Module 2 - Solutions and Their Properties

MODULE 2SOLUTIONS AND THEIR PROPERTIES

Types of Intermolecular Forces of Attraction:

1. van der Waals forces (Johannes van der Waals, Dutch physicist)a. Dipole-dipole forces – attractive forces between polar molecules. That is between

molecules that possess dipole moments.b. Dispersion forces or London forces (Fritz London, 1930) – attractive forces that arise

as a result of temporary dipoles induced in atom or moleculeb1. ion-induced dipole forces – attraction between an ion and the induced dipole.b2. dipole-induced dipole forces – attraction between a polar molecule and the induced dipole.

2. Ion-dipole forces – attractive forces between an ion (either a cation or an anion) and a polar molecule.

3. Hydrogen bonding – special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N-H, O-H, or F-H and an electronegative atom O, N, or F atom.

Types of Interactions and the Solution Process1. solvent-solvent interaction2. solute-solute interaction3. solvent-solute interaction

Types of solution:1. According to final physical state

a. solid solutionb. liquid solutionc. gaseous solution

2. According to amount of solute dissolveda. unsaturated solutionb. saturated solutionc. supersaturated solution

Solubility – maximum amount of solute that will dissolve in a given amount of solvent at a specific temperature.

3. According to ratio of solute to solventa. concentrated solutionb. dilute solution

Concentration of a solution is the amount of solute present in a given amount of solvent or solution.

Types of Concentration Units 1. Percent by mass

The percent by mass is defined as

percent by mass of solute =

Module 2: Solutions and their Properties Page 1 of 10

or

percent by mass solute =

Practice Exercise 1.1: The dehydrated form of Epsom salt is magnesium sulfate. What is the percent MgSO4 by mass in a solution made from 16.0 g MgSO4 and 100 ml of H2O at 25oC. The density of water at 25oC is 0.997 g/ml.

Practice Exercise 1.2: An aqueous solution contains 167 g CaSO4 in 820 ml of solution. The density of the solution is 1.195 g/ml. Calculate the percent CaSO4 in the solution.

2. Molarity (M)Molarity is defined as

molarity =

Practice Exercise 1.3: What is the molarity of the MgSO4 solution made in Practice Exercise 1.1? Assume that the density remain unchanged upon addition of MgSO4 to the water.


Practice Execise 1.4: An aqueous solution contains 167 g CaSO4 in 820 ml of solution. The density of the solution is 1.195 g/ml. Calculate the molarity of the solution.

3. Molality (m) Molality is defined as the number of moles of solute per mass of solvent (in kg)

molality =

Practice Exercise 1.5: What is the molality of the MgSO4 solution made in Practice Exercise 1.1?

Practice Exercise 1.6: An aqueous solution contains 167 g CaSO4 in 820 ml of solution. The density of the solution is 1.195 g/ml. Calculate the molality of the solution.

4. Mole fraction


mole fraction expresses the ratio of the number of moles of one component to the number of moles of all component present

mole fraction of solute =

Practice Exercise 1.7: What is the mole fraction of MgSO4 in the MgSO4 solution made in Practice Exercise 1.1?

Practice Exercise 1.8: An aqueous solution contains 167 g CaSO4 in 820 ml of solution. The density of the solution is 1.195 g/ml. Calculate the mole fraction of CaSO4 .

Factors Affecting Solubility:

Solubilty is defined as the maximum amount of a solute that will dissolve in a given quantity of solvent at a specific temperature.

1. Nature of Solute and SolventPolar or ionic solute will be dissolved in a polar solvent. Non-polar solute will be dissolved in a non-polar solvent.

2. Temperaturea. For solids in liquids, in general an increase in temperature means an increase in

solubilityb. For gases in liquids, increase in temperature means a decrease in solubility

3. Pressure. For gases in liquids, the higher pressure, the higher the solubility. ( Henry’s law)

c P c = kP

where, c – is the molar concentration of the dissolved gas


k - is a constant for a given gas that depends only on the temperature.P – is the pressure, in atmospheres of the gas over the solution.

Practice Exercise 1.9: What is the concentration of O2 at 25oC in water that is saturated with air at an atmospheric pressure of 645 mmHg? The Henry’s constant (k) for oxygen is 3.5 x 10-4 mol / L – atm. Assume that the mole fraction of oxygen in air is 0.209.

Practice Exercise 1.10: The Henry’s Law constant for CO is 9.73 x 10-4 mol/L-atm at 25oC. What is the concentration of dissolve CO in water, if the partial pressure of CO in the air is 0.015 mmHg.?

Colligatve Properties of Nonelectrolytes

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

1. Vapor Pressure Lowering, Raoults lawIf a solute is nonvolatile, the vapor pressure of its solution is always less than that of the pure solvent, Raoult’s law quantifies this relationship by stating that the partial pressure

of a solvent over a solution, P1, is given by the vapor pressure of the pure solvent, ,

times the mole fraction of the solvent in the equation, X1:

P1 = X1

If the solution contains only one solute, X1 = 1 – X2, where X2 is the mole fraction of the solute. Substituting for X1

P1 = (1 – X2)

P1 = -X2

- P1 = X2

ΔP = X2


Practice Exercise 1.11: Calculate the vapor pressure of an aqueous solution at 30oC made from 100 g sucrose (C12H22O11) and 100 g water. The vapor pressure of pure water at 30oC is 31.8 mmHg.

Practice Exercise 1.12: At 25oC, the vapor pressure of pure water is 23.76 mmHG and that of an aqueous sucrose (C12H22O11) is 23.28 mmHg. Calculate the molality of the solution.

2. Boiling-point elevationThe boiling point of a solution is the temperature at which its vapor pressure equals the external atmospheric pressure. From Raoult’s law, a nonvolatile solute always decreases the vapor pressure of the solution relative to the pure solvent. Consequently, the boiling point of the solution is higher than the pure solvent, because more energy in the from of heat must be added to raise the vapor pressure of the solution to the external atmospheric pressure. The change in boiling point is proportional to the concentration of solute.

ΔTb = Kbmwhere,

ΔTb = Tb - (where Tb is the boiling point of the solution and is the

boiling point of the pure solvent)m is the molal concentrationKb is the molal boiling-point elevation constant of the solvent with units of

oC/m

Practice Exercise 1.13: What is the boiling point of an “antifreeze/coolant” solution made from a 50-50 mixture (by volume) of ethylene glycol, C2H6O2 and water? Assume the density of water is 1.00 g/ml and the density of ethylene glycol is 1.11 g/ml?


Practice Exercise 1.14: Benzene boils at 80.1oC. When 2.11 g of naphthalene, C10H8, is added to 100 g of benzene, the solution boils at 80.5oC. Calculate the boiling-point elevation constant (kb) for benzene.

3. Freezing-point depressionThe freezing point of a liquid (or the melting point of a solid) is the temperature at which the solid and liquid phases coexist in equilibrium. To melt a solid, the intermolecular forces holding the solid molecules together must be overcome. Adding another solid substance to a pure solid disrups the intermolecular forces of the formerly pure solid. Hence, it is easier to overcome the intermolecular forces and the mixture melts at a lower temperature than the pure solid. The melting point (or the freezing point) is depressed. The depression of freezing point can be represented by the following equation.

ΔTf = Kfmwhere,

ΔTf = - Tf (where Tf is the freezing point of the solution and is the

freezing point of the pure solvent)m is the molal concentrationKf is the molal freezing-point depression constant of the solvent with units of

oC/m

Practice Exercise 1.15: How many grams of isopropyl alcohol, C3H7OH, should be added to 1.0 L of water to give a solution that will freeze at -16oC. Assume the density of water is 1.00 g/ml.


Practice Exercise 1.16: What is the freezing point of an aqueous solution that boils at 102.5oC?

4. Osmotic pressureOsmosis is the net movement of solvent molecules through a semipermeale membrane from a pure solvent or from a dilute solution to a more concentrated solution. The osmotic pressure (Π) of a solution is the pressure required to stop osmosis. The osmotic pressure of a solution is given by:

Π = MRTwhere:

M is the molarity of the solutionR is the gas constant (0.0821 L-atm/K-mol)T is the absolute temperature in K

Practice Exercise 1.17: The average osmotic pressure of seawater is about 30.0 atm at 25oC. Calculate the molar concentration of an aqueous solution of urea (NH2CONH2) that is isotonic with sea water.

Practice Exercise 1.18: The walls of red blood cells are semipermeable membranes, and the solution of NaCl within those walls exerts an osmotic pressure of 7.82 atm at 37oC. What concentration of NaCl must a surrounding solution have so that this pressure is balanced and cell rapture (hemolysis) is prevented?


Determining Molar Mass Using Colligative Properties

Any of the four colligative properties discussed can be used to calculate the molar mass of the solute. However, in practice, only freezing-point depression and osmotic pressure are used because they show the most pronounced changes.

Practice Exercise 1.19: Benzene has a normal freezing point of 5.51oC. The solution of 1.25 g of an unknown compound to 85.0 g of benzene produces a solution with a freezing point of 5.52oC. What is the molar mass of the unknown compound?

Practice Exercise 1.20: 30 g of sucrose is dissolved in water making 100 ml of solution. The solution has an osmotic pressure of 20.8 atm at 16.0oC. What is the molar mass of sucrose?

Colligative Properties of Electrolytes

Electrolytes – dissociate into ions in solution and so one unit of an electrolyte compound separates into two or more particles when it dissolves.

To account for the dissociation of an electrolyte into ions, the equations for colligative properties must be modified as follows:

Tb = i KbmTf = i Kfm = i MRT

where,i is the van’t Hoff factor which is defined as


i =

Practice Exercise 1.21: Calculate the value of i for an electrolyte that should dissociate into 2 ions, if a 1.0 m aqueous solution of the electrolyte freezes at -3.28oC. Why is the value of I less than 2?

Solve Problems: 13.9, 13.22, 13.35, 13.49, 13.55, 13.62, 13.64, 13.74,13.79 pages. 448 - 449, General Chemistry: The Essential Concepts, 5th edition by Raymond Chang

References:

General Chemistry: The Essential Concepts, 5th edition by Raymond ChangChemistry : 8th edition by Whitten, Davis, Peck and StanleyCHEMISTRY: The Molecular Nature of Matter and Change , 4th ed. by Martin Silberberg

Chemistry: The General Science, 10th ed, by Brown, Le May and Bursten.


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Module 2 - Solutions and Their Properties