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Courk
Daedalian Member

 Posted: Fri Nov 19, 2004 10:04 pm    Post subject: 1 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
Courk
Daedalian Member

 Posted: Fri Nov 19, 2004 10:11 pm    Post subject: 2 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.
impossibleroot
Hi-Keeba!

Posted: Fri Nov 19, 2004 10:30 pm    Post subject: 3

 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!

+1

 Posted: Fri Nov 19, 2004 10:42 pm    Post subject: 4 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.
Guest

Posted: Fri Nov 19, 2004 11:09 pm    Post subject: 5

 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])
austinap
Daedalian Member

 Posted: Fri Nov 19, 2004 11:12 pm    Post subject: 6 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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