When an excess of copper (II) iodate, Cu(IO₃)₂, is added to water at 298 K, the equilibrium concentration of copper ions is 3.27 *10^-3 mol dm^-3. Determine the solubility constant at this temperature.

Question

Quizplus · Accepted Answer

The solubility product constant (Ksp) for copper (II) iodate, Cu(IO3)2, can be calculated using the equilibrium concentration of copper ions. Given that the formula of copper (II) iodate suggests for every mole of Cu(IO3)2 that dissolves, 1 mole of Cu2+ ions and 2 moles of IO3- ions are produced. If the equilibrium concentration of Cu2+ is 3.27×10^-3 mol dm^-3, then the concentration of IO3- ions would be twice that, or 6.54×10^-3 mol dm^-3. The Ksp can be calculated as [Cu2+][IO3-]^2 = (3.27×10^-3)(6.54×10^-3)^2 = 1.40×10^-7.

When an Excess of Copper (II) Iodate, Cu(IO3)2, Is Added

When an Excess of Copper (II) Iodate, Cu(IO₃)₂, Is Added