Question

The solubility of Ag3PO4 at 25°C is 0.0178 g/L.

a. Calculate the molar solubility of Ag3PO4. (ans in ml/L)

b. Calculate Ksp.

Answer 1

a)

Molar mass of Ag3PO4,

MM = 3*MM(Ag) + 1*MM(P) + 4*MM(O)

= 3*107.9 + 1*30.97 + 4*16.0

= 418.67 g/mol

Molar mass of Ag3PO4= 418.67 g/mol

s = 1.78*10^-2 g/L

To covert it to mol/L, divide it by molar mass

s = 1.78*10^-2 g/L / 418.67 g/mol

s = 4.252*10^-5 mol/L

Answer: 4.25*10^-5 mol/L

b)

At equilibrium:

Ag3PO4 <----> 3 Ag+ + PO43-

   3s s

Ksp = [Ag+]^3[PO43-]

Ksp = (3s)^3*(s)

Ksp = 27(s)^4

Ksp = 27(4.252*10^-5)^4

Ksp = 8.822*10^-17

Answer: 8.82*10^-17