Question

The solubility of magnesium arsenate, Mg3(AsO4)2 was determined experimentally to be 1.6 x 10-3 g/100 g of water. What is the Ksp for Mg3(AsO4)2? a. 2.3 x 10-20 b. 2.3 x 10-30 c. 9.2 x 10-5 d. 4.6 x 10-5 e. 1.4 x 10-4

Answer 1

Lets calculate the solubility in molar first

Molar mass of Mg3(AsO4)2,

MM = 3*MM(Mg) + 2*MM(As) + 8*MM(O)

= 3*24.31 + 2*74.92 + 8*16.0

= 350.77 g/mol

mass(Mg3(AsO4)2)= 1.6*10^-3 g

use:

number of mol of Mg3(AsO4)2,

n = mass of Mg3(AsO4)2/molar mass of Mg3(AsO4)2

=(1.6*10^-3 g)/(3.508*10^2 g/mol)

= 4.561*10^-6 mol

since density of water is 1 g/mL and mass of water is 100 g,

volume , V = 1*10^2 mL

= 0.1 L

use:

Molarity,

M = number of mol / volume in L

= 4.561*10^-6/0.1

= 4.561*10^-5 M

At equilibrium:

Mg3(AsO4)2<----> 3 Mg2+ + 2 AsO4-

   3s 2s

Ksp = [Mg2+]^3[AsO4-]^2

Ksp = (3s)^3*(2s)^2

Ksp = 108(s)^5

Ksp = 108(4.561*10^-5)^5

Ksp = 2.1*10^-20

Answer: a