Lisachem I beg to differ...

[kenhung123]

Ok for the 3rd error I would think that the % N is too high because we would calculate n(NaOH) that reacted from finding how much is reacted subtract from the amount added. If we in fact added less NaOH it means there will be less excess and we would subtract less from the amount added. This results in more NaOH reacted. We relate the amount of NaOH reacted to the amount of NH4+. If there is more NaoH there is more NH4+ and therefore increase %?

[kyzoo]

Err what is the 3rd error? I remember doing thsi question and getting the same answer as solutions (%N too low) but I forget the specific error.

[kenhung123]

sorry

[m@tty]

It definitely results in a lower percentage. *After* the original sample has been weighed some is left behind. This reduces the amount of ammonium is lower, which creates less precipitate, which lowers the percentage determined.

[kenhung123]

There's not precipitate involved they are adding excess NaOH to NH4+ and want to find how much of the NaOH reacted to determine the amount NH4+ reacted (present). We think that the amount of NaOH is all transferred (i.e. higher than normal in the initial amount). But after back titrating, we find that there is quite less excess. So NaOH that we THINK reacted is a lot because only little is left behind. As a result we think there is lots of NH4+ since there is lots of NaOH reacting with it initially.


Tell me if I'm thinking a bit weird lol

[m@tty]

Sorry dude. I meant the amount of OH^- reacted reduces and hence more HCl is required. And going backwards: You find that more HCl reacts, this then implies that there was more OH excess from the first reaction which implies that there was less ammonium. And as the calculated mass is lower then the percentage will also be lower.

[kenhung123]

But we have less NaOH and same amount of ammonium. So say we have 10 molecule of ammonium and 11 OH- ions but we think we added 12 in. So using HCl back titration we find that we only have 1 OH- left. So we think "oh there is only 1 OH- left so that means 11 reacted!" But in fact 10 reacted and 1 was left not 11 reacted. So we think there is 11 NH4+ since 11OH- reacted. (But again, in actual fact 10 reacted and 1 was excess).

[kenhung123]

can someone confirm whether this is right or wrong?

[shinny]

It's right. What do you mean we have the same amount of ammonium ions, but less NaOH? It's the other way around. Since some fertiliser, and hence ammonium, is sitting in the container still, we're clearly going to underestimate the amount of ammonium in the fertiliser. On the other hand, the amount of NaOH we pour in is still the same; 100 mL of 0.108 M.

[kenhung123]

Oh damn, I thought the error occured in the NaOH preparation...sorry about that

[m@tty]

In practice exams I have made far too many errors from misinterpreting a question... What I do to try and expunge this petty error is to re-read every question after I have done it.

[chansthename]

^^ although that problem is when you have a longer test you end up with a small margin of time being the difference between finishing and not (2009 VCAA for example)

[m@tty]

Just a quick 5 second re-read of both the question and your response. Shouldn't impose on the completion of the rest of the exam.

Just lost a mark in VCAA 09 for this same kind of thing; addressed only one part of a two part question

[shinny]

It's faster to take longer in reading the question and getting it right the first time around than having to rub/cross it all out and repeat it. But yeh, this is why I'd spend my time reading short answer rather than doing MCQs.