Question

(a) If the molar solubility of PbF2 at 25 °C is 0.00202 mol/L, what is the Kgn at this temperature? Ksp (b) It is found that 1.66e-06 g of Ca2(PO4)2 dissolves per 100 mL of aqueous solution at 25 °C. Calculate the solubility-product constant for Ca3(PO)2 Ksp (c) The Ksn of Cu3(PO4)2 at 25 °C is 1.40e-37. What is the molar solubility of Cu3(PO4)2? solubility mol/L

Answer 1

a)

At equilibrium:
PbF2      <---->     Pb2+     +         2 F-    

                     s                  2s      

Ksp = [Pb2+][F-]^2
Ksp = (s)*(2s)^2
Ksp = 4(s)^3
Ksp = 4(2.02*10^-3)^3
Ksp = 3.297*10^-8
Answer: 3.30*10^-8

b)
S = 1.66*10^-6 g / 100 mL
= 1.66*10^-6 g / 0.1 L
= 1.66*10^-5 g / L

Molar mass of Ca3(PO4)2,
MM = 3*MM(Ca) + 2*MM(P) + 8*MM(O)
= 3*40.08 + 2*30.97 + 8*16.0
= 310.18 g/mol


Molar mass of Ca3(PO4)2= 310.18 g/mol
s = 1.66*10^-5 g/L
To covert it to mol/L, divide it by molar mass
s = 1.66*10^-5 g/L / 310.18 g/mol
s = 5.352*10^-8 mol/L

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

                     3s                 2s      

Ksp = [Ca2+]^3[PO43-]^2
Ksp = (3s)^3*(2s)^2
Ksp = 108(s)^5
Ksp = 108(5.352*10^-8)^5
Ksp = 4.741*10^-35
Answer: 4.74*10^-35

C)

At equilibrium:
Cu3(PO4)2 <---->    3 Cu+     +         2 PO43-

                     3s                 2s      

Ksp = [Cu+]^3[PO43-]^2
1.4*10^-37=(3s)^3*(2s)^2
1.4*10^-37= 108(s)^5
s = 1.669*10^-8 M
Answer: 1.67*10^-8 M