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Ksp, Solubility Question

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    The Ksp of Al(OH)3 is 1.0 x 10-33. What is the solubility of Al(OH)3 in 0.0010 M Al(NO3)3? Give your answer using scientific notation and to 2 significant figures (i.e., one decimal place).

    2. Relevant equations

    3. The attempt at a solution

    I can solve easier questions than this but I'm confused on how the .0010 M Al(NO3)3 will dissociate into and how it will effect the Ksp.

    From what I gather AL(NO3)3 will dissociate into 4 ions, AL, and 3(NO3). Since it's at .001 M the 3(NO3) has a value of .003M.

    1.8E-33 = (x)(.003) = 6E-31. Which I know isn't even close. The correct answer is 3.3e-11 but I can only guess on how to get that answer.

    Any help would be appreciated.
  2. jcsd
  3. Nov 12, 2012 #2


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    Staff: Mentor

    Not bad, with one important mistake.

    Ignoring everything else, how dos the formula for Ksp of Al(OH)3 look like?
  4. Nov 12, 2012 #3
    Ksp = ( Al^3+) ( 3OH)^3?
  5. Nov 12, 2012 #4


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    Staff: Mentor

    Why do you throw this 3 in every possible place? Apparently you are not sure where it should be, aren't you?

    [tex]K_{sp} = [Al^{3+}][OH^-]^3[/tex]

    Now, if concentration of dissolved Al(OH) is x, and there were already 0.0010 M of Al3+ present, what are concentrations of both Al3+ and OH-?

    Note: it may turn out we will need an additional step, but it will become obvious later.
  6. Nov 12, 2012 #5
    Al would be .001 and OH would be .001^3 ?
  7. Nov 12, 2012 #6


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    Staff: Mentor

    No. When the Al(OH)3 is dissolved it adds to the ions already present in the solution.

    In some cases the difference in the concentration will be negligible, but let's start by getting these things right, we will think about approximations and negligibility later.
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