The solubility product of CU(IO3)2 is 1.08×10^-3 assuming that neither ions react appreciably with water to form H+ and OH- . What is the solubility of this salt?

Answer 1

To find the solubility of CU(IO3)2, we need to first write out the equilibrium expression for the dissolution of the salt:

CU(IO3)2 (s) ⇌ Cu^2+ (aq) + 2IO3- (aq)

The solubility product constant (Ksp) for this equilibrium is given as 1.08×10^-3.

Let x be the solubility of CU(IO3)2 in mol/L. Then the equilibrium concentrations are [Cu^2+] = x and [IO3-] = 2x.

Substitute these values into the Ksp expression:

Ksp = [Cu^2+][IO3-]^2
1.08×10^-3 = x(2x)^2
1.08×10^-3 = 4x^3

Solve for x:

x = (1.08×10^-3 / 4)^(1/3)
x ≈ 0.158 mol/L

Therefore, the solubility of CU(IO3)2 is approximately 0.158 mol/L.