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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.

  • chemistryyyy - ,

    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^-

    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) = ?

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