VCE Chemistry Question Thread

[lzxnl]

I would be saying ethanol, as the C=O is normally a sharp intense peak @ 1700.  Also the acid is a much broader OH peak which over laps some of the C-H peaks, this OH shows clear definition between the OH and CH. 

If you google them have a look at the C=O peaks and become familiar with their shapes.

Interesting. I did a quick google search of ethanol IR and I managed to find the IR given, so it does seem to be ethanol. However, other infra red spectra seem to differ in that they don't have the peak at 1700 cm-1. It's likely to cause confusion in VCE chemistry because students are taught to think carbonyl when they see that peak.

It still doesn't explain the origin of that peak.

Also, in practice, you commonly get a peak at ~3200 cm-1 solely because when you're making the infra red sample, you use potassium bromide which unfortunately attracts water from the atmosphere quite well so you end up with an OH absorption that means nothing.

[Brunette15]

Can someone please help me understand 3e to this question, I don't really like the vcaa explanation...

[Yacoubb]

Is the answer acid I has a greater pH change than acid IV?

[Professor_Oak]

Are we meant to round off our answers with each part of a question? I did the attached question on VCAA 2009 and ended up being 0.1 off on the second part because I didn't round my previous answer in the calculation.

[Robert123]

Can someone please help me understand 3e to this question, I don't really like the vcaa explanation...

Ok, we have two acids, one is strong  (I), the other is weak (IV). Now what is going to happen when we dilute the solution? The first one is going to act logically and has a pH increase of 1, however, the second one is going to act more interestingly since it is a weak acid.
The acid equilibrium constant for the weak acid is going to be equal to [h+][A-]/[HA]
Now when the solution is diluted by a factor or 10, there is going to be a net forward reaction so that the concentration of H+ will partially increase. Therefore, it will have less change in pH
So really, this question was a 'hidden' le chatelier's principle.
Does that clarify it for you?

[Brunette15]

Ok, we have two acids, one is strong  (I), the other is weak (IV). Now what is going to happen when we dilute the solution? The first one is going to act logically and has a pH increase of 1, however, the second one is going to act more interestingly since it is a weak acid.
The acid equilibrium constant for the weak acid is going to be equal to [h+][A-]/[HA]
Now when the solution is diluted by a factor or 10, there is going to be a net forward reaction so that the concentration of H+ will partially increase. Therefore, it will have less change in pH
So really, this question was a 'hidden' le chatelier's principle.
Does that clarify it for you?

Yeah that makes much more sense...thankyou!

[Valyria]

Hey,

For the solutions, shouldn't their very first line be "In NaF(aq) the oxidant..."? If not, why?

[Robert123]

Hey,

For the solutions, shouldn't their very first line be "In NaF(aq) the oxidant..."? If not, why?

They are talking about molten NaF, not NaF dissolve in water. The reason molten is used is because if you look at the chemical series, Na+ is below H20, so when electrolysis occur in an aqueous solution, the water would be reduced, not Na+. Therefore, they use liquid NaF (this is done at a very hot temperature) so that Na would react.
Does that clarify it for you?

[Valyria]

They are talking about molten NaF, not NaF dissolve in water. The reason molten is used is because if you look at the chemical series, Na+ is below H20, so when electrolysis occur in an aqueous solution, the water would be reduced, not Na+. Therefore, they use liquid NaF (this is done at a very hot temperature) so that Na would react.
Does that clarify it for you?

The ECS doesn't have Na(l) or F(l), so in the solution with molten NaF, shouldn't water be oxidised and reduced whilst in the aqueous solution, F^-1 gets oxidsed and Na^+1 gets reduced?

[lzxnl]

The ECS doesn't have Na(l) or F(l), so in the solution with molten NaF, shouldn't water be oxidised and reduced whilst in the aqueous solution, F^-1 gets oxidsed and Na^+1 gets reduced?

Molten NaF isn't a solution. It's pure sodium and fluoride ions.

Also, just because something isn't listed in your ECS doesn't mean it doesn't react. There are stronger reductants than lithium metal out there; they're just not mentioned because you don't need to know about them.

I think you mean in NaF (aq), water would be reduced and oxidised.

[Robert123]

The ECS doesn't have Na(l) or F(l), so in the solution with molten NaF, shouldn't water be oxidised and reduced whilst in the aqueous solution, F^-1 gets oxidsed and Na^+1 gets reduced?

Molten NaF does not contain water, that is the point of using molten. The ECS technically does not apply for this reaction since it won't be in standard conditions, however, they are not Na(l) and F(l) form due to still being in ionic form.
For the aqueous solution, the strongest oxidant gets reduced and the strongest reductant get oxidised. Since water is both a strongest oxidant and reductant than both Na+ and F-, it will react with itself.
When predicting  redox reaction, it is always best to start circling what you have in your solution. In this case, we have H20, Na+ and F-. Since H20 is the highest reactant on the left side ( at -0.83V) and the lowest in the right side (+1.23V), it would undergo a redox reaction by itself.

You should have covered similar stuff in class using molten NaCl(l) which behaves in the same way.
Does any of that help any more?

[lucypevensie]

I have a question about sig figs

When doing pH questions, does giving your answer as 2.13 mean 2 sig figs? I remember one of the vcaa exams mentioned sig figs for pH only includes the decimal places. Can someone confirm if this is correct? We've never being taught this at school so I'm not too sure what's the correct way.

Also, in the 2012 Unit 4 exam, Question 9) b. i. the answer is given to 2 sig figs, but all the data is given to 3. Is it only 2 because when you convert minutes to seconds you multiply by 60?

[AngelWings]

Does anyone have ¾ trials that they can share pls

I can share one, if you could send a message, stating your e-mail and roughly which company/ year you'd like. I'll try to get what I can for you. I owe you anyway.

I have a question about sig figs

When doing pH questions, does giving your answer as 2.13 mean 2 sig figs? I remember one of the vcaa exams mentioned sig figs for pH only includes the decimal places. Can someone confirm if this is correct? We've never being taught this at school so I'm not too sure what's the correct way.

For pH, I'm pretty sure it's after the decimal place, so: 2.13 = 2 sig. figs. Someone please confirm that I've not been fed the wrong information all year, please?

[psyxwar]

clarification re; homogenous equilibrium with all compounds being liquids:

Can we still write an equilibrium expression for this reaction?

Apparently for aA (l) + bB (l) <-> cC(l) + dD (l), K=[C]^c[D]^d/[A]^a[B ]^b as usual. I think this is what the chief assessor said in a lecture (might be misremembering), but I can't find any evidence to support this. The activity of pure liquids and solids should be 1 right, so how you do even write an equilibrium expression for this?

[lzxnl]

clarification re; homogenous equilibrium with all compounds being liquids:

Can we still write an equilibrium expression for this reaction?

Apparently for aA (l) + bB (l) <-> cC(l) + dD (l), K=[C]^c[D]^d/[A]^a[B ]^b as usual. I think this is what the chief assessor said in a lecture (might be misremembering), but I can't find any evidence to support this. The activity of pure liquids and solids should be 1 right, so how you do even write an equilibrium expression for this?

To put it simply, you can't. The rate of reaction of these liquids isn't dependent on how much liquid you have; rather, it's dependent on the area of contact between the liquids. The fact that you've written individual liquid species suggests these liquids do not dissolve in each other and so you have several different phases. Reactions only occur on the boundary between two phases and as such, an equilibrium isn't well defined.

In solution, however, ALL of the solutes are capable of reacting. This isn't the case with liquids and solids.