The solubility equilibrium of copper iodate may be represented as :
`Cu(IO_(3))_(2) hArr Cu^(2+) (aq) +2IO_(3)^(-)(aq)`
`Cu^(2+)(aq)` are to be provided from copper chlorate solution while `IO_(3)^(-) (aq)` ions from sodium iodate solutions
`Cu(CIO_(3))_(2)(s) overset((aq))(to) Cu^(2+) (aq) +2CIO_(3)^(-) (aq)`
`NaIO_(3)(s)overset((aq))(to) Na^(+) (aq) +IO_(3)^(-) (aq)`
Since equal volumes of the solutions have been mixed therefore , concentration of both `Cu^(2+) (aq)` and `IO_(3)^(-)(aq)` ions in solution after mixing will be reduced to half i.e.,
`(0.002M)/(2) =0.001 M`
The ionic product `=[Cu^(2+)][IO_(3)^(-)]^(2) =(0.001) xx (0.001)^(2) = 1.0 xx 10^(-9)`
`K_(sp) " value of " Cu(IO_(3))_(2) =7.4 xx 10^(-8)` (given)
Since the ionic product is less than the `K_(sp)` value copper iodate will not be precipitated.