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Chapter 11

Properties of Solutions

Shaun Williams, PhD

Solution Composition

Various Types of Solutions

Example State of Solution State of Solute State of Solvent
Air, natural gas Gas Gas Gas
Vodka, antifreeze Liquid Liquid Liquid
Brass Solid Solid Solid
Carbonated water Liquid Gas Liquid
Seawater, sugar solution Liquid Solid Liquid
Hydrogen in platinum Solid Gas Solid

Solution Composition

$$ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} $$ $$ \text{Mass (weight) percent} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\% $$ $$ \text{Mole fraction}\,(\chi_\text{A}) = \frac{\text{moles}_\text{A}}{\text{total moles of solution}} $$ $$ \text{Molarity}\,(m) = \frac{\text{moles of solute}}{\text{kilogram of solvent}} $$

Exercise 1

You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

Exercise 1 - Answer

You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

$$ 8.00\,\text{M} $$

Exercise 2

You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?

Exercise 2 - Answer

You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?

$$ 0.200\,\text{L} $$

Exercise 3

A solution of phosphoric acid was made by dissolving 8.00 g of \(\chem{H_3PO_4}\) in 100.0 mL of water. Calculate the mole fraction of \(\chem{H_3PO_4}\). (Assume water has a density of 1.00 g/mL.)

Exercise 3 - Answer

A solution of phosphoric acid was made by dissolving 8.00 g of \(\chem{H_3PO_4}\) in 100.0 mL of water. Calculate the mole fraction of \(\chem{H_3PO_4}\). (Assume water has a density of 1.00 g/mL.)

$$ 0.0145 $$

Exercise 4

A solution of phosphoric acid was made by dissolving 8.00 g of \(\chem{H_3PO_4}\) in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)

Exercise 4 - Answer

A solution of phosphoric acid was made by dissolving 8.00 g of \(\chem{H_3PO_4}\) in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)

$$ 0.816\,\text{m} $$

The Energies of Solution Formation

Formation of a Liquid Solution

  1. Separating the solute into its individual components (expanding the solute).
  2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).
  3. Allowing the solute and solvent to interact to form the solution.

Steps in the Dissolving Process

Energy is required to separate the solute molecules. Energy is required to separate the solvent molecules. Energy is released in mixing the solute and the solvent.

  1. Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent.
  2. Step 3 usually releases energy.
  3. Steps 1 and 2 are endothermic, and step 3 is often exothermic.

Enthalpy (Heat) of Solution

Enthalpy (Heat) of Solution

If the energy released by mixing is larger than the sum of the energy going in, the process is exothermic overall. Otherwise, the mixing process is endothermic.

The Energy Terms for Various Types of Solutes and Solvents

\(\Delta H_1\) \(\Delta H_2\) \(\Delta H_3\) \(\Delta H_\text{soln}\) Outcome
Polar solute, polar solvent Large Large Large, negative Small Solution forms
Nonpolar solute, polar solvent Small Large Small Large, positive No solution forms
Nonpolar solute, nonpolar solvent Small Small Small Small Solution forms
Polar solute, nonpolar solvent Large Small Small Large, positive No solution forms

In General

Factors Affecting Solubility

Structure Effects

Pressure Effects

A Gaseous Solute

As the pressure on a gas above a liquid increases, more gas particles are forced to dissolve into the liquid.

Temperature Effects (for Aqueous Solutions)

The Solubilities of Several Solids as a Function of Temperature

For most solids, as the temperature increases, the solid becomes more soluble in water.

The Solubilities of Several Gases in Water

For all gases, as the temperature increases, the gas becomes less soluble in water.

The Vapor Pressure of Solutions

An Aqueous Solution and Pure Water in a Closed Environment

When a pure water beaker and a beaker containing an aqueous solution are placed together in a closed container, the water ends up getting tranfered from the pure water beaker to the aqueous beaker due to vapor pressure differences.

Vapor Pressures of Solutions

A Solution Obeying Raoult's Law

For a solution obeying Raoult's Law will show a linear relationship between mole fraction of solventand the vapor pressure of the solution.

Nonideal Solutions

Vapor Pressure for a Solution of Two Volatile Liquids

When the solution contains a volatile liquid, the vapor pressure curves stop being linear and start curving away from the ideal straight line.

Summary of the Behavior of Various Types of Solutions

Interative Forces Between Solute (A) and Solvent (B) Particles \(\Delta H_\text{soln}\) \(\Delta T\) for Solution Formation Deviation from Raoult's Law Example
\(\chem{A\leftrightarrow A}, \chem{B\leftrightarrow B}\equiv \chem{A\leftrightarrow B}\) Zero Zero None (ideal solution) Benzene-toluene
\(\chem{A\leftrightarrow A}, \chem{B\leftrightarrow B} \lt \chem{A\leftrightarrow B}\) Negative (exothermic) Positive Negative Acetone-water
\(\chem{A\leftrightarrow A}, \chem{B\leftrightarrow B} \gt \chem{A\leftrightarrow B}\) Positive (endothermic) Negative Positive Ethanol-hexane

Boiling-Point Elevation and Freezing-Point Depression

Colligative Properties

Boiling-Point Elevation

Freezing-Point Depression

Changes in Boiling Point and Freezing Point of Water

When adding a solutions, the pahse diagram lines shift to higher temperature and higher pressure causing the effects observed.

Exercise 5

A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution.

Exercise 5 - Answer

A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution.

$$ 100.35^\circ \text{C} $$

Osmotice Pressure

Equilibrium in Osmosis System

When a pure medium is exposed to a solution, at first the pure solvent flows into the solution. Over time the rate of solvent flowing in equals the rate flowing out and equilibrium is established.

Measuring Osmotic Pressure

A u-shaped glass tube is fitted with a semipermeable membrane at the bend. Pure solvent is poured in one side while solution is poured in the other side to the same level. Solvent flows from pure solvent into the solution to increase the volume of the solution side. Eventually gravity overcomes the fource of the osmosis that level stops rising. The difference in the liquid levels is a measure of the osmotic pressure.

Exercise 6

When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.

Exercise 6 - Answer

When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.

$$ 111 \, \bfrac{\text{g}}{\text{mol}} $$

Colligative Properties of Electrolyte Solutions

van't Hoff Factor, \(i\)

Examples

Ion Pairing

Colloids

Types of Colloids

Examples Dispersing Medium Dispersed Substance Colloid Type
Fog, aerosol sprays Gas Liquid Aerosol
Smoke, airborne bacteria Gas Solid Aerosol
Whipped cream, soap suds Liquid Gas Foam
Milk, mayonnaise Liquid Liquid Emulsion
Paint, clays, gelatin Liquid Solid Sol
Marshmallow, polystyrene foam Solid Gas Solid foam
Butter, cheese Solid Liquid Solid emulsion
Ruby glass Solid Solid Solid sol

Coagulation

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