Copper(II) iodate precipitates from iodate solution. The formula of the precipitate is CU(IO3)2. A saturated solution of this substance was made in 0.01 M KLO3. The copper concentration was found to be 0.0013 M. Determine the [IO3-] and Ksp of the substance.
Ksp = (Cu^+2)(IO3^-)
(Cu^+2) = 0.0013
(IO3^-) = 0.01
(0.0013)(0.01)^2 = ??
The Internet shows Ksp = 1.4 x 10^-7. Your result is quite good.
To determine the concentration of IO3- and the Ksp of the substance, we can use the concept of solubility products and the given information.
First, let's identify the balanced chemical equation for the dissolution of CU(IO3)2:
CU(IO3)2(s) ⇌ CU2+(aq) + 2IO3-(aq)
From the equation, we can see that 1 mol of CU(IO3)2 produces 2 mol of IO3- ions when it dissolves.
Given:
Copper concentration (CU2+) = 0.0013 M
Using the ratio from the chemical equation, the concentration of IO3- (denoted as [IO3-]) is twice the copper concentration:
[IO3-] = 2 * 0.0013 M
[IO3-] = 0.0026 M
Next, let's determine the Ksp (solubility product constant) of CU(IO3)2. The Ksp expression for this compound can be written as:
Ksp = [CU2+][IO3-]^2
Substituting the given values:
Ksp = (0.0013 M)(0.0026 M)^2
Ksp = 8.058 x 10^-9 M^3
Therefore, the concentration of IO3- is 0.0026 M, and the Ksp of CU(IO3)2 is 8.058 x 10^-9 M^3.