Question #ea97e
1 Answer
Here's what I got.
Explanation:
The van't Hoff factor,
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,
The solubility product constant,
#K_(sp) = 2.07 * 10^(-33)#
Keep in mind that you do have a dissociation equilibrium going on in aqueous solution
#"Ca"_ 3("PO"_ 4)_ (2(s)) rightleftharpoons 3"Ca"_ ((aq))^(2+) + 2"PO"_ (4(aq))^(3-)#
If you take
#K_(sp) = ["Ca"^(2+)]^3 * ["PO"_4^(3-)]^2#
#K_(sp)= (3s)^3 * (2s)^2 = 108s^5#
This will get you
#s = root(5)(K_(sp)/108) = root(5)((2.07 * 10^(-33))/108) = 1.14 * 10^(-7)"M"#
This means that in order to get a saturated solution of calcium phosphate at
In a saturated solution of calcium phosphate, every mole of calcium phosphate that dissociates produces
- three moles of calcium cations,
#3 xx "Ca"^(2+)# - two moles of phosphate anions,
#2 xx "PO"_4^(3-)#
This means that the van't Hoff factor will be equal to
Therefore, you can say that calcium phosphate has a van't Hoff factor equal to