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

This discussion is closed.
FoOtYdUdE
Badges: 0
Rep:
?
#581
Report 8 years ago
#581
(Original post by FoOtYdUdE)
dyu knw wen ur working out delta G

erm.. it sez in the textbook that wen delta H is -ve and delta s is +ve delta G is always -ve and the reaaction is feasible

ok...


but i thought the formula also includes T...so what if T is -ve?? eh??? am i making sense? is T always positive then?
just went thru a ppt presentation and it says..T is always positive...that's cleared that up then.
0
CHILLIN-DRAGON
Badges: 10
Rep:
?
#582
Report 8 years ago
#582
(Original post by TwirlGirl)
Any one with the OCR A endorsed text book:

Pg 210 - Fig 2. Is says it's the isomerism for colbalt, but Ni is in the diagram. Would Ni do the same thing? I presume it's a typo...
i asked my teacher and she said although its Ni in the diagram either Ni or colbalt would be okay to have in the centre
0
cyborg
Badges: 0
Rep:
?
#583
Report 8 years ago
#583
(Original post by FoOtYdUdE)
just went thru a ppt presentation and it says..T is always positive...that's cleared that up then.
Yeah T is always the temperature in Kelvin - so it must always be positive.
0
steph_v
Badges: 13
Rep:
?
#584
Report 8 years ago
#584
(Original post by Pedus)
Oh, I haven't even started revising for that one yet. Been focusing on my other exams... I got 9 days gap though to revise that so tis not the end of the world for me!



I am so tempted to say that after reading this post, it felt as if I've already said this before in my head... :hmmmm:

deja vu :proud:

I've seen you around once or twice... You visited the OCR Chemistry Thread??

Click here :pieree:
:yep: We have spoken before. You sent me some good revision packs
0
matt^
Badges: 0
Rep:
?
#585
Report 8 years ago
#585
(Original post by student92)
How do you work out the pH of diprotic acid given the Ka value? Could someone explain step by step - i tried searchin in google, but its confused me!
I'd like to know this aswell
Can anyone do this?
Ive forgotten how to
0
cyborg
Badges: 0
Rep:
?
#586
Report 8 years ago
#586
For more "light hearted" revision, check out Khan Academy on Youtube.

Most of our spec is covered in the videos - and they are great !
0
andy892
Badges: 0
Rep:
?
#587
Report 8 years ago
#587
(Original post by matt^)
I'd like to know this aswell
Can anyone do this?
Ive forgotten how to
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.
0
andy892
Badges: 0
Rep:
?
#588
Report 8 years ago
#588
(Original post by cyborg)
For more "light hearted" revision, check out Khan Academy on Youtube.

Most of our spec is covered in the videos - and they are great !
Yeah they are good actually
Thanks for posting that
0
Markh1000
Badges: 1
Rep:
?
#589
Report 8 years ago
#589
(Original post by 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.
It depends on how the strong acid loses it's successive protons. For example, H2SO4 loses the first proton as a strong acid, but the second as a weak acid (HSO4-) - so it's not as simple as doubling the conc. (This is shown on pg 136 of the OCR textbook.)

You work out the first [H+] using strong acid method, and use the Ka value given to work out the second [H+] and then add together to take negative log.
0
matt^
Badges: 0
Rep:
?
#590
Report 8 years ago
#590
(Original post by 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.
(Original post by Markh1000)
It depends on how the strong acid loses it's successive protons. For example, H2SO4 loses the first proton as a strong acid, but the second as a weak acid (HSO4-) - so it's not as simple as doubling the conc. (This is shown on pg 136 of the OCR textbook.)

You work out the first [H+] using strong acid method, and use the Ka value given to work out the second [H+] and then add together to take negative log.
Ahh, thanks to you both.
I'll make sure that i learn that.
i fear that i havent a different textbook as my page 136 doesnt have that on it.
i dont think diprotic acids are mentioned in the specification so hopefully i wont need to know it anyway.
0
_Andrew_
Badges: 9
Rep:
?
#591
Report 8 years ago
#591
Guys, I really don't think calculating the pH of a diprotic acid solution will come up, it does say specifically on the spec that it's only monoprotic acids|!
0
cyborg
Badges: 0
Rep:
?
#592
Report 8 years ago
#592
(Original post by _Andrew_)
Guys, I really don't think calculating the pH of a diprotic acid solution will come up, it does say specifically on the spec that it's only monoprotic acids|!
:yep: The wording of the spec is very clear.

I hope this exam has a lot of maths in it though; I'm better at Maths than Chemistry :p:
0
andy892
Badges: 0
Rep:
?
#593
Report 8 years ago
#593
(Original post by Markh1000)
It depends on how the strong acid loses it's successive protons. For example, H2SO4 loses the first proton as a strong acid, but the second as a weak acid (HSO4-) - so it's not as simple as doubling the conc. (This is shown on pg 136 of the OCR textbook.)

You work out the first [H+] using strong acid method, and use the Ka value given to work out the second [H+] and then add together to take negative log.
Thanks for that, very useful Wasn't aware that it needed to be calculated in that manner. But yes I think it may not come up too.
0
andy892
Badges: 0
Rep:
?
#594
Report 8 years ago
#594
(Original post by matt^)
Ahh, thanks to you both.
I'll make sure that i learn that.
i fear that i havent a different textbook as my page 136 doesnt have that on it.
i dont think diprotic acids are mentioned in the specification so hopefully i wont need to know it anyway.
No it is on that page.
It shows the first proton donation as being a strong acid - i.e. no equilibrium sign. But it shows the second proton donation as being weak - i.e. it has an equilibrium sign.
0
cyborg
Badges: 0
Rep:
?
#595
Report 8 years ago
#595
Are you guys all revising organic chemistry as well (briefly) for any synoptic elements that might come up ?

There were a few questions in the specimen with the phenol (although admittedly it was related to buffers).
0
danhirons
Badges: 15
Rep:
?
#596
Report 8 years ago
#596
(Original post by cyborg)
Are you guys all revising organic chemistry as well (briefly) for any synoptic elements that might come up ?

There were a few questions in the specimen with the phenol (although admittedly it was related to buffers).
Nope, there's enough in this section of chemistry to keep me more than busy
0
cyborg
Badges: 0
Rep:
?
#597
Report 8 years ago
#597
(Original post by danhirons)
Nope, there's enough in this section of chemistry to keep me more than busy
True - especially for transition metals
0
danhirons
Badges: 15
Rep:
?
#598
Report 8 years ago
#598
Yeah, I think we'll be really unlucky if some hard organic chemistry comes up on the paper tbh! I HOPE they're not going to be that cruel
0
andy892
Badges: 0
Rep:
?
#599
Report 8 years ago
#599
(Original post by danhirons)
Nope, there's enough in this section of chemistry to keep me more than busy
Snap
If I don't remember the organic stuff, it isn't getting revised. Especially since I'm abandoning Chemistry after this exam too
0
_Andrew_
Badges: 9
Rep:
?
#600
Report 8 years ago
#600
I have so many past papers to do for chemistry this week :emo:!
0
X
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (264)
38.94%
No - but I will (46)
6.78%
No - I don't want to (51)
7.52%
No - I can't vote (<18, not in UK, etc) (317)
46.76%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise