Chemistry Examination Discussion

[Jeggz]

Oh cool, you've made my night! (but our crappy calculators get 80.1 as an answer lol)

I got 80.1 too

[thushan]

OOPS. Misread calc - I meant 80.1%.

[thepikanation]

OOPS. Misread calc - I meant 80.1%.

Night is officially MADE!

[andyse7en17]

For the CH3COOH and NaOH question. The part with "using exactly half the volume of NaOH needed", what did everyone tick?
I consider the first ([CH3COOH]<[CH3COO-]) and second (equal concentrations) choices... If what VCAA asks for is literally "BEST DESCRIBE" then the first box should be the answer and the last part becomes [H3O+]<Ka, not [H3O+]=Ka.

My only possible reasoning is that at half the volume needed, there're still half the original mole of [CH3COOH] in the solution, this amount will still undergo ionisation in water and thus a very small amount of [CH3COO-] is produced while a very small amount of [CH3COOH] is decreased, thus the first choice is the "BEST" answer... Idk why they put the equilibrium question in between, which confused me @_@ But yeah, I chose the first box, then wrote the explanation below... Do you guys think I'll still get penalized if my explanation is correct?

[joey7]

For the CH3COOH and NaOH question. The part with "using exactly half the volume of NaOH needed", what did everyone tick?
I consider the first ([CH3COOH]<[CH3COO-]) and second (equal concentrations) choices... If what VCAA asks for is literally "BEST DESCRIBE" then the first box should be the answer and the last part becomes [H3O+]<Ka, not [H3O+]=Ka.

My only possible reasoning is that at half the volume needed, there're still half the original mole of [CH3COOH] in the solution, this amount will still undergo ionisation in water and thus a very small amount of [CH3COO-] is produced while a very small amount of [CH3COOH] is decreased, thus the first choice is the "BEST" answer... Idk why they put the equilibrium question in between, which confused me @_@ But yeah, I chose the first box, then wrote the explanation below... Do you guys think I'll still get penalized if my explanation is correct?

I did the first box then in the second part I wrote they are approximately the same

[Stick]

It's 80.1% for the copper purity sample.

I'm writing up solutions here - this is a tricky question, but my reasoning is here:

"This question is actually fairly tricky. What I gather is that the mass of copper in the impure electrode is more accurately reflected by the amount of copper that has been deposited on the cathode. The reason for this is because the copper deposited on the cathode is a result of the amount of electrons pulled away at the anode by the current. Metals that are more reactive than copper would all dissolve in the H2SO4, and that would account for some of the mass loss in the impure sample. Metals that are less reactive than copper will be, as stated in the following part, not oxidised and will form a sludge."

If you check the electrochemical series, copper is the most reactive element that does not react with H+.

@a-rod: If there is copper left in the impure sample after electrolysis, then determining the percentage purity of copper in the original impure sample would be inaccurate.

Another question incorrect. >_<

[Sentar]

For the CH3COOH and NaOH question. The part with "using exactly half the volume of NaOH needed", what did everyone tick?
I consider the first ([CH3COOH]<[CH3COO-]) and second (equal concentrations) choices... If what VCAA asks for is literally "BEST DESCRIBE" then the first box should be the answer and the last part becomes [H3O+]<Ka, not [H3O+]=Ka.

My only possible reasoning is that at half the volume needed, there're still half the original mole of [CH3COOH] in the solution, this amount will still undergo ionisation in water and thus a very small amount of [CH3COO-] is produced while a very small amount of [CH3COOH] is decreased, thus the first choice is the "BEST" answer... Idk why they put the equilibrium question in between, which confused me @_@ But yeah, I chose the first box, then wrote the explanation below... Do you guys think I'll still get penalized if my explanation is correct?

I ticked the first box for the same reason.
For the Relationship I said (Ka/10)=[H30+]^2 cause I didn't understand the question, reckon I can get a mark? because I did give the actual relationship

[andyse7en17]

I did the first box then in the second part I wrote they are approximately the same

I actually did the second box, then first, then second, then first again... After all the trials I decided to show the whole working underneath to show that it is less than lol. I hope VCAA really did put that question to be the trickiest...

[lzxnl]

For the CH3COOH and NaOH question. The part with "using exactly half the volume of NaOH needed", what did everyone tick?
I consider the first ([CH3COOH]<[CH3COO-]) and second (equal concentrations) choices... If what VCAA asks for is literally "BEST DESCRIBE" then the first box should be the answer and the last part becomes [H3O+]<Ka, not [H3O+]=Ka.

My only possible reasoning is that at half the volume needed, there're still half the original mole of [CH3COOH] in the solution, this amount will still undergo ionisation in water and thus a very small amount of [CH3COO-] is produced while a very small amount of [CH3COOH] is decreased, thus the first choice is the "BEST" answer... Idk why they put the equilibrium question in between, which confused me @_@ But yeah, I chose the first box, then wrote the explanation below... Do you guys think I'll still get penalized if my explanation is correct?

Let's make an analogy.
Pretend we have 1 mol of HA, a weak acid
And we add 0.5 mol of NaOH. Essentially, you can then let all of the OH- react, leaving 0.5 mol HA and 0.5 mol A-. As these are weak acids and bases, they won't ionise much further, so you'll be left with that. I think you're forgetting that adding base to HA forms A-.

I ticked the first box for the same reason.
For the Relationship I said (Ka/10)=[H30+]^2 cause I didn't understand the question, reckon I can get a mark? because I did give the actual relationship

Ka=[H+] though?

[Emily C]

I got so confused with that ethanoic acid question , i thought of it as being a weak acid so there would be more CH3COOH than the ethanoate ions - so i ticked box 3 but i think im wrong in saying that, confusing question

[thushan]

The answer was meant to be that the concentrations of the substances are equal, and that Ka = [H+].

I didn't like this question though - I think the analysis for this is too complex for Year 12 level, especially for those who try to think deeply through this.

So, if we took reaction:

CH3COOH --> CH3COO- + H+

and we added exactly half the amount of OH- to neutralise the CH3COOH, we are effectively removing reactant and adding product, and their concentrations will be equal at this point. And you are right by saying that by Le Chatelier's Principle, some of the CH3COO- will react with the H+ to form CH3COOH subsequently. So technically, [CH3COOH] will be ever so slightly higher than [CH3COO-] and hence Ka > [H+].

However, the idea was that so little of the CH3COO- will have reacted since its concentration is so much higher than that of H+, so it doesn't take measurable amounts of it to react with the H+. Hence, we can say that [CH3COOH] = [CH3COO-] and Ka = [H+].

If I was the chief examiner (which I am most certainly not ) I would pay both answers because I think it's unfair for students to be expected to perform that kind of reasoning.

[Stick]

FML I ticked the first box thinking that water would drive the ionisation forwards a bit more. I think I underestimated this exam. XD

[andyse7en17]

Let's make an analogy.
Pretend we have 1 mol of HA, a weak acid
And we add 0.5 mol of NaOH. Essentially, you can then let all of the OH- react, leaving 0.5 mol HA and 0.5 mol A-. As these are weak acids and bases, they won't ionise much further, so you'll be left with that. I think you're forgetting that adding base to HA forms A-.

Ka=[H+] though?

Nope. I did put the extra 0.5 mol A- in my working for the equilibrium and it showed that the ionisation process still occurs to produce a little bit more CH3COO-, not much though. You can try the calculation, someone already posted the paper I think.

If we think it this way I must be correct. Say Putting more NaOH in will give us 1 mol CH3COOH and 1 mol CH3COO- . Now they're in the same solution, same V so the concentrations should be the same initially... Assuming 1M each. Initially Q=(1 x 10^-7)/1 < Ka (1.7 x 10^-5), so the equilibrium will shift to the right, which will indeed produce more CH3COO- and thus the first choice.

[thushan]

andyse7en17 - check my previous post.

[Sentar]

Ka=[H+] though?

Ah, made the assumption that [CH3COO-]=[H3O+], not in this case...