Physical Properties of Solutions
Types of problems
1. Concentration units
- % by mass
- Molarity
- Molality
2. Henry’s Law: Pressure and solubility
3. Colligative Properties
- Raoult’s Law: Vapor pressure lowering
- Boiling-point elevation
- Freezing-point depression
- Osmotic pressure
4. Colligative properties of electrolytes
Problem Type 1: Concentration Units
% by mass
More precisely expressed, “mass percent of solute in a solution.”
Thus,
Example 1: Dehydrated Epsom salt is magnesium sulfate. What is the % by mass of magnesium sulfate in a solution made from 16.0 g of the salt and 100.0 mL of water at 25ºC? The density of water at 25ºC is 0.997 g/mL.
mass of solute = 16.0 g
mass of solvent =
Molarity
Molarity is defined as: it has the units mol/L = M
Note that M is used both as a variable (M=molarity) and as a dimension (M=mol/L or molar). Thus, we can say that M (molarity) of a solution is 2.0 M (molar) or 2.0 M (mol/L). Confusing?
Molality
Molality is defined as: it has the units mol/Kg = m
Note that m is used both as a variable (m=molality) and as a dimension (m=mol/Kg or molal). Thus, we can say that m (molality) of a solution is 2.0 m (molal) or 2.0 m (mol/Kg). Confused again?
Working with molarity has an advantage in laboratory work: It is easy to dispense solutions by volume, allowing the chemist to easily dispense mole quantities by adjusting volumes.
Working with molality has an advantage when temperatures change: The molality of a solution does not change with temperature, but molarity does since temperature affects the density of the solution, and thus its volume.
Also, don’t forget that M = molarity, while m = molality. Some textbooks will refer to them as CM and Cm, respectively, where C stands for concentration.
Problem Type 2: Henry’s Law: Pressure and Solubility
Solubility is defined as the maximum amount of solute that will dissolve is a given amount of solvent at a specific temperature. External pressure usually has a negligible effect on the solubility of liquids and solids, however it has a dramatic effect on the solubility of gases. Henry’s Law describes this nearly linear, directly proportional relationship as: where c = concentration in molarity
P = pressure in atmospheres
k = proportionality constant that
is specific to a gas at a given
temperature in mol/L•atm
Kinetic Theory explains this by noting that increasing the pressure of a gas over a liquid increases the frequency of gas particle collision with the surface, thus increasing the number of particle that dissolve in the liquid. Alternatively, this can be viewed as a consequence of Le Chatelier’s Principle, in which increasing the pressure of a gas over a liquid increases the concentration of one reactant in the equilibrium expression, A(g) + H2O(l) → A(aq), thus creating a stress that can be minimized by an equilibrium shift to the right.
Example 2: What is the concentration of O2 at 25 ºC in water that is saturated with air at an atmospheric pressure of 645 mmHg? The Henry’s Law constant (k) of oxygen at 25 ºC is 3.5 x 10-4 mol/L•atm. Assume that the mole fraction of O2 in air is 0.209.
Henry’s Law:
c = ?
k = 3.5 x 10-4 mol/L•atm
PO2 =
Problem Type 3: Colligative Properties
Colligative properties of solutions are those properties that depend on the number of solute particles in solution, but are otherwise independent of the nature of the solute particles.
Raoult’s Law: Vapor Pressure Lowering
Caution: Raoult’s Law applies only to non-volatile solutes!
Adding a non-volatile solute to a solvent lowers the vapor pressure of the solvent in direct proportion to the mole fraction of the solute according to Raoult’s Law:
P1 = X1 Pº1 where P1 = partial pressure of the solvent over a solution
X1 = mole fraction of the solvent in the solution
Pº1 = vapor pressure of the pure solvent
A little algebra will yield a different version of Raoult’s Law for a one solute system:
ΔP = X2 Pº1 where ΔP = amount the vapor pressure is lowered
X2 = mole fraction of the solute in the solution
Pº1 = vapor pressure of the pure solvent
The phenomenon of vapor pressure lowering can be understood by analogy. Evaporation only occurs at the surface of a liquid. Imagine the surface as a checkerboard with one particle of liquid per square. Let the probability that a particle leaves the surface and enter the gas phase be 10%. Over time and equilibrium is established between the liquid phase and gas phase, with particles transferring in both directions at an equal rate. We can shift the equilibrium by reducing the probability of a particle leaving the surface. Now, imagine replacing 1/10 of the liquid particles in the squares with another particle that cannot escape. That is, fewer particles are available to enter the gas phase, while the probability of gas particles re-entering the liquid phase remains unchanged. Thus, the equilibrium is shifted away from the gas phase and the vapor pressure is depressed.
Example 3: Calculate the vapor pressure of an aqueous solution at 30 ºC made from 1.00 x 102 g of sucrose (C12H22O11) and 1.00 x 102 g of water. The vapor pressure of pure water at 30 ºC is 31.8 mmHg.
Raoult’s Law: P1 = X1 Pº1
Pº1 = 31.8 mmHg
Boiling Point Elevation
The boiling point of a liquid is the temperature at which the vapor pressure equals the external atmospheric pressure. For water at 1 atmosphere pressure, the vapor pressure equals 1 atm when the temperature of the water reaches 100 ºC. However, as discussed under Raoult’s Law, a non-volatile solute always lowers the vapor pressure of a solution, so at the normal boiling point, the vapor pressure of the solution is less than atmospheric pressure, and it does not boil. The vapor pressure must be increased by raising the temperature of the solution. Thus, the addition of a solute raises the boiling point of a solution. The change in boiling point is directly proportional to the concentration of the solute and is given by: ΔTb = kbm where ΔTb = Tb – Tbº (where Tb is the boiling point of the
solution and Tbº is boiling point of the pure solvent)
m = the molal concentration of the solute
kb = the molal boiling-point elevation constant of the
solvent with units ºC/m
Remember: molality must be used because we are dealing with a system not at constant temperature. Molarity varies with temperature, while molality does not.
Example 4: What is the boiling point of an antifreeze solution made from a 50-50 mixture (by volume) of ethylene glycol, C2H6O2 and water? Assume the density of water is 1.00 g/mL and the density of ethylene glycol is 1.11 g/mL. From a table, kb = 0.52 ºC/m for water.
Tb = Tbº + ΔTb = Tbº + kbm
Tbº = 100.0 ºC
kb = 0.52 ºC/m
assume 100 mL solution:
Freezing Point Depression
Analogous to boiling point elevation, the freezing point of a solution is depressed in direct proportion to the solute concentration according to the equation,
ΔTf = kfm where ΔTf = Tfº – Tf (where Tf is the freezing point of the
solution and Tfº is freezing point of the pure solvent)
m = the molal concentration of the solute
kf = the molal freezing-point depression constant of the
solvent with units ºC/m
Example 5: How many grams of isopropyl alcohol, C3H7OH, should be added to 1.0 L of water to give a solution that will not freeze above -16ºC. From a table, kf (water) is 1.86ºC/m.
ΔTf = Tfº – Tf = 0-(-16) = 16ºC
kf = 1.86ºC/m
kg (solvent) = 1.0 L x 1 kg/L = 1.0 kg
Van’t Hoff Equation: Osmotic Pressure
Osmosis is the net movement of solvent molecules through a semi-permeable membrane from a pure (or less dilute) solvent to a more concentrated solution. The osmotic pressure (π) is the pressure required to stop osmosis. The osmotic pressure of a solution is directly proportional to the concentration of the solute, and is given by the Van’t Hoff equation:
π = MRT where M = molarity of the solution
R = Ideal gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin
Osmotic pressure is easily perceived by examining Figure 1 to the right, where the osmotic pressure is the pressure necessary to exactly balance the flow of water caused by osmosis.
Example 6: The average osmotic pressure of seawater is about 30.0 atm at 25ºC. Calculate the molar concentration of an aqueous solution of urea (NH2CONH2) that is isotonic with seawater.
rearranging the Van’t Hoff Equation:
To be isotonic with seawater, it must have the same osmotic pressure = 30.0 atm. So,
π = 30.0 atm
R = 0.0821 L•atm/mol•K
T = 25 + 273 = 298 K
Problem Type 4: Colligative Properties of Electrolytes
Remember: Colligative properties of solutions are those properties that depend on the number of solute particles in solution. Also, Electrolytes dissociate into ions in solution. Thus, the number of particles in a solution will be different than the number of particles placed into the solution. For example, one “particle” of NaCl dissolved in water becomes one “particle” of Na+ and one “particle” of Cl-, or two particles. Thus, the concentration used for colligative property calculations is twice what might be otherwise expected.
As a result, we have to modify each of the equations previously discussed to account for the dissociation of electrolytes:
ΔTb = ikbm
ΔTf = ikfm
π = iMRT
where
The value of i should be 1 for all non-electrolytes. It should be 2 for strong electrolytes such as KCl and BaSO4, and it should be 3 for strong electrolytes such as CaCl2 and Na2CO3. In reality, however, the value of i is usually less than simple stoichiometry predicts. This is because at higher concentrations, the ions tend to pair due to electrostatic attractions (although they are still dissolved as pairs, like polar molecules), thus reducing the number of solute particles in solution. Van’t Hoff factors (i) can be looked up in tables.