VCE Chemistry Question Thread

[lzxnl]

Neutralise = n(acid) = n(base)
As both bases have the same volumes, concentration considerations are the same as mole considerations

As for base strength, you'd need to show us the actual graphs as you haven't given all the necessary information yet about them.

[Eiffel]

with questions that ask like to calc the conc of acid 1 that reacted in the 1L flask after the student added NaOH.

Is the way these questions work is that we get the moles of NaOH and the mols of HCL that reacted the use the initial number of mols - reacted = left over (which is in the flask).

Once HCl has reacted (in any context - or like this one) is it right to say it is not HCl anymore? and now is like water / carbon dioxde etc (but not HCl right?))

Neutralise = n(acid) = n(base)
As both bases have the same volumes, concentration considerations are the same as mole considerations

As for base strength, you'd need to show us the actual graphs as you haven't given all the necessary information yet about them.

i under stand that to neutralise it is all about moles, but if something is more concentrated why doesnt the pH increase (for base) and decrease (for acid).

i cant show a graph since it is those checkpoint improvised questions and not an actual VCAA one.

Image attached, need help with a,b
here https://drive.google.com/file/d/0B4jDYm8MQMIFcW5Ud21HNm1XUFU/view?usp=sharing

[Champ101]

Hi all!

Could someone please explain why the lower the concentration is in a conical flask the higher the titre volume must be in the burette and vice versa.

Thanks in advance!

[Redoxify]

Hi all!

Could someone please explain why the lower the concentration is in a conical flask the higher the titre volume must be in the burette and vice versa.

Thanks in advance!

could you repeat the question, if it's in a titration the number of mole is usually the same for both the burette and the concial flask

[Redoxify]

Hi all!

Could someone please explain why the lower the concentration is in a conical flask the higher the titre volume must be in the burette and vice versa.

Thanks in advance!

could you repeat the question, if it's in a titration the number of mole is usually the same for both the burette and the concial flask

[Champ101]

could you repeat the question, if it's in a titration the number of mole is usually the same for both the burette and the concial flask

Not necessarily, it depends on the mole ratio. I'm talking about titration for volumetric analysis

[keltingmeith]

Not necessarily, it depends on the mole ratio. I'm talking about titration for volumetric analysis

"titration for volumetric analysis"
Erm... They're two different terms for the exact same thing. You've basically said "titration for titration".

Anyway, for a titration, we have two moles - , for the acid, and , for the base. They can be calculated like so:



Now, once we hit the equivalence point, these moles are equal, so we can simplify to:



So, now we get to your question.
Let's say we lower the concentration of the acid - this means, we need one of two things:

a) The numerator to also lower, since that'll give us a smaller number.
b) The denominator to raise, since that will also give us a smaller number.

To lower the numerator, we either use a less concentrated base, or we use less base in the burette. (this assumes the acid is in the conical flask, and the base is in the burette) Similarly, we could also put more acid in the conical flask.

Does that explain it at all?

[cosine]

Can someone please explain the process of titration to me, in simple terms. I kind of understand it but I think if I read it in another perspective it'll click in. Thanks

Oh and what's the purpose of it too, and is it effective in terms of expense, time consuming and accuracy?

[RazzMeTazz]

What would be the reduction and oxidation half equations for this overall equation:

2HCHO(aq) + O2(g) --> 2HCOOH(aq)

Thanks

[lzxnl]

What would be the reduction and oxidation half equations for this overall equation:

2HCHO(aq) + O2(g) --> 2HCOOH(aq)

Thanks

You first need to attempt to balance HCHO => HCOOH
This gets you HCHO + H2O => 2H+ + 2e- + HCOOH

Now reduce oxygen to water
You get O2 + 4H+ + 4e- => 2H2O
Double the first equation and add the second
You'll find that the protons and waters cancel as needed

[RazzMeTazz]

Thankyou so much lzxnl!
Is there any explanation behind knowing why O2 needs to be reduced into H2O?

[RazzMeTazz]

Would it be correct to say that when combining oxidation and reduction half equations to give the full ionic equation, you cant always cancel out the H+ ions and H2O from either sides of the equation, as in some cases it makes the equation unbalanced?

This has happened to me a few times and Im not sure if Im doing something wrong!

Thanks

[lzxnl]

Thankyou so much lzxnl!
Is there any explanation behind knowing why O2 needs to be reduced into H2O?

Would it be correct to say that when combining oxidation and reduction half equations to give the full ionic equation, you cant always cancel out the H+ ions and H2O from either sides of the equation, as in some cases it makes the equation unbalanced?

This has happened to me a few times and Im not sure if Im doing something wrong!

Thanks

Scale each half reaction to cancel the electrons only.

As for oxygen reducing to form water, if there are any protons available in water, the reduction potential of the reduction of oxygen going to water is the highest out of what oxygen can do. It's most stable that way.

[scarletmoon]

I have an extended experimental investigation SAC that combines a volumetric, gravimetric and colorimetry prac together. What kind of questions would be on this SAC? Because all the calculations for each of the pracs has already been completed so what would they ask on the SAC? 

[The Living Legend... almost]

I have an extended experimental investigation SAC that combines a volumetric, gravimetric and colorimetry prac together. What kind of questions would be on this SAC? Because all the calculations for each of the pracs has already been completed so what would they ask on the SAC? 

possible errors in the prac itself , if the aim was met, how you would do the prac again to avoid errors

these are some possible q's but ask your teacher if he/she can give you any clues to the questions which could be asked.