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# ocr a f325 revision thread watch

Do we have to know the indicators?? It says on the spec guess I will have to cram it...
nope we dont where one spec does it say that?
2. (Original post by DudeBoy)
June 2010
7b.
You times by 34 as that is the Ar of H2O2
2(1)+2(16)
thanks don't know why I didnt find that lol

(Original post by DudeBoy)
Jan 2011
7ai. Use CV/1000 to get S2O3(2-) used.
=2.46x10^-5

Look at the equations, 2 moles of S2O3 form 2 moles of I(-)
So I(-) = 2.46x10^-5
2 mole of I(-) reacts 2 moles of Mn(OH)3 so =2.46x10^-5.
4 moles of Mn(OH)3 is formed by 1 mole of O2, so O2 moles = mole of O2 = 2.46x10^-5/4.
=6.15x10^-6

Use CV/100 to get conc of O2.
This method doesn't work because
If I go to OH- instead of Mn(OH)3, this is what happens:
2 mole of I(-) reacts 2 moles of OH- so =2.46x10^-5.
8 moles of OH- is formed by 1 mole of O2, so O2 moles = mole of O2 = 2.46x10^-5/8.
=3.08x10^-6

3. (Original post by Namod)
nope we dont where one spec does it say that?
'(p) for acid–base titration pH curves for strong
and weak acids and bases:
(i) interpret, or sketch, their shapes,
(ii) explain the choice of suitable indicators
for acid–base titrations, '

In case they may not give you indicators, hence you may need to name one closest to equivalence point
'(p) for acid–base titration pH curves for strong
and weak acids and bases:
(i) interpret, or sketch, their shapes,
(ii) explain the choice of suitable indicators
for acid–base titrations, '

In case they may not give you indicators, hence you may need to name one closest to equivalence point
"explain the choice" "State the indicator that must be used"
Big difference mate.

GL
5. after in depth revision, began ppq... i got a low b at first mark, a B it still is lool... definitely more area for improvements...hws it going?
6. (Original post by MathsNerd1)
What challenging questions could actually be asked on transition metals/elements as I'm struggling to see where they become challenging Same with all the calculation questions too.
Try jan 13 last q.
7. (Original post by DudeBoy)
It is D. AgOH⇌Ag(+) + OH(-)
If Ag(+) is added it reacts with OH(-) thus decreasing the Kc value.
pH may also decrease, but I am not sure because there will be less OH(-) ions.
Thats wrong. chiggy guy is right. kc not affected by changesa in conc
8. (Original post by Gulzar)
But I just looked in the book and they carried out a enthalpy of neutralisation calculation pg 156...using degrees C and not Kelvin

(Original post by Better)
Go in the internet and search SI Units. Degrees C, is not an SI Unit.

The Examples in the book are just Examples. All of these Papers I have done work on Kelvin.

If you convert to Degrees C AFTER you've done the calculation they will give you a mark but you will get it wrong if you do it before, as you will use the wrong number.
Change temp for neutralisation so wouldnt matter k or c
9. (Original post by MedMed12)
YAAAY really? :O I hope I get that now means I can get a B in unit 4 -_-
I doubt we'll get papers like the old ones though heard they are toughening up!
the boundaries are nice for this paper!
my friend had a theory that everyone should enter for all exams (only work for 4 subjects they are sitting) and everyone else in the room who doesn't need/study chem should just sleep and get Us and it'll lower the grade boundaries for chem-takers! Clever thing huh?
Waste of money..
10. (Original post by Theafricanlegend)
Thats wrong. chiggy guy is right. kc not affected by changesa in conc
if the concentration changes, the equilibrium shift to decease the effects, so Kc will remain constant, only thing that change the Kc is change in temperature

if reaction exothermic= *increase the temp-> equilibrium shifts to left hence Kc decrease
*decrease the temp -> equilibrium shifts to right, hence Kc increases

if reaction endothermic= *increase the temp-> equilibrium shifts to the right, Kc increases
*decrease the temp -> equilibrium shifts to the left, Kc decrease

Kc=[products]/[reactants]
11. (Original post by Theafricanlegend)
Try jan 13 last q.
Hi, do you have F325 jan 13, can you please give me the link, or upload it here, many thanks
12. (Original post by amin666)
Hi, do you have F325 jan 13, can you please give me the link, or upload it here, many thanks
Here ya go
Attached Images
13. F325 Jan 2013 QP.pdf (279.9 KB, 57 views)
14. F325 Jan 2013 MS.pdf (564.4 KB, 58 views)
15. For a standard hydrogen electrode. What is the actual electrode, is it Platinum wire? or is it Platinum foil at the bottem of the gas imput thingy
16. (Original post by amin666)
if the concentration changes, the equilibrium shift to decease the effects, so Kc will remain constant, only thing that change the Kc is change in temperature

if reaction exothermic= *increase the temp-> equilibrium shifts to left hence Kc decrease
*decrease the temp -> equilibrium shifts to right, hence Kc increases

if reaction endothermic= *increase the temp-> equilibrium shifts to the right, Kc increases
*decrease the temp -> equilibrium shifts to the left, Kc decrease

Kc=[products]/[reactants]
i know that.
You didnt need to write it.
17. (Original post by Theafricanlegend)
i know that.
You didnt need to write it.
no need to be rude!
18. Hi, doing some extra questions but surely this answer that it says on the mark scheme is wrong?

This is the question: Calculate the pH of a buffer solution that is 0.100 mol dm–3 with respect to sodium methanoate and 0.200 mol dm–3 with respect to methanoic acid. The Ka of methanoic acid is 1.8 × 10–4 mol dm–3.

I did Ka x Acid/Salt and then -log10 of the H+ conc to give me a pH of 3.44 from a H+ of 3.6 x10-4

However, the mark scheme says this:

[HCOOH(aq)]eqm ≈ 0.100 mol dm–3 and [HCOO–(aq)]eqm ≈ 0.200 mol dm–3
Ka = [HCOO–(aq)]eqm[H+(aq)]
[HCOOH(aq)]eqm
So [H+(aq)] = 9.0 × 10–5 mol dm–3
pH = 4.05

^^ is this not wrong? If not can someone explain where I am wrong please. Thank you.
19. Does ionisation constant mean the same as Ka?
20. (Original post by ofudge)
Hi, doing some extra questions but surely this answer that it says on the mark scheme is wrong?

This is the question: Calculate the pH of a buffer solution that is 0.100 mol dm–3 with respect to sodium methanoate and 0.200 mol dm–3 with respect to methanoic acid. The Ka of methanoic acid is 1.8 × 10–4 mol dm–3.

I did Ka x Acid/Salt and then -log10 of the H+ conc to give me a pH of 3.44 from a H+ of 3.6 x10-4

However, the mark scheme says this:

[HCOOH(aq)]eqm ≈ 0.100 mol dm–3 and [HCOO–(aq)]eqm ≈ 0.200 mol dm–3
Ka = [HCOO–(aq)]eqm[H+(aq)]
[HCOOH(aq)]eqm
So [H+(aq)] = 9.0 × 10–5 mol dm–3
pH = 4.05

^^ is this not wrong? If not can someone explain where I am wrong please. Thank you.
u're right. where is this q from/?
21. (Original post by otrivine)
Does ionisation constant mean the same as Ka?
You talking about the ionic product of water?

If you are, it's similar. ka= [OH-][H+]/[h2O]
As H2O is constant, they move it to other side and KAx[H2O]=Kw=[OH-][H+]
22. (Original post by otrivine)
Does ionisation constant mean the same as Ka?
http://lmgtfy.com/?q=ionisation+constant

heehee

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