Understanding Freezing Point Depression with Ionic Dissociation

When a solute dissolves and dissociates in a solvent, it affects the colligative properties like freezing point depression. The observed depression is greater than expected if dissociation occurs because the number of particles increases.

Step 1: Write the dissociation reaction

For MX₂:

$\mathrm{MX}2\rightleftharpoons M^{2+}+2X^{-}$

Step 2: Define the van't Hoff factor (i)

The van't Hoff factor accounts for dissociation and is given by:

$i=1+\alpha (n-1)$

where α is the degree of dissociation (0.5) and n is the number of ions produced per formula unit (3 for MX₂).

Step 3: Calculate i

$i=1+0.5(3-1)=1+0.5\times 2=2$

Step 4: Relate to freezing point depression

The freezing point depression is given by ΔT_f = i K_f m, where K_f is the cryoscopic constant and m is molality.

Observed depression (with dissociation): ΔT_f(obs) = i K_f m = 2 K_f m

Depression without dissociation: ΔT_{f(no diss)} = K_f m (since i=1 for no dissociation)

Step 5: Find the ratio

Ratio = ΔT_f(obs) / ΔT_{f(no diss)} = (2 K_f m) / (K_f m) = 2

Final Answer

The ratio is $2$.

Related Topics

• Colligative Properties
• Van't Hoff Factor
• Degree of Dissociation
• Electrolytic Solutions

Key Formulae

Van't Hoff factor: $i=1+\alpha (n-1)$

Freezing point depression: $\Delta T_f=i{K}_{f}m$

For dissociation reactions, always calculate the total ions (n) correctly from the balanced equation.