Suggested Solutions - DONE

[Synesthetic]

Actually I thought it was electrolytic initially but I was second guessing myself there.

b) cell operates at 24.0 hours at 25.5 A
i) [1 mark]
Q = 2.20 x 10^6 A

ii) [3 marks]
12.2 g of methanol

2.20 x 10^6 C

Hmm... I got 122 g for m(CH3OH)

[lwine]

To Mao:

Question 6 b) ii.

I got 122g, and so did Synesthetic.

Operating a fuel cell for 24 hours at 35.5A with 12.2 grams of methanol feels just a little unrealistic, but I can't remember the specifics of the question.

I was hoping you could check that.

Thanks.

[We Want Peace.]

I think i have several different ans as Mao.
some MC .

[Mao]

To Mao:

Question 6 b) ii.

I got 122g, and so did Synesthetic.

Operating a fuel cell for 24 hours at 35.5A with 12.2 grams of methanol feels just a little unrealistic, but I can't remember the specifics of the question.

I was hoping you could check that.

Thanks.

arghhhh i hate not being able to read my own scribbles. [very messy indeed]
fix'd

[We Want Peace.]

To Mao:

Question 6 b) ii.

I got 122g, and so did Synesthetic.

Operating a fuel cell for 24 hours at 35.5A with 12.2 grams of methanol feels just a little unrealistic, but I can't remember the specifics of the question.

I was hoping you could check that.

Thanks.

me too.

[riadnicolas]

6c...
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.

so you'd have a net gain of CO2 actually.

My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.

wasnt it just about methanol production in fuel cell

Oh - yeah, it WAS a fuel cell wasn't it...

i thought it was getting rid of CO2

CO2 was on the reactants side of the equation producing methanol

wat did the question say exactly?

what's this about fuel cell...? i don't see it on that page at all

just "electrolytic cell"

my bad, and also for the methanol question i f we said it reacted with oxygen and therefore lost some mass is that correct

[We Want Peace.]

Why mc q1 is A?.
why increase the pressure??. ( 2 particles on left. 4 particles on right?)

[Synesthetic]

it was about volume not pressure

[We Want Peace.]

oh my god..my bad..

[Synesthetic]

f)
i) [2 marks]
Concentration of methanoic acid solution that has a pH the same as IV (2.1)
using Ka found from data booklet, we find the equilibrium concentration of methanoic acid to be 0.350M
however, if we add the concentration of H+ at equilibrium, we will have 0.358M for initial concentration
so it's either 0.35M or 0.36M.
Which one VCAA accept/want will be a completely different matter. It's only 2 marks.

Right...so they wanted a value from the data booklet for 2 marks??

I calculated [H+] from pH of Acid IV, and then let that equal [HCOOH] = 7.9x10^-3 M ...since it is monoprotic
In supposition this is completely off, but in the exam I could not imagine the question to be asking us to look up that figure in the data booklet!

*edit* Well, now I see that you added [H+] to that value, took me a while

---
While I'm at it I'd appreciate elucidation on this:

e) [2 marks]
diluting I and IV by a factor of 10. We note that IV is a weak acid (incomplete ionisation), hence dilution will increase the percentage ionisation. Also note that I is a strong acid, hence dilution will not increase the number of moles of H+. Hence the change in pH will be greater for I than IV.

So, we needed to observe Ka [equilibrium] principles? I thought dilution would affect each acid in the same way, adding 1 to the pH value of both.

*edit* Is it incorrect to assume pH=2.1 => strong acid? Although, I am aware that acidity strength relates to Ka

It almost seems...unreasonable to me that they should expect us to take that into consideration. I thought they were testing us on the formula for interchanging pH / concentration, hence in my solution:

e) Same change in pH, +1 (tick the third box)
Since dilution by a factor of 10 <=> divide [H+] by 10,
new pH = -log10{[H+]/10} = old pH - log10(.1) = old pH 1 + 1 => same change in pH for both acids (+1)

[205ism]

I'm such a spastic, 67 or 68 /79

There goes 40, damn

[username]

Oh I'm so pleased. Today's a good day. ^_^

[cosmicgate]

I'm such a spastic, 67 or 68 /79

There goes 40, damn

wat u mean? last year u could drop 11 marks and get an A+
u should have a good chance at 40+ if u had A+ mid year, A+ sacs and HIGH A or A+ end of year... otherwise no idea.

[Synesthetic]

*referring to my post above*

It might just be me but the ambiguity of these acid equilibria questions is terrible...I'd be surprised if the majority of students were able to get most of the marks available.

It's these types of questions that make me really frustrated with chem...

[Mao]

This is not in the course, but fuck, the course is over, so I'll just do it anyways:

for acid IV,

hence, when it is diluted to 0.01M, assuming the H+ from self ionisation from water is insignificant



solving this quadratic, , pH=2.64

so change in pH for IV is about 0.5, whereas it is 1.0 for I.