What is the solubility-product expression for the equilibrium established when calcium phosphate, ca3(po4)2(s), is added to pure water at 25°c?

Ks = [Ca2+]^3 . [PO4 3+]^2

Ks (solubility product) is the product of calcium ion concentration multiplied by phosphate ion concentration

Explanation: when a salt is added to water it dissolves and the molecules dissociates into ions. It can be represented in the equation:

Ca3(PO4)2(s) <--> 3Ca2+ (aq) + 2PO4 3+ (aq)

Where s and aq indicate the state in which the molecule is. S for solid and aq for aqueous.

This solution reach and equilibrium, that means that salt is dissociated at the same rate that ions reunite to form a molecule.

In chemical equilibrium there is a constant that reflects this rate and is generally expressed as: K = [D]^d .[C]^c/ [A]^a .[B]^b

D, C represent products and A, B reagents. Lower case letters are the stoichiometric coefficients.

Another thing to remind is that only gases and aqueous molecules are taking into account in the constant, solids and liquids do not affect it and values are replace in the equation by 1.

Now, Ca3(PO4)2 is in solid state so it doesn't affect Ks, in that case only concentrations of Ca and PO4 are useful. Since these ions are products, the expression of Ks is

Ks= [Ca2+]^3 . [PO4 3+]^2

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"