ocr a f325 revision thread

Original post by Emma2129

What's a precipitation reaction?

i think its where 2 soluble substances react to form an insoluble salt? anyon correct me here :s-smilie:

Reply 1402

Weasleys_jumper

Original post by otrivine

Hi

Okay what you do is this:-

1) You can use the formula for buffers, I use [H+]=Ka x [HA]/[A-]
2) We have Ka, thats fine , we need to get the concentrations of HA and A-
3) Before we do that, we need to get the moles of the butaneoic acid and the NaOH
4) The moles of butanoic acid is 0.0125 and the moles of NaOH is 2.5x10-3
5) now using your knowledge of buffers, we know that some acid remains in solution and hence, the acid is in excess and so we substract the moles of butanoic acid and NaOH to get the moles of butanoic acid that has reacted with NaOH.
6) so the new mole of butanoic acid is 0.01 and the moles of NaOH is the same 2.5 x 10-3, but we have to get the concentration so we divide by the volume which would be 100cm3(0.1dm3) , this is because there is 50cm3 of butanoic acid and 50cm3 of NaOH , so we add the volumes.
7) Now that you got the concentration just sub the values into the Ph buffer formula and you get the value of [H+] and then do the -log[H+] which should give 4.22 correct to 2 decimal place

please tell me if you are unsure :smile:

aah thank you for this brilliant step-by-step explanation, i remember you on the AS threads for last year :smile:

May i ask what you've applied to do at uni? Was it med or pharmacy? :smile:

x

Reply 1403

Original post by Member427

ahh that makes so much sense now thank you so much, so if a question says for example potassium is oxidised we assume it turns into k^1+ ? And also will it be the same for chlorine as well if it says cl is reduced it goes to cl^1- ?

You're welcome, that's what I'd do, so yes :smile:

Reply 1404

Pride

21

Original post by master y

i think its where 2 soluble substances react to form an insoluble salt? anyon correct me here :s-smilie:

wouldn't produce a salt, 'compound' would probably do. I would say where two solutions react to form an insoluble compound.

Reply 1405

Original post by otrivine

Hi

Okay what you do is this:-

1) You can use the formula for buffers, I use [H+]=Ka x [HA]/[A-]
2) We have Ka, thats fine , we need to get the concentrations of HA and A-
3) Before we do that, we need to get the moles of the butaneoic acid and the NaOH
4) The moles of butanoic acid is 0.0125 and the moles of NaOH is 2.5x10-3
5) now using your knowledge of buffers, we know that some acid remains in solution and hence, the acid is in excess and so we substract the moles of butanoic acid and NaOH to get the moles of butanoic acid that has reacted with NaOH.
6) so the new mole of butanoic acid is 0.01 and the moles of NaOH is the same 2.5 x 10-3, but we have to get the concentration so we divide by the volume which would be 100cm3(0.1dm3) , this is because there is 50cm3 of butanoic acid and 50cm3 of NaOH , so we add the volumes.
7) Now that you got the concentration just sub the values into the Ph buffer formula and you get the value of [H+] and then do the -log[H+] which should give 4.22 correct to 2 decimal place

please tell me if you are unsure :smile:

omg thank you, finally makes sense :smile:

I never knew what to do after I found the first set of number of moles :biggrin:

Reply 1406

Better

8

Could someone please help on the

Specimen Paper - My Mistakes - Any advice appreciated

Q1aiii - Deduce the intial rate of FORMATION of O2 ?
They do a division by 4 but I don't know why? :s-smilie:

Q3ii Suggest why methanol is added to petrol
Their Answer: Adds Oxygen?
I don't understand why

Q6d- AS MOLAR Equations :frown:

This is the most important
AS Molar Equations I have a difficulty with these could anyone please break this down for me - it would be much appreciated or give me any tips!

Reply 1407

Original post by Pride

wouldn't produce a salt, 'compound' would probably do. I would say where two solutions react to form an insoluble compound.

Examples you might need to know:

Fe^2+ (aq) + 2OH- (aq) ------> Fe(OH)2 (s)
green ------> green precipitate

Co^2+ (aq) + 2OH- (aq) ------> Co(OH)2 (s)
pink ------> light blue ppt

Fe^3+ (aq) + 3OH- (aq) --------> Fe(OH)3 (s)
yellow -------> brown ppt

Cu^2+ (aq) + 2OH- (aq) -------> Cu(OH)2 (s)
light blue -------> dark blue ppt

:biggrin:

(edited 10 years ago)

Reply 1408

chignesh10

15

Original post by ofudge

4Ci) jan 13
"how does S2082- and I- catalysed by Fe2+?
- I got the first one as Fe2+ has higher E theta than S208

but in the mark scheme, the second one is 2Fe3+ + 2I- --> 2Fe2+ + I2

Surely ^^ is being catalysed by fe3+ and not fe2+?

no because fe2+ is regenerated showing that fe2+ is the catalyst

Reply 1409

H0ls

18

I want someone to revise with. I can't not get an A on this exam.

Reply 1410

Original post by H0ls

I want someone to revise with. I can't not get an A on this exam.

how have you been doing on past papers?

Reply 1411

H0ls

18

Original post by master y

how have you been doing on past papers?

I haven't done any for a couple of weeks. I'm getting between 70-80. So I need to really crack on.

Reply 1412

Original post by H0ls

I haven't done any for a couple of weeks. I'm getting between 70-80. So I need to really crack on.

They' re good marks, that's an A!

Reply 1413

Original post by otrivine

m

please can you explain clock reactions to me?

Reply 1414

H0ls

18

Original post by master y

They' re good marks, that's an A!

I know, but I don't want to slip up in the exam. I always seem to drop a couple of marks due to the pressure. So I want to be hitting in the 90's so I can not give myself panic attacks until results day haha.

Reply 1415