June 04, 2020, 12:49:08 AM
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Topic: Acid-Base Equilibrium  (Read 12647 times)

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Offline Kryolith

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Re: Acid-Base Equilibrium
« Reply #15 on: February 03, 2008, 04:18:40 PM »
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?

Offline MitchTwitchita

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Re: Acid-Base Equilibrium
« Reply #16 on: February 03, 2008, 04:29:20 PM »
BH+ would donate the proton and Ka would equal [OH-]/[H30] ?

Offline Kryolith

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Re: Acid-Base Equilibrium
« Reply #17 on: February 03, 2008, 04:37:13 PM »
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 ;)

Offline MitchTwitchita

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Re: Acid-Base Equilibrium
« Reply #18 on: February 03, 2008, 04:45:31 PM »
H2O ---> H+ + OH-

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

Offline Kryolith

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Re: Acid-Base Equilibrium
« Reply #19 on: February 03, 2008, 04:52:28 PM »
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 Ka 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.

Offline MitchTwitchita

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Re: Acid-Base Equilibrium
« Reply #20 on: February 03, 2008, 05:02:00 PM »
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

Offline Kryolith

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Re: Acid-Base Equilibrium
« Reply #21 on: February 03, 2008, 05:03:59 PM »
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.

Offline MitchTwitchita

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Re: Acid-Base Equilibrium
« Reply #22 on: February 03, 2008, 05:12:42 PM »
Thanks Kryolith!

Offline Kryolith

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Re: Acid-Base Equilibrium
« Reply #23 on: February 03, 2008, 05:15:33 PM »
You're welcome.

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