Question

a saturated solution cu(io3)2 was prepared in 0.015 m klo3. The copper concentration was found to be 0.0010 M. Set up an ice table and caunlate the [IO3-] and the Ksp.

Answer 1

IO3- + 5I- + 6H+ <--> 3I2 + 3H2O

2S2O3^2- + I2 <---> 2I- + S4O6^2-

Cu(IO3)2(s) --> Cu2+ + 2IO3-

3 moles I₂   / 1mole IO₃^- x 2 moles (S₂O₃)^-2 / 1 mole I₂ = 6 moles (S₂O₃)^-2 / 1mole IO₃^-
thus, 1 mole of IO₃^- will require 6 moles of (S₂O₃)^-2

0.015 m klo3

0.0010 M =  copper concentration )

We should GET 2 MOLE IO₃^- from 1 mle Cu(IO3)2

since we have 0.0010 M Cu(IO3)2 we will get --0.0020 M IO₃^-

we allready have 0.015 m klo3 so totall we have 0.0170M IO₃^-

Ksp = [Cu+2][IO3-}^2

= (0.0010)(0.0170)^2
= 2.89 x 10^-7