Colligative Properties

It turns out that for dilute solutions of nonvolatile solutes the Tfp and the Tbp are proportional to the molality of the solution.

to get an equality we add a constant and

The magnitude of the constants are different for a particular solvent and also vary with different solvents. The units on 'k', the freezing point constant or the boiling point constant, are degrees C m-1. For water the freezing point constant is 1.86 degrees C m-1 and the boiling point constant is 0.512 degrees C m-1. A 1 molal aqueous solution of any nonvolatile nonelectrolyte boils at 100.512 degrees C and melts at 1.86 degrees C. So if we have a concentration of a nonvolatile solution other than 1 molal we can use either equation to calculate the new boiling or freezing point. It must be remembered that when solving for T that if the freezing point is expected the T must be subtracted from 0 degrees C (freezing point depression) and added to 100 degrees C (boiling point elevation).

Another sample problem;

We can also use the expression to determine the molecular weight of an unknown solute. If we recognize that;

and that;

substituting into the freezing point expression we have;

rearranging and solving for MW we have;

Another more interesting example;

We can also use the colligative properties to help understand the behavior of strong electrolytes as well as nonelectrolytes. If we look at some experimental freezing point data some interesting information can be obtained.

What is it about the second group of compounds that would explain the observed freezing point depressions? They are all ionic compounds! They are what we have called strong electrolytes, because we understand these compounds to completely dissociate in aqueous solution. But do they?

Experimentally the Tfp for NaOH is 3.44 degrees C. Ideally we would expect that all of the NaOH would dissolve according to the reaction:

NaOH(s) H2O--> Na+(aq) + OH-(aq)

When an ionic compound dissolves in water the ionic compound exists as ions in solution. 1 mol of NaOH dissolves to form 2 moles of ions (particles). Inorder to deal with this behavior we need a variation on the freezing point expression. It is;

where;

Ideally we would predict that NaOH would completely dissociate so that 'i' would have a value of 2 and the theoretical freezing point would be 3.72 degrees C. We experimentally observe that all the NaOH dissolves yet the freezing point depression data suggest that not all of the NaOH has dissociated. Some of the NaOH remains associated. Based on the experimental change in freezing point the experimental value of 'i' is 1.85. In general, the more concentrated the solution the greater the observed deviation from ideality. As the solution becomes more dilute the experimental 'i' value approaches the theoretical value. The difference between the ideal value of 'i' and the observed is due to something we call 'ion-pairing'. In an aqueous solution the ions of the solute are surrounded by water molecules. If the the solution is very dilute the large number of water molecules that surround the ions effectively prevent the ions from seeing each other. All an individual ion sees is other water molecules. At higher concentrations of ions the number of water molecules available for solvation is lowered and there are fewer water molecules surrounding any particular ion. The result is the ions can 'see' each other and they feel some attraction and they will 'pair up' in solution. When they pair there are fewer particles in the solution and 'i' drops from it ideal value.

We can use experimental 'i' values obtained by measuring the freezing point depression of electrolytes to determine whether the substance is strong, weak or a nonelectrolyte. For example, when the freezing point of a 0.100 molal solution of HCl is measured the observed value is 0.353 degrees C. Is HCl an electrolyte? If it were not the expected change in temperature would be 0.186 degrees C. Based on our experimental data HCl is in fact a strong electrolyte and is 90 % dissociated. On the other hand a 0.100 molal solution of HC2H3O2 has a Tf of 0.195 degrees C, very close to 0.186 degrees C. Acetic acid is considered to be a weak electrolyte, while dextrose and urea are considered to be nonelectrolytes. With more understanding of this kind of problem we would be able to calculate the percent dissociation of the compound.

Sample Problem:

Calculate the freezing point and the boiling point of a saturated solution of Li2CO3. The solubility of lithium carbonate is 0.72 g per 100 g of water at 100 degrees C.

Answer:

Show Me!

Try another sample problem:

2.57 g of an ionic compound with the formula KX are dissolved in 120 g of water. The freezing point of the solution was lowered by 1.37 degrees C. Determine the formula weight of X.

Answer: