 Topic - Any chemists out there?.........
|
|---|
 by - blackreaver  on - 10 Nov 2012 |
I'm unsure of an assumption that I've made whilst doing some calculations for a chemistry uni lab report.
We made up 3 buffer solutions of 0.1M ethanoic acid (EA) and 0.1M sodium ethanoate (SE)
Solution 1: 25ml EA, 25ml SE, pH was measured to be 4.46
Solution 2: 5ml EA, 50ml SE, pH was measured to be 5.80
Solution 3: 50ml EA, 5ml SE, pH was measured to be 3.52
I have to calculate the pka (acid dissociation constant) for each solution.
Would pka just be the same as the pH as:
pKa = pH - log([SE]/[EA])
Am I correct in saying that "[SE]/[EA]" would just be 1 as the concentrations of EA and SE are the same. (and log1=0)
Thanks in advance,
BR |
... not showing 15 replies to this topic ...
Register as a New User or login to gain full access to the forums. Registration is quick and completely free.
If you are definitely logged on, press Ctrl+F5 keys to reload this page [read more]
Unregistered users can only read messages in the most recent topics. |
|
