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Homework Help: Determining Molar Solubility

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the molar solubility of Mg(OH)2 (Ksp1.8*10-11) in8.62*10-2M MgCl2 (aq)?



    2. Relevant equations Chemistry is awwweeesssommmmeee....



    3. The attempt at a solution

    Okay. Here goes. But mind you, I have not put my all into this course, so I will probably need a good slap in the face every now and again when I make a silly assumption or ask a stupid question

    The first thing I need to do is write the reaction, which I am guessing is

    [itex]Mg(OH)_2\rightarrow 2OH+Mg[/itex] but I am not entirely sure why. That is, don't I need to include the water or something? Or do we just assume it does not participate?



    Next I will construct an ice table.. arrg what's that array command again...

    [tex]\left[\begin{array}{ccc}Mg(OH)_2 & 2OH & Mg \\8.62*10^{-2} & 0 & 0\\ -s & +2s & +s\\ 8.62*10^{-2}-s & 2s & s\end{array}\right][/tex]

    This may be wrong (probably). Maybe I did need to include the water. I assumed that the concentration of the Mg(OH)2 was equal to that of the MgCl2

    I don't think that is correct. Can I get some guidance from here?

    Thanks,
    Casey
     
  2. jcsd
  3. Dec 11, 2008 #2

    Borek

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    Staff: Mentor

    You starting concentration of Mg(OH)2 is 0.

    However, you already have dissolved magnesium in the solution, so initial concentration of Mg2+ is not 0.

    Using ICE table is not a bad idea, but it will be tricky. Your first column - the one with dissolved solid - will remain unchanged. You will have only two columns - one for magnesium and one for OH-. That's the easy part. It will yield third degree polynomial, that's the hard part.

    However, you can start assuming that you know concentration of Mg2+ - after all it should not change substantially. Use it to calculate concentration of OH- in saturated solution, that will give you information about amount of dissolved Mg(OH)2. Finally check, if your initial assumption (that concentration of Mg2+ won't change substantially) was valid.
     
  4. Dec 11, 2008 #3
    So then my reaction is not written correctly then? That is what is most difficult for me, is setting up the reaction.

    If the [Mg(OH)2]0=0 then the rxn [itex]Mg(OH)_2\rightarrow 2OH+Mg[/itex] can have no meaning.

    So what is the correct rxn ?
     
  5. Dec 11, 2008 #4

    Borek

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    Staff: Mentor

    Mg(OH)2(s) <-> Mg2+(aq) + 2 OH-(aq)

    You start with a solid hydroxide, so its concentration is 0.

    Don't ignore charges when dealing with ions.
     
  6. Dec 11, 2008 #5
    I do not see how this reaction makes sense (but that's why I suck at chemistry).

    Why isn't the MgCl2 included in the rxn ?
     
  7. Dec 11, 2008 #6

    Borek

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    Staff: Mentor

    Because it doesn't react, it is just dissolved.

    However, solubility product tells you when Mg(OH)2 will start to precipitate - when [Mg2+][OH-]2 will be higher than the given value. So [Mg2+][OH-]2 product can't be never higher than Kso. When there was already Mg2+ there is the same limit - just now there exist some initial concentration of Mg2+, so when Mg(OH)2 dissolves you should take this original concentration into account.
     
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