Solubility product?

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Starlight

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Solubility product?

« on: May 03, 2012, 11:05:34 am »

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Anyone know how to do this question?

Calcium phosphate (Ca3(Po4)2) has a solubility product Ksp= 2.07 x 10^-33 at 25 degrees celsius. What is the concentrtion of Po43- ions in a saturated solution of Ca3(Po4)2?

(Answer is 1.1 x 10^-7 M)

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Mao

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Re: Solubility product?

« Reply #1 on: May 04, 2012, 03:08:05 pm »

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$Ca_3(PO_4)_2 (s) \leftrightharpoons 3Ca^{2+}(aq) + 2PO_4^{3-} (aq)$

$K_{sp}=[PO_4^{3-}]^2[Ca^{2+}]^3 = 2.07\times 10^{-33}$

Assume we start from pure water. At equilibrium, $[Ca^{2+}] = \frac{3}{2} [PO_4^{3-}]$ from mole ratio.

Thus, $K_{sp}=\left( \frac{3}{2}\right)^3 [PO_4^{3-}]^5 = 2.07\times 10^{-33}$

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Starlight

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Re: Solubility product?

« Reply #2 on: May 04, 2012, 05:11:29 pm »

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Thanks!

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