ap chemistry notes solutions & colligative properties

AP Chemistry Notes

Solutions & Colligative Properties

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…

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…

Three types of interactions in the solution process: solvent-solvent; solute-solute; and solvent-solute

Hsoln = H1 + H2 + H3

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 solutemass of solution

Concentration…

Mole Fraction (X)

XA = moles of A

sum of moles of all components

Concentration…

Molarity:

Molality:

M =moles of solute

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 ethanol927 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

Solubility…

Solid solubility and

temperature

Solubility…

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:

low P

low c

high P

high c

The solubility of a gas in a liquid is proportional to the The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (pressure of the gas over the solution (Henry’s lawHenry’s law).).

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 10

P 10 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

If the solution contains only one solute:

X1 = 1 – X2

P 10 - P1 = P = X2 P 1

0

X2 = mole fraction of the solute


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.


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


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.


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


PB = XB P B0

PT = PA + PB

PT = XA P A0 + XB P B

0

PA = XA P A0

Ideal Solution


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.


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…

Tb = Tb – T b0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb > T b0 Tb > 0

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m)

Freezing Point Depression…

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

Constants…

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.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf water = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 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 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

Colligative Properties

Electrolytic 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.

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 be1

2

3

Nonelectrolytes

NaCl

CaCl2


Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT


At what temperature will a 5.4 molal solution of NaCl At what temperature will a 5.4 molal solution of NaCl freeze?freeze?

Solution:Solution:

∆∆TTFPFP = K = Kff • m • i • m • i

∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 5.4 m • 2C/molal) • 5.4 m • 2

∆ ∆TTFP FP = 20.1= 20.1 ooCC

FP = 0 – 20.1 = -20.1FP = 0 – 20.1 = -20.1 ooCC


Osmotic Pressure ():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 moreconcentrated


HighP

LowP

= 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

ap chemistry notes solutions & colligative properties

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