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Finding Equilibrium concentrations with added solutes?

  1. Nov 19, 2014 #1
    HF(aq) ↔ H+ + F-

    Determine equilibirum concentrations with Initital concentrations and Kc.

    Kc = 7.2 x 10-4
    [HF] = 1.00M
    [NaF] = 0.50M

    I understand how to do an ICE table and put together questions, but what do you do with [NaF] when it is not part of the chemical equation? How to I factor it into it? Thanks!
     
  2. jcsd
  3. Nov 19, 2014 #2
    What you have to remember is that ionic salts dissociate into their ions when in solution.

    Consider the dissociation reaction of the NaF:

    NaF → F- + Na+

    Hopefully that should help you understand what to do.
     
  4. Nov 19, 2014 #3
    So the if it all dissociates then the initial concentration for F- should be 0.5M. I use that molarity but it still doesn't seem to work out. I can't seem to get anything to equal the equilibrium constant.
     
  5. Nov 19, 2014 #4
    It is possible that I am misunderstanding. You were asked to find the equilibrium concentrations correct?

    so you should have an ice table similar to this:
    HF(aq) ↔ H+ + F-
    i 1M 0M 0.5M
    c -x +x +x
    e 1-x x 0.5+x

    from here you should be able to create your equilibrium expression of Kc = [products]e/[reactants]e

    then if substitute in your expressions from ice table and the equilibrium constant you should be able to calculate x and then determine all of your equilibrium concentrations.

    You should also note that it is probably acceptable to assume that 1-x ≈ 1 and that 0.5+x ≈ 0.5. this makes your calculations much simpler.

    You were already given the equilibrium constant. you are solving for the equilibrium concentrations are you not?
     
  6. Nov 19, 2014 #5
    I can setup my ice table like that fine, but where does [NaF] come into play, do I just add 0.50M to F- when I have equilibrium concentrations??
     
  7. Nov 19, 2014 #6
    thank you for the help btw :) !
     
  8. Nov 19, 2014 #7

    Borek

    User Avatar

    Staff: Mentor

    Have you checked the ICE table Ryan Schmidt gave you? 0.5 is already in the I (initial) line, taken care of.

    @Ryan Schmidt - please don't go too far with help, by forum rules we guide people to the answer, not give it to them.
     
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