Calculate the molar solubility of copper(ii) arsenate (cu3(aso4)2) in water. use 7.6 x 10-36 as the solubility product constant of cu3(aso4)2.
9.1 x 10-4 m
3.4 x 10-2 m
3.7 x 10-8 m
8.7 x 10-2 m

Ionization equilibrium: 

Cu3(AsO4)2(s) <> 3 Cu2+(aq) + 2 AsO42-(aq) 

Ksp = [Cu2+]^3 [AsO43-]^2 = 7.6X10^-36 

Now, let s = molar solubility 
When 1 mole of the compound dissolves, you produce 3 moles Cu2+. So, in terms of molar solubility, [Cu2+] = 3s 
Likewise, [AsO43-] = 2s 

Plugging those into the expression for Ksp gives: 

7.6X10^-36 = (3s)^3 (2s)^2 = 27s^3 X 4s^2 = 108s^5 

Solving for s gives: 
s = 3.7X10^-8 mol/L This is the molar solubility.