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Homework Help: K value

  1. Jan 26, 2006 #1
    How would you determine K for the following reaction:
    [tex] HOCl_{aq} + CN^{-} \rightarrow HCN_{aq} + OCN^{-} [/tex]?

    You dont do [tex] \frac{1}{K_{a}} [/tex] or [tex] \frac{1}{K_{b}} [/tex]

  2. jcsd
  3. Jan 26, 2006 #2
    If you mean K as in the equilibrium constant, then I think you would do the following:

    [tex]K_c = \frac{[HCN][OCN^{-}]}{[HOCl][CN^{-}]}[/tex], where [] indicate the concentration in moles per liter.

    If you're talking about another K, then I can't help you off the top of my head. We just started learning about equilibrium on Monday in AP Chemistry, so you may not want to trust me on this.
    Last edited: Jan 26, 2006
  4. Jan 27, 2006 #3


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    Perchloric acid is a strong acid, the reaction will go towards completion.
  5. Jan 28, 2006 #4


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    HOCl is not a strong perchloric acid. HOCl is a pretty weak hypochlorous acid.

    There is something wrong with the reaction - where is Cl on the right side?
  6. Jan 28, 2006 #5
    I think it should be:

    [tex] HOCl_{(aq)} + CN^{-} \rightarrow HCN_{(aq)} + OCl^{-} [/tex]
    Last edited: Jan 28, 2006
  7. Jan 29, 2006 #6


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    That was my first idea too, but HOCN does exist (it is cyanic acid) - so perhaps the question was not acid/base related, but redox related - with cyanide being oxidized to cyanate by hypochlorous acid.

    No idea what is halfpotential for cyanide/cyanate oxidation.
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