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 K_(sp)=7.4xx10^(-8) )

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

1.	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 K_(sp)=7.4xx10^(-8) )
Answer» Solution :Sodium Iodate `(NaIO_3)` : It is completely ionized `{:(NaIO_3 to Na_((aq))^(+)+ IO_(3(aq))^(-)),(0.002M to 0.002 M + 0.002 M):}` `THEREFORE` Concentrationof iodate before mixing `[IO_3^-]`= 0.002 M "When equal volume of two solution mix together then the concentration of mixture is half" `therefore` In mixture `IO_3^(-)=("INITIAL" [IO_3^-])/2=0.002/2`=0.001 M By taking the complete ionization of Cupric chlorate `CU(ClO_3)_2` `{:(Cu(ClO_3)_2 hArr, Cu_((aq))^(2+) + , 2ClO_(3(aq))^(-)),(0.002 M, 0.002M , 2(0.002)M):}` `therefore [Cu^(2+)]`= 0.002 M before mixing "In equal volume of solutionwhen mix, the concentration become half " `therefore` (Concentration of copper ion in mixture ) = `1/2` (Initial concentration of `Cu^(2+)` ion ) `=1/2(0.002)`=0.001 M Equilibrium of `Cu(IO_3)_2` : `{:(Cu(IO_3)_2 hArr, Cu_((aq))^(2+) +, 2IO_(3)^(-)),("Concentration in mixture:", 0.001 M, 0.001 M):}` Thus, `Q_(SP)` of `Cu(IO_3)_2 = [Cu^(2+)][IO_3^-]^2` `=(0.001)(0.001)^2` `=1.0xx10^(-9)` `Cu(IO_3)_2` of `{:((Q_(sp)),(1.0xx10^9)) lt ((K_(sp)),(7.4xx10^(-8))):}` The value of `Q_(sp)` is less than `K_(sp)` so, precipitation of copper iodate does not occurs.