The solubility product of Cu(IO3)2 is 1.08 * 10-7. Assuming that neither ions react appreciable with water to form H+ and OH-, what is the solubility of this salt?
On dissociation, Cu(IO\(_3\))\(_2\) ⇌ Cu\(^{2+}\)(aq) + 2IO\(_3\)\(^-\)(aq).
Solubility product constant = Ksp
The formula is : Ksp = [Cu\(^{2+}\)] [IO\(_3\)\(^-\)]\(^2\)
Let the concentrations of Cu\(^{2+}\) and IO\(_3\)\(^-\) be expressed as d.
[Cu\(^{2+}\)] = d ; [ IO\(_3\)\(^-\)] = 2d (because 2IO\(_3\)\(^-\) )
Substituting these expressions into the Ksp
Ksp = [Cu\(^{2+}\)] [IO\(_3\)\(^-\)]\(^2\)
Ksp = (d) x (2d)\(^2\)
Ksp = d.4d\(^2\)
Ksp = 4d\(^3\)
Recall that Ksp = 1.08 x 10\(^{-7}\)
4d\(^3\) = 1.08 x 10\(^{-7}\)
Divide both sides by 4
d\(^3\) = \(\frac{1.08 x 10^{-7}}{4}\)
d\(^3\) = 0.000000027
d = ∛0.000 000 027
d = 0.003
d = 3 x 10\(^{-3}\) moldm\(^{-3}\)
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