OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

Original post by Kamara7

I literally just did this paper and finished marking it XD I also got the exact same answer as you :lol:

Did you get to the 4th step in @itsConnor_'s working out above? If so, then you don't need to divide by two afterwards because you're finding the % mass of chromium and not the chromium-containing compound (Fe(CrO₂)₂) and as you can see in the equation, the moles of Cr are the same on both sides of the equation (4 moles on each side), so you don't need to multiply/divide by anything as you already got the moles of Cr (0.034)! I don't know if you did the same mistake as me though, so it could be something else.

lol I literally just finished the paper and marked it today as he posted it too! Got 79 :s 87 for A* :O

Reply 2704

pineneedles

14

Original post by uk_shahj

Why do I multiply the moles of Mg by 2? Lost 1 mark on this because I didn't multiply by 2

Question from June 2015?

2C2H5COOH + Mg2+ > (C2H5COO-)2Mg2+ + H2
One mole of magnesium produces two moles of the salt, and reacts with two moles of the acid. Hence the moles of weak acid decreases by twice the moles of magnesium and the moles of the salt increases by twice the moles of magnesium.

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Reply 2705

uk_shahj

1

Original post by pineneedles

2C2H5COOH + Mg2+ > (C2H5COO-)2Mg2+ + H2
One mole of magnesium produces two moles of the salt, and reacts with two moles of the acid. Hence the moles of weak acid decreases by twice the moles of magnesium and the moles of the salt increases by twice the moles of magnesium.

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Thanks, I see why we x2 with that equation. I didn't write the equation out when I did this one lol

Reply 2706

Original post by HFancy1997

June 14, 6d. The answer uses Ka of the acid as the Ka for the buffer solution

the ka you use in buffer solution calculation is the ka of the acid as ka is the acid dissociation constant - it's always like that

Reply 2707

Original post by RayMasterio

Can someone please explain why the answer to June 2014 question 3b(i) is 0.504.

I don't understand why they used the molar ratio for c2h2 and not the ch4

please please pleaseeeeeeeeeeeeeeee

because they are the products -
the products originally was a 0 moles whilst CH4 initial moles was already something at the start of the reaction
So you would directly compare the mole ratios of the products as they started off as the same intial moles of 0 rather than the moles of the reactant, so 0.168x3 in the 1:3 ratio would be 0.504 moles for H2
Hope that sort of makes sense

Reply 2708

Dentistry101

2

Can someone help me with June2010- last part of question6 and last part of question7 please

Reply 2709

username2640523

8

Original post by tcameron

the ka you use in buffer solution calculation is the ka of the acid as ka is the acid dissociation constant - it's always like that

Can you explain why? If the acid and buffer have different ch3Coo- Ion concentrations it shouldnt be the same?

Reply 2710

Original post by HFancy1997

Can you explain why? If the acid and buffer have different ch3Coo- Ion concentrations it shouldnt be the same?

The ch3coo- you are assuming to use in the buffer is from the ch3coona fully dissociating and not from the ch3cooh partial dissociation because you need the ch3coo- and ch3cooh to be much larger than the H+ concentration
so the ch3coo- in the equation is from the salt and not from the acid but the ch3cooh is from the acid and so you use the ka of the acid
Does that make sense :/ not sure how to really explain it
but don't think you need to know as all the buffer calculations use the ka from the acid as that's the only place it can come from

Reply 2711

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Original post by Dentistry101

Can someone help me with June2010- last part of question6 and last part of question7 please

attached question 6:

with q7:
moles KMnO4- = 0.02x0.0234 = 4.69x-4 moles
moles H202 is therefore 4.69x-4 x 5/2 due to the 5:2 ratio in the redox equation so = 1.1725x-3 moles
need to x10 to get this in 250cm3 from 25cm3 so = 0.011725moles
to find mass (because the answer is needed in gdm-3) need to times the moles by the Mr of H202 which is 0.011725x34 = 0.39865g
concentration will then be 0.39865/0.025 to get is as 15.9 gdm-3
moles of oxygen will be the diluted moels of H202 so is 1.1725x10-3 x24 = 0.028dm3 for the volume of oxygen

Reply 2712

Professor L

14

Original post by itsConnor_

lol I literally just finished the paper and marked it today as he posted it too! Got 79 :s 87 for A* :O

I'm terrified that I'll make a mistake on questions like where they added Mg. Or get a last redox titration one wrong because I'll be rushing it because of time :afraid:

Reply 2713

Dentistry101

2

Original post by tcameron

attached question 6:

with q7:
moles KMnO4- = 0.02x0.0234 = 4.69x-4 moles
moles H202 is therefore 4.69x-4 x 5/2 due to the 5:2 ratio in the redox equation so = 1.1725x-3 moles
need to x10 to get this in 250cm3 from 25cm3 so = 0.011725moles
to find mass (because the answer is needed in gdm-3) need to times the moles by the Mr of H202 which is 0.011725x34 = 0.39865g
concentration will then be 0.39865/0.025 to get is as 15.9 gdm-3
moles of oxygen will be the diluted moels of H202 so is 1.1725x10-3 x24 = 0.028dm3 for the volume of oxygen

Thank you so much. I was wondering though- so can you use gdm-3 as moles? Cuz in both questions that's what it seems has happened???

Reply 2714

If it asked to draw a diagram of how you would measure the standard electrode potential of this Cl2 half cell would this be right??

EDIT: and would the electromotive force be just 1.36V?
EDIT2: and would overall equation be Cl2 + H2 --> 2H+ +2Cl-?
EDIT3: and would pH of solution in the hydrogen one DECREASE as the reaction progressed as conc. of H+ decreased sicne its electropotential is more negative than chlorine?
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(edited 7 years ago)

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Reply 2715

Original post by Dentistry101

Thank you so much. I was wondering though- so can you use gdm-3 as moles? Cuz in both questions that's what it seems has happened???

no they're different

Reply 2716

username2703735

18

Original post by itsConnor_

If it asked to draw a diagram of how you would measure the standard electrode potential of this Cl2 half cell would this be right??

EDIT: and would the electromotive force be just 1.36V?
EDIT2: and would overall equation be Cl2 + H2 --> 2H+ +2Cl-?
EDIT3: and would pH of solution in the hydrogen one DECREASE as the reaction progressed as conc. of H+ decreased sicne its electropotential is more negative than chlorine?
Posted from TSR Mobile

All correct apart from the pH value of the hydrogen would decrease due to an increase in H+ concentration. The higher the concentration of H+ ions = a lower pH.

The standard hydrogen half cell is 2H+ + 2e- <=> H2 and as this is less positive than the chlorine half cell it is reduced. Therefore H+ ion concentration would increase.

(edited 7 years ago)

Reply 2717

Original post by VMD100

All correct apart from the pH value of the hydrogen would decrease due to an increase in H+ concentration. The higher the concentration of H+ ions = a lower pH.

how do we know H+ would decrease exactly though? I know I said it just don't rly understand :smile:

Also if we EVER have a gas in an electropotential, we draw it as a hydrogen half cell BUT with the gas going in instead of the hydrogen? Never seen an OCR A question like this before :s

Reply 2718

username2703735

18

Original post by itsConnor_

how do we know H+ would decrease exactly though? I know I said it just don't rly understand :smile:

Also if we EVER have a gas in an electropotential, we draw it as a hydrogen half cell BUT with the gas going in instead of the hydrogen? Never seen an OCR A question like this before :s

I am not sure what you mean by the first part? The H+ concentration is increasing not decreasing?
And while I have never seen an OCR question with another gas other than hydrogen yes if it is gaseous you would need the 1 atmosphere/100kPa, which is really 101kPa :wink:

and for it to be bubbled into the aqueous solution of its ions.

Reply 2719