[Babinimeister]

How did you do it? It was a 6 mark question.

[leah2323]

(Original post by Babinimeister)
How did you do it? It was a 6 mark question.

Well i wrote out an equation like
HX + NaOH goes to XNa + H20

then worked out the moles of HX to be 0.00261
moles of NaOH to be 0.00125
then did 0.00261-0.00125 which gave 0.00136
then used the equation H+ = Ka multiplied by Acid over Salt
which gave me a H+ concentration of 0.000032748 then - logged that and got 4.4848...

[Babinimeister]

(Original post by leah2323)
Well i wrote out an equation like
HX + NaOH goes to XNa + H20

then worked out the moles of HX to be 0.00261
moles of NaOH to be 0.00125
then did 0.00261-0.00125 which gave 0.00136
then used the equation H+ = Ka multiplied by Acid over Salt
which gave me a H+ concentration of 0.000032748 then - logged that and got 4.4848...

Thankyou so much for your help, but I'm not fully understanding why you would take away the two values for the number of moles.

[Abbieastoria]

(Original post by leah2323)
Well i wrote out an equation like
HX + NaOH goes to XNa + H20

then worked out the moles of HX to be 0.00261
moles of NaOH to be 0.00125
then did 0.00261-0.00125 which gave 0.00136
then used the equation H+ = Ka multiplied by Acid over Salt
which gave me a H+ concentration of 0.000032748 then - logged that and got 4.4848...

I did exactly the same and wrote
Ph= 4.4848
Therefore ph=4.49

I rounded up to 4.485 then 4.49 do you think this will matter or will they have a small range you can fit into?


This was posted from The Student Room's iPhone/iPad App

[Babinimeister]

(Original post by Abbieastoria)
I did exactly the same and wrote
Ph= 4.4848
Therefore ph=4.49

I rounded up to 4.485 then 4.49 do you think this will matter or will they have a small range you can fit into?


This was posted from The Student Room's iPhone/iPad App

They should allow for this range if the rounding is correct, which it was, so yeah.

[leah2323]

(Original post by Babinimeister)
Thankyou so much for your help, but I'm not fully understanding why you would take away the two values for the number of moles.

If you look at the equation initially there was 0.00261 moles of acid
they then react with the NaOH which is 0.00125 since the acid mole number is higher which is assumed as it is an acidic buffer all the NaOH is used up meaning 0.00125 moles of salt is created and therefore 0.00125 moles of NaOH reacted with 0.00261 HX to produce this salt therefore the amount of HX left after reaction is initial moles of HX 0.00261 take away the moles of NaOH that have reacted 0.00125

im not too good at explaining but there ye go ahah

[leah2323]

(Original post by Abbieastoria)
I did exactly the same and wrote
Ph= 4.4848
Therefore ph=4.49

I rounded up to 4.485 then 4.49 do you think this will matter or will they have a small range you can fit into?


This was posted from The Student Room's iPhone/iPad App

oh god i dont know i hate it when they get all technical with that

[Babinimeister]

(Original post by leah2323)
If you look at the equation initially there was 0.00261 moles of acid
they then react with the NaOH which is 0.00125 since the acid mole number is higher which is assumed as it is an acidic buffer all the NaOH is used up meaning 0.00125 moles of salt is created and therefore 0.00125 moles of NaOH reacted with 0.00261 HX to produce this salt therefore the amount of HX left after reaction is initial moles of HX 0.00261 take away the moles of NaOH that have reacted 0.00125

im not too good at explaining but there ye go ahah

Thank you. I've been waiting for 2 hours for a reply to this question, and don't worry, I understand now.

Lost 6 marks either way.

[leah2323]

(Original post by Babinimeister)
Thank you. I've been waiting for 2 hours for a reply to this question, and don't worry, I understand now.

Lost 6 marks either way.

Oh dear me happy to help anyway im sure you did fine on the rest i found myself losing stupid marks later on

[Strangeclouds]

(Original post by zoeaw)
I got BBAABA...

Agreed

(Original post by leah2323)
I got 4.48 or something like that! D:

Agreed

[leah2323]

This will sound stupid but what was the first question.. like moles of hydrogen and iodine

[Abbieastoria]

(Original post by leah2323)
This will sound stupid but what was the first question.. like moles of hydrogen and iodine

Doesn't sound stupid at all those questions are there to trip people up!

It was H2 + I2 >>> 2HI

It said there was 1.2 hydrogen and 1.7 iodine initially and 2.06 HI at equilibrium

So we divide 2.06 by two to get 1 mole that's 1.03 then take that away from each reactant so 0.17 h2 and 0.67 i2

I think those were the right numbers not sure, but that is deffo the right process of how to work it out


This was posted from The Student Room's iPhone/iPad App

[Picture~Perfect]

(Original post by leah2323)
What did people get for the 6mark ph question..

I got 4.48 from what I can tell that seems to be what most people got

[leah2323]

(Original post by Abbieastoria)
Doesn't sound stupid at all those questions are there to trip people up!

It was H2 + I2 >>> 2HI

It said there was 1.2 hydrogen and 1.7 iodine initially and 2.06 HI at equilibrium

So we divide 2.06 by two to get 1 mole that's 1.03 then take that away from each reactant so 0.17 h2 and 0.67 i2

I think those were the right numbers not sure, but that is deffo the right process of how to work it out


This was posted from The Student Room's iPhone/iPad App

I am an idiot ahahh thanks!

[small'un]

(Original post by leah2323)
i got 4.48 or something like that! D:

i got this :d

[Josh1993]

Unofficial mark scheme.

1a)

i)0.47,0.170
ii)sraightforward.
iii)equal amount of mols on both sides.
iv)52.1

b)

i) D
ii)B
iii)A
iv)C

2) 4.5*10^-4
4.5*10^-3
4.33*10^-2

b) rate = 4.44 mol^-2dm^6s^-1

a) proton donor.

b) i) ii) iii) BBAABA in that order.

c) 37.8 cm^3

d) i) 4.52
ii) standard
iii)2.64
iv)4.48.

4 a) mechanism as standard (methylpropanoate)
b) i)1,5-pentanediol
ii) simple enough.
iii) base hydrolysis (base is usually ignored). acts as nucleophile blah blah

c i) ethanoic acid
ii)not sure.nucleophyllic addition-eleimination.
iii) no HCl , not as vigourous etc....

d) can't draw

e) i ) ester
ii) i count 10 but think i may be wrong. it is 12
iii) look at formula sheet.

f) i) 2nd one and last one( not sure again)
ii) pink ------> colourless

5) a)Nuclephyllic addition.
B)2-Bromobutanenitrile
c) NH3 , Excess , nucleophyllic substitiution.
d i) zwitterion
ii) stronger forces of attraction , Vdw in the other molecule etc....

e) i) NH2CH2COOCH2CH3
ii) COH(CH3)2CONH2
iii) NH2CH2CH2CH2COOH
iii) CH3CH2CH(N(CH3)2)COOH hard without drawing

6 a) AlCl3 + CH3COCl -----------------> [AlCl4]- + CH3CO+
AlCl4 - + H+ ---------------> AlCl3 + HCl
Phenylethanone

b) C6H5CO+
c) Inductive effects?? I should really know this but oh well.

7 a) propan-1-ol conc H2SO4 orange-----> green
propanal , fehlings/tollens , brick red PPT, silver mirror
propanoic acid , Na2CO3 , Effervesence(CO2)
Chloropropane , dil HNO3 , NaOH , AgNO3, white PPT

b) oxidation, state wavenumber of c=o,diffrent affinites/solubilities.

8) 1,4 dinitrobenzene
1,2 dinitrobenzene

double bond with 4 methyl groups around it.
cyclocompund (cyclohexane)

c=o on benzene
aldehyde on benzene.


3 ethyl groups as bonded to N.
c(CH3)3NHCH3

Feel free to amend and add to this as some questions I dont know (only a couple). This is very rushed so likely to be mistakes.

[WrongSideUp]

(Original post by leah2323)
Well i wrote out an equation like
HX + NaOH goes to XNa + H20

then worked out the moles of HX to be 0.00261
moles of NaOH to be 0.00125
then did 0.00261-0.00125 which gave 0.00136
then used the equation H+ = Ka multiplied by Acid over Salt
which gave me a H+ concentration of 0.000032748 then - logged that and got 4.4848...

For Acid/salt... Which value did you use for the acid and how did you get it?

[lamalas600]

Guys you can't change your exam paper

[WrongSideUp]

(Original post by lamalas600)
Guys you can't change your exam paper

Well I can't speak for the others but if I've got something wrong I want to know how to get it right. Maybe I'm weird.

[lamalas600]

(Original post by WrongSideUp)
Well I can't speak for the others but if I've got something wrong I want to know how to get it right. Maybe I'm weird.

You've done the exam now, so how will it help?