Solutions & Colligative Properties AP Chemistry Notes Solutions & Colligative Properties

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.

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. nonelectrolyte weak electrolyte strong electrolyte

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.

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

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)

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

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

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

Concentration… What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? m = moles of solute mass of solvent (kg) M = moles of solute liters of solution 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 m = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 kg solvent = 8.92 m

Solid solubility and temperature

Gas solubility and temperature

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

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). high P low P high c low c

Colligative Properties NON-Electrolytic Solutions

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

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

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.

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

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.

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

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

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.

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

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

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

Constants…

DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m 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 = Kf m Kf water = 1.86 0C/m = 3.202 kg solvent 478 g x 1 mol 62.01 g m = moles of solute mass of solvent (kg) = 2.41 m DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C DTf = T f – Tf Tf = T f – DTf = 0.00 0C – 4.48 0C = -4.48 0C

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. Vapor-Pressure Lowering P1 = X1 P 1 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT

Colligative Properties Electrolytic Solutions

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 van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be 1 2 3 Nonelectrolytes NaCl CaCl2

Colligative Properties… Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT

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 = 20.1 oC FP = 0 – 20.1 = -20.1 oC

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

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)

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