AP Chemistry Notes

1. AP Chemistry Notes Solutions & Colligative Properties

2. Solutions… • A solution is a homogenous mixture of two or more substances. • The solute is(are) the substance(s) present in the smaller amount(s). • The solvent is the substance present in the larger amount.

3. nonelectrolyte strong electrolyte weak electrolyte Solutions… • An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. • A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

4. Solutions… • A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. • An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. • A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

5. Solutions… • Three types of interactions in the solution process: • solvent-solvent; solute-solute; and solvent-solute DHsoln = DH1 + DH2 + DH3

6. Solutions… • Two substances with similar intermolecular forces are likely to be soluble in each other. • ‘LIKE DISSOLVES LIKE’ • Non-polar molecules are soluble in non-polar solvents • CCl4 in C6H6 • Polar molecules are soluble in polar solvents • C2H5OH in H2O • Ionic compounds are more soluble in polar solvents • NaCl in H2O or NH3 (l)

7. mass of solute x 100% = mass of solution Concentration… • The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. • Percent by Mass:

8. moles of A XA = sum of moles of all components Concentration… • Mole Fraction(X)

9. moles of solute liters of solution moles of solute m = mass of solvent (kg) M = Concentration… • Molarity: • Molality:

10. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent Concentration… What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? • Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol • 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m

11. Solubility… Solid solubility and temperature

12. Solubility… Gas solubility and temperature

13. Solubility… • Pressure and Solubility of Gases • The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). • Mathematically expressed: 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/L•atm) that depends only on temperature

14. low P high P low c high c Henry’s Law: The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

15. Colligative Properties NON-Electrolytic Solutions

16. Colligative Properties… • 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 • Boiling Point Elevation • Freezing Point Depression • Osmosis

17. = vapor pressure of pure solvent 0 P1 = X1 P 1 0 0 0 P 1 P 1 P 1 - P1 = DP = X2 Vapor Pressure Lowering… • Raoult’s law: X1= mole fraction of the solvent If the solution contains only one solute: X1 = 1 – X2 X2= mole fraction of the solute

18. Vapor Pressure Lowering… • At a given temperature water has a vapor pressure of 22.80 mmHg. Calculate the vapor pressure above a solution of 90.40 g of sucrose (C12H22O11) in 350.0 mL of water, assuming the water to have a density of 1.000 g/mL.

19. Vapor Pressure Lowering… • 90.40 g C12H22O11 0.26 mol • 350.0 ml H2O  19.4 mol • X = (19.4/(19.4+0.26)) = 0.9868 • VP = XP • VP = (0.9868)(22.80) • VP = 22.50 mmHg

20. Vapor Pressure Lowering… • 23.00 g of an unknown substance was added to 120.0 g of water. The vapor pressure above the solution was found to be 21.34 mmHg. Given that the vapor pressure of pure water at this temperature is 22.96 mmHg, calculate the Molar Mass of the unknown.

21. Vapor Pressure Lowering… • 23.0 g X • 120.0 g H2O  6.65 mol H2O • VP = 21.34 • P = 22.96 • VP = XP • 21.34 = X (22.96) • X = 0.929 • X = mol H2O / total mol • 0.929 = 6.65/(6.65 + x) • X mol = 0.506 • M = g/mol = 23.0 / 0.506 • M = 45.46 g/mol

22. 0 0 0 0 PB = XB P B PA = XA P A PT = XA P A +XB P B Ideal Solution Vapor Pressure Lowering… PT = PA + PB

23. Vapor Pressure Lowering… • At 20.0 oC the vapor pressures of methanol (CH3OH)and ethanol (C2H5OH) are 95.0 and 45.0 mmHg respectively. An ideal solution contains 16.1 g of methanol and 92.1 g of ethanol. Calculate the vapor pressure.

24. Vapor Pressure Lowering… • 16.1 g CH3OH  0.5 mol • 92.1 g C2H5OH  2 mol • Total mol = 2.5 • VP = XaPa + XbPb • VP = (0.5/2.5)(95) + (2/2.5)(45) • VP = 55 mmHg

25. 0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b DTb = Kbm Boiling Point Elevation… T b is the boiling point of the solution DTb > 0 m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m)

26. 0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf DTf = Kfm Freezing Point Depression… T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m)

27. Constants…

28. 0 DTf = T f – Tf moles of solute m= mass of solvent (kg) = 3.202 kg solvent 1 mol 62.01 g 478 g x 0 Tf = T f – DTf What is the freezing point of a solution containing 478g of ethylene glycol (antifreeze) in 3202 g of water? GIVEN: The molar mass of ethylene glycol is 62.01 g. DTf = Kfm Kf water = 1.86 0C/m = 2.41 m DTf = Kfm = 1.86 0C/m x 2.41 m = 4.48 0C = 0.00 0C – 4.48 0C = -4.48 0C

29. Vapor-Pressure Lowering Boiling-Point Elevation DTb = Kbm 0 P1 = X1 P 1 Freezing-Point Depression DTf = Kfm p = MRT Osmotic Pressure (p) Summary… • Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

30. Colligative Properties Electrolytic Solutions

31. actual number of particles in soln after dissociation van’t Hoff factor (i) = number of formula units initially dissolved in soln Colligative Properties… • 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 solution 0.1 m Na+ ions & 0.1 m Cl- ions • 0.1 m NaCl solution 0.2 m ions in solution i should be 1 2 3 Nonelectrolytes NaCl CaCl2

32. Boiling-Point Elevation DTb = iKbm Freezing-Point Depression DTf = i Kfm p = iMRT Osmotic Pressure (p) Colligative Properties…

33. Colligative Properties… At what temperature will a 5.4 molal solution of NaCl freeze? Solution: ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1oC FP = 0 – 20.1 = -20.1oC

34. Colligative Properties… Osmotic Pressure (p): 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 (p) is the pressure required to stop osmosis. more concentrated dilute

35. Colligative Properties… High P Low P p = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K)

36. Colloids… • A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. • Colloid versus solution: • Collodial particles are much larger than solute molecules • Collodial suspension is not as homogeneous as a solution