# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.

Message body

 Emoticons View more Emoticons
Options
HTML is OFF
BBCode is ON
Smilies are ON
 Disable BBCode in this post Disable Smilies in this post

 All times are GMT
 Jump to: Select a forum Puzzles and Games----------------Grey Labyrinth PuzzlesVisitor Submitted PuzzlesVisitor GamesMafia Games Miscellaneous----------------Off-TopicVisitor Submitted NewsScience, Art, and CulturePoll Tournaments Administration----------------Grey Labyrinth NewsFeature Requests / Site Problems
 Topic review
Author Message
austinap
Posted: Fri Nov 19, 2004 11:12 pm    Post subject: 1

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?
Guest
Posted: Fri Nov 19, 2004 11:09 pm    Post subject: 0

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])
Posted: Fri Nov 19, 2004 10:42 pm    Post subject: -1

Interesting stuff.

What does K a represent?
impossibleroot
Posted: Fri Nov 19, 2004 10:30 pm    Post subject: -2

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!

Courk
Posted: Fri Nov 19, 2004 10:11 pm    Post subject: -3

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.
Courk
Posted: Fri Nov 19, 2004 10:04 pm    Post subject: -4

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.