Question

The Ksp (solubility product) of calcium phosphate is 1.2 x 10^-26. Find the solubility in moles per litre of the following.

(a) Calcium phosphate in water.

(b) Calcium phosphate in 0.010 M CaCl₂ solution.

Please show all steps. Thank you.

Answer 1

a)

At equilibrium:

Ca3(PO4)2 <----> 3 Ca2+ + 2 PO43-

   3s 2s

Ksp = [Ca2+]^3[PO43-]^2

1.2*10^-26=(3s)^3*(2s)^2

1.2*10^-26= 108(s)^5

s = 2.565*10^-6 M

Answer: 2.56*10^-6 mol/L

b)

CaCl2 here is Strong electrolyte

It will dissociate completely to give [Ca2+] = 0.01 M

At equilibrium:

Ca3(PO4)2 <----> 3 Ca2+ + 2 PO43-

   0.01 +3s 2s

Ksp = [Ca2+]^3[PO43-]^2

1.2*10^-26=(1*10^-2 + 3 s)^3*(2s)^2

Since Ksp is small, s can be ignored as compared to 1*10^-2

Above expression thus becomes:

1.2*10^-26=(1*10^-2)^3*(2s)^2

1.2*10^-26= 1*10^-6 * 4(s)^2

s = 5.477*10^-11 M

Answer: 5.48*10^-11 mol/L