Question

A solution is made by dissolving 0.0150 mol of HF in enough water to make 1.00 L of solution. At 26 °C, the osmotic pressure of the solution is 0.449 atm. What is the percent ionization of this acid?

Answer 1

Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm

Required:
percent ionization

Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367

Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x

Adding all the concentration and equating to iM
x +x + 0.0150 - x = 1.219367 (0.0150)
x = 3.2905 x 10^-3

percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%

Also,
percent dissociation = (i -1) (100) = (1.219367 * 1) (100) = 21.94%