

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In General for M(s) + n X2(g)  MX2n(s)

M+(g) + X(g) M+(g) + X(g)

EA1 + EA2 etc IE1 + IE2 etc M+(g) + X-(g)

M(g) + X(g)

n ΔHBE M(g) + n X2(g) ΔHlast = -Ulattice

ΔHsub M(s) + n X2(g)

ΔHf MX(s)

Calculating Ulattice

° ΔHf = ΔHsub + n ΔHBE +  IE +  EA + ΔHlast

Rearranging and solving for ΔHlast -

ΔHlast = ΔHf − ΔHsub − n ΔHBE −  IE −  EA

Ulattice = -ΔHlast

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Calculating Ulattice for NaCl(s)

Na+(g) + Cl(g) Na+(g) + Cl(g)

EA1 IE1 Na+(g) + Cl-(g)

Na(g) + Cl(g)

½ ΔHBE Na(g) + ½ Cl2(g) ΔHlast = -Ulattice

ΔHsub Na(s) + ½ Cl2(g)

ΔHf NaCl(s)

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ΔHlast = ΔHf − ΔHsub − ½ ΔHBE − IE1 − EA1

Sample Exercise 11.1 Calculating Lattice Energy

Calcium fluoride occurs in nature as the mineral fluorite, which is the principle source of the world’s supply of fluorine. Use the following data to calculate the lattice energy of CaF2.

ΔHsub,Ca = 168 kJ/mol

BEF2 = 155 kJ/mol

EAF = -328 kJ/mol

IE1,Ca = 590 kJ/mol

IE2,Ca = 1145 kJ/mol

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Chapter Outline

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Vapor Pressure Pressure exerted by a gas in equilibrium with its liquid

Where are we going with this? Colligative Properties – boiling point elevation, freezing point depression, osmosis

Vapor Pressure of Solutions

• Vapor Pressure: – Pressure exerted by a gas in equilibrium with liquid – Rates of evaporation/condensation are equal

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The tendency for a liquid to evaporate increases as -

1. the temperature rises 2. the surface area increases 3. the intermolecular forces decrease

Boiling Point

“If you have a beaker of water open to the atmosphere, the mass of the atmosphere is pressing down on the surface. As heat is added, more and more water evaporates, pushing the molecules of the atmosphere

aside (wsystem < 0). If enough heat is added, a temperature is eventually reached at which the vapor pressure of the liquid equals the atmospheric pressure, and the liquid boils.”

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Normal boiling point

the temperature at which the vapor pressure of a liquid is equal to the external atmospheric pressure of 1 atm.

Increasing the external atmospheric pressure increases the boiling point

Decreasing the external atmospheric pressure decreases the boiling point

o Location Elevation (ft) b.p H2O ( C) San Francisco sea level 100.0 Salt Lake City 4400 95.6 Denver 5280 95.0 Mt. Everest 29,028 76.5

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Intermolecular Forces and Vapor Pressure

Diethyl ether - dipole-dipole interactions

Water - hydrogen bonding (stronger)

Clausius-Clapeyron Equation Vapor Pressure vs Temperature

H vap  1  lnPvap       C R  T 

liquid  solid

gas Pressure

Temperature 

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Clausius-Clapeyron Equation Vapor Pressure vs Temperature

Plot of ln(P) vs 1/T yields straight line:

• Slope = −ΔHvap/R • Intercept = constant

Sample Exercise 11.2:

Calculating Vapor Pressure Using Hvap

The compound with the common name isooctane has the structure o shown below. Its normal boiling point is 99 C, and Hvap = 35.2 kJ/mol. What is the vapor pressure at 25 oC?

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Practice Exercise, p. 473

Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into Styrofoam, which is used in coffee cups and other products that have good thermal insulation properties. The normal boiling point of pentane is 36 oC; its vapor pressure at 25 oC is 505 torr. What is the enthalpy of vaporization of pentane?

Chapter Outline

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Fractional Distillation

. A method of separating a mixture of compounds based on their different boiling points . Vapor phase enriched in the more volatile component

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Solutions of Volatile Compounds

. Raoult’s Law: • Total vapor pressure of an ideal solution depends on how much the partial pressure of each volatile component contributes to total vapor pressure of solution

• Ptotal = 1P1 + 2P2 + 3P3 + …

where i = mole fraction of component i, and

Pi = equilibrium vapor pressure of pure volatile component at a given temperature

Sample Exercise 11.3 (simplified): Calculating the Vapor Pressure of a Solution

What is the vapor pressure of a solution prepared by dissolving 13 g o of heptane (C7H16) in 87 g of octane (C8H18) at 25 C?

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Real vs. Ideal Solutions

. Deviations from Raoult’s Law: Due to differences in solute–solvent and solvent–solvent interactions (dashed lines = ideal behavior)

Negative deviations Positive deviations

Concept Test, p. 478

Which of the following solutions is least likely to follow Raoult’s Law: (a) acetone/ethanol, (b) pentane/hexane, (c) pentanol/water?

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Chapter Outline

Colligative Properties of Solutions

• Colligative Properties: – Set of properties of a solution relative to the pure solvent – Due to solute–solvent interactions – Depend on concentration of solute particles, not the identity of particles – Include lowering of vapor pressure, boiling point elevation, freezing point depression, osmosis and osmotic pressure

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Vapor Pressure of Solutions

• Raoult’s Law: • Vapor pressure of solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of solvent in the solution

• Psolution = solvent·P solvent • Vapor pressure lowering: • A colligative property of solutions (Section 11.5) • Ideal Solution: • One that obeys Raoult’s law

Sample Exercise 11.4: Calculating the Vapor Pressure of a Solution of One or More Nonvolatile Substances

Most of the world’s supply of maple syrup is produced in quebec, Ontario, where the sap from maple trees is evaporated until the concentration of sugar (mostly sucrose) in the sap reaches at least 66% by weight. What is the vapor pressure in atm of a 66% aqueous solution of sucrose (MW=342) at 100 oC? Assume the solution obeys Raoults’s Law.

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Colligative Properties of Solutions

• Colligative Properties: Vapor pressure lowering

Consequences:  in b.p. and f.p. of solution relative to pure solvent

Solute Concentration: Molality

• Changes in boiling point/freezing point of solutions depends on molality: n m  solute kg of solvent – Preferred concentration unit for properties involving temperature changes because it is independent of temperature – For typical solutions: molality > molarity

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Calculating Molality

Starting with: a) Mass of solute b) Volume of solvent, or c) Volume of solution

Sample Exercise 11.5: Calculating the Molality of a Solution

A popular recipe for preparing corned beef calls for preparing a seasoned brine (salt solution) that contains 0.50 lbs of kosher salt (NaCl) dissolved in 3.0 qt of water. Assuming the density of water = 1.00 g/mL, what is the molality of NaCl in this solution for “corning” beef?

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Chapter Outline

Colligative Properties of Solutions

. Colligative Properties: • Solution properties that depend on concentration of solute particles, not the identity of particles.

Previous example: vapor pressure lowering.

Consequences: change in b.p. and f.p. of solution.

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Boiling-Point Elevation and Freezing-Point Depression

. Boiling Point Elevation (ΔTb):

• ΔTb = Kb∙m

• Kb = boiling point elevation constant of solvent; m = molality.

. Freezing Point Depression (ΔTf):

• ΔTf = Kf∙m

• Kf = freezing-point depression constant; m = molality.

Sample Exercise 11.7: Calculating the Freezing Point Depression of a Solution

What is the freezing point of an automobile radiator fluid prepared by mixing equal volumes of ethylene glycol (MW=62.07) and water at a temperature where the density of ethylene glycol is 1.114 g/mL and the density of water is 1.000 g/mL? o The freezing point depression constant of water, Kf = 1.86 C/m.

Tf = Kf  m Assume 1.00 L of each = 1000 mL

1.00 g 1 kg kg H2O (solvent) = 1000 mL x x =1.00 kg H O mL 1000 g 2 1.114 g x 1 mol mol E.G. (solute) = 1000 mL x =17.95 mol E.G. ml 62.07 g 17.95 mol m = = 17.95 m 1.00 kg

o o Tf = Kf  m = (1.86 C/m)(17.95 m) = 33.4 C

New f.p. = 0.0 oC – 33.4 oC = -33.4 oC

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The van’t Hoff Factor

. Solutions of Electrolytes: • Need to correct for number of particles formed when ionic substance dissolves. . van’t Hoff Factor (i): • number of ions in formula unit. • e.g., NaCl, i = 2

. ΔTb = i∙Kb∙m & ΔTf = i∙Kf∙m . Deviations from theoretical value due to ion pair formation.

Values of van’t Hoff Factors

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Sample Exercise 11.9: Using the van’t Hoff Factor

Some Thanksgiving dinner chefs “brine” their turkeys prior to roasting them to help the birds retain moisture and to season the meat. Brining involves completely immersing a turkey for about 6 hours in a brine (salt solution) prepared by dissolving 910 g of salt (NaCl,

MW=58.44) in 7.6 L of water (dH2O = 1.00 g/mL). Suppose a turkey soaking in such a solution is left for 6 hours on an unheated porch on a day when the air temperature is -8 oC. o Are the brine (and turkey) in danger of freezing? Kf = 1.86 C/m; assume that i = 1.85 for this NaCl solution.

Tf = i  Kf  m

1.00 g 1 kg kg H O (solvent) = 7.6 x 103 mL x 푥 = 7.57 kg H O 2 mL 1000 g 2 1 mol mol NaCl (solute) = 910 g x = 15.52 mol NaCl 58.44 g 15.52 mol m = = 2.05 m 7.57 kg

o o Tf = i  Kf  m = (1.85)(1.86 C/m)(2.05 m) = 7.0 C

New f.p. = 0.0 oC – 7.0 oC = -7.0 oC BUT the air T is -8 oC so the brine will freeze

Chapter Outline

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Molar Mass from Colligative Properties

ΔTf = Kfm and ΔTb = Kbm

Step 1: ΔT or ΔT m = ΔT /K solve for molality f b f f

kg of solvent Step 2: molality moles solve for moles

Step 3: MW = g sample/moles calc. MW

Sample Exercise 11.12: Using Freezing Point Depression to Determine Molar Mass

Eicosene is a hydrocarbon used as an additive in the high-pressure fracturing of rock layers to enhance recovery of petroleum and natural gas, a process sometimes called “fracking”. The freezing point of a solution prepared by dissolving 100 mg of eicosene in 1.00 g of benzene is 1.75 oC lower that the freezing point of pure benzene. What is the molar mass of eicosene? o (Kf for benzene is 4.90 C/m)

Step 1: ∆T 1.75 oC m = = = 0.357 m solve for molality Kf 4.90 oC/m

0.357 mol Step 2: moles solute = X 0.00100 kg benzene solve for moles kg benzene

= 3.57 x 10-4 mol eicosene

−3 Step 3: 100 x 10 g eicosene MW = = 280 g/mol calc. MW 3.57 x 10−4 mol