stonecold's chem questions :)

[stonecold]

erm, i'd take a punt and say just like the normal salt bridge the H+ goes to the anode and vice versa for SO42-... but its weird since both half equations need SO4..

STAV says SO42- to the positive electrode, but this confused me, because I was under the impression that cations always flow to the cathode and anions flow to the anode.

Uuugh, so many damn holes in my learning.

I just think of it this way: the SO42- is negatively charged, and so will be attracted to the positively charged electrode (cathode) and the H+, being positive, will be attracted to the negatively charged anode.

The reason that it does not work like a standard galvanic cell, I think, is because there is no real loss/gain of charge around each electrode (like the production of a metal ion).

I think. 

Gotta hate it when the rule you would bet your house on fails you.

That is how I used to remember it, the positive to negative and negatve to positive thing.  Then I learnt the other definition and went along with that. 

I'll go back to the old one now. 

Sorry for the dumb buffers question, but I don't even remember learning about these other than being told weak acids and maintain stable PH.

[Greggler]

yeah i stuffed this question up to because i always thought positive to positive etc.

[stonecold]

yeah i stuffed this question up to because i always thought positive to positive etc.

So how do you remember it now...

What fady said?  If no ions are produced, then opposite charges attract?

[Greggler]

You just have to look at the specific electrode. I forget but for that one even though it was a positive electrode, i think it was producing positive ions or something.

My teacher did say that it was a rare case/dodgy and unlikely question; but still she basically said that you just have to think logically, see what ions are produced at the specific electrode and see what environment it is in. If it is getting more positive, then give it some negative ions to negate this effect.

I cant remember exactly because i dont have the question in front of me, but im pretty sure that this is along the lines of what she said. ... maybe was it in a solution of positive ions or something, or surrounded by positive ions...

[kenhung123]

Which question is this?

[stonecold]

You just have to look at the specific electrode. I forget but for that one even though it was a positive electrode, i think it was producing positive ions or something.

My teacher did say that it was a rare case/dodgy and unlikely question; but still she basically said that you just have to think logically, see what ions are produced at the specific electrode and see what environment it is in. If it is getting more positive, then give it some negative ions to negate this effect.

I cant remember exactly because i dont have the question in front of me, but im pretty sure that this is along the lines of what she said. ... maybe was it in a solution of positive ions or something, or surrounded by positive ions...

Are people sure stav is right?

I am looking at this question again, and the reduction half equation isn't producing any +ive ions... :S

[kenhung123]

what question is this

[stonecold]

what question is this

STAV 09 Q6c.

[stonecold]

And furthermore, seeing as H+ is a reactant for the reduction reaction, doesn't that further indicate that H+ would migrate towards the +ive electrode/cathode?

[kenhung123]

Yea I would have thought both?

[stonecold]

Yeah, I am pretty sure they screwed it.  The circle of negative charge is the best way to go about this questions I reckon.

Can someone clarify on STAV 2010 Q4d.

So it is possible to have HX <---> H+ X-  where

[H+] =/= [X-]    ?

I always thought that they had to be equal...

[kyzoo]

Yeah, I am pretty sure they screwed it.  The circle of negative charge is the best way to go about this questions I reckon.

Can someone clarify on STAV 2010 Q4d.

So it is possible to have HX <---> H+ X-  where

[H+] =/= [X-]    ?

I always thought that they had to be equal...

That's only when you start with only HX(aq), if you have X- or H+ at the start as well then [H+] does not neccesarily equal [X-]

[m@tty]

^That's also assuming that no further reactions take place in solution.

[Mao]

Yeah, I am pretty sure they screwed it.  The circle of negative charge is the best way to go about this questions I reckon.

Can someone clarify on STAV 2010 Q4d.

So it is possible to have HX <---> H+ X-  where

[H+] =/= [X-]    ?

I always thought that they had to be equal...

Try this to test your understanding:

Starting with a 0.1M vinegar solution, allow this to reach equilibrium.
We then add some sodium ethanoate to this solution. The pH increases to 5 when equilibrium is re-established. Find:

i) the concentration of ethanoate at equilibrium
ii) the mass of sodium ethanoate added if we have 500mL of vinegar to start with.


Solution:
Mechanism:
Step 1: At the first equilibrium, [H+]=[CH3COO-]=0.0013 M
Step 2: Addition of [CH3COO-] drives the first equilibrium almost completely to the left ([H+]=0.00001M when equilibrium is re-established), thus almost all [H+] and [CH3COO-] from the initial ionisation of CH3COOH will convert back to CH3COOH in this back reaction.
Step 3: At this new equilibrium, essentially all the CH3COOH is converted back, essentially all CH3COO- comes from the addition of CH3COONa, and [H+] is very small.

Calculations:
i) Using Ka from data booklet, [CH3COO-] = K_a [CH3COOH]/[H+] = 0.17M
ii) m(CH3COONa) = Mr(CH3COONa) * 0.17 * 0.5 = blahh

[stonecold]

Yeah, I am pretty sure they screwed it.  The circle of negative charge is the best way to go about this questions I reckon.

Can someone clarify on STAV 2010 Q4d.

So it is possible to have HX <---> H+ X-  where

[H+] =/= [X-]    ?

I always thought that they had to be equal...

Try this to test your understanding:

Starting with a 0.1M vinegar solution, allow this to reach equilibrium.
We then add some sodium ethanoate to this solution. The pH increases to 5 when equilibrium is re-established. Find:

i) the concentration of ethanoate at equilibrium
ii) the mass of sodium ethanoate added if we have 500mL of vinegar to start with.


Solution:
Mechanism:
Step 1: At the first equilibrium, [H+]=[CH3COO-]=0.0013 M
Step 2: Addition of [CH3COO-] drives the first equilibrium almost completely to the left ([H+]=0.00001M when equilibrium is re-established), thus almost all [H+] and [CH3COO-] from the initial ionisation of CH3COOH will convert back to CH3COOH in this back reaction.
Step 3: At this new equilibrium, essentially all the CH3COOH is converted back, essentially all CH3COO- comes from the addition of CH3COONa, and [H+] is very small.

Calculations:
i) Using Ka from data booklet, [CH3COO-] = K_a [CH3COOH]/[H+] = 0.17M
ii) m(CH3COONa) = Mr(CH3COONa) * 0.17 * 0.5 = blahh

Thanks Mao.

I think I got it.

I got the answer to part a right, and then for part b I got 6.91g, your way says 6.97g.

Is that good enough, or have i made a mistake.

I subtracted the initial mol n(CH3COO-) from the 0.17M thing after converting it to mol.

I think this is why I got a different answer.