June 01, 2020, 10:27:49 AM
Forum Rules: Read This Before Posting


Topic: Acid-Base Equilibrium  (Read 12643 times)

0 Members and 1 Guest are viewing this topic.

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Acid-Base Equilibrium
« on: February 02, 2008, 09:38:34 PM »
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?

Offline Kryolith

  • Full Member
  • ****
  • Posts: 269
  • Mole Snacks: +19/-4
  • Gender: Male
Re: Acid-Base Equilibrium
« Reply #1 on: February 03, 2008, 03:20:49 AM »
BH+ = 10^-11.08 = 8.32 x 10^-12

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

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #2 on: February 03, 2008, 01:46:34 PM »
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. 

Offline Kryolith

  • Full Member
  • ****
  • Posts: 269
  • Mole Snacks: +19/-4
  • Gender: Male
Re: Acid-Base Equilibrium
« Reply #3 on: February 03, 2008, 02:12:24 PM »
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?

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #4 on: February 03, 2008, 02:15:48 PM »
1

Offline Kryolith

  • Full Member
  • ****
  • Posts: 269
  • Mole Snacks: +19/-4
  • Gender: Male
Re: Acid-Base Equilibrium
« Reply #5 on: February 03, 2008, 02:42:25 PM »
correct  :)

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #6 on: February 03, 2008, 02:52:26 PM »
I'm afraid I don't understand where you're leading me.  Surely, BH+ doesn't have the same concentration as OH-?

Offline Kryolith

  • Full Member
  • ****
  • Posts: 269
  • Mole Snacks: +19/-4
  • Gender: Male
Re: Acid-Base Equilibrium
« Reply #7 on: February 03, 2008, 03:06:57 PM »
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.


Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #8 on: February 03, 2008, 03:12:47 PM »
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? 

Offline Kryolith

  • Full Member
  • ****
  • Posts: 269
  • Mole Snacks: +19/-4
  • Gender: Male
Re: Acid-Base Equilibrium
« Reply #9 on: February 03, 2008, 03:19:31 PM »
neutral means [OH-]=[H+]=10-7 mol/l
we have [OH-]=[BH+]=1.2 * 10-3 mol/l

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #10 on: February 03, 2008, 03:31:39 PM »
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?


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 25776
  • Mole Snacks: +1686/-400
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Acid-Base Equilibrium
« Reply #11 on: February 03, 2008, 03:56:13 PM »
BH+ is not the same thing as H+.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Kryolith

  • Full Member
  • ****
  • Posts: 269
  • Mole Snacks: +19/-4
  • Gender: Male
Re: Acid-Base Equilibrium
« Reply #12 on: February 03, 2008, 03:56:22 PM »
Ok, I'll start from the beginning.

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)

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+]

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #13 on: February 03, 2008, 04:01:25 PM »
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*

Offline MitchTwitchita

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +1/-1
Re: Acid-Base Equilibrium
« Reply #14 on: February 03, 2008, 04:03:18 PM »
Why did the last post say *delete me* for help?  Anyhoo, the rest of the answers were right though?  Thanks again!

Sponsored Links