/ Any chemists out there?.........
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,
Correct - each of the above will give you the same value. For example, solution 1 -
pKa = 4.46 - log(25/50) = 4.46 - log 1/2 = 4.46 - (-0.5) = 4.76
You are most welcome and thank you for not calling me on my poor arithmetic skills
log of 0.5 is of course -0.3
4.46 - (-0.3) = 4.76
Phew. Glad I caught that before the rest of UKC.
pKa= pH -log (salt)/(acid)
Solution 1 pKa = 4.46 -log 1/1 = 4.46
Solution 2 pKa = 5.8 - log 10/1 = 4.8
Solution 3 pKa = 3.52 - log 1/10 = 4.52
No - you've not allowed for the cation in the salt. The pKa can't vary all over the place, either
Copying and pasting from the link 'cos I'm too lazy to type -
pH = pKa + log([A-]/[HA])
pH = pKa+log([conjugate base]/[undissociated acid])
Solution 1 4.76
Solution 2 6.84
Solution 3 3.56
> No - you've not allowed for the cation in the salt.
Could you explain please.
I'm assuming complete dissociation of the salt and negligible dissociation of the acid.
The pKa should be the same in each case - it's a property of the ethanoic acid
In each case, the base has been diluted - so the salt concentration in (1) is 0.05 M
The cation concentration is the same - 0.1M. So the salt/acid part of the equation is 1/2. Log of this is -0.3. So pKa is pH + 0.3.
I calculate the pKa's to be 4.46, 4.8, 4.52, which doesn't look right. The calculated pH values, based on a pKa for ethanoic acid of 4.76 should be 4.76, 5.76, 3.76, versus the OPs values.
It has been 40 years since I learnt this - I may have taught this ~ 20 years ago.....
This is lab data - is it possible that the OP measured his solutions incorrectly (not too likely)? Or that the pH meter wasn't calibrated correctly (quite possible)?
Like you, I've been retired for a few years and like you suspect that the original question may be from experimental data. In which case the answers are in the right ballpark.
Thanks everyone for their help.
It did require the blowing out of several substantial cobwebs! Good exercise.
The hole digging?
I enjoyed it too.
Sorry. Couldn't resist it and I was on my way out the door without the time to do a smiley. :-)
To Blackreaver Best of luck with the lab report.
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