Question 22.7: A 1.0 × 10^−3 M NaIO3(aq) solution is made 0.010 M in Cu^2+(......

General Chemistry [1096620]

A $1.0 × 10^{−3}\ M\ NaIO_{3}(aq)$ solution is made 0.010 M in $Cu^{2+}(aq)$ by dissolving the soluble salt $Cu(ClO_{4})_{2}(s)$ . Does $Cu(IO_{3})_{2}(s)$ ( $K_{sp} = 7.4 × 10^{−8}\ M^3$ ) precipitate from the solution at 25°C?

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The chemical equation for the solubility equilibrium is

Cu(IO_{3})_{2}(s) ⇋ Cu^{2+}(aq) + 2\ IO_{3}^−(aq)

The value of $Q_{sp}$ is

Q_{sp} = [Cu^{2+}]_{0}[IO_{3}^−]^{2}_{0} = (0.010\ M)(1.0 × 10^{−3}\ M)^2 = 1.0 × 10^{−8}\ M^3

Because $Q_{sp} \lt K_{sp}$ , no precipitate of $Cu(IO_{3})_{2}(s)$ forms.

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