[Kryolith]

Oh man, I got it now!  my malfunction was confusing BH+ with H+. 

I am pleased.

The Ka would be [BH+][OH-]/H2O.

No. Kb is for the base B. The conjugated acid of B is BH⁺. How does it react with water? What's the formula for Ka?

[MitchTwitchita]

BH+ would donate the proton and Ka would equal [OH-]/[H30] ?

[Kryolith]

BH+ would donate the proton

Yes

and Ka would equal [B ][OH-]/[H30] ?

No. Please post the equilibrium you want to describe and then set up the acid dissociation constant.

P.S.
B is interpreted as a bold tag

[MitchTwitchita]

H2O ---> H+ + OH-

Ka = [H+][OH-]/[H2O], is this what you mean?

[Kryolith]

This is K (not Ka!) for the self-ionization of water. I'll tell you what I meant:

You have the equilibrium
B + H₂O BH⁺ + OH^-



The value of K_a is not of interest here. Just for you to understand, try to set up the equation for Ka. Which equilibrium is described by Ka? (remember conjugated acid/base pairs)

BH⁺ + H₂O B + H₃O⁺



Ka*Kb = 10^-14 for conjugated acid/base pairs.

[MitchTwitchita]

O.K., so my Kb should be [BH+][OH-]/
=(1.2 x 10^-3)^2/0.44 - 1.2 x 10^-3
=3.28 x 10^-6

[Kryolith]

OK.

It is not 100 % clear whether 0.44 mol/l means [B ] or [B ]₀, but because of [B ]~[B ]₀ you don't have to care.

[MitchTwitchita]

Thanks Kryolith!

[Kryolith]

You're welcome.