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Hydroxylapatite, Ca1o(PO4)6(OH)2, has a solubility constant of Ksp 2.34 x106, and dissociates according to Solid hydroxylapat

Hydroxylapatite, \(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6}(\mathrm{OH})_{2}\), has a solubility constant of \(K_{\mathrm{sp}}=2.34 \times 10^{\mathrm{F}}\), and dissociates according to

$$ \mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6}(\mathrm{OH})_{2}(s) \rightleftharpoons 10 \mathrm{Ca}^{2+}(a q)+6 \mathrm{PO}_{4}^{3-}(a q)+20 \mathrm{H}^{-}(a q) $$

Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of \(\mathrm{Ca}^{2+}\) in this solution if \(\left[\mathrm{OH}^{-}\right]\) is somehow fixed at \(6.50 \times 10^{-6} \mathrm{M}\) ?

Let \(\left[\mathrm{Ca}^{2+}\right]=x .\) For every \(10 \mathrm{~mol}\) of \(\mathrm{Ca}^{2+}\) produced there are \(6 \mathrm{~mol}\) of \(\mathrm{PO}_{4}{ }^{3-}\), so \(\left[\mathrm{PO}_{4}{ }^{3-}\right]=(6 / 10) x=(3 / 5) x\). The solubility product expression is

$$ \boldsymbol{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{10}\left[\mathrm{P} \mathrm{O}_{4}^{3-}\right]^{6}\left[\mathrm{OH}^{-}\right]^{2} $$

Substitute the expressions found above for \(\left[\mathrm{Ca}^{2+}\right]\) and \(\left[\mathrm{PO}_{4}{ }^{3-}\right]\), and the numeric values for \(\left[\mathrm{OH}^{-}\right]\) and \(K_{\mathrm{sp}}\), and solve for \(x\)

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