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Homework Help: Dissociation Constant Homework Check

  1. Jul 3, 2006 #1
    Question:
    A solution of hydrofluoric acid contains 2.0g of HF per litre and has a pH of 2.2. What is the dissociation constant for HF?

    My Answer:
    HF + H2O <-> H3O+ + F-

    HF = 2.0g
    H30+ = 10^-2.2 = .0063g
    F- = H3O = .0063

    Ka = [H3O+][F-] / HF
    = (.0063)(.0063) / 2.0
    =1.98 x 10^-5

    Therefore the dissociation constant for HF is 2.0 x 10^-5

    Can anyone tell me if I am on the right track?
    Thanks
     
  2. jcsd
  3. Jul 3, 2006 #2
    To find [tex]K_a[/tex] you need the concentration of [tex]HF[/tex] expressed in [tex]mol.dm^{-3}[/tex]. Convert [tex]2.0 g.dm^{-3}[/tex] to [tex]mol.dm^{-3}[/tex] using molar mass.


    The concentration of [tex]H_3O^+[/tex] that you calculated using that expression is molar concentration, so the units are [tex]mol.dm^{-3}[/tex].
     
  4. Jul 3, 2006 #3

    My Text mentions nothing about [tex]mol.dm^{-3}[/tex]
    it does say to calculate [tex][H_3O^{+}] = 10^{-x}[/tex]
    So thats how I got .0063 mol/L

    I also have, from the text the formula for calculating the acid dissociation constant [tex]K_{a}[/tex]

    Ka = [H30+][X-] / [HX]
     
  5. Jul 4, 2006 #4
    That's right you got [tex]0.0063 mol/L[/tex]! But you indicated on your previous post:

    I am just correcting the units. It is not grams but mol/dm3 or mol/L, since 1 L = 1 dm3.
    You, maybe, ask why the units are those. Because you find the pH of a solution from the molar concentration of [tex]H_3O^+[/tex]. The expression is:
    [tex]pH = - \log\left[ H_3O^{+}_{(aq)} ][/tex]

    Then, using that expression, knowing the value of pH, you can just find the molar concentration of [tex]H_3O^+[/tex].
    --------
    The formula is correct.
     
    Last edited by a moderator: Jul 4, 2006
  6. Jul 4, 2006 #5
    Hmm... yes I seem to mess up on the units sometimes.

    I would like to do this all in mol/L if possible. (we have been taught to this point only in mol/L)

    So does this look better?
    HF + H2O <-> H3O+ + F-

    HF = 1+19=20 g/mol
    2g HF x 1 mol HF / 20g HF = .1 mol HF
    So HF = .1 mol/L

    H30+ = 10^-2.2 = .0063 mol/L
    F- = H3O = .0063 mol/L

    Ka = [H3O+][F-] / HF
    = (.0063)(.0063) / .1
    =4.0 x 10^-4

    Therefore the dissociation constant for HF is 4.0 x 10^-4

    Can you please give me your feedback on this?

    Thank you very much for all your help. This forum and people who help have been a valuable resource.
     
  7. Jul 5, 2006 #6

    Hootenanny

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    Looks good to me :smile:
     
  8. Jul 5, 2006 #7
    Thank you PPonte and Hootenanny for taking the time to review my work and help me out.

    Thanks Again.
     
  9. Jul 5, 2006 #8
    My pleasure! :approve:
     
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