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Courk
Daedalian Member
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Posted: Fri Nov 19, 2004 10:04 pm Post subject: 1 |
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I just learned this today in my Chem class and need a bit of help.
Equation as given by the teacher:
Suppose we have a solution of 0.110 M of NaC
2
H
3
O
2
and 0.090 M of CH
3
COOH. In this buffer system, CH
3
COOH is the acid, and NaC
2
H
3
O
2
is its conjugate base. For acid concentration, my teacher used the abbreviation [HA], and for conjugate base concentration, she used [A - ]. So, she wrote the Henderson-Hasselbalch equation as:
pH = pK
a
- log ([A - ]/[HA]) Equation 1
We were told K
a
= 1.8*10 -5
That means the whole equation (with numbers plugged in) should be:
pH = -log(1.8*10 -5 ) - log(0.110/0.090), which, when entered exactly like that in my calculator, ~ equals 4.66. She said the answer is ~ 4.83.
My book has the equation pH = pK
a
+ log ([anion]/[acid]), which, using her symbols, is equal to
pH = pK
a
+ log ([A - ]/[HA]) Equation 2
Using this equation and the same numbers I get
pH = -log(1.8*10 -5 ) + log(0.110/0.090), which, when entered exactly like that in my calculator, gives me ~ 4.83 (her answer). It appears equation 2 is the right equation.
I'm not done yet, please bear with me.
She gave us another problem to work on, this time [HA] = 0.110 M and [A - ] = 0.090 M.
We were already told pK
a
= 7.00.
Using the original equation pH = pK
a
- log([A - ]/[HA])
I get pH = 7 - log(0.090/0.110) ~ 7.09 (which is her answer).
Using the second equation pH = pK
a
+ log ([A - ]/[HA])
I get 7 + log(0.090/0.110) ~ 6.91 (not her answer).
So... one way equation 1 works, the other way equation 2 works. Is she wrong in the second example, and the answer really isn't 7.09, but is actually 6.91? I want to use equation 2 (the one from the book), but if I'm doing something wrong and her equation (equation 1) should work, I want to know what I'm doing wrong.
BTW, if you want me to, I can make the equations more readable by making an image out of them.
Last edited by Courk on Sat Nov 20, 2004 12:21 am; edited 1 time in total |
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Courk
Daedalian Member
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Posted: Fri Nov 19, 2004 10:11 pm Post subject: 2 |
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| Oh bah. She just e-mailed the whole class to correct herself. It really is pH = pK a + log ([A - ]/[HA]). I put waaaay too much work into the sup's and sub's to delete this now. |
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impossibleroot
Hi-Keeba!
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Posted: Fri Nov 19, 2004 10:30 pm Post subject: 3 |
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| Quote: |
| Suppose we have a solution of 0.110 M of NaC 2 H 3 O 2 and 0.090 M of CH 3 COOH. In this buffer system, CH 3 COOH is the acid, and NaC 2 H 3 O 2 is it's conjugate base. |
'Its' shouldn't have an apostrophe!
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Samadhi
+1
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Posted: Fri Nov 19, 2004 10:42 pm Post subject: 4 |
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Interesting stuff.
What does K
a
represent? _________________ And he lived happily ever after. Except for the dieing at the end and the heartbreak in between. |
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Guest
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Posted: Fri Nov 19, 2004 11:09 pm Post subject: 5 |
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| Samadhi wrote: |
Interesting stuff.
What does K
a
represent? |
The dissociation constant of an acid (its an equilibrium constant).
Your professor just did a step only half way. She switched the log around but didn't change the sign.
Deriving the equation from scratch, it starts with
[H
3
O + ] = K
a
([HA]/[A - ])
taking the -log of each side:
-log([H
3
O + ] ) = -log(K
a
([HA]/[A - ]))
=> pH = -log(K
a
([HA]/[A - ]))
= -log(K
a
) -log([HA]/[A - ])
= pK
a
- log([HA]/[A - ])
and for where she got messed up:
= pK
a
+ log([A - ]/[HA]) |
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austinap
Daedalian Member
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Posted: Fri Nov 19, 2004 11:12 pm Post subject: 6 |
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Sorry that was me. And those sub's and sup's take far too much time!
Any way you can enable html in your posts? |
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