CHM 123 Chapter 11 – Properties of Solution
11.2 – Energy changes and the solution process
11.4 - Factors that affects solubility
• when there is an attraction between the particles of the solute and solvent.
• when a polar solvent such as water dissolves polar solutes such as sugar, and ionic solutes such as NaCl.
• when a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease.
Effect of Temperature on Solubility
· Saturated Solution: A solution containing the maximum possible amount of dissolved solute at equilibrium.
· Supersaturated Solution: A solution containing a greater-than-equilibrium amount of solute.
Example: The solubility of NaNO3 in water at 50oC is 110g/100g of water. In a laboratory, a student use 230.0 g of NaNO3 with 200 g of water at the same temperature
o How many grams of NaNO3 will dissovle?
o Is the solution saturated or unsaturated?
o What is the mass, in grams, of any solid NaNO3 on the bottom of the container?
Gases in Solution
• In general, the solubility of gases in water increases with increasing mass as the attraction between the gas and the solvent molecule is mainly dispersion forces.
• Larger molecules have stronger dispersion forces.
Henry’s law states
• the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.
• at higher pressures, more gas molecules dissolve in the liquid.
Example: Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this temperature is 3.1 x 10–2 mol/L-atm.
Answer: 0.12M
11.3 – Units of Concentration
Example: A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (% m/m) of the solution.
Answer: 6.0%
Example: What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, C27H46O, in 40.0 g chloroform, CHCl3?
Answer: 0.0249 m
11.5 – 11.10 – Physical Behavior of Solutions: Colligative Properties
Colligative Properties: Properties that depend on the amount of a dissolved solute but not on its chemical identity.
• Vapor-Pressure Lowering Boiling-Point Elevation
• Freezing-Point Depression Osmotic Pressure
Solutions of ionic substances often have a vapor pressure significantly lower than predicted, because the ion-dipole forces between the dissolved ions and polar water molecules are so strong.
Vapor-Pressure Lowering of Solutions: Raoult’s Law
Boiling-Point Elevation and Freezing-Point Depression
The change in boiling point Tb for a solution is (BPsolution – BPsolvent) = ΔTb = m∙Kb
The freezing-point depression for a solution relative to that of a pure solvent depends on the concentration of solute particles, just as boiling-point elevation does.
(FPsolvent – FPsolution) = ΔTf = m∙Kf
ΔTf = Kf • m
For ionic substances: ΔTb = Kb • m • i
ΔTf = Kf • m • i
Example: What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2? Kf = 1.86 oC/m
Answer: -3.2oC
Osmosis and Osmotic Pressure
Osmosis: The passage of solvent through a semipermeable membrane from the less concentrated side to the more concentrated side.
- Solvent flows from a high solvent concentration to a low solvent concentration. Or, solvent flows from a low solute concentration to a high solute concentration.
- the level of the solution with the higher solute concentration rises.
- the concentrations of the two solutions become equal with time.
Suppose a semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens?
Example: What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25°C?
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