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K values with acid and base reactions

  1. Mar 2, 2014 #1
    Given an aqueous solution (I'm omitting the subscripts):

    ## NH_3 + H_2O ←→ NH_4^+ + OH^- K_b ##


    Then we can also say:

    ## NH_3 + H^+ ←→ NH_4^+ K_b ## as well since they're equivalent reactions, right? They both have the same Kb value as a result, right?

    So then, ## K_b = \frac {[NH_4^+][OH^-]}{[NH_3]} = \frac{[NH_4^+]}{[NH_3][H^+]} ##

    I would just like to clarify, is this true? It looks like ##[OH^-] = \frac{1}{[H^+]}##

    Is this true? Does it take into account the constant "concentration" of the water in this case?
     
  2. jcsd
  3. Mar 2, 2014 #2
    [tex][h^+][oh^-]=10^{-14}[/tex]
     
  4. Mar 2, 2014 #3
    Ok. So are the two K expressions not equivalent? I just don't understand why since they're representing the same reaction, are they not?
     
  5. Mar 3, 2014 #4
    There are actually two interacting reactions involved. One is NH3 + H2O, and the other one is OH+H going to H2O. If you're only interested in finding the OH concentration, and the original solution is pH=7, then you can solve for the new OH concentration using the first equilibrium constant. However, if you also want to find the new H+ concentration, you need to use the second equilibrium constant (OH+H). Your second reaction expression is not in a form that you can use to solve directly and uniquely for the product concentrations (and, of course, the K has to be modified).

    Chet
     
  6. Mar 3, 2014 #5
    Oh okay! Thank you! So would I just divide Kb by Kw to get the second reaction I specified?
     
  7. Mar 3, 2014 #6

    Borek

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    Staff: Mentor

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