
# 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.

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?

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.)

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.)

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).

### 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

• 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.

### 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 Vapor Pressure of Solutions

### 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

### 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.

### 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

• 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

### 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.

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

• 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)

### 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.

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

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

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

/