[Dolphinsiu]

1. What is the solubilty of Ba(IO3)2 in a solution of 0.0200M Ba(NO3)2
[Ksp of Ba(IO3)2 = 1.5 x 10-9 mol3L-3]

I know that Ksp of Ba(IO3)2 = [Ba2+][IO3-]

and Ecell = E' - 0.0592/2 [1/Ba2+] for Ba2+ + 2e --> 2Ba

Then I start get stucked. Can anyone help?

2. In a potentiometric measurement using calcium ion-selective electrode and a standard hydrogen elcetrode (SHE), and the cell potential Ecell was found to be -2.988 V in a 1.00 x 10-4 M Ca2+ ion. The electrode potential was -2.986 V when it was immersed in the solution contains 1.00 x 10-4 M Ca2+ and 1.0 x 10-4 M Mg2+ ions. Calculate the selectivity coefficient KCa2+,Mg2+ of the electrode.

-2.988 = E' - 0.0592/2 log [1/1x10-4]

-2.986 = E' - 0.0592/2 log [1/1x10-4]

Then I get stucked. Please help ! Thank you

[Borek]

I know that Ksp of Ba(IO3)2 = [Ba2+][IO3-]

Close, but wrong. You don't need anything else, there is no redox reaction going on. Just dissolution/precipitation equilibrium.

[Dolphinsiu]

Just 1.5 x 10-9 / 0.02 = Solubility . Am I right?

How about Q.2?