Question

What concentration of of CN- is needed to produce the same POH as 3.5*10^-5M solution of BA(OH)2?

Answer 1

.........Ba(OH)2 ==> Ba^2+ + 2OH^-
I.......3.5E-4........0.......0
C......-3.5E-4.....3.5E-4..2*3.5E-4
E........0.........3.5E-4...7.0E-4

So the (OH^-) from Ba(OH)2 = 7.0E-4. We want to know how much CN^- it will take to produce OH^- = 7.0E-4 M.

.........CN^- + HOH ==> HCN + OH^-
I........x...............0.....0
C......-7.0E-4........7.0E-4..7.0E-4
E.....x-7.0E-4........7.0E-4..7.0E-4

Kb for CN^- = (Kw/Ka for HCN) = (7.0E-4)^2/(x-7.0E-4)
Solve for x = molar concn of CN^- required. Kw is 1E-14. You will look up Ka for HCN.

Answer 2

I got 2*10^-5 for the Kb, is that right?

Answer 3

When you ask a question like this you should tell us what you used for Ka. Our tables may not give us the same value as your table(s). Therefore, we don't know if you've done it right or not. I used 6.2E-10 for Ka for HCN and using that value Kb for CN^- is 1.6E-5.