Question:
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 Ksp = 7.4 x 10-8 ).
Answer:
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together, then, molar concentration is reduces by half i.e, 0.001 M
NaIO3 → Na+ + IO-3
0.001 M 0.001 M
Cu(ClO3) → Cu+ + 2ClO-3
0.001 M 0.001 M
Now, equilibrium can be represents as:
Cu(IO3) → Cu+ + 2IO-3
Ionic product of copper iodate = [Cu+ ][IO-3] = [0.001] x [0.001] = 1 x 10-9
Ionic product of copper iodate is less than its Ksp = 7.4 x 10-8, then precipiation will not occur.
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