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Solubility Problem~

  1. Oct 10, 2007 #1
    The question is:

    Calculate the solubility, in g/100mL, of Ca5(PO4)3OH.

    when it asks for solubility, is that the same as asking for the Ksp of Ca5(PO4)3OH??

    so basically, i have to find the concentration of Ca2+, PO4^3- and OH- and then find the Ksp ?

    BUT, i thought Ksp have no units...... how can it be in g/100mL then????

    any one help please?
     
  2. jcsd
  3. Oct 11, 2007 #2

    HallsofIvy

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    What it is asking for is "how many grams of [itex]Ca_5(PO_4)_3OH[/itex] can you dissolve in 100 mL of water at standard room temperature?"
     
  4. Oct 11, 2007 #3

    chemisttree

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    Try writing down the expression for Ksp as a first step.
     
  5. Oct 11, 2007 #4
    Ca5(PO4)3OH <----> 5Ca2+ + 3PO4^3- + OH-

    ksp = [Ca2+]^5 [PO4)3-]^3 [OH-]
    ksp = (5X)^5 (3x)^3 (x)
    6.80E-37 = 84375x^9
    x = 2.72E-5 M = [Ca5(PO4)3OH ]

    (2.72E-5 mol/L)* (502.307 g/mol) = 0.01366275g/L = 0.0136627g/1000mL

    divide the top and bottom by 10 to get per 100mL:
    0.0136627g/1000mL

    1.37 E-3 g/100mL <---- my answer

    am i doing it right? thanks
     
  6. Oct 11, 2007 #5

    chemisttree

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    That should do it.
     
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