chapter 11 properties of solutions. important vocabulary homogeneous means there is only one phase...

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, H2CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the molarity.

• Answer: 1.87 M H2CO

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, H2CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the mass percent.

• Answer: 5.52 % H2CO, 94.48% H2O

Mole Fraction• Represented by X• Moles of part/moles of solution X 100%

Mole Frac. A = XA = nA/(nA+nB)

• Example: A solution was prepared by adding 5.84 g of formaldehyde, H2CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the mole fraction.

• Answer: XH2CO = 0.0338, XH2O = 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, H2CO, to 100.0 g of water. The final volume of the solution was 104.0 mL. Calculate the molality.

• Answer: 1.94 m H2CO

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. Structure2. Pressure3. 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 O2 is 2.2 X 10-4 M at 0C and 0.10 atm. Calculate the solubility of O2 at 0C and 0.35 atm.

• Answer: 7.7 X 10-4 M O2

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

Psoln = XsolventP0solvent

• Psoln = observed vapor pressure of solution

• Xsolvent = mole fraction of solvent

• P0solvent = vapor pressure of the pure solvent

Example

• Glycerin, C3H8O3, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g of glycerin to 338 mL of H2O 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 CuCl2, 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 solution1. Boiling-Point Elevation2. Freezing-Point Depression3. 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 ∆Tb is the

difference between the boiling point of the solution and the pure solvent

• Unit: °C/m• Calculated using

∆Tb = Kb X msolute

Kb 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 (C12H22O11) 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 ∆Tf is the difference between the freezing point of the solution and the pure solvent

• Unit: °C/m• Calculated using

∆Tf = Kf X msolute

Kf 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 (CH3OH) to 250.0 g of water.

• Answer: -6.39°C

Example

• Find the boiling point of a 1.50m solution of calcium chloride, CaCl2 and water.

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