Question

The molar solubility of Cd3(PO4)2 is 1.19×10-7 mol/L at 25°C.

Calculate the solubility of Cd3(PO4)2 at 25°C._____g/L.

Calculate Ksp at 25°C._______

Answer 1

1)

Molar mass of Cd3(PO4)2 = 3*MM(Cd) + 2*MM(P) + 8*MM(O)

= 3*112.4 + 2*30.97 + 8*16.0

= 527.14 g/mol

Molar mass of Cd3(PO4)2= 527.14 g/mol

s = 1.19*10^-7 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.19*10^-7 mol/L * 527.14 g/mol

s = 6.273*10^-5 g/L

Answer: s = 6.27*10^-5 g/L

2)

The salt dissolves as:

Cd3(PO4)2 <----> Cd2+ + 2 PO43-

   s 2s

Ksp = [Cd2+][PO43-]^2

Ksp = (s)*(2s)^2

Ksp = 4(s)^3

Ksp = 4(1.19*10^-7)^3

Ksp = 6.741*10^-21

Answer: Ksp = 6.74*10^-21