Question

5.0 g of calcium phosphate is mixed into 1 L of solution

a) what is the concentration of Ca^2+ and PO4^3-? calculate

b) explain how you could dissolve will calcium phosphate

c) what is the molarity of the solutions, if you were able to dissolve all of it?

Answer 1

Mass of Ca3(PO4)2 = 5 g

Moles of Ca3(PO4)2 =

= 0.016 mol

c)

Suppose if Ca3(PO4)2 is completely soluble

Molarity of Ca3(PO4)2 = 0.016 mol/1 L = 0.016 M

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a)

Finding concentration of ions

First checking solubility of calcium phosphate in solution

Ca3(PO4)2 ---> 3Ca²⁺ + 2PO₄^3- (if completely soluble)

Concentration of Ca²⁺ =

= 0.048 M [Ca²⁺]

Concentration of PO₄^3- =

= 0.032 M [PO₄^3-]

and ionic product = [Ca²⁺]³[PO₄^3-]²

= (0.048)³(0.032)²

Ionic product =1.13 x 10^-7

Solubility product =  2.0 x 10^-33

Ionic product > solubility product

Calcium phosphate is a sparingle soluble in water

Ca₃(PO₄)₂ <===> 3Ca²⁺ + 2PO₄^3-

at Equilibrium 3s and 2s are the concentrations of ions

Ksp = [Ca²⁺]³[PO₄^3-]²

Ksp = (3s)³ * (2s)²

Ksp = 108 ( s⁵)

But Ksp of Ca₃(PO₄)₂ = 2.0 x 10^-33

2.0 x 10^-33 = 108 (s⁵)

s⁵ = 0.0185 x 10^-33

s = 1.13 x 10^-7

Therfore, molarity of Ca²⁺ = 3s = 3 x 1.13 x 10^-7 = 3.39 x 10^-7 M

Molarity of PO₄^3- ions = 2s = 2 x 1.13 x 10^-7 = 2.26 x 10^-7 M

b)

We can add nitric acid or HCl to shift the equilibrium to product side

The chloride ions or nitrate ion react with calcium ions so the equilibrium shift to the product side and solubility increases

It is just like removing product and shifting the reaction to product side.