Solute Effects on Solvents
Dissolving a solute in a solvent has several important effects: first, the vapor pressure of the solvent is lowered; second, the boiling point of the solvent is elevated; and third, the freezing point is depressed. These effects are known as the colligative properties of a solution.
Because the number of solvent molecules at the surface of a liquid is reduced when a solute is added (because some solute molecules are at the surface as well), the vapor pressure of the solvent is reduced. It follows that more solute molecules will further reduce the vapor pressure. Francois Raoult developed an equation relating the vapor pressure of a solvent with the mole fraction of solvent times the normal vapor pressure:
Psolvent = Xsolvent * P°solvent
Psolvent is the corrected vapor pressure in mm Hg, Xsolvent is the mole fraction of the solvent (the number of moles of solvent over the total number of moles of solvent and solute), and P°solvent is the vapor pressure of the pure solvent. As with the Ideal Gas Law, Raoult's Law is an approximation of ideal behavior, but it's sufficiently accurate under most conditions.
For example, what is that vapor pressure of 5.5 moles of water if 20 grams of salt are dissolved at 25 °C? First, find the number of moles of salt: 20 grams of NaCl, which has a molar mass of 58.5, is equal to 0.342 moles. The normal vapor pressure of water at this temperature is Now, using Raoult's Law, we write:
Psolvent = (5.5 / (0.342 + 5.5)) * 23.8 mm Hg
Some simple math gives a new vapor pressure of 22.41 mm Hg. Raoult's Law can also be used to find the molar mass of an unknown solute, using the principle of mole fractions:
Psolvent = (moles solvent / (moles solvent + grams solute / molar mass solute)) * P°solvent
Another important effect of solutes is the elevation of the solvent's boiling point. Because the vapor pressure is lowered, the temperature must be raised above normal to equal atmospheric pressure. The elevation, or change, in boiling point temperature is represented by the symbol Δtbp:
Δtbp = Kbp * msolute
Where m is the molality of the solute. For a definition of molality, see the Moles, Molarity, and Molality page in the "Background Information" chapter. Kbp is a boiling point elevation constant that is unique to each solvent. The constant for water is 0.5121; other constants will be given in a problem.
For example, let's find the boiling point elevation of 2 liters of water when 100 grams of salt are added. 100 grams of NaCl is equivalent to 1.709 moles, and 2 liters of water weighs 2 kilograms. However, since NaCl dissociates into two solutes, Na+ and Cl-; therefore, another factor needs to be introduced into our equation above to account for the two ions in the solution. This symbol is called the van't Hoff factor, and its symbol is i. The van't Hoff factor is equal to the number of ions produced when a solid dissolves, in whole numbers, for dilute solutions; in more concentrated solutions, the van't Hoff factor becomes smaller than expected. The corrected equation becomes:
Δtbp = Kbp * msolute * i
For salt, the van't Hoff factor is about two, and substituting the other factors into the equation results in:
Δtbp = (0.5121)(1.709 mol / 2 kg)(2) = +.876 degrees Celsius
In this case, the boiling point of water is elevated by .876 degrees to 100.876 °C.
Freezing points of solvents are depressed by dissolved solutes, and as a bonus for chemists, the equation for freezing point is just the same as above; the constant is the only thing that changes.
Δtfp = Kfp * msolute * i
For freezing point depression, the constant for water is -1.86. Using the equation above, let's find the freezing point depression caused by the addition of salt in the previous example problem:
Δtfp = (-1.86)(1.709 mol / 2 kg)(2) = -3.178 degrees Celsius
The freezing point of water is depressed to -3.178 degrees Celsius. The effect of dissolved solutes is put to use in antifreeze, in which a soluble liquid is dissolved in water to keep a car's cooling fluid from boiling or freezing at normal temperatures.
As with Raoult's Law, these equations can be used to find the molar mass of an unknown substance; since molality is defined as moles of solute per liter of solvent, the moles component of the equations can be solved for molar mass:
Δtbp = Kbp * ((grams solute / molar mass) / kilograms
solvent) * i, or:
Δtfp = Kfp * ((grams solute / molar mass) / kilograms
solvent) * i