Updates Midterms are Feb. 08 and Mar. 15 at 7pm… anyone with a night class or other midterm will write it at 6 pm (notify me at least 1 week prior if you have one of these conflicts) Assignment 01 is posted and is due (in class) Wed., January 24

Physical Properties of Solutions Chapter 13

Effect of pressure on gas solubility The solubilities of solids and liquids are not affected appreciably by pressure When the pressure of a gas is increased, as in (b), the rate at which gas molecules enter the solution increases The concentration of solute molecules at equilibrium increases in proportion to the pressure So the solubility of a gas increases with pressure

Pressure and Solubility of Gases 13.5 Solubility decreases as pressure decreases Soft drink bottled under CO 2 pressure greater than 1 atm When the bottle is opened, partial pressure of CO 2 above the solution decreases Solubility of CO 2 decreases -> bubbles

Pressure and Solubility of Gases 13.5 Solubility decreases as pressure decreases In many consumer beverages such as soft drinks, carbonation is used to give "bite". Contrary to popular belief, the fizzy taste is caused by the dilute carbonic acid inducing a slight burning sensation, and is not caused by the presence of bubbles. This can be shown by drinking a fizzy drink in a hyperbaric chamber at the same pressure as the beverage. This gives much the same taste, but the bubbles are completely absent.

Pressure and Solubility of Gases 13.5 The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c = kP c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution k is a constant (mol/Latm) that depends only on temperature low P low c high P high c

SAMPLE EXERCISE A Henry’s Law Calculation Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO 2 in water at this temperature is 3.1  10 –2 mol/L-atm. PRACTICE EXERCISE Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0  10 –4 atm. Solution Analyze: We are given the partial pressure of CO 2, and the Henry’s law constant, k, and asked to calculate the concentration of CO 2 in the solution. Plan: With the information given, we can use Henry’s law to calculate the solubility. Solve: Check: The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO 2 and the value of Henry’s constant. 2

Blood gases and deep sea diving Solubility increases as pressure increases Divers who use compressed gases must be concerned about the solubility of the gases in their blood At depth, the blood contains higher concentrations of dissolved gases Ascension, if too rapid, will cause the blood to fizz similar to 7-UP when opened! This is called decompression sickness, or “the bends”, which is painful and can be fatal because the bubbles affect things like nerve impulses

Temperature and Solubility 13.4 Solubility of most solid solutes in water increases with increasing temperature In contrast, solubility of gases in water decreases with increasing temperature

Temperature and Solubility 13.4 In contrast, solubility of gases in water decreases with increasing temperature Carbonated beverages go “flat” as they warm due to a decreased solubility of dissolved CO 2 Bubbles form on the inside wall of a cooking pot when water is heated even though the temperature is well below boiling Thermal pollution of lakes and streams causes low oxygen levels in deeper layers

Colligative Properties Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. Among colligative properties are  Vapor pressure lowering  Boiling point elevation  Melting point depression  Osmotic pressure

Vapor Pressure Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

Colligative Properties of Nonelectrolyte Solutions Vapor-Pressure Lowering Raoult’s law If the solution contains only one solute: X 1 = 1 – X 2 P P 1 =  P = X 2 P = vapor pressure of pure solvent X 1 = mole fraction of the solvent X 2 = mole fraction of the solute 13.6 P 1 = X 1 P 1 0 Nonvolatile solutes lower the vapor pressure of a solvent by an amount proportional to the solute mole fraction.

SAMPLE EXERCISE Calculation of Vapor-Pressure Lowering Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to mL of water. The vapor pressure of pure water at 25°C is 23.8 torr. Solution Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute. Plan: We can use Raoult’s law to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, X A, is the ratio of the number of moles of solvent (H 2 O) to total solution (moles C 3 H 8 O 3 + moles H 2 O). Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C 3 H 8 O 3 and H 2 O:

PRACTICE EXERCISE The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: SAMPLE EXERCISE continued The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water. We now use Raoult’s law to calculate the vapor pressure of water for the solution:

P A = X A P A 0 P B = X B P B 0 P T = P A + P B P T = X A P A 0 + X B P B If both components of the solution are volatile, the vapor pressure of the solution is the sum of the individual partial pressures. Pure toluenePure benzene 1:1 mix

Boiling-Point Elevation  T b = T b – T b 0 T b > T b 0  T b > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution  T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m) 13.6

Freezing-Point Depression  T f = T f – T f 0 T f > T f 0  T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution  T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m) 13.6

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g.  T f = K f m m =m = moles of solute mass of solvent (kg) = 2.41 m = kg solvent 478 g x 1 mol g K f water = C/m  T f = K f m = C/m x 2.41 m = C  T f = T f – T f 0 T f = T f –  T f 0 = C – C = C 13.6

Osmotic Pressure (  ) 13.6 Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (  ) is the pressure required to stop osmosis. dilute more concentrated

High P Low P Osmotic Pressure (  )  = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 13.6

A cell in an: isotonic solution hypotonic solution hypertonic solution 13.6

Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles Vapor-Pressure Lowering P 1 = X 1 P 1 0 Boiling-Point Elevation  T b = K b m Freezing-Point Depression  T f = K f m Osmotic Pressure (  )  = MRT

Colligative Properties of Electrolyte Solutions m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl 2 i should be 1 2 3

Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = iMRT Colligative Properties of Electrolyte Solutions 13.7