Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: MitchTwitchita on February 02, 2008, 09:38:34 PM
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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?
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BH+ = 10^-11.08 = 8.32 x 10^-12
This is [H+] not [BH+]. Now what can you tell about [BH+]?
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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.
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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?
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1
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correct :)
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I'm afraid I don't understand where you're leading me. Surely, BH+ doesn't have the same concentration as OH-?
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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.
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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?
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neutral means [OH-]=[H+]=10-7 mol/l
we have [OH-]=[BH+]=1.2 * 10-3 mol/l
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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?
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BH+ is not the same thing as H+.
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Ok, I'll start from the beginning.
You have the equilibrium
B + H2O (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%5Crightleftharpoons&hash=157e2a14e3d1ec4923b187ed514226fe1ec08b02) BH+ + OH-
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3FK_b%3D%5Cfrac%7B%5BBH%5E%2B%5D%2A%5BOH%5E-%5D%7D%7B%5BB%5D%7D&hash=7cf029a8f3b3bf2936b7a98da2d517e077ab52c8)
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+]
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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*
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Why did the last post say *delete me* for help? Anyhoo, the rest of the answers were right though? Thanks again!
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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?
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BH+ would donate the proton and Ka would equal [OH-]/[H30] ?
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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 ;)
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H2O ---> H+ + OH-
Ka = [H+][OH-]/[H2O], is this what you mean?
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This is K (not Ka!) for the self-ionization of water. I'll tell you what I meant:
You have the equilibrium
B + H2O (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%5Crightleftharpoons&hash=157e2a14e3d1ec4923b187ed514226fe1ec08b02) BH+ + OH-
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3FK_b%3D%5Cfrac%7B%5BBH%5E%2B%5D%2A%5BOH%5E-%5D%7D%7B%5BB%5D%7D&hash=7cf029a8f3b3bf2936b7a98da2d517e077ab52c8)
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 (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%5Crightleftharpoons&hash=157e2a14e3d1ec4923b187ed514226fe1ec08b02) B + H3O+
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3FK_a%3D%5Cfrac%7B%5BB%5D%2A%5BH%5E%2B%5D%7D%7B%5BBH%5E%2B%5D%7D&hash=edb7dc2b35eda8eaae5e261528eda70604373d4c)
Ka*Kb = 10-14 for conjugated acid/base pairs.
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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
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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.
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Thanks Kryolith!
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You're welcome.