if 2.0 x 10-5 mol of sparingly soluble Cu(IO3)2 can dissolve in 2L of a very soluble NaIO3 sol'n, what is the molar concentration of the NaIO3 sol'n?
(Ksp=1.4x10-5 for Cu(IO3)2)
To find the molar concentration of the NaIO3 solution, we need to use the concept of solubility product constant (Ksp) and the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the dissociation of Cu(IO3)2:
Cu(IO3)2 (s) ⇌ Cu2+ (aq) + 2IO3- (aq)
According to the stoichiometry of the reaction, the concentration of Cu2+ ions will be equal to half the concentration of Cu(IO3)2, while the concentration of IO3- ions will be equal to twice the concentration of Cu(IO3)2.
Given that 2.0 x 10^-5 mol of Cu(IO3)2 can dissolve in 2 L of NaIO3 solution, the molar concentration of Cu(IO3)2 can be calculated as follows:
Molar concentration of Cu(IO3)2 = (2.0 x 10^-5 mol) / (2 L) = 1.0 x 10^-5 M
Since the concentration of Cu2+ ions is equal to half the concentration of Cu(IO3)2, the concentration of Cu2+ ions is:
Concentration of Cu2+ ions = 0.5 x (1.0 x 10^-5 M) = 5.0 x 10^-6 M
Using the solubility product constant (Ksp = 1.4 x 10^-5), we can set up an equation to find the concentration of IO3- ions:
Ksp = [Cu2+] * [IO3-]^2
Substituting the concentration of Cu2+ ions:
1.4 x 10^-5 = (5.0 x 10^-6 M) * [IO3-]^2
Rearranging the equation to solve for [IO3-]^2:
[IO3-]^2 = (1.4 x 10^-5) / (5.0 x 10^-6)
[IO3-]^2 = 2.8
Taking the square root of both sides:
[IO3-] = √2.8
[IO3-] = 1.67 M
Therefore, the molar concentration of the NaIO3 solution is 1.67 M.