Chemistry Examination Discussion

[Stick]

I think everyone saying that this exam was straightforward has underestimated it a bit - certainly not impossible, but I think there's enough challenging material in there. Never underestimate the amount of detail they require for written answers, too.

It was a lot like the Biology exam, in that sense.

[Michaeljohns]

I got Ka = ([H3O+]^2)/2; where [H3O+]=[CH3COO-].
Worked out when I calculated it, but even when I wrote it down I felt that the answer seemed a bit dodgy.
We'll see when the official answers come out.

In my opinion, time was not a factor for this exam, but rather attention to detail was the key to doing well.

[DefyingGravity]

For the question with how many days it took to plate the copper or whatever,
if you got 12.odd (whatever it was LOL), should you round up or down to the nearest 2 sig figs. i.e. should it be 12 or 13? I mean, it was closer to 12 but it takes longer than 12...

[Stick]

Why wasn't the first box correct? Couldn't the excess water drive the ionisation of ethanoic acid, meaning its concentration is slightly lower than the ethanoate ion? I just want to make sure I learn from my mistakes. :S

[joey7]

Why wasn't the first box correct? Couldn't the excess water drive the ionisation of ethanoic acid, meaning its concentration is slightly lower than the ethanoate ion? I just want to make sure I learn from my mistakes. :S

I think it's because the solution was already in equilibrium when the NAOH was added hence the removal of a product drives equilibrium backwards, thats just my understanding Thushan is the man with the answers

[Stick]

Urgh... applied Le Chatelier's principle to the wrong equilibrium system then. +_+

[joey7]

Urgh... applied Le Chatelier's principle to the wrong equilibrium system then. +_+

Yeah its a killer hey, I did exactly the same thing

[andyse7en17]

Why wasn't the first box correct? Couldn't the excess water drive the ionisation of ethanoic acid, meaning its concentration is slightly lower than the ethanoate ion? I just want to make sure I learn from my mistakes. :S

Most likely not... I don't think VCAA will be that kind.
It will take a whole page to explain the first choice as we discussed
I'm confused now. As we add NaOH, the OH- can react with H+ and drives the equilibrium to the right, but the equation given was between OH- and CH3COOH, which is the opposite.
But yeah, if it takes so much effort to explain just for 1-2 marks, I don't think the first choice is correct anymore...

[presto]

Hey presto (hehe nice pun) - "if it's below 10^-2, then there are generally more reactants than products" - that's not something to be religiously followed, you could have relatively more product if the concentration if one of your products is excruciatingly small compared to that of the other products and reactants.

Ok, I learnt something there  . So was my reasoning just a lucky guess? Because [CH3COOH]=0.2M is quite high relative to the order of concentration for the CH3COO- right?

[Stick]

The thing was that I completely ignored the given equilibrium system since the NaOH was no longer present... Assumed that the concentrations of CH3COOH and CH3COO- would initially be equal, but the water would drive the reaction CH3COOH(aq) + H2O(l) <---> CH3COO-(aq) + H3O+(aq) forward. I'm pretty sure I'm wrong and my reasoning is flawed - I just want to know how. Thanks.

[Mafioso]

So were you supposed to tick the last box?

[simba]

Just wait, for the percentage purity of copper I think I get 87%
I thought we were meant to subtract the final amount of impure copper by the initial? How is that wrong? :|

[Cazalinko]

Just wait, for the percentage purity of copper I think I get 87%
I thought we were meant to subtract the final amount of impure copper by the initial? How is that wrong? :|

I'm pretty sure I did that as well, I'm not sure why you can't do that. It did seem a little simple for 3 marks though now that i think about it....

[12AM]

Just wait, for the percentage purity of copper I think I get 87%
I thought we were meant to subtract the final amount of impure copper by the initial? How is that wrong? :|

I did the exact same thing. :/

[simba]

I'm pretty sure I did that as well, I'm not sure why you can't do that. It did seem a little simple for 3 marks though now that i think about it....

yeah, but I just worked out on my calc to get the 80.1% purity you used the initial mass of the impure copper rather than subtracting the 0.855kg

but I don't understand why you don't subtract the 0.855kg. An explanation would be super!