Electrolytes and Colligative Properties

Learning Objectives

Perform calculations involving the effects of electrolytes on colligative properties.

What effect do ions have?

The presence of ions in solutions affect the properties of that solution

Ionic solution measurement. Image from Wikimedia.

The addition of ions creates significant changes in properties of solutions. Water molecules surround the ions and are somewhat tightly bound to them. Colligative properties are affected because the solvent properties are no longer the same as those in the pure solvent.

Ionic compounds are electrolytes and dissociate into two or more ions as they dissolve. This must be taken into account when calculating the freezing and boiling points of electrolyte solutions. The sample problem below demonstrates how to calculate the freezing point and boiling point of a solution of calcium chloride. Calcium chloride dissociates into three ions according to the equation:

$text{CaCl}_2 (s) rightarrow text{Ca}^{2+} (aq)+2text{Cl}^-(aq)$

The values of the freezing point depression and the boiling point elevation for a solution of CaCl ₂ will be three times greater than they would be for an equal molality of a nonelectrolyte.

Sample Problem: Freezing and Boiling Point of an Electrolyte

Determine the freezing and boiling point of a solution prepared by dissolving 82.20 g of calcium chloride into 400.g of water.

Step 1: List the known quantities and plan the problem.

Known

mass CaCl ₂ = 82.20 g
molar mass CaCl ₂ = 110.98 g/mol
mass H ₂ O = 400. g = 0.400 kg
$K_f(text{H}_2text{O})=-1.86^circ text{C}/ m$
$K_b(text{H}_2text{O})=0.512^circ text{C}/ m$
CaCl ₂ dissociates into 3 ions

Unknown

$T_f=? ^circ text{C}$
$T_b=? ^circ text{C}$

The moles of CaCl ₂ is first calculated, followed by the molality of the solution. The freezing and boiling points are then determined, including multiplying by 3 for the three ions.

Step 2: Solve.

$82.20 text{ g CaCl}_2 times frac{1 text{ mol CaCl}_2}{110.98 text{ g CaCl}_2} &= 0.7407 text{ mol CaCl}_2\frac{0.7407 text{ mol CaCl}_2}{0.400 text{ kg H}_2text{O}} &= 1.85 m text{CaCl}_2$

$Delta T_f &= K_f times m times 3=-1.86^circ text{C}/m times 1.85 m times 3=-10.3^circ text{C} && T_f=-10.3^circ text{C}\Delta T_b &= K_b times m times 3=0.512^circ text{C}/m times 1.85 m times 3=2.84^circ text{C} && T_b=102.84^circ text{C}$

Step 3: Think about your result.

Since the normal boiling point of water is 100.00°C, the calculated result for $Delta T_b$ must be added to 100.00 to find the new boiling point.

Summary

The effect of ionization on colligative properties is described.

Practice

Do the practice and homework problems dealing with ionic solutions toward the end of the section on the link below:

http://avon-chemistry.com/solution_lec_p2.html

Review

Why do ionic materials change the colligative properties of a solution?
Would HCl be expected to alter colligative properties?
Calcium carbonate is ionic, but insoluble in water. What effect would it have on the boiling point of water?

Solutions