Buffer Calculations Help

1) Calculate the pH of a buffer which is 0.2 moldm-3 with respect to ammonium sulphate and 0.1 moldm-3 with respect to ammonia. (Ka of NH4+ = 5.6 x 10-10 moldm-3)
The answer is 8.95, but I keep getting 7.95
-log(5.6x10^-10 x 0.2/ 0.1^2) =7.95

Reply 1

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Reply 2

Protoxylic

I definitely keep getting 8.65 not 8.95. The way I did it was treat the ammonium salt as the associated version of the acid so (NH4)2SO4, this dissociates (assuming) into 2 moles of ammonium ions, therefore the concentration of NH4+ is 0.4M. So I get Ka={[H+][NH3]/2[(NH4)2SO4]} rearranging to find pH I get -log((Ka x 2[(NH4)2SO4]/[NH3])) which gives me 8.649751982 which is roughly 8.65.

Right I see where the answer comes from, the question assumes that the salt acts to dissociate to give 1 mole of ammonium ions rather than 2. So, pH=-log{[Ka x [(NH4)2(SO4)]/[NH3]} = 8.950781977 = 8.95 2sf

(edited 9 years ago)

Reply 3

username1445490

Neither of the answers in the original post is correct.
You have to consider what the concentration of NH4+ ions is in a 0.2 solution of (NH4)2SO4. It isn't 0.2!
For some reason you seem to be squaring in your equation which should not be there.

Reply 4

Protoxylic

Original post by Madasahatter

8.95 is definitely the correct answer. The concentration would be 0.2 because it only mono dissociates. So [NH4+]=[(NH4)2SO4]

(edited 9 years ago)

Reply 5

username1445490

Original post by Protoxylic

it only mono dissociates. So [NH4+]=[(NH4)2SO4]

Where do you get that information from ?

Reply 6

Protoxylic

Original post by Madasahatter

Where do you get that information from ?

The answer is 8.95, which is the answer you get if you assume a mono dissociation. If you assume a di dissociation you get my other answer, but I am assuming the book teaches mono dissociation hence why the answer is so.

Reply 7

username1445490

Original post by Protoxylic

The answer is wrong.
It doesn't mono dissociate (why would it?). The book, like so many textbooks, has an error.
When you've been in the game for a while you soon learn that textbooks are full of errors especially when it comes to answers to problems.
The people who write textbooks are human and make mistakes. A friend of mine just worked his way through all the physics problems in a new edition of a physics textbook. The author had made loads of errors with the new problems and he found a load of wrong answers from problems held over from the previous edition.
Never assume the book is right!

(edited 9 years ago)

Reply 8

Protoxylic

Original post by Madasahatter

The OP hasn't really stated the full question, for all we know the question could have stated explicitly to assume a mono dissociation, which in this case, seems to be what the question is doing. If the question did state that, then the question isn't technically wrong, but otherwise you don't even know the dissociation constants.

Reply 9

username1445490

Original post by Protoxylic

This is because ammonium sulphate fully ionises in water and so does not have a meaningful dissociation constant.

Reply 10

Protoxylic

Original post by Madasahatter

This is because ammonium sulphate fully ionises in water and so does not have a meaningful dissociation constant.

This is in ammonia, not water.

Reply 11

username1445490

Original post by Protoxylic

This is in ammonia, not water.

Ammonia is a gas.
These are aqueous solutions forming a buffer.
The ammonium sulphate and ammonia are dissolved in water forming dilute solutions. This is almost entirely water.

Reply 12

Protoxylic

Original post by Madasahatter

Ammonia is a gas.
These are aqueous solutions forming a buffer.
The ammonium sulphate and ammonia are dissolved in water forming dilute solutions. This is almost entirely water.

You're making assumptions that it is aqueous. Ammonia isn't a gas at all temperatures. This question requires assumptions for almost all parts of it, that is why you get 2-3 different answers that are all correct provided the assumptions match up.

Reply 13

username1445490

Now you are getting silly!
If ammonia was a liquid here then this is not going to act as a buffer. All buffers are aqueous. pH is meaningless without water.
If ammonia was the solvent then how could it have a concentration of 0.1 moldm-3. What is the dm-3 of? It is the solvent that the ammonia is dissolved in ie WATER!

Just spotted you are an A2 student. I guess that explains the lack of understanding.

I think we should stop this here before the OP gets confused with foolish suggestions.

This is an aqueous solution. There is one correct answer (7.95 is completely wrong because of the squaring error) and no assumptions needed and the book has made an error by not accounting for the full ionisation of the ammonium sulphate. Believe me I know as I have had to deal with this sort of stuff for over 20 years.

(edited 9 years ago)

Reply 14

Protoxylic

Original post by Madasahatter

Alright, I went off track. But that doesn't stop the fact that the question could have assumed a mono dissociation. I do understand that some textbooks have errors within them, however, the OP hasn't stated the question fully, only fragments.

Reply 15

username1445490

Original post by Protoxylic

In that case the book would be even more wrong as ammonium sulphate is an ionic salt and fully ionises in aqueous solution.
(NH₄)₂SO_{4 (s)} +aq --> 2NH₄⁺_(aq) + SO₄^2-_(aq)

0.2M --> 0.4M + 0.2M

This is a bit like a question asking you to calculate the pH of 0.04 moldm-3 solution of HCl and someone saying they haven't given the full question because it doesn't say HCl is not a weak acid and maybe the question is assuming HCl does not fully ionise but they haven't given the Ka. This would be ridiculous.

ALL textbooks have errors in them. Set yourself the challenge of finding them and report them to the publisher.