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Complexation Equilibria problem

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data
    How much AgBr could dissolve in 1.0 L of 0.40 M NH3? Assume that Ag(NH3)2+ is the only complex formed. [Kf=1*108 ; Ksp=5*10-13]

    2. Relevant equations



    3. The attempt at a solution
    Let Ag+ and Br- formed from AgBr be x. Since Ag+ form a complex with NH3, let the amount of Ag+ remaining be x-y, hence amount of NH3 remaining is 0.4-2y and amount of complex formed is y.

    Therefore
    [tex]K_{sp}=(x-y)x[/tex]
    [tex]K_f=\frac{y}{(x-y)(0.4-2y)^2}[/tex]
    I hope my equations are correct.

    Now the trouble is solving the equations. I can easily plug them in a calculator and get the answer but I want to know if there is a way to do this problem without using one.

    Since Kf is so large, I think that ammonia will be almost completely used and hence, y=0.2 M but I don't see if this helps or if my assumption is correct.

    Any help is appreciated. Thanks!
     
    Last edited: Jul 28, 2013
  2. jcsd
  3. Jul 28, 2013 #2
    I think you can solve the problem this way:

    reaction 1:
    [tex]AgBr(s) => Ag^+ + Br^-,\,\,\,K_{sp}[/tex]
    reaction 2:
    [tex]Ag^+ + 2NH_3=> Ag(NH_3)_2^+,\,\,\,K_f[/tex]
    ===
    the total reaction:

    [tex]AgBr(s) + 2NH_3=> Ag(NH_3)_2^+ + Br^-[/tex]


    [tex]K=K_f*K_{sp}=5*10^{-5}=\frac{[Br^-]*[Ag(NH_3)_2^+]}{[NH_3]^2}[/tex]

    [tex]K=5*10^{-5}=\frac{x*x}{(0,4-2x)^2}[/tex]

    that's another equation, I believe...

    where
    x >> sqrt(Ksp(AgBr))
     
  4. Jul 28, 2013 #3
    I was expecting someone to post this. I have already seen this way to solve the problem and it does give the right answer. But what's wrong with my approach?
     
  5. Jul 28, 2013 #4
    from the first equation you have assumed [Ag^+] not equal [Br^-]
    that's wrong...you said above that equation: "Let Ag+ and Br- formed from AgBr be x"
    i.e:

    [Ag^+] equal [Br^-] = x
     
  6. Jul 28, 2013 #5
    Yes, I did say that but Ag+ gets consumed when it forms complex with ammonia. I assumed that the amount of Ag+ getting consumed be y.
     
  7. Jul 29, 2013 #6
    Anyone?
     
  8. Jul 30, 2013 #7

    epenguin

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    Gold Member

    These things sound simple then when you start they seem to become quite confusing, hard to get in focus. It is good IMHO to do a complete treatment as Pranav tries, but also good to look at all approximations and then at end compare.

    (Premiss - I don't know what the official definition of these constants are, and whether molarity is built into them so they are dimensionless. I am always happier to see units, like M-2 for this Kf but maybe that's not today's textbook.)

    For approximations, I would say that the complex is pretty tight, and AgBr is pretty insoluble, and I would expect therefore the ammonia to near 100% complex the Ag+, so it would be an 0.2M solution of complexed Ag, 0.2M Br- to first approximation. To a second approximation at first sight the amount of free Ag+ would be a very insignificant amount more. I mean it would be much less even than in solution of saturated AgBr. Because then you have only a tiny amount of Br- in solution. But here you have around 0.2M Br- held in solution as it were by complexing of the Ag+ which is still charged. So the solubility product will give an utterly minuscule amount of Ag+ in solution, much less even than AgBr on its own. Easily calculated from the Ksp equation putting [Br-] = 0.2 (M?) giving free [Ag+] = 2.5X10-12M (as compared with a whopping 7X10-7M for ordinary saturated AgBr solution) if I am not mistaken. So I'm thinking incomplete complexation might be the more important effect at second approximation.

    In connection with what I just said, if I understand, janhaa would be wrong to say [Ag^+] equal [Br^-] if by [Ag^+] he means the free Ag+ concentration. The Ag+(NH3)2 is still charged and this nearly = [Br-]. However I agree it is a good idea to multiply Ksp by Kf and get rid of at least one nuisance factor. I mean I think you get

    Ksp.Kf = (x - y)x/[0.4 - 2y]2 which is no worse than a quadratic.

    The OP said he has has done the complete calclations so it would be interesting to know the results and the comparisons with the approximations I sketched.
     
    Last edited: Jul 30, 2013
  9. Jul 31, 2013 #8
    I too agree that [Br-] should be 0.2 M as Ag+ gets used up again and again in forming the complex but janhaa's method gives a very small value for [Br-] So do we have to use Kf here?

    Why do you get (x-y) in the numerator? If we consider the net reaction, K comes out to be very small, so it is safe to assume that y is much smaller than 0.4 . If 2y amount of NH3 is used up, we get y amount of the complex and the Br- ion. Hence
    [tex]K_{sp}.K_f=\frac{y\cdot y}{(0.4-2y)^2} =\frac{y^2}{0.16}[/tex]
    Solving this, y=2.8*10^(-3). This is the amount of AgBr which dissolves in the ammonia solution.
     
  10. Aug 4, 2013 #9

    epenguin

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    From your first comment that they could be easily plugged in etc. I thought you had done the exact calculations. If so please give us the results, so we see how they compare with mine below.

    I thought "ammonia will be almost completely used" too, and it is a mistake, in fact rather little is used, from which there are mistakes in my comments in earlier posts, in particular the eventual free [Ag+] is very low but not nearly as low as I estimated. Almost all the [Ag+] is complexed, true, but there can be very little free [Ag+] in solution in equilibrium with it because it precipitates - two factors acting opposite ways.

    If you don't mind I found your notation confusing so will use explicit chemical notation.

    I start from your equations

    Ksp = [Ag+][Br-] = 5.10-13 M (1)

    Kf = [Ag+(NH2)2]/[Ag+][NH3]2 = 108 (2)

    The total AgBr is

    [Ag+] + [Ag+(NH3)2] = [Br-] (3)

    But because of the tightness of the complex we can approximate

    [Ag+(NH3)2] ≈ [Br-] (4)

    As before we have

    KspKf = [Ag+(NH3)2][Br-]/[NH3]2

    Using (4) plus other things above, we have to very good approximation

    5.10-5 = [Br-]2/(0.4 - 2[Br-])2

    so we have only a square root to take instead of solving a quadratic equation. Then

    0.00707 = [Br-]/(0.4 - 2[Br-])

    Only now can I see that little of the ammonia (about 1%) is complexed. I don't know if anyone can see a way I could have seen that before.

    I finally find the answer [AgBr] = [Br-] = 0.00279

    From which I can also calculate for the free silver ions [Ag+] = 1.79.10-10 , much more than I said before, but more importantly still more than 100 tines less than in plain saturated AgBr solution for the reason I said.

    Now I see you seem to have followed this same idea and have the same result. Any corrections by not making the approximation must be vanishingly small.
     
    Last edited: Aug 4, 2013
  11. Aug 6, 2013 #10
    I am not sure if I understand your reply completely but it looks good to me, I will spend some more time on it.

    I plugged the equations I posted in Wolfram|Alpha, I have attached the image of the solutions. Sorry, I cannot post the link because Wolfram asks you to continue with additional time and you have to go through the trouble of signing-in.

    The solutions by wolfram alpha match your answer (x is the [Br-]). I guess I understand most of this now. I liked your explanation about complex being formed in a very less amount.

    Thanks a lot for your effort and time epenguin. It was a great learning experience for me. :smile:
     

    Attached Files:

  12. Aug 7, 2013 #11

    DrDu

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    I didn't check the correctness of all the equations you wrote down. Just one point.
    You want to know whether it is permissible to neglect 2y in comparison with 0.4.
    To find the answer just do the calculation with 0.4 in place of (0.4-2y), which presumably can be done by hand, and then check whether 2y is really much smaller than 0.4.
    That's called check for self-consistency.
     
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