Chapter 11

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SOLUTIONS

concentration of acetic acid molecules of water/molecules of acid

1 M 50/1

6 M 6/1

In most solutions there is a lot more __________ than ___________.

CONCENTRATION UNITS

% (w/w) =

Ax =(mole fraction)

M =(molarity)

m =(molality)

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3

3Example: The density of 6 M HNO (MM = 63.0 g/mol) is 1.19 g/mL. Calculate the%(w/w), molality, and moles fractions of both nitric acid and water. What doesthe ratio of the mole fraction of water to the mole fraction of nitric acid indicate?

When you do not have enough information to calculate you may choose a value for any one(1)property that is determined by the sample size. Choices are any of the following variables in thechart:

Some choices may be easier to work with than others.

INTERMOLECULAR FORCES AND SOLUBILITY

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Why in most solutions do the number of solvent molecules outnumber the solute molecules? The solvent serves as "molecular chaperones" to keep the solute particles separated. The moresimilar the _________________ between solute and solvent, the better able the solvent is to keepthe solute apart and the _______ soluble the solute will be. The stronger the IMF between soluteparticles the _______solvent molecules required to keep them apart.

ENERGETICS OF SOLUTION FORMATION

solutioncompound ÄT sign of ÄH

KCl + water

2CaCl + water

acetone + water

acetone + 2-propanol

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ENTHALPY OF SOLUTION

heat of solution:

lattice energy:

Always ________. Increases in magnitude with increasing ________ and _________ size of theions.

hydration energy:

Always ________. Increases in magnitude with increasing ________ and _________ size of theions.

solution compound lattice energy (kJ/mol) hydration energy (kJ/mol) ÄH (kJ/mol)

NaCl -778 -774

LiCl -846 -884

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f solution substance ÄH (kJ/mol) ÄH (kJ/mol)o 298K o 298K

(aq)Ca -5432+

2 (s)CaCl -795

(aq)Cl -167-

FACTORS THAT EFFECT SOLUBILITY

1. Type of IMF - the more similar in strength the IMF are in the solute and solvent the morelikely the solute will ___________. “Like dissolves like.”

IMF 1 2 3 4

1**

2

3

4

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Line Structures (used extensively in organic chemistry)**

a. The end of a line or the intersection of two or more lines represents a _____ atomunless labeled with another symbol.

b. Carbon always has ______ bonds.c. Any “missing” bonds are assumed to be between C and _____.

Benzaldehyde:

Example: Within each IMF type there is a wide variation in actual strengths. Which

2 2 3 6molecule probably has the stronger dipole forces, CH Cl or C H O ?

3Demonstration: Add potassium chromate to water and CH OH. Which solvent must havestronger H-bonds?

2 3Example: Which would be more soluble in water: CO or NH ?

DIELECTRIC CONSTANT

A more quantitative measure of a solvent’s polarity is called the dielectric constant. Thelarger the dielectric constant the better able the solvent can dissolve ions. Generally the largerthe dipole moment of the molecule the higher its dielectric constant. However, the dielectricconstant is also influenced by the ease with which a dipole can be induced in the molecule.

compound dipole moment (D) dielectric constant

water 1.85 79

methanol 1.70 33

acetone 2.89 21

methylene chloride 1.57 9

hexane 0 2

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2. Pressure - only a significant effect when the solute is a gas. The higher the partialpressure of the gas the ___________ the solubility.

Henry’s Law:

M = molarityk = constant that varies with the gas used and the temperatureP = the partial pressure of the gas that is being dissolved.

Example: Air is about 21 mole % oxygen. What is the molarity of dissolved oxygen at 25C in Rexburg where the normal atmospheric pressure is 640 torr. The Henryo

constant for oxygen at 25 C is 1.3 x 10 M/atm.o -3

3. Temperature

solutioncompound ÄH (kJ/mol) graph symbol

(s)NaCl 4 square

(s)KCl 17 circle

2 (g)SO -21 diamond

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As temperature increases, solutes with an endothermic heat of solution usually_________in solubility and solutes with an exothermic heat of solution usually __________in

solutionsolubility. The smaller the magnitude of ÄH the _______effect temperature has on thesolubility

VAPOR PRESSURE LOWERING

compound relative vapor pressure

acetone

methyl salicylate

acetone + methylsalicylate

When a solute with a low vapor pressure is dissolved in a volatile solvent, the vapor pressure ofthe solvent ________.

RAOULT'S LAW

AP = vapor pressure produced when the liquid solvent evaporates.

Ax = mole fraction of solvent = moles solvent/total moles (if the solute is an electrolyte be sureto include the moles of ALL the dissolved particles

AP = vapor pressure of pure solvento

Example: Calculate the mole fraction of acetone in the above solution.

WARNING: Be careful that you do not confuse this equation with Dalton’s Law from Chem105.

Raoult’s Law is applied when you have a liquid. Dalton’s Law is used when you have onlygases.

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Example: Calculate the vapor pressure of a solution that contains 20 grams of a non-volatile,non-electrolyte (MM = 111 g/mol) in 100 grams of water at a temperature of 100 deg C .Since water boils at 100 deg when the atmospheric pressure is 760 mm, the vapor pressure ofwater at 100 deg C must be ____mm.

VOLATILE SOLUTES

Raoult’s Law assumes that the solute is non-volatile and the only vapor pressure is due to____________. When both solute and solvent have significant vapor pressures, the observedvapor pressure of the solution is expected to have a value between the vapor pressures of the twochemicals. The vapor pressure can be estimated from:

A Bx = mole fraction of liquid A x = mole fraction of liquid B

A BP = vapor pressure of pure liquid A P = vapor pressure of pure liquid Bo o

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Example: Acetone has a vapor pressure of 400 mm at 40 C. 2-propanol has a vaporo

pressure of 100 mm at the same temperature. Calculate the expected vaporpressure of a solution made by mixing 0.15 moles of acetone with 0.25 moles of2-propanol. Using Dalton’s Law, find the composition of the vapor above thesolution in terms of mole fractions of acetone and 2-propanol.

IDEAL SOLUTIONS

solutionThe Raoult’s equation for volatile solutes is only valid if the ÄH = 0. Such a solution is

solutioncalled an ______ solution. If the ÄH is endothermic the interaction between solute andsolvent molecules is weaker when in solution relative to the pure liquids. This causes themolecules to have a greater tendency to escape in the solution and the vapor pressure is usuallyhigher than calculated. (Positive deviation from Raoult’s Law) Exothermic heats of solutionimply that the interaction between the solute and solvent molecules in the solution is strongerthan in the pure liquids, there is a reduced tendency to evaporate in the solution and the vaporpressure is usually lower than calculated. (Negative deviation from Raoult’s Law). Thedirection of deviation is usually __________as the sign of the enthalpy of solution.

solutionliquid solution Ä H deviation

acetone/2-propanol

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i-factor

The ionization factor (i-factor) of a solute is the number of free particles the solute forms whendissolved.

compound expected i-factor in aqueous solution

4KMnO

3 4Na PO

3CH OH

3HNO

OSMOTIC PRESSURE

Since the vp of the solvent in a solution decreases as the concentration of the dissolved solute____________, if solutions of differing concentration are placed on either side of a solventpermeable membrane, a pressure difference will exist due to the differing vapor pressures thatcauses the solvent to move. The amount of pressure required to stop this movement is calledosmotic pressure.

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ð = osmotic pressure

i = number of particles the solute forms when dissolved in the solvent

M = molarity of the solution

R = gas constant (0.0821 L atm/mol K or 62.4 L torr/mol K)

T = Kelvin temperature

Example: What is the calculated osmotic pressure in atm of a solution that contains 20

2grams of CaCl (MM = 111 g/mol) dissolved in 100 mL of solution at 25 C? o

BOILING POINT CHANGE

Since the solution has a ________vp, in order to boil (vp = Patm) the temperature must be above the boiling point of the solvent. This is boiling point _________. Allnon-volatile solutes that dissolves in water will the boiling point of water. The moresolute dissolved, the the boiling point. What would happen to the boiling point ofwater if a solute with a higher vapor pressure were used? ____________________________

atm P vp

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Demonstration: Measure the boiling temperature of water and salt water. Add more saltand note the change in boiling temperature.

liquid boiling temperature ( C)o

pure water

salt water

more salt

BOILING POINT ELEVATION:

bÄT = change in the boiling temperature

i = number of particles the solute forms when dissolved in the solvent

bK = constant for the solvent = 0.52 deg/molal for water

m = molality of the solution

Example: What is the boiling point of a solution that contains 20 grams of a non-electrolyte

2(MM = 111 g/mol), dissolved in 100 g of water? If the compound had been CaCl (MM = 111g/mol), what would the boiling point have been?

FREEZING POINT DEPRESSION

compound change in temperature? ( C) lowers freezing point of water?o

KCl

2CaCl

sand

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A substance will lower the freezing point of a solvent if . The volatility of thesolute has no influence on the result.

FREEZING POINT DEPRESSION WHEN SOLUTE IS A LIQUID

In addition to lowering the freezing point of water, antifreeze will also help prevent the waterfrom boiling when the engine get hot. Do they add something special to it to get it to raise theboiling point of the water? FREEZING POINT DEPRESSION:

fÄT = change in freezing temperature i = number of particles the solute forms when dissolved in the solvent

fK = constant for the solvent = 1.86 deg/molal for waterm = molality of solution

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Example: Calculate the freezing temperature of the solution used previously that contains 20grams of non-electrolyte (MM = 111 g/mol) dissolved in 100 g of water? What would the result

2be if the compound had been CaCl ?

Summary: A non-volatile solute that dissolves in a solvent will the boilingpoint and the vapor pressure of the solvent. Any solutethat dissolves in a solvent will the freezing point.

EXPERIMENTAL VALUE OF i

In very dilute solutions, the experimental values of i are usually close to the value predicted fromthe number of ions the solute should produce. As solutions become more concentrated, theexperimental i values are often different than expected due to a number of factors. In ionicsolutes, for example, ions may associate in groups (“ion-pairs”) rather than existing as free,independent ions in solution. This reduces the actual number of solute particles and causes the ifactor to appear to decrease.

Example: Calculate the apparent i factors for the following

f experimental expectedsolute molality (mol/kg) experimental T ( C) i io

acetone 0.262 0.48

HCl 0.277 0.98

2MgCl 0.269 1.32