OCR Chemistry A - Equilibria, Energetics and Elements (F325).

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Reply 600

TwirlGirl

I can't wait til this exam is over! I havent even looked at C4 or Biology Unit 5 yet... :sad:

Reply 601

Pedus

steph_v

We have spoken before. You sent me some good revision packs :smile:

Oh yes, I remember youu :yep:

Bahh to do some Buffers and Transitions revision now :emo:

Reply 602

TwirlGirl

Pedus

Oh yes, I remember youu :yep:

Bahh to do some Buffers and Transitions revision now :emo:

How do you revise at this time? :eek:

I'm so tired, i wouldn't be able to concentrate :yawn:

Reply 603

student92

andy892

It really is simple.

You, of course, don't know need to know the Ka if working it out for a strong acid.
You simply take the acid's concentration and multiply it by 2, then find the negative log of that.
I.e. 0.2 mol H2SO4 would be negative log of 0.4.

For a weak acid it's basically the same as you would normally, but you do it for each [H+], then add the 2 together. Then take the negative log of that. Again, simple.
For example, if it has a concentration of 0.1 mol, and the Ka for the first [H+] is 7 x 10-4 and the Ka for the second [H+] is 6x10-10, then you just use the formula for working out [H+] for each of the Ka values, add together and take negative log.

the formula for weak acid being: Ka is H+ squared / conc of reactants... then rearrange to get conc of H+? and the 2 eqns for H2SO4 would be H2SO4 >> H+ + HSO4- anddd HSO4- >> H+ + SO42- so for the first equation wud u use strong acid formula - ka is conc of products / conc of reactants and for the 2nd equation use the weak acid formula???

Reply 604

Pedus

TwirlGirl

How do you revise at this time? :eek:

I'm so tired, i wouldn't be able to concentrate :yawn:

I've got my page open for buffers but haven't even started reading it yet :teehee:

andy892

student92

I am told that the first proton dissociation is a simple calculation, as it's the same as the conc of the acid.

To work out the concentration of the second one, you use this (because this is a weak acid):
[H+] = SQRT (Ka x [HA-]). The conc of the HA- should be the same as the initial value too.

Then you add the two [H+] values together.
THen do -log(Ans)

And it should all work out fine.

But as someone said earlier, I wouldn't worry too much about this! Not technically on spec.

Reply 607

Inter-viewer

cyborg

For the rate equation, the value of k increases as the temperature increases - but how come it does not depend on the change in the equilibrium position when the temp increases/decreases, as this will change the concentrations of the reactants :confused:

Cyborg, i used to get confused regarding this also, but there's a really easy way of remembering it!

Think logically!

When you increase the temperature, according to Le chets principle the system tries to decrease it (and vice versa)

So if the forward reaction is endo
And the temperature is increased
System tries to decrease it, therefore forward endo reaction is favoured
Because forward reaction is favoured, theres increased conc of Products
So equilibruim lies on the RHS
When equilibrium lies on the RHS, Kc> 1
So kc is increased!

Now let's apply it when forward reaction is exo
Increase in temp
System tries to decrease it, so reverse endo reaction is favoured
So there's an increase in conc of Reactants
Equilibrium lies on the LHS
when equilibrium lies on the LHS kc<1
So kc is decreased!

Does that make sense?

(if i'm wrong someone tell me!!!)

Reply 608

student92

andy892

So basically for the 1st equation - Use Ka = [Products] / [Reactants]

2nd equation - Use Ka = [H+]^2 / [Reactants]

Then add both the values for [H+] and antilog the value to get pH

Is that right?

Reply 609

TwirlGirl

student92

So basically for the 1st equation - Use Ka = [Products] / [Reactants]

2nd equation - Use Ka = [H+]^2 / [Reactants]

Then add both the values for [H+] and antilog the value to get pH

Is that right?

Pretty sure it is.

Reply 610

student92

TwirlGirl

Pretty sure it is.

Great... thanks :biggrin:

Reply 611

cyborg

Thanks to Inter-viewer by the way, for explaining that :smile:

Reply 612

_Andrew_

student92

So basically for the 1st equation - Use Ka = [Products] / [Reactants]

2nd equation - Use Ka = [H+]^2 / [Reactants]

Then add both the values for [H+] and antilog the value to get pH

Is that right?

No, that is not. You don't need a ka expression as technically a strong acid doesn't equilibrate, as it completely dissociates. conc of acid= [H+], and then -log that.

Reply 613

Motorbiker

I worried about how much this exam sucked when i only got 75% on the mock, then i realised i need 46% to get an A overall. Bliss...

Want the A* now though, only need 86%...

Reply 614

dpd5

MHorman

I worried about how much this exam sucked when i only got 75% on the mock, then i realised i need 46% to get an A overall. Bliss...

Want the A* now though, only need 86%...

i thought you needed 90% in every A2 module to get an A*? or am i mistaken?

Reply 615

dpd5

sorry i just checked, 90% overall in A2 units...

Reply 616

dpd5

i have to get exactly 150UMS...thats 100% then

Reply 617

cyborg

I need around 79/150 for an A - but 138/150 for an A* :p:

(Of course, this depends on how the coursework this year is scaled as well)

At least the content in this module is far more interesting than the january organic chemistry.

Reply 618

CHILLIN-DRAGON

can anyone help i think i got this question right but i dont know.

basically in the question it said it was a weak acid and told me to work out the PH they gave me the volume and the mass.

i worked out moles from mass over M, then the concentration using the volume.

because it is a week acid i used K=(H+)squared/[HA]
instead of [H+][A-]/[HA]

they gave the value of K so i did K X [HA]=[H+]squared

and worked out [H+] and did -log of it to fnd the ph

my teacher said i was right but the ph was really low something like 2.4

could someone explain why the ph is low even though the acid was weak :biggrin:

Reply 619

andy892

How can we all be predicting what we need to get an A or an A* when we do not yet know the F326 coursework boundaries!
Need to be entirely sure about what they are first. Last year they were incredibly high, so people ended up with lower UMS than they may have expected.
What are you lot basing the UMS marks for that module on?