A solution is made by dissolving 0.500 mol of HF in enough water to make 1.00 L of solution. At 21 C, the osmotic pressure of the solution is 1.35 atm. What is the percent ionization of this acid?

Question

This is what i did.. is it correct?
(1.35 atm) = (2)(M)(0.08206)(294 K)
M = 0.0279
0.0279/0.500 = 0.0559 X 100 = 5.59%? is that correct...? it doesn't seem right please help!

Answer 1

Eric, will you please look at the problem to be sure you have posted it correctly. I have struggled with this because the numbers didn't look right to me and finally figured out why I had doubts. IF (capitalized IF) HF did not ionize at all, then i would be 1. But if we try to calculate i we get
1.35 = i*0.5*0.08206*294
i = about 0.1 and that just can't be. It can't be less than 1 give or take a little to make up for non-ideal solutions and that kind of thing. A 0.05M soln would come closer

Answer 2

I'm sorry! I did post it incorrectly! it was 0.05 M of solution not 0.500!

Answer 3

so would it be.. 1.35 = 1.206i
i = 1.119?

is that when you plug it in with osmotic pressure = i(m)(r)(t)?

i'm sorry for my mistake!

Answer 4

Please be careful. I spent about an hour trying to figure out what was wrong before I suddenly realized that the post had to be wrong.
Yes, solve for i and I get 1.119 also, then 1.119 x 0.05 = about 0.0560 (so it is 0.05 but appears to be 0.0560). A few of the HF molecules break up but only some of them.
.............HF ==>H^+ + F^-
initial.....0.05...0.....0
change.......-x....x.....x
equil......0.05-x...x.....x

We add all of the ions and make it add up to 0.0560
x + x+ 0.05-x = 0.0560
Solve for x = about 0.006
Then %ion = (0.006/0.05)*100 = ?

Answer 5

I am very sorry Dr. Bob =/ thank you so much for the help though I understand now