[MitchTwitchita]

A 0.44 M solution of a weak base B has a pH of 11.08.  The equation for ionization is B(aq) + H2O ---> BH+ + OH- 
What are the BH+, OH-, and B concentrations at equilibrium?  Calculate Kb for the base.

O.k. so, would 0.44 be the concentration for B?

BH+ = 10^-11.08 = 8.32 x 10^-12

OH- = 1 x 10^-14/8.32 x 10^-12 = 1.20 x 10^-3

Kb = [BH+][OH-]/B
= (8.32 x 10^-12)^2/0.12 - 8.32 x 10^-12
=5.77 x 10^-22 ?

Somehow I think I'm doing this problem wrong.  Can anybody please show me what I'm doing wrong?

[Kryolith]

BH+ = 10^-11.08 = 8.32 x 10^-12

This is [H⁺] not [BH⁺]. Now what can you tell about [BH⁺]?

[MitchTwitchita]

I think you would have to add the B in order to get the concentration of BH+ but the number 8.32 x 10^-12 is so nominal it doesn't even contribute. 

[Kryolith]

I think you would have to add the B in order to get the concentration of BH+ but the number 8.32 x 10^-12 is so nominal it doesn't even contribute. 

 

Your calculation of [OH^-] is correct. For 1 mol OH^- how many moles BH⁺ are formed?

[MitchTwitchita]

1

[Kryolith]

correct 

[MitchTwitchita]

I'm afraid I don't understand where you're leading me.  Surely, BH+ doesn't have the same concentration as OH-?

[Kryolith]

Why not? I confess, I neglected the self-ionization of water and assumed that the pH was not changed by adding a different acid/base.

[MitchTwitchita]

The autoionization of water is assumed to be negligible.  If the concentrations of BH+ and OH- were the same, wouldn't that make the solution neutral? 

[Kryolith]

neutral means [OH^-]=[H⁺]=10^-7 mol/l
we have [OH^-]=[BH⁺]=1.2 * 10^-3 mol/l

[MitchTwitchita]

Kw = (Ka)(Kb)

1 x 10^-14 = (BH+)(1.20 x 10^-3)
BH+ = 1 x 10^-14/1.20 x 10^-3
=8.33 x 10^-12

wouldn't this be the way that BH+ would be calculated?

[Borek]

BH⁺ is not the same thing as H⁺.

[Kryolith]

Ok, I'll start from the beginning.

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)

Ok. Let's go on. We know that every OH^- comes from the dissociation of the base (neglecting self-ionization of water). From the reaction equation we see that for every OH^- ion we also get one BH⁺ ion. So [OH^-] must be the same as [BH⁺]. The pH is given, so we can calculate [OH^-] resp. [BH⁺]

[MitchTwitchita]

Oh man, I got it now!  my malfunction was confusing BH+ with H+.  The Ka would be [BH+][OH-]/H2O.  Thanks a lot, you've been a very big *delete me*

[MitchTwitchita]

Why did the last post say *delete me* for help?  Anyhoo, the rest of the answers were right though?  Thanks again!