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

Equal volumes of 0.002 M solutions of sodium iodate and copper chlorat

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