Given:

Ks_p of cupric iodate = 7.4 × 10^–8

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together.

Ionization of sodium iodate:

2NaIO₃ + CuCrO₄ Na₂CrO₄ + Cu(IO₃)₂

After mixing, [NaIO₃] = [IO₃^-] = = 10^-3M

[CuCrO₄] = [Cu²⁺] = = 10^-3M

For cupric iodate, the ionization is:

Cu(IO₃)_{2 ↔} [Cu²⁺][IO₃^-]

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Cu²⁺]

⇒[B^-] = [IO₃^-]

∴ K_sp = [Cu²⁺][IO₃^-]

⇒K_sp = (10^-3)(10^-3)

⇒K_sp = 10^-9

As we know that precipitation only occur when Ks_p < [A⁺][B^-]

As ionic product 10^-9 is less than the K_sp (7.4 × 10^–8), hence no precipitation will occur.