# Question ea97e

Jan 31, 2017

Here's what I got.

#### Explanation:

The van't Hoff factor, $i$, tells you how many moles of particles of solute you can expect to find in solution for every mole of solute dissolved in solution.

In other words, the van't Hoff factor tells you the extent to which the solute dissociates in solution to produce particles.

Now, calcium phosphate, "Ca"("PO"_4)_3#, is an insoluble ionic compound, which basically means that only small amounts dissociate in aqueous solution to produce calcium cations, ${\text{Ca}}^{2 +}$, and phosphate anions, ${\text{PO}}_{4}^{3 -}$.

The solubility product constant, ${K}_{s p}$, for calcium phosphate is equal to

${K}_{s p} = 2.07 \cdot {10}^{- 33}$

Keep in mind that you do have a dissociation equilibrium going on in aqueous solution

${\text{Ca"_ 3("PO"_ 4)_ (2(s)) rightleftharpoons 3"Ca"_ ((aq))^(2+) + 2"PO}}_{4 \left(a q\right)}^{3 -}$

If you take $s$ to be the molar solubility of calcium phosphate in water at ${25}^{\circ} \text{C}$, you can say that

${K}_{s p} = {\left[{\text{Ca"^(2+)]^3 * ["PO}}_{4}^{3 -}\right]}^{2}$

${K}_{s p} = {\left(3 s\right)}^{3} \cdot {\left(2 s\right)}^{2} = 108 {s}^{5}$

This will get you

$s = \sqrt[5]{{K}_{s p} / 108} = \sqrt[5]{\frac{2.07 \cdot {10}^{- 33}}{108}} = 1.14 \cdot {10}^{- 7} \text{M}$

This means that in order to get a saturated solution of calcium phosphate at ${25}^{\circ} \text{C}$, you must dissolve $1.14 \cdot {10}^{- 7}$ moles of calcium phosphate for every liter of water.

In a saturated solution of calcium phosphate, every mole of calcium phosphate that dissociates produces

• three moles of calcium cations, $3 \times {\text{Ca}}^{2 +}$
• two moles of phosphate anions, $2 \times {\text{PO}}_{4}^{3 -}$

This means that the van't Hoff factor will be equal to $5$, since $1$ mole of dissolved calcium phosphate produces $5$ moles of ions in solution.

Therefore, you can say that calcium phosphate has a van't Hoff factor equal to $5$ in saturated solution.