Question

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 $x$

Answer 1