Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, K_sp = 7.4 × 10^-8)

To determine if mixing equal volumes of 0.002 M sodium iodate and cupric chlorate will result in the precipitation of copper iodate, we need to assess the solubility product constant (Ksp) of cupric iodate.

Understanding the Reaction

When sodium iodate (NaIO3) and cupric chlorate (Cu(ClO3)2) are mixed, the relevant reaction is:

• Cu²⁺ + 2IO₃^- ⇌ Cu(IO₃)₂

Calculating Ion Concentrations

After mixing equal volumes, the concentration of Cu²⁺ and IO₃^- will each be halved:

• Cu²⁺: 0.002 M / 2 = 0.001 M
• IO₃^-: 0.002 M / 2 = 0.001 M

Finding the Ion Product

The ion product (Q) for the precipitation of copper iodate can be calculated as follows:

• Q = [Cu²⁺][IO₃^-]²
• Q = (0.001)(0.001)² = 1 × 10^-9

Comparing Q and Ksp

Now, we compare the calculated ion product (Q) with the Ksp of cupric iodate:

• Ksp = 7.4 × 10^-8
• Q = 1 × 10^-9

Conclusion on Precipitation

Since Q (1 × 10^-9) is less than Ksp (7.4 × 10^-8), the solution is unsaturated with respect to copper iodate. Therefore, no precipitation will occur when these two solutions are mixed.