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  • 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 ).

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

Answers (1)

best_answer

We have,
solubility product (K_{sp}) of cupric iodate =7.4 \times 10^{-8}

When equal volumes of sodium iodate and cupric chlorate are mixed together, the molar concentration of both solutions becomes half (= 0.001)

The ionisation of cupric iodate is:

Cu(IO_3)_2\rightleftharpoons Cu^{2+}+2IO_3^-
 0.001 M                                 0.001 M

So, K_{sp} can be calculated as:

K_{sp}= [Cu^{2+}][IO_3^-]^2
           = (0.001)^3 = 10^{-9}

Since the ionic product is less than the K_{sp}, so, no precipitation occurs.

Posted by

manish

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