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

` [IO_3^(-) ]= 0.002 M and [Cu^(2+) ] = 0.002M`
` " " Cu^(2+) + 2IO_3^(-) to Cu(IO_3) _2`
` K_(sp) Cu(IO_3) _2= [Cu^(2+) ] [ IO_3^(-) ]^(2) = 7.4 xx 10 ^(-8) ` ( given ) .
On mixing equal volumes , the concentration gets halved .
` [Cu^(2+) ] [IO_3^(-) ] = ((0.002)/(2)) xx ((0.002)/(2))^(2) = 1.0 xx 10 ^(-9)`
Since , the ionic product is less than the solubility product, precipitations will not take place.