Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If $K_f$ for water is $ 1.86^{\circ} C/m$, the freezing point of the solution will be -

• A. $-0.24^{\circ} C$
• B. $-0.18^{\circ} C$
• C. $-0.54^{\circ} C$
• D. $-0.36^{\circ} C$

Answer 1

The Correct Option is A

To find the freezing point of the 0.1 molal aqueous solution of the weak acid which is 30% ionized, we need to use the concept of freezing point depression.

The formula for freezing point depression is:

\Delta T_f = i \cdot K_f \cdot m

Where:

• \Delta T_f is the change in freezing point
• i is the van't Hoff factor
• K_f is the freezing point depression constant
• m is the molality of the solution

For a weak acid HA that is 30% ionized, the ionization can be represented as:

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\

Initially, concentration of HA is 0.1 molal. Since it is 30% ionized:

• Concentration of ions (H⁺ and A^-) will be 0.03 molal each.
• Remaining concentration of HA will be 0.07 molal.

The van’t Hoff factor, i, is the measure of the number of particles after dissociation:

i = (1 - \text{ionization fraction}) + (\text{ionization fraction} \times \text{Number of particles after dissociation})

i = 0.7 + (0.3 \times 2) = 0.7 + 0.6 = 1.3

Now substitute the values into the freezing point depression formula:

\Delta T_f = 1.3 \times 1.86 \times 0.1 = 0.2418

The freezing point of pure water is 0°C. Thus, the freezing point of the solution will be:

0 - \Delta T_f = 0 - 0.2418 \approx -0.24^\circ C

Therefore, the freezing point of the solution is approximately -0.24^\circ C.