Question

A solution is made by dissolving 0.0100 mol of HF in enough water to make 1.00 L of solution. At 23 °C, the osmotic pressure of the solution is 0.308 atm. What is the percent ionization of this acid?
Please help.

Answer 1

I have seen another problem and I doubted it completely. This one I don't believe but at least it's possible. I will assume the author is just making up problems.
pi = iMRT
Plug in all the numbers ans solve for i = van't Hoff factor.
Then i*0.01 = concn the 0.01 appears. I think it's about 0.013 but you need to clean up the numbers.
The ionization of HF is
.............HF == H^+ + F^-
initial....0.01....x.....x
change......-x.....+x....+x
equil.....0.01-x....x......x

So what do we have in solution?
We have x + x + 0.01-x = ? and it APPEARS to be 0.013
Solve for x which will give you the ion concn, then %ionization = 100*(x/(0.01) = ?