Chapter 11

Properties of Solutions

Important Vocabulary

•

Homogeneous

means there is only one phase (compositions do not vary) • • Ex: Kool Aid, air, steel

Solute

: Gets dissolved •

Solvent

: Does the dissolving •

Solution

: Homogeneous mixture consisting of a solute and solvent

Dilute vs. Concentrated

• Can’t be used in calculations • Molarity, mass percent, and mole fraction can be used to show solution concentrations

Molarity

• Moles of solute/liters of solution • Represented by

M

• Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the molarity.

• Answer: 1.87 M H 2 CO

Mass Percent

• Percent by mass of the solute in the solution • Mass Percent = (mass of solute/mass of solution) X 100% • Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the mass percent.

• Answer: 5.52 % H 2 CO, 94.48% H 2 O

Mole Fraction

• Represented by

X

• Moles of part/moles of solution

X

100% Mole Frac. A =

X

A = n A /(n A +n B ) • Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the mole fraction.

• Answer: X H2CO = 0.0338, X H2O = 0.9662

Molality

• Represented by

m

• Moles of solute per kilogram of solvent Molality = moles of solute/kilogram of solvent • Example: A solution was prepared by adding 5.84 g of formaldehyde, H 2 CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the molality.

• Answer: 1.94

m

H 2 CO

Solubility

• Shows what will dissolve in what • “Like dissolves like” = polar solvents will dissolve polar/ionic solutes and nonpolar solvents will dissolve nonpolar solutes

Factors Affecting Solubility

1. Structure 2. Pressure 3. Temperature

1. Structure Effects

• Polarity of solute/solvent (like dissolves like) • Example: vitamins are fat-soluble and water soluble – Fat-soluble = nonpolar, hydrophobic (water-fearing), build up/stored in fatty tissue, too much = hypervitaminosis – Water-soluble = polar, hydrophilic (water-loving), extra are excreted by the body

2. Pressure Effects

• Doesn’t affect liquids/solids, but has a large affect on gases • Gas solubility increases as the partial pressure of the gas above the solution increases

Henry’s Law

• Shows relationship between gas pressure and concentration of dissolved gas: C = kP • C = concentration of dissolved gas • K = constant for particular solution • P = partial pressure of gas above solution • Works best with gases that don’t dissociate in/react with solvent

Henry’s Law Example

• The solubility of O 2 is 2.2 X 10 -4 M at 0C and 0.10 atm. Calculate the solubility of O 2 at 0C and 0.35 atm.

• Answer: 7.7 X 10 -4 M O 2

3. Temperature Effects

• For most solids, solubility

increases

as temperature increases • For most gases, solubility

decreases

as temperature increases –Thermal Pollution in lakes: increase in temp. lowers dissolved oxygen concentrations

Vapor Pressure of Solutions

• If a solution contains a nonvolitile (not easily vaporized) solute, its vapor pressure is LOWER than the pure solvent.

• Shells of water solvation make it so it’s harder for the solvent to vaporize

• Molecules that do not dissociate (break up) in water (solvent) have higher vapor pressures than ionic compounds that do dissociate • The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution.

Answer This…

• Which compound affects the vapor pressure of a solution the least: glucose, sodium chloride, or calcium chloride?

• Solutions with covalent compounds > Solutions with ionic compounds

Raoult’s Law

• Calculates the expected vapor pressure of a solution based on the solute/solvent

P soln = X solvent P 0 solvent

• P soln = observed vapor pressure of solution • X solvent = mole fraction of solvent • P 0 solvent = vapor pressure of the pure solvent

Example

• Glycerin, C 3 H 8 O 3 , is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g of glycerin to 338 mL of H 2 O at 39.8C? The vapor pressure of pure water at 39.8C is 54.74 torr and its density is 0.992 g/mL.

• Answer: 50.0 torr

Example #2

• What is the vapor pressure of a solution made by adding 52.9 g of CuCl 2 , a strong electrolyte, to 800.0 mL of water at 52.0C? The vapor pressure of water is 102.1 torr, and its density is 0.987 g/mL.

• Answer: 99.4 torr

Colligative Properties

• Depend on the number of solute particles, NOT their identity in an ideal solution 1. Boiling-Point Elevation 2. Freezing-Point Depression 3. Osmotic Pressure

1. Boiling-Point Elevation

• Review boiling point definition: when vapor pressure = atmospheric pressure • When solute is added to solvent, it lowers the vapor pressure.

• More kinetic energy must be added to bring the solution to boiling • Boiling point is HIGHER in solutions than in pure solvents

• Antifreeze in car engines (solute) makes it so car engines don’t boil in high temperatures • The more solute particles dissolved, the higher the boiling point (identity doesn’t matter)

Boiling-Point Elevation Calculation

• Change in boiling point ∆T b is the difference between the boiling point of the solution and the pure solvent • Unit: °C/

m

• Calculated using K b ∆T b = K b X

m solute

is a molal boiling-point elevation constant of the solvent found on pg. 517

Example

• What is the boiling point of a solution containing 96.7g of sucrose (C 12 H 22 O 11 ) in 250.0g water at 1atm?

• Answer: 100.579°C

2. Freezing-Point Depression

• When solute is present, the normal molecular freezing pattern is disrupted • This makes it so the solution has to lose more kinetic energy (get colder) in order to solidify • Freezing point of the solution is LOWER than that of the pure solvent

• The more solute particles dissolved, the more the freezing point decreases (identity doesn’t matter) • Sidewalk salt and car antifreeze work this way

Freezing-Point Depression Calculation

• Change in freezing point ∆T f is the difference between the freezing point of the solution and the pure solvent • Unit: °C/

m

• Calculated using ∆T f = K f X

m solute

K f is a molal freezing-point depression constant of the solvent found on pg. 517

Example

• Determine the freezing point of a solution made by adding 27.5 g of methanol (CH 3 OH) to 250.0 g of water.

• Answer: -6.39°C

Example

• Find the boiling point of a 1.50

m

solution of calcium chloride, CaCl 2 and water.

• Answer: 2.30°C so the new boiling point would be 102.30

°C.