Announcements Dec. 18 Hour Exam 1 C-109 Start time 6PM Coverage is Chapter 12 and 13. 10-multiple choice 3-fairly short problems 3-longer problem solving 100 point Exam Lecture slides updated and homework solutions posted. It is critical that you are keeping up. Ask or see me if you need help.
These are must know units of concentration. Molarity (M) = moles solute liters solution Molality (m) = Mole Fraction (XA) = moles solute kg solvent moles of A total number of moles (solvent + solute) % by mass = mass of solute x 100 mass of solution % by volume = volume of solute volume of solution mass of solute ppm = x 10 mass of solution 6 x 100
Suppose: 5.00 g of NaCl(s) is dissolved in 100 g of pure water at 25 C and 1 atm pressure. Dissolve a solute in a pure solvent Property Vapor- Pressure Pure Water 4.8% w/w NaCl (aq) 23.8 23.1 Boiling Point 100 C 100.8 C Freezing Point Osmotic Pressure 0 C -3.0 C 0 atm 33 atm These are the Colligative Properties.
Colligative properties are physical properties of solutions that arise because of the number of solute molecules dissolved in solution (and not on the kind of solute particles dissolved in solution). Pure Solvent Solutes added to a pure solvent reduce the fraction of molecules of fraction of solvent molecules at the surface thereby reduce the vapor pressure relative to a pure solvent. Pure Liquid Solvent Molecules Pure Liquid with solute Added Solute Molecules
There are four colligative properties that we need to understand and memorize the equations for. The Four-Colligative Properties Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure (π) ΔP soln = X solute P solv ΔT bp = K b m ΔT fp = K f m π = MRT
The colligative properties are used in many applications. adding antifreeze to a car s radiator to lower FP de-iceing airplanes spreading salt on ice to melt it (lower FP). IV with saline solution. Dehydration drinks and contacts lens solutions
There are four colligative properties that result when we add non-volatile solute to solvents. 1. Vapor-Pressure Lowering P solv - Psoln = ΔP = P solvent (χ Solute ) 2. Boiling-Point Elevation Tbp - T bp = ΔTb = Kb m 3. Freezing-Point Depression T fp - Tfpsln = ΔTf = Kf m 4. Phenomena of Osmotic Pressure π = M R T
Raoult s Law: The vapor pressure of a pure solvent in the presence of a non-ionic solute is proportional to the mole fraction of the solvent. P solvent α P solvent P solution = χ solvent P solvent Vapor pressure of solution Mole Fraction of Solvent Vapor Pressure of Pure Solvent a) pure solvent A with a vapor pressure of P b) after solute is added the fraction of escaping molecules at the surface and is reduced thereby reducing the vapor pressure of the pure solvent.
Raoult s Law can be expressed in terms of mole fraction of the solute and vapor pressure lowering. P Soln = χ solvent P solvent (1) χ Solute + χ Solvent = 1 (2) χ Solvent = 1 - χ Solute (3) P Soln = (1 - χ Solute ) P solvent (4) P Soln = P solvent - P solvent (χ Solute ) (5) substituting 3 into 1 expanding 4 rearranging 5 ΔP = (P solvent - Psoln) = P solvent (χ Solute )
Calculate the vapor pressure lowering, ΔP, when 10.0 ml of glycerol (C 3 H 8 O 3 ) is added to 500. ml of water at 50. o C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/ml. The density of glycerol is 1.26 g/ml. PLAN: Find the mol fraction, χ, of glycerol in solution and multiply by the vapor pressure of water. ΔP = (P solvent - Psolvent) = P solvent (χ Solute ) SOLUTION: 10.0 ml C 3 H 8 O 1.26 g C 3 H 8 O 3 mol C 3 H 8 O 3 3 x ml C 3 H 8 O x = 0.137 mol C 3 92.09 g C 3 H 8 O 3 H 8 O 3 3 0.988 g H 500.0 ml H 2 O 2 O mol H 2 O x x = 27.4 mol H ml H 2 O 18.02 g H 2 O 2 O χ = 0.00498 ΔP = 0.137 mol C 3 H 8 O 3 x 92.5 torr = 0.461 torr 0.137 mol C 3 H 8 O 3 + 27.4 mol H 2 O
What is the vapor pressure, at 100ºC, of a 50/50 % (v/v) solution of ethylene glycol, C 2 H 6 O 2, in water at 1 atm? (MM C 2 H 6 O 2 = 62.06 g/mol d(c 2 H 6 O 2 ) = 1.1155 g/ml, d(h 2 O) = 1.0000 g/ml, d(50/50) = 1.069 g/ml P H2O = χ H2O P H2O Moles C 2 H 6 O 2 = 500. ml X 1.1155 g C 2 H 6 O 2 X 1 mol = 8.99 mol ml 62.06 g Moles H 2 O = 500. ml X 1.000 g H 2 O X 1 mol = 27.8 mol ml 18.02 g Mole Fraction H 2 O = 27.8 mol/ (27.8 mol + 8.99 mol) = 0.7556 P H2O = χ H2O P H2O =.7556 X 760. torr = 574. torr
2. Boiling Point Elevation: The addition of a nonvolatile non-ionic solute to a pure solvent increases the boiling the point of a solution. Tbp - T bp = ΔTb = Kb m Molality (mol/kg) Solution boiling point Pure solvent boiling point Change in boiling point of solution (+) Molal boiling point elevation constant ( C/m)
3. Freezing Point Depression: The addition of a non-volatile non-ionic solute to a pure solvent decreases the freezing point of the solution. ΔTf = T f - Tfp = Kf m Freezing point change (+) Freezing point solution Molal freezing point depression constant ( C/m) Solution Molality (mol/kg) Freezing point pure solvent
Kb and Kf are constants of proportionality unique to any given pure solvent measured by experiment and tabulated in handbooks.
We can compare the freezing and boiling pts of a pure solvent and a solution in a phase diagram. 0 Pure Solvent T b 0 Solution Curve Pressure SOLID LIQUID VAPOR ΔT b T b Temperature
You add 1.00 kg of ethylene glycol antifreeze (C 2 H 6 O 2 ) to 4450 g and add it to your car s radiator. What are the boiling and freezing points of the resulting solution?
You add 1.00 kg of ethylene glycol antifreeze (C 2 H 6 O 2 ) to 4450 g and add it to your car s radiator. What are the boiling and freezing points of the resulting solution? PLAN: Find the number of mols of ethylene glycol and m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water. SOLUTION: mol C 2 H 6 O 2 = 1.00 x 10 3 g C 2 H 6 O x mol C 2 H 6 O 2 2 = 16.1 mol C 2 H 6 O 2 62.07 g C 2 H 6 O 2 m C 2 H 6 O 2 = 16.1 mol C 2 H 6 O 2 4.450 kg H 2 O = 3.62 m C 2 H 6 O 2 ΔT b = 0.512 o C/m x 3.62 m = 1.85 o C ΔT f = 1.86 o C/m x 3.62 m = 6.73 C BP = 100 C + 1.85 C = 101.85 o C FP = -6.73 o C
4. Osmosis is the selective passage of solvent molecules through a semipermeable membrane from a dilute solution to a more concentrated one. dilute solution more concentrated solution
Osmosis is the passage of a solvent (frequently water) through a semi-permeable membrane, from a solution of low solute concentration (high-water potential) to a solution with high solute concentration (low water potential). pure solvent semipermeable membrane solution osmotic pressure An applied pressure is needed to prevent volume increase; this pressure is the osmotic pressure!
An pressure difference results from the net movement of solvent from a less-concentrated (hypotonic) to the more-concentrated (hypertonic) solution. π = M R T Osmotic Pressure (atm) Solution molarity R is the gas constant temperature in Kelvin
Suppose we have a 0.020 molar solution of table sugar (sucrose) and a semi-permeable membrane not permeable to sucrose. Calculate the osmotic pressure in mm Hg and to what height could this pressure support a column of water (density Hg =13.6 g/ml and water = 1g/mL? π = M R T π = 0.02 M x 0.0821 L atm/mol K x 298K π =.49 atm x 760 torr/1 atm h = 371 mm Hg x 13.6 = 5.0 meters
Biochemists have discovered more than 400 mutant varieties of hemoglobin (Hb), the blood protein that carries oxygen throughout the body. A physician studying a form of Hb associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.0 o C to make 1.50 ml of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this Hb mutant? PLAN: We know Π as well as R and T. Convert Π to atm and T to Kelvin. Use the Π equation to find the molarity M and then the amount and volume of the sample to calculate M. SOLUTION: # mol = g/m M = Π RT 2.08 x 10-4 mol L 21.5 mg x x g 10 3 mg = 3.61 torr x atm 760 torr (0.0821 L. atm/mol. K)(278.15 K) L 1.50 ml x = 3.12 x 10-7 mol 10 3 ml x 1 3.12 x 10-7 mol = 2.08 x 10-4 M = 6.89 x 10 4 g/mol
When there is a concentration difference across a semi-permeable membrane there will be an osmotic pressure across the membrane. Water will move from the dilute solute side to the concentrated solute side When Concentrations Are the Same: No Osmotic Pressure isotonic solution hypotonic solution hypertonic solution
When there is a concentration difference across a semi-permeable membrane there will be an osmotic pressure across the membrane. Normal Onion Cell Plasmolyzed cell (cell membrane has shrunk from the cell wall.
Because colligative properties depend only on the number of solute particles in the solution, ionic solutes have a greater impact on the colligative properties differently than do non-ionic solutes. 0.1 m nonelectrolytes solution 0.1 m in solution 0.1 m NaCl solution 0.2 m ions in solution 0.1 m CaCl2 solution 0.3 m ions in solution
van t Hoff Factors for Electrolytes Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure ΔT b = i K b m ΔT f = i K f m π = i M R T van t Hoff Factors
Reverse osmosis is used to purify water by overcoming forcing dirty water through a semipermeable membrane. Pressure Memb
Now this is wild: Anology to Osmosis In a closed container the solution with the highest vapor pressure will completely transfer to the container of lower vapor pressure until the mole fractions of solvent are equal in both! Cool... Pure High Vapor Pressure Low Vapor Pressure