# Equal volumes of

Given:

Ksp 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,

Ksp = [A+] [B-]

Where A and B are the ions dissolved

In the above reaction,

[A+] = [Cu2+]

[B-] = [IO3-]

Ksp = [Cu2+][IO3-]

Ksp = (10-3)(10-3)

Ksp = 10-9

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

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

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