Off Belay forum


View latest messages in My Forums
This topic has been archived, and won't accept reply postings.

Topic - Any chemists out there?.........

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.

This topic has been archived, and won't accept reply postings.

My Forums