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Moles of ions

  1. Apr 19, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    ##300.0 mL## of ##1.50 \times 10^{-10} M## ##Na_2S_{(aq)}## is combined with ##200.0 mL## of ##1.50 \times 10^{-10} M## ##{Co(NO_3)_2}_{(aq)}##.

    Determine what precipitates.
    Determine how many moles precipitate.


    2. Relevant equations



    3. The attempt at a solution

    The first question was easy. I simply found the moles of each substance before mixing, which are equal to the moles of the ions ##Co^{+2}, S^{-2}##. Then I divided the moles of ions by the new total volume of ##500.0 \times 10^{-3} L## to get the new ion concentrations after mixing.

    I found ##K_{sp} = 4.0 \times 10^{-21}## in my reference material and I calculated the ion product to be ##Q = 5.40 \times 10^{-21}##. Since ##Q > K_{sp}##, the solution is supersaturated and so ##CoS_{(s)}## precipitate will form until ##Q = K_{sp}##.

    When the question asks how many moles precipitate, is it simply asking me to find the limiting reagant that produces fewer moles of ##CoS##?
     
  2. jcsd
  3. Apr 19, 2014 #2

    Borek

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    No, you need to check how much can precipitate before Q=Ksp.
     
  4. Apr 19, 2014 #3

    Zondrina

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    EDIT: Thinking...

    ##Q > K_{sp}## is telling me the ion concentration is too high and the reaction will shift to produce more ##CoS##... So that means:

    ##CoS ⇌ Co^{+2}+S^{-2}##

    Will shift to the left.

    Okay I think I got the answer now. I'm getting ##n_{CoS} = 3.00 \times 10^{-11} mol##. Does that sound reasonable?
     
    Last edited: Apr 19, 2014
  5. Apr 20, 2014 #4

    Borek

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    That's not what I got.

    After precipitating 3e-11 moles of CoS there would be no Co left in the solution, and solution would be not saturated.

    Assume x moles precipitated. What is concentration of Co2+ now? What is concentration of S2- now? What should their product be equal to?
     
  6. Apr 20, 2014 #5

    Zondrina

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    Bah! My original idea about using the ICE strategy with concentrations was the way to go then.

    So at the moment the two solutions are mixed, ##Q > K_{sp}## and some of the ions are going to be used up to form a precipitate.

    Considering the equilibrium: ##Co^{+2}+S^{-2} ⇌ CoS##

    Initial: ##9.00 \times 10^{-11}, 6.00 \times 10^{-11}, 0##
    Change: ##-x, -x, +x##

    So at equilibrium:

    ##K_{sp} = \frac{1}{[Co^{+2}][S^{-2}]}##
    ##4.0 \times 10^{-21} = \frac{1}{(9.00 \times 10^{-11} - x)(6.00 \times 10^{-11} - x)}##

    This does not make sense though, if I use the quadratic formula from here I get an absurd value. What am I ignoring?

    EDIT: If the reactant terms could be moved to the numerator... i get a reasonable answer, but it's always supposed to be [products]/[reactants].

    If I solve it this way (which obviously makes more sense logically):

    ##4.0 \times 10^{-21} = (9.00 \times 10^{-11} - x)(6.00 \times 10^{-11} - x)##

    I get ##x = 1.0 \times 10^{-11}## and ##x = 1.4 \times 10^{-10}##

    Though I'm not certain which value I would pick.

    Any advice on this one?
     
    Last edited: Apr 20, 2014
  7. Apr 20, 2014 #6
    Can you show how did you got those values for initial moles? I don't think that's right.
    It looks as if you plugged in the moles instead of the concentrations of ions.
     
  8. Apr 20, 2014 #7

    Borek

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    That's the correct approach.

    As you were already told, these are wrong.

    This is wrong too. What is the definition of Ksp?
     
  9. Apr 20, 2014 #8

    Zondrina

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    Here is the working of the first question:

    I know now that I want the ion product concentration minus 'x' from each ion. That's like asking 'how much precipitates before ##K_{sp}## is reached'.

    EDIT: I got the answer now thanks guys.
     
    Last edited: Apr 20, 2014
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