Question

A 0.15 m aqueous solution of a weak acid has a freezing point of -0.31 °C. What is the percent ionization of this weak acid at this concentration? The molal freezing-point-depression constant of water is 1.86 °C/m.

Answer 1

Answer:

11%

Explanation:

1) Calculate van 't Hoff factor:

Δt = i Kf m

0.31 = i (1.86) (0.15)

i = 1.111

2) Calculate value for [H+]:

CCl3COOH ⇌ H+ + CCl3COO¯

total concentration of all ions in solution equals:

(1.11) (0.15) = 0.1665 m

This is a molality, but we will act as if it a molarity since we will assume the density of the solution is 1.00 g/cm3, which makes the molarity equal to the molality.

0.1665 = (0.15 − x) + x + x

x = 0.0165 M

3) Calculate the percent dissociation:

0.0165/ 0.15 = 11 %