TODAY•1. Review- Concentrations

Influences on SolubilityTemperature, Pressure

•2. Vapor Pressure of Solutions- Raoult’s Law

•3. Consequences of Vapor Pressure LoweringColligative Properties

Importance: Way to get molar mass!

Quiz #17 (8)Given that we want to make 25.0 mL of 0.40 M

NaOH (40.0 g/mol).

1.(2) How many mol of NaOH are needed?

2. (2) How many grams of NaOH are needed?

3. (1) How much water is needed?

Influences on Solubility

•Temperature

•For gases- Decrease solubility with increase in T (Fig.11.7)

• For solids- Often (but not always) increase with increase in T (Fig. 11.6)

Figure 11.7 The Solubilities of Several Gases in Water

(Solubility Decreases with Increase in T)

Figure 11.6The Solubilities of Several Solids as a Function of Temperature

Influences on Solubility (cont.)

•Pressure

•For gases- Increase solubility with increase in P (Fig.11.5)

• For solids- Small effect

Figure 11.5 A Gaseous Solute

(Increase in Pressure Increases solubility)

Henry’s Law

P = kCP = partial pressure of gaseous solute above

the solutionC = concentration of dissolved gask = a constant

The amount of a gas dissolved in a solution is The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas directly proportional to the pressure of the gas above the solution.above the solution.

Quiz #18• Given: N2 solubility in water is 8.21*10 -4

mol/L at 0 C and 0.790 atm.

• Find: Concentration of N2 in water at 0 C when P of nitrogen is 1.10 atm.

Vapor Pressure of Solutions

Presence of non-volatile solute affects vapor pressure of solvent.

•Lowers it (Figs. 11.9,10)

• Raoult’s Law - Ideal SolnPA = XA PA° 0 < X <1So PA < PA° (Words?)

Figure 11.9 An Aqueous Solution and Pure Water in a Closed Environment

Figure 11.10 Presence of a Nonvolatile Solute Inhibits Escape of Solvent Molecules from Liquid

Raoult’s Law(“Ideal” Soln.)

Psoln = χsolventP°solvent

Psoln = vapor pressure of the solutionχsolvent = mole fraction of the solventP°solvent = vapor pressure of the pure solvent

The presence of a nonvolatile solute lowers The presence of a nonvolatile solute lowers the vapor pressure of a solvent.the vapor pressure of a solvent.

Figure 11.11A Solution Obeying Raoult’s Law

Influences on SolubilityPressure - solutions, Raoult’s Law

11.48, 50•11.48•Given: 53.6 g glycerin (C3H8O3) and 133.7 gethanol (C2H5OH) solution has vapor pressure of 113 Torr at 40 C.

Glycerin is “not” volatile (has very low vapor pressure)

Find: Vapor pressure of pure ethanol at 40 C Soln:

Note:

Even if solution does not obey Raoult’sLaw ( if not ideal soln.) the solution vapor pressure still obeys Dalton’s Law.

Ptotal = PA + PB + …..

Fig. 11.12 Total Vapor Pressure of a Solution(Sum of vapor pressure of each component)

Figure 11.13Vapor Pressure for a Solution of Two Volatile

Liquids (Nonideal)

TODAY•1. Review-

Colligative Properties- ∆Tb

• 2. Colligative Properties- (Cont.)∆Tf , Osmotic pressure

•3. Electrolyte solutions - more particles

•4. Colloidal Dispersions

•5. Reaction Rates

Colligative Properties

•Colligative:Means depends on number of

particles, NOT kind. (Collective)

•Importance: Can estimate Molar Mass.

Colligative Properties(Cont.)

•Based on Vapor Pressure Lowering of solvent because of solute.

(Next Slide, Fig. 11.14)

Figure 11.14 Development of Colligative Properties, BP Elevation, MP Depression and Osmotic Pressure

Colligative Properties (Cont.)• Boiling Point Elevation

∆ Tb = Kb m

m = molality of solute particles

Kb = molal boiling point constant

(deg*kg solvent/mol particles)

∆ Tb = Tbsoln - Tb° (positive number)

Tb° = boiling point pure solvent

Colligative Properties (Cont.)•Ex: 11.60, p. 555•Given: 2.00 g of X in 15.0 g CCl4.

B. pt. of soln = 77.85 deg. Kb for CCl4 = 5.03 deg*kg/mol . Tb° = 76.50 deg

• Find: Mm of X, a biomolecule.•Soln: Use ∆ Tb = Kb m

Colligative Properties (Cont.)•Freezing Point Depression

∆ Tf = - Kf m

m = molality of solute particles

Kf = molal freezing point constant

(deg*kg solvent/mol particles)

∆ Tf = Tfsoln - Tf° (negative number)

Tf ° = freezing point pure solvent

Colligative Properties(Cont.)Ex:11.66, p. 555•Given: Anthraquinone, emp. form. C7H4O.∆ Tf = - 22.3 deg when 1.32 g anth. dissolved in 11.4 g camphor.Kf = 40. deg*kg solvent/mol particles (Table 11.5)Find: Molecular formula.Soln: Find molar mass using ∆ Tf = -Kf m

Colligative Properties(Cont.)•Osmotic Pressure

•Another result of vapor pressure lowering of solvent because of solute.

•Osmosis (Fig. 11.16)Flow of solvent through semipermeable membrane into a solution.(because of different vapor pressures of solvent in solution)

Colligative Properties(Cont.)•Osmotic Pressure

Natural tendency is to try to even out concentrations.

Solute can’t go through semipermeablemembrane so solvent moves.

Figure 11.15 Ice in Equilibrium

Figure 11.16

Osmotic Pressure

Name? TA NameQuiz #19 (8)

• Given: 0.255 g of a compound,Y, is dissolved in 11.12 g benzene. The solution starts boiling at 80.26 deg. Pure benzene boils at 80.10 deg and the boiling point constant for benzene is 2.53 deg*kg/mol.

• Find: Estimate of molar mass of Y

Colligative Properties(Cont.)•Osmotic Pressure(Cont.)

•Pressure which must be supplied to surface of solution to stop osmosis.

π = M R T

π = osmotic pressureM = molarity of particlesT = Temperature in Kelvin

Colligative Properties(Cont.)Ex:11.68, p. 556•Given: 20.0 mg of a protein dissolved in water to make 25.0 mL soln. Osmotic pressure is 0.56 Torr at 25 C. •Find: Molar mass.•Soln: Find molar mass using π = M R T

Colligative Properties(Cont.)Electrolyte Solutions?Remember that Colligative Properties depend on number of (solute) particles in solution.

i = van’t Hoff factor

= moles particles in solutionmoles of solute dissolved

[ion pairing keeps i from having ideal value]

Colligative Properties(Cont.)Electrolyte Solutions?Remember that Colligative Properties depend on number of (solute) particles in solution.Expected values:0.050 m NaCl i = 2

m = 2 (ions/mol)* 0.050 mol/kg water

Actual values: i= 1.9, Table 11.6

m = 1.9 (ions/mol)* 0.050 mol/kg water

Colligative Properties(Cont.)Electrolyte Solutions?Remember that Colligative Properties depend on number of (solute) particles in solution.Expected values:0.050 m MgCl2 i = 3

m = 3 (ions/mol)* 0.050 mol/kg water

Actual values: i = 2.7

m = 2.7 (ions/mol)* 0.050 mol/kg water

Colligative Properties(Cont.)Electrolyte Solutions?Remember that Colligative Properties depend on number of (solute) particles in solution.Expected values:0.050 m FeCl3 i = 4

m = 4 (ions/mol)* 0.050 mol/kg water

Actual values: i = 3.4, Table 11.6

m = 3.4 (ions/mol)* 0.050 mol/kg water

ColloidsColloidal Dispersion:

Beam of light visible passing through.Tyndall Effect , Fig. 11.23Particle Size: 1 to 1000 nmAround size of light wavelengths.

True Solution:Beam of light NOT visible passing through.Particle Size: < 1 nmLess than size of light wavelengths.