Question

Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to precipitation of copper iodate ? For copper iodate, `K_(sp)=7.4xx10^(-8)`.

Answer 1

`2 NaIO_(3)+CuCrO_(4) rarr Na_(2)CrO_(4)+Cu(IO_(3))_(2)`
After mixing, `[NaIO_(3)]=[IO_(3)^(-)]=(2xx10^(3))/(2)=10^(-3)M`
`[CuCrO_(4)]=[Cu^(2+)]=(2xx10^(-3))/(2) = 10^(-3)M`
Ionic product of `Cu(IO_(3))_(2)=[Cu^(2+)][IO_(3)^(-)]^(2)=(10^(-3))(10^(-3))^(2)=10^(-9)`
As ionic product is less than `K_(sp)`, no precipitation will occur.