16.15: Electrolytes and Colligative Properties

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.

Electrolytes and Colligative Properties

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:

$\ce{CaCl_2} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right)$

The values of the freezing point depression and the boiling point elevation for a solution of $$\ce{CaCl_2}$$ will be three times greater than they would be for an equal molality of a nonelectrolyte.

Example 16.15.1

Determine the freezing and boiling point of a solution prepared by dissolving $$82.20 \: \text{g}$$ of calcium chloride into $$400. \: \text{g}$$ of water.

Solution:

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

Known

• Mass $$\ce{CaCl_2} = 82.20 \: \text{g}$$
• Molar mass $$\ce{CaCl_2} = 110.98 \: \text{g/mol}$$
• Mass $$\ce{H_2O} = 400. \: \text{g} = 0.400 \: \text{kg}$$
• $$K_f \left( \ce{H_2O} \right) = -1.86^\text{o} \text{C}/\textit{m}$$
• $$K_b \left( \ce{H_2O} \right) = 0.512^\text{o} \text{C}/\textit{m}$$
• $$\ce{CaCl_2}$$ dissociates into 3 ions

Unknown

• $$T_f = ? \: ^\text{o} \text{C}$$
• $$T_b = ? \: ^\text{o} \text{C}$$

The moles of $$\ce{CaCl_2}$$ 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.

\begin{align} 82.20 \: \text{g} \: \ce{CaCl_2} \times \frac{1 \: \text{mol} \: \ce{CaCl_2}}{110.98 \: \text{g} \: \ce{CaCl_2}} &= 0.7407 \: \text{mol} \: \ce{CaCl_2} \\ \frac{0.7407 \: \text{mol} \: \ce{CaCl_2}}{0.400 \: \text{kg} \: \ce{H_2O}} &= 1.85 \: \textit{m} \: \ce{CaCl_2} \end{align}

$\begin{array}{ll} \Delta T_f = K_f \times \textit{m} \times 3 = -1.86^\text{o} \text{C}/\textit{m} \times 1.85 \: \textit{m} \times 3 = -10.3^\text{o} \text{C} & T_f = -10.3^\text{o} \text{C} \\ \Delta T_b = K_b \times \textit{m} \times 3 = 0.512^\text{o} \text{C}/\textit{m} \times 1.85 \: \textit{m} \times 3 = 2.84^\text{o} \text{C} & T_b = 102.84^\text{o} \text{C} \end{array}$

Since the normal boiling point of water is $$100.00^\text{o} \text{C}$$, the calculated result for $$\Delta T_b$$ must be added to 100.00 to find the new boiling point.