Question

For an aqueous solution of HF, determine the van't Hoff factor assuming 0% ionization. For the same solution, determine the van't Hoff factor assuming 100% ionization. A solution is made by dissolving 0.0300 mol HF in 1.00 kg of water. The solution was found to freeze at -0.0644°C. Calculate the value of i and estimate the percent ionization of HF in this solution.

Answer 1

van't Hoff factor assuming 0% ionization : i = 1

van't Hoff factor assuming 100% ionization : i = 2

solution made by dissolving 0.0300 mol HF in 1.00 kg water : i = 1.154

percent ionization = 15.4 %

Explanation

moles of HF = 0.0300 mol

mass of water = 1.00 kg

molality of HF = (moles of HF) / (mass of water in kg)

molality of HF = (0.0300 mol) / (1.00 kg)

molality of HF = 0.0300 m

freezing point of solution = -0.0644 ^oC

decrease in freezing point = (freezing point of pure water) - (freezing point of solution)

decrease in freezing point = (0.0 ^oC) - (-0.0644 ^oC)

decrease in freezing point = 0.0644 ^oC

van't Hoff factor = (decrease in freezing point) / [(K_f) * (molality of HF)]

van't Hoff factor = (0.0644 ^oC) / [(1.86 ^oC/m) * (0.0300 m)]

van't Hoff factor = 1.154