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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 precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10-8 ).

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M manish

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 the solution becomes half (= 0.001)

Ionization 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}

Sinc eionic product is less than the K_{sp} so no precipitation occurs.

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