# Question #3447f

Jun 11, 2015

This reaction produces hydroxyapatite and hydroxide anions.

#### Explanation:

When calcium hydroxide, $C a {\left(O H\right)}_{2}$, reacts with phosphate ions, $P {O}_{3}^{3 -}$, the reaction produces hydroxyapatite, $C {a}_{5} {\left(P {O}_{4}\right)}_{3} O H$, which precipitates out of the solution, and hydroxide anions, $O {H}^{-}$.

The balanced chemica lequation for this reaction looks like this

$5 C a {\left(O H\right)}_{2 \left(a q\right)} + 3 P {O}_{\textrm{4 \left(a q\right]}}^{3 -} \to C {a}_{5} {\left(P {O}_{4}\right)}_{3} O {H}_{\textrm{\left(s\right]}} \downarrow + 9 O {H}_{\left(a q\right)}^{-}$

The complete ionic equation looks like this

$5 C {a}_{\left(a q\right)}^{2 +} + 10 O {H}_{\left(a q\right)}^{-} + 3 P {O}_{4 \left(a q\right)}^{3 -} \to C {a}_{5} {\left(P {O}_{4}\right)}_{3} O {H}_{\textrm{\left(s\right]}} + 9 O {H}_{\left(a q\right)}^{-}$

If you remove spectator ions, which are ions present on both sides of the equation, you get the net ionic equation

$5 C {a}_{\left(a q\right)}^{2 +} + O {H}_{\left(a q\right)}^{-} + 3 P {O}_{4 \left(a q\right)}^{3 -} \to C {a}_{5} {\left(P {O}_{4}\right)}_{3} O {H}_{\textrm{\left(s\right]}}$

This reaction is actually used to remove phosphate ions from various solutions.

Here's how the structure of hydroxyapatite looks like: 