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 ).

Asked by Pragya Singh | 1 year ago |  153

##### Solution :-

Cupric chlorate and sodium iodate having equal volume are mixed together, then the molar concentration of cupric chlorate and sodium iodate will reduce to half. So, the molar concentration of cupric chlorate and sodium iodate in a mixture is 0.001M.

$$Na(IO_3)_2 → Na^{+} + IO_{3}^{-}$$

0.0001M           0.001M

Cu(ClO3)2→ Cu2+ + $$2CIO_{3}^{-}$$

0.001M                0.001M

The Solubility for Cu(IO3)2 ⇒ Cu2+ (aq) + 2IO3–  (aq)

Now, the ionic product of the copper iodate is:

=$$[Cu^{2+}] [IO_{3}^{-}]^{2}$$

$$= (0.001)(0.001)^2$$

$$= 1.0\times 10^{-9}$$

As the value of Ksp is more than Ionic product. Thus, precipitation will not occur.

Answered by Abhisek | 1 year ago

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