A 0.2 molal aqueous solution of a weak monobasic acid is 20% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be

To solve this problem, we need to use the formula for the freezing point depression of a solution:

ΔTf = Kf * m * i

where:

• ΔTf is the change in freezing point,
• Kf is the cryoscopic constant of the solvent (water in this case),
• m is the molality of the solution,
• i is the van't Hoff factor, which represents the number of particles the solute splits into in solution.

Given in the problem:

• Kf for water is 1.86°C/m,
• m (molality) is 0.2 mol/kg,
• The weak monobasic acid is 20% ionized, which means i = 1 + α = 1 + 0.2 = 1.2 (since for weak monobasic acid, i = 1 + degree of ionization).

Substituting these values into the formula, we get:

ΔTf = 1.86°C/m * 0.2 mol/kg * 1.2 = 0.4464°C

The freezing point of a solution is lower than that of the pure solvent. The freezing point of pure water is 0°C, so the freezing point of the solution is:

0°C - 0.4464°C = -0.4464°C

So, the freezing point of the solution will be -0.4464°C.