ch14

Jeffrey MackCalifornia State University,

Sacramento

Chapter 14

Solutions and Their Behavior

• A solution is a homogeneous mixture of two or more substances in a single phase.

• By convention, the component present in largest amount is identified as the solvent and the other component(s) as the solute(s)

Solutions

• Solutions can be classified as saturated or unsaturated.

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

Solutions

• An unsaturated solution can still take on more solute at a given temperature.

• SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.

Solutions

M =moles solute

L of solution

Molarity (M)

Concentration Units

%=

mass solute

Total mass of solution

Weight (mass) %

100

Concentration Units

% by mass = x 100%mass of solutemass of solute + mass of solvent

Xi =moles solute (i)

Total moles in solution

Mole Fraction (X)

Concentration Units

Mole Fraction (X)

• Where “i” is the moles of one component of the solute.

• Total moles are all species:

mols solute (i) + mols solvent.• The sum of all of the mole fractions for each

component are 1 exactly.

iX 1

Concentration Units

Concentration Units

M =moles of solute

liters of solution

m =moles of solute

mass of solvent (kg)

Molarity (M)

Molality (m)

m =mg solute

kg solution

Parts Per Million (ppm)

Concentration Units

• Parts per million (ppm): grams of solute/grams of solution (then multiplied by 106 or 1 million)

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• Molarity: moles of solute/liter of solution• Percent by mass: grams of solute/grams of solution (then multiplied

by 100%)• Percent by volume: milliliters of solute/milliliters of solution (then

multiplied by 100%)• Mass/volume percent: grams of solute/milliliters of solution (then

multiplied by 100%)

Most concentration units are expressed as:

Solution Concentration

Amount of solvent or solution

Amount of solute

Problem:62.1 g (1.00 mol) of ethylene glycol is dissolved in 250. g of water.

Calculate mol fraction, molality, and weight % of the solution.

Calculating Concentrations

2 6 22 6 2 2 6 2

22 2

1 mol C H O62.1 g C H O 1.00 mol C H O

62.1 g

1 mol H O250. g H O 13.9 mol H O

18.02 g



Mole Fraction:


Mole Fraction:

2 6 2

2 6 22 6 2 2 6 2

22 2

C H O

1 mol C H O62.1 g C H O 1.00 mol C H O

62.1 g

1 mol H O250. g H O 13.9 mol H O

18.02 g

1.00 molX 0.0671

1.00 mol 13.9 mol


2 6 2

2 6 22 6 2

C H O

2 3

1 mol C H O62.1 g C H O

62.1 gm 4.00 mol / kg

1 kg250. g H O

10 g


Molality:


2 6 22 6 2

2 6 2 2

62.1 g C H O% C H O 100 19.9%

62.1 g C H O 250. g H O



Wt. %:

17

What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solution

Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

• Solutes dissolve in solvents by a process called solvation.

• Polar solvent dissolve polar solutes, non-polar solvents dissolve non-polar solutes. (aka: “like dissolves like”.

• If two liquids mix to an appreciable extent to form a solution, they are said to be miscible.

• In contrast, immiscible liquids do not mix to form a solution; they exist in contact with each other as separate layers.

The Solution Process

The Solution Process

When a cation exists in solution, it is surrounded by the negative dipole ends of water molecules.

When as anion exists in solution, it is surrounded by the positive dipole ends of water molecules.

+

Solvation of Ions

Energy must be supplied to separate the ions from their attractive forces. (an endothermic process)

latticeH

Energetics of the Solution Process: solutionH



latticeH

Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation.



latticeH

Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation.

This process, referred to as the Energy of Hydration when water is the solvent, is strongly exothermic.

We can therefore represent the process of dissolving KF in terms of these chemical equations:

Step 1: KF(s) → K+(g) + F(g) = latticeH

Step 2: K+(g) + F (g) → K+(aq) +F(aq) =hydrationH


The overall reaction is the sum of these two steps. The enthalpy of the overall reaction, called the enthalpy of solution (solnH), is the sum of the two enthalpies.

Overall:KF(s)→ K+(aq) + F(aq)

solnH = latticeH + hydrationH



If the enthalpy of formation of the solution is more negative than that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic!

Energetics of the Solution Process

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

• The enthalpy of solution for sodium acetate is ENDOthermic.

Supersaturated Sodium Acetate

Sodium acetate has an ENDOthermic heat of solution.

NaCH3CO2 (s) + heat f Na+(aq) + CH3CO2

-(aq)

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

Na+(aq) + CH3CO2-(aq) f

NaCH3CO2 (s) + heat

Supersaturated Sodium Acetate

30

Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

Gas solubility (mol/L)

Sg =KHPg

Factors Affecting: Solubility Pressure & Temperature— “Henry’s Law”

32

Temperature and Solubility

O2 gas solubility and temperature

solubility usually decreases with increasing temperature

Do you like your coke hot or cold? Why?

33

Chemistry In Action: The Killer Lake

Lake Nyos, West Africa

8/21/86CO2 Cloud Released1700 Casualties

Trigger?

• earthquake

• landslide

• strong Winds

• Relative to a pure solvent, a solution has:–Lower vapor pressure–Higher boiling point–Lower freezing point–A higher osmotic pressure

• Example:Pure water: b.p.=100°C f.p.= 0°C1.00 m NaCl (aq) b.p.= 101°C f.p. = -3.7 °C

Colligative Properties

Upon adding a solute to a solvent, the properties of the solvent are affected:

–Vapor pressure decreases–Melting point decreases–Boiling point increases

• Osmosis is possible (osmotic pressure)

Collectively these changes are called COLLIGATIVE PROPERTIES.

They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

Colligative properties

To understand colligative properties, one must consider the LIQUID-VAPOR EQUILIBRIUM for a solution.

Understanding Colligative Properties

LIQUID-VAPOR EQUILIBRIUM

Understanding Colligative Properties

• Vapor pressure: pressure exerted by the vapor when a liquid in a closed container is at equilibrium with the vapor.

• Vapor pressure of solution is decreased by the presence of a solute.

• More solute particles, lower vapor pressure of solution.

Changes in Vapor Pressure: Raoult’s Law

Psolution = Xsolvent Po

Psolution = the vapor pressure of a mixture of solute and solventPo = the vapor pressure of the pure solvent Xsolvent = the mole fraction of the solvent.

The expression can also be written in the form:

osolvent soluteP X P

P is the change to the vapor pressure of the pure solvent.

Raoult’s Law

40

PA = XA P A0

PB = XB P B0

PT = PA + PB

PT = XA P A0 + XB P B

0

Ideal Solution

Special case of 2 Liquids

41

PT is greater thanpredicted by Raoults’s law

PT is less thanpredicted by Raoults’s law

ForceA-B

ForceA-A

ForceB-B< &

ForceA-B

ForceA-A

ForceB-B> &

Vapor Pressure Lowering of Benzene + non volatile solute

Problem:Pure iodine (105 g) is dissolved in 325 g of CCl4 at

65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

Raoult’s Law

22 2

44 4

1 mol I105g I = 0.414 mol I

253.8 g

1mol CCl325g CCl = 2.11 mol CCl

153.8 g

Raoult’s Law



4

4 4 4

2.11 0.836

2.11 0.414

º 0.836 531 444

CCl

CCl CCl CCl

molX

mol mol

P X P mm Hg mm Hg

Raoult’s Law



The temperature of the normal boiling point of a solution is increased by:

The temperature of the normal freezing point of a solution is decreased by:

Where the K’s are the respective boiling and freezing point constants and msolute is the molality of the solution.

b b soluteT K m

ff soluteT K m

Molecular Compounds

Boiling Point Elevation & Freezing Point Depression

b b soluteT K m

ff soluteT K m

For boiling point elevation: Kb > 0 (positive)For freezing point depression: Kf < 0 (negative)


Molecular Compounds


Boiling and Freezing point effects involving ions:

When solutions containing ions are involved, the total concentration of solute particles must be considered.The change in b.p. or f.p. is given by the equation:

soluteT K m i Where m is the calculated molaity based on formula wt.“i” = the number of ions (van’t Hoff factor)

compound Type iCH3OH molecular 1NaCl strong electrolyte 2Ba(NO3)2 strong electrolyte 3HNO2 weak electrolyte 1 - 2

van’t Hoff Factor

The Boiling Point of a Solution is Higher Than That of a Pure Solvent

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

Pure water Water solution

Change in Freezing Point

Water with and without antifreeze

When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

Lowering the Freezing Point

Problem:If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

Freezing Point Depression

soluteT K m i

2 3

1 52.5 LiF

25.94 6.61

1306

10

LiF

mol LiFg

gm m

kgg H O

g



1.86 C6.61 m (2)

24.6 C

0 C ( 24.6 C) 24.6 C

fp

fp

fp

Tm

T

T


soluteT K m i


Problem:Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate?

Molar Mass By Boiling Point Elevation

1. Solution: The molality of the solution can be found from the freezing point depression.

2. Knowing molality and mass of solvent, one can find the moles of solute.

3. Knowing mass and moles of solute, the molar mass of the compound can be determined.



f f solute

fsolute

f

T K m

Tm

K



61.82 C 61.70 C 0.12 C

0.12 C 0.033

3.63 C

bp

bpbenzyl acetate

bp

T

Tm m

Km



–43 3

3

–4

0.033 mol benzyl acetate 1 kg 25.0 g CHCl 8.3 10 mol

1 kg CHCl 10 g

0.125 g benzyl acetate 150 g/mol

8.3 10 mol benzyl acetate



Dissolving the shell in vinegar Egg in corn syrupEgg in pure water

Osmosis

• A semipermeable membrane allows only the movement of solvent molecules.

• Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute.

Osmosis

• Osmosis of solvent from one solution to another will occur as they try to equalize one another’s concentration.

• At the point where they have equal osmotic pressures, they are said to be Isotonic.

Osmosis

Osmosis at the Particulate Level

Process of Osmosis

Equilibrium is reached when the internal pressure of the apparatus equals the external pressure of the open tube.

Since the internal pressure, Pint is greater then the atmospheric pressure, a column of liquid rises:

Patm +Pcol = Pint

Recall that:

Pcol = g d h

Osmotic Pressure,

Osmotic pressure

Pcol is referred to as the Osmotic Pressure, ∏.

= cRT

c = concentration of solute (mols/L)

T = the absolute temperature (K)

L atmR 0.08206

mol K

is in units of atm.

Osmotic Pressure,

Osmosis & Living Cells

Water desalination plant

Reverse Osmosis: Water Desalination

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Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure (p) p = CRT

Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure (p) p = iCRT

Electrolyte Solutions

Colloids

• In a solution, dispersed particles are molecules, atoms, or ions (roughly 0.1 nm in size). Solute particles do not “settle out” of solution.

• In a suspension (e.g., sand in water) the dispersed particles are relatively large, and will settle from suspension.

• In a colloid, the dispersed particles are on the order of 1–1000 nm in size.

• Although they are larger than molecules/atoms/ions, colloidal particles are small enough to remain dispersed indefinitely.

A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

• Colloidal dispersions scatter light, a phenomenon known as the “The Tyndall effect”

Colloids

• A hydrophobic colloid is stabilized by positive ions absorbed onto each particle and a secondary layer of negative ions.

• Because the particles bear similar charges, they repel one another, and precipitation is prevented.

Hydrophobic Colloids

Soap molecules interact with water through the charged, hydrophilic end of the molecule.

The long, hydrophobic end of the molecule binds through dispersion forces with nonpolar hydrocarbons and other non polar substances.

Soap

• An emulsifier (also known as an emulgent) is a substance which stabilizes an emulsion by increasing its kinetic stability.

• One class of emulsifiers is known as surface active substances, or surfactants.

• Examples of food emulsifiers are egg yolk (where the main emulsifying agent is lecithin) and honey.

• In some cases, particles can stabilize emulsions as well through a mechanism called Pickering stabilization.

• Both mayonnaise and Hollandaise sauce are oil-in-water emulsions that are stabilized with egg yolk lecithin or other types of food additives such as Sodium stearoyl lactylate.

Emulsifiers & Surfactants

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Documents

vant hoff

expected freezing

freezing point

individual

boiling point

mole fraction

colligative

osmotic pressure