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Concentration at Equilibrium

  1. Nov 29, 2016 #1
    Equal Volumes of 3.60 M A2+ and 6.80 M B- are mixed. After the reaction, equilibrium is established with [B-] = 0.40 M.
    What is the [A2+] at equilibrium?

    2. Relevant equations
    Ice table, algebra

    3. The attempt at a solution
    This is probably a really easy problem, but I seem to be getting the wrong answer. Probably just a small lapse, but anyways, here's my work.

    Equilibrium constant not given.

    Ice Table:
    A2+ (aq) + 3B- ==><== AB3-
    I 3.60 6.80 0
    C +x -3x +x
    E 3.60 - x 0.40 +x

    By subtracting 0.40 from 6.80, I determined 3x to be 6.4.
    Using the ratios, I divided 3x by 3 to get 2.13(3 repeating).
    Next I subtracted x from 3.60, and got my answer or 1.47.

    The answer key says I am wrong, and that [A2+] at equilibrium is 0.80 M.

    So am I going about this wrong, or is the book wrong (it has been in the past)?
  2. jcsd
  3. Nov 30, 2016 #2
    Shouldn't it be $$A^{2+} + 2B^{-} \rightleftharpoons AB_2$$

    \text{ }& A^{2+} & 2B{-}&AB_2\\
    \text{Initial concentration}& 3.60 & 6.80 & 0\\
    \text{Change}& -x&-2x&x\\
    \text{Equilibrium concentration }& 3.6 -x & 6.8 -2x=(0.40) & x\\

    Anyhow answer in books seems incorrect.
    Last edited: Nov 30, 2016
  4. Nov 30, 2016 #3


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    Staff: Mentor

    As long as the reaction is not given in the problem statement it can be everything.
  5. Nov 30, 2016 #4


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    Science Advisor

    It says "equal volumes of.......are mixed". So immediately after mixing the initial concentrations are 1.8M and 3.4M. But as Borek says, if they do not specify the reaction there is no unique answer.
  6. Nov 30, 2016 #5
    The reaction was specified in this case, and it is the one I posted above. This text has a lot of errors, so I'm not too surprised it's wrong this time. Good to double check though.
  7. Dec 1, 2016 #6


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    Staff: Mentor

    Then the answer given is the correct one. See mjc123 post.

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