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

1. Can anyone explain where I'm going wrong in qu7 of the June 2015 paper? I ended up with a value that was half what it should have been. Is this because the ratio in the ore was fe(CrO)2 as in 2 moles of chromium to one mile of ore, so I should have multiplied by 2?
2. (Original post by maisym00)
Can anyone explain where I'm going wrong in qu7 of the June 2015 paper? I ended up with a value that was half what it should have been. Is this because the ratio in the ore was fe(CrO)2 as in 2 moles of chromium to one mile of ore, so I should have multiplied by 2?
You got a pic of your workings?
Step by step:
• Work out moles of S2O3^2- = (25.5/1000)x0.1 = 2.55x10^-3 mol
• Work out moles of I2 = (2.55x10^-2)/2 = 1.275x10^-3 mol
• Work out moles of CrO4^2- = (1.275x10^-3)/1.5 = 8.5x10^-4 mol
• Work out moles of CrO4^2- in 1dm^-3 = 40x(8.5x10^-4) = 0.034 mol
• Work out mass of Cr(Cr^3+) = 0.034 x 52 = 1.768g
• Work out percentage by mass = 1.768/5.25 = 33.7%
3. Can I double check my answer to this question with anyone, I think the mark scheme is wrong:

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 methanoicacid is 1.8 × 10–4 mol dm–3.
4. Do we also need to know about the effect of volume on equilibrium?
5. (Original post by RetroSpectro)
Can I double check my answer to this question with anyone, I think the mark scheme is wrong:

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 methanoicacid is 1.8 × 10–4 mol dm–3.
i got ph=3.44
6. (Original post by Arima)
i got ph=3.44
Yep thats what I got, the mark scheme says 4.05 but I think they switched the concentrations for the acid and the salt
7. (Original post by maisym00)
Can anyone explain where I'm going wrong in qu7 of the June 2015 paper? I ended up with a value that was half what it should have been. Is this because the ratio in the ore was fe(CrO)2 as in 2 moles of chromium to one mile of ore, so I should have multiplied by 2?
I literally just did this paper and finished marking it XD I also got the exact same answer as you 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.
8. (Original post by RetroSpectro)
Yep thats what I got, the mark scheme says 4.05 but I think they switched the concentrations for the acid and the salt
which question paper are you doing might i ask? if it's an official past paper id be really worried LOL
9. (Original post by Arima)
which question paper are you doing might i ask? if it's an official past paper id be really worried LOL
Its from the Stretch and challenge questions from Exam Cafe

Im not surprised the mark scheme was wrong, there are so many mistakes in the book itself
10. 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

11. 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?
12. (Original post by Kamara7)
I literally just did this paper and finished marking it XD I also got the exact same answer as you 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
13. (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.

Posted from TSR Mobile
14. (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.

Posted from TSR Mobile
Thanks, I see why we x2 with that equation. I didn't write the equation out when I did this one lol
15. (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
16. (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

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
17. Can someone help me with June2010- last part of question6 and last part of question7 please
18. (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?
19. (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
20. (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
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