The solubility of copper(II)iodate Cu(IO₃)₂ is reported as 0.12 g per 100 mL.What is the solubility product constant for this salt? A)9.8 × 10^-8 B)8.6 × 10^-7 C)7.3 × 10^-6 D)6.4 × 10^-5 E)1.2 × 10^-4

Question

Quizplus · Accepted Answer

The solubility product constant (Ksp) for copper(II) iodate, Cu(IO3)2, can be calculated from its solubility. Given the solubility is 0.12 g per 100 mL, we first convert this to moles per liter (molarity, M) for the calculation. The molar mass of Cu(IO3)2 is approximately 413.38 g/mol. Thus, 0.12 g per 100 mL is equivalent to 0.12 g / 413.38 g/mol = 2.9 × 10^-4 mol in 0.1 L, or 2.9 × 10^-3 M. The dissociation of Cu(IO3)2 in water is represented by the equation: Cu(IO3)2(s) ⇌ Cu^2+(aq) + 2IO3^-(aq). From this, the Ksp expression is Ksp = [Cu^2+][IO3^-]^2. Given the stoichiometry of the dissolution, [Cu^2+] = s and [IO3^-] = 2s, where s is the solubility in mol/L (2.9 × 10^-3 M). Substituting these values into the Ksp expression gives Ksp = (2.9 × 10^-3)(2 × 2.9 × 10^-3)^2 = 9.8 × 10^-8, matching choice A.

The Solubility of Copper(II)iodate Cu(IO3)2 Is Reported as 0

The Solubility of Copper(II)iodate Cu(IO₃)₂ Is Reported as 0