Chemistry - Posted for Rudy

Posted by Rudy on Sunday, January 26, 2014 at 7:24pm.


Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10-59, and dissociates according to

Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of Ca2 in this solution if [OH–] is somehow fixed at 4.30 × 10-6

I understand to set this question up as such:
2.34e-59 = (10x)^10(6x)^6(4.3e-6+2x)

But I don't understand the next step, can anyone help? you don't have to give the answer just the next few steps

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Ms. Sue

  1. wouldn't the Ca ion cncentration be 5 times the OH conc?

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    posted by bobpursley

  2. First, I think you left a squared term off the OH^-.
    Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

    (Ca^2+) = 10x
    (PO4^3-) = 6x
    (OH^-) = 2x

    Ksp = (Ca^2+)^10(PO4^3-)^6(OH^-)^2

    Substituting the x values we have
    Ksp = (10x)^10(6x)^6(4.3e-6 + 2x)^2
    I would make the assumption that
    4.3e-6 + 2x = 4.3e-6 so we have
    Ksp = (10x)^10(6x)^6(4.3e-6)^2
    2.34e-59 = 10^10*x^10 (6x)^6*(1.85e-11)
    2.34e-59 = 10^10*x^10*4.66e4*x^6*(1.85e-11)
    2.34e-59/[(10)^10*(4.66e4)*(1.85e-11) = x^16.

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    posted by DrBob222

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