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

In reply to blackreaver:

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

Should be-

4.46 - (-0.3) = 4.76

Phew. Glad I caught that before the rest of UKC.

In reply to blackreaver: It's getting late but I would have said

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

In reply to JimSh:

No - you've not allowed for the cation in the salt. The pKa can't vary all over the place, either

In reply to pneame:

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])

In reply to pneame: So the equation is {pH}-log([EA]/[EA+SE])

Making:

Solution 1 4.76

Solution 2 6.84

Solution 3 3.56

BR

In reply to pneame:

> (In reply to JimSh)

> 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.

In reply to JimSh:

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.

Hmmmm....

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)?

> 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.

>

Funny that. These are the same as my answers above.

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.

In reply to blackreaver: They're from experimental data so that is probably why they are a little off. My lab partner made up the solutions so I'm not sure how accurate they are and the pH meter may have not been calibrated properly.

Thanks everyone for their help.

BR

In reply to JimSh:

It did require the blowing out of several substantial cobwebs! Good exercise.

In reply to pneame:

> Good exercise.

The hole digging?

I enjoyed it too.

In reply to JimSh: To pneame.

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.