Chapter 11

Solutions and Colloids

Shaun Williams, PhD

Solution Composition

Various Types of Solutions

Solution Composition

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

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

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 change associated with the formation of the solution is the sum of the ΔH values for the steps: $$ \Delta H_\text{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 $$
• \(\Delta H_\text{soln}\) may have a positive sign (energy absorbed) or a negative sign (energy released).

Enthalpy (Heat) of Solution

The Energy Terms for Various Types of Solutes and Solvents

In General

• One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed.
• Processes that require large amounts of energy tend not to occur.
• Overall, remember that "like dissolves like".

Factors Affecting Solubility

• Structure Effects:
  • Polarity
• Pressure Effects:
  • Henry’s law
• Temperature Effects:
  • Affecting aqueous solutions

Structure Effects

• Hydrophobic (water fearing)
  • Non-polar substances
• Hydrophilic (water loving)
  • Polar substances

Pressure Effects

• Little effect on solubility of solids or liquids
• Henry’s law: $$ C = kP $$
  • \(C\) - concentration of dissolved gas
  • \(k\) - constant
  • \(P\) - partial pressure of gas solute above the solution
• Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

A Gaseous Solute

Temperature Effects (for Aqueous Solutions)

• Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature.
• Predicting temperature dependence of solubility is very difficult.
• Solubility of a gas in solvent typically decreases with increasing temperature.

The Solubilities of Several Solids as a Function of Temperature

The Solubilities of Several Gases in Water

The Vapor Pressure of Solutions

An Aqueous Solution and Pure Water in a Closed Environment

Vapor Pressures of Solutions

• Nonvolatile solute lowers the vapor pressure of a solvent.
• Raoult’s Law: $$ P_\text{soln} = \chi_\text{solv} P_\text{solv}^\circ $$
  • \(P_\text{soln}\) - observed vapor pressure of solution
  • \(\chi_\text{solv}\) - mole fraction of solvent
  • \(P_\text{solv}^\circ\) - vapor pressure of pure solvent

A Solution Obeying Raoult's Law

Nonideal Solutions

• Liquid-liquid solutions where both components are volatile.
• Modified Raoult’s Law: $$ P_\text{total} = \chi_\text{A} P_\text{A}^\circ + \chi_\text{B} P_\text{B}^\circ $$
• Nonideal solutions behave ideally as the mole fractions approach 0 and 1.

Vapor Pressure for a Solution of Two Volatile Liquids

Summary of the Behavior of Various Types of Solutions

Boiling-Point Elevation and Freezing-Point Depression

Colligative Properties

• Depend only on the number, not on the identity, of the solute particles in an ideal solution:
  • Boiling-point elevation
  • Freezing-point depression
  • Osmotic pressure

Boiling-Point Elevation

• Nonvolatile solute elevates the boiling point of the solvent.
• \(\Delta T = K_bm_\text{solute}\)
  • \(\Delta T\) - boiling-point elevation
  • \(K_b\) - molal boiling-point elevation constant
  • \(m_\text{solute}\) - molality of solute

Freezing-Point Depression

• When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.
• \(\Delta T = K_fm_\text{solute}\)
  • \(\Delta T\) - boiling-point elevation
  • \(K_f\) - molal freezing-point depression constant
  • \(m_\text{solute}\) - molality of solute

Changes in Boiling Point and Freezing Point of Water

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.

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

Osmotic Pressure

• Osmosis – flow of solvent into the solution through a semipermeable membrane.
• \( \Pi = MRT \)
  • \(\Pi\) - osmotic pressure (atm)
  • \(M\) - molarity of the solution
  • \(R\) - gas law constant
  • \(T\) - temperature (Kelvin)

Equilibrium in Osmosis System

Measuring 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, \(i\)

• The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: $$ i=\frac{\text{moles of particles in solution}}{\text{moles of solute dissolved}} $$
• Modified equations $$ \Delta T = imK $$ $$ \Pi = iMRT $$

Examples

• The expected value for \(i\) can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur).
  • \(\chem{NaCl}\): \(i = 2\)
  • \(\chem{KNO_3}\): \(i = 2\)
  • \(\chem{Na_3PO_4}\): \(i = 4\)

Ion Pairing

• At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.
• Ion pairing is most important in concentrated solutions.
• As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs.
• Ion pairing occurs to some extent in all electrolyte solutions.
• Ion pairing is most important for highly charged ions.

Colloids

• A suspension of tiny particles in some medium.
• Tyndall effect – scattering of light by particles.
• Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.

Types of Colloids

Coagulation

• Destruction of a colloid.
• Usually accomplished either by heating or by adding an electrolyte.