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)` ).

Explanation

Concentration of `[Cu^(2+)]` before mixing `=0.002 M`
Conc. of `[Cu^(2+)]` after mixing `=0.002/2=0.001 M`
(since equal volume, `:.` Volume becomes double after mixing and hence conc. Is half)
Similarly conc. Of `[IO_(3)^(Θ)]=0.001 M`
`Q_(sp)=[Ca^(2+)][IO_(3)^(Θ)]^(2)=0.001xx(0.001)^(2)`
`=10^(-3)xx(10^(-3))^(2)=10^(-9)`
Since `Q_(sp)lt K_(sp)`
`10^(-9) lt 7.4xx10^(-8)`
`:. Cu(IO_(3))_(2)` will not precipitate.