Talk:Buffer solution
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[edit] removed section
I removed a section "Illustration of a buffer solution in action" because, to me at least, it was redundant with the following section, and the less well presented of the two. I hope that doesn't cause consternation. Olin 03:35, 9 March 2006 (UTC)
I did some further cleanup, although the article still needs some help. Olin 04:10, 9 March 2006 (UTC)
I have changed the equation of calculating the PH. The HA and A- was wrong way round. This is further confirmed through text book.
I changed it back, per Henderson-Hasselbalch equation. I'll look it up when I get a chance. Tom Harrison Talk 14:19, 6 June 2006 (UTC)
Deduction:
![\mathrm{K_a = \frac{[H^+][A^-]}{[HA]}}](../../../math/a/0/2/a026fcfe3468355857a6b832c680db8c.png)
![\mathrm{[H^+] = {K_a} \frac{[HA]}{[A^-]}}](../../../math/2/d/9/2d9fe1b2ff65f75f7a7cad9803ab359c.png)
![\mathrm{log_{10}{[H^+]} = log_{10}{(K_a)} + log_{10} \frac{[HA]}{[A^-]}}](../../../math/5/b/b/5bb4b6b99645d7d9a557d70060af3f28.png)
![\mathrm{-log_{10}{[H^+]} = -log_{10}{(K_a)} + -log_{10} \frac{[HA]}{[A^-]}}](../../../math/8/a/e/8ae674733eb43fbb0f26329fc9e04608.png)
![\mathrm{-log_{10}{[H^+]} = -log_{10}{(K_a)} + log_{10} \frac{[A^-]}{[HA]}}](../../../math/a/3/d/a3d2f907b739eb9888b231ffe6325854.png)
![pH=pK_a+log_{10}\frac{[A^-]}{[HA]}](../../../math/6/7/2/67262074271899856d78d6415b3ae320.png)
QED. IMHO, the formula stated is the correct one. --Dirk Beetstra 15:34, 6 June 2006 (UTC)
ah ha....sorry, forgotten about the -log. thanks for reverting it back.
[edit] more buffers @ sigma-aldrich
Spam link! I do not see and kind of scientific value!
http://en.wikipedia.org/wiki/User_talk:213.188.227.119
Best regards
- Look, Scotty has beamed them to a more appropriate position, just as this remark of you! --Dirk Beetstra 16:21, 29 May 2006 (UTC)
