[TheStudent.]

Hi,

Does anyone know how to answer questions 5.f.i and ii?

As well as question 1.c.iii. If you go by the n+1 rule then you'd surely have a splitting value of 4 and then 2? How could it be zero both times?

I cant seem to figure it out :s

Here's a copy of the paper and mark scheme if you don't have it:
http://www.thestudentroom.co.uk/show...3#post37931943

Thank you

[Mathlete 4 the win]

(Original post by TheStudent.)
Hi,

Does anyone know how to answer questions 5.f.i and ii?

As well as question 1.c.iii. If you go by the n+1 rule then you'd surely have a splitting value of 4 and then 2? How could it be zero both times?

I cant seem to figure it out :s

Here's a copy of the paper and mark scheme if you don't have it:
http://www.thestudentroom.co.uk/show...3#post37931943

Thank you

I've just done this paper so will write up some solutions in a second.

[Mathlete 4 the win]

For 5) f) i)
A student sets out to make a buffer solution. The student measures out 27 cm3 of
0.050 mol dm–3 HA solution and reacts it with one-third of the volume of 0.10 mol dm–3 sodium
hydroxide needed for complete neutralisation.

HA + NaOH --------> NaA + H₂O

So first of all the moles of Acid HA.
n = C X V / 1000
n = (0.05 X 27) / 1000
n = 1.35X10^-3 moles

As the volume added was a 1/3 of the require volume to neutralise, the amount of moles in the volume actually added will be the moles required / 3.

So...
actual n = (1.35X10^-3) / 3
actual n = 4.5X10^-4

V added = (n X 1000) / C where C is the concentration of sodium hydroxide

V added = ( 4.5X10^-4 X 1000) / 0.1
V added = 4.5 cm³

5))ii)
Here you use a lot of the values you calculated above ^^^
moles of HA added to the solution = 1.35X10^-3 moles
moles of A^- added to the solution = 4.5X10^-4 moles

For part 2 I have 0.2 less than the mark scheme so going to have a quick check of the ol' solutions before I post this.


And for 1)c) if you draw out the structure you'll find the there is in fact an oxygen between the CH₃ and the H bonded to the oxygen. This there is no splitting.

[TheStudent.]

(Original post by Mathlete 4 the win)
For 5) f) i)
A student sets out to make a buffer solution. The student measures out 27 cm3 of
0.050 mol dm–3 HA solution and reacts it with one-third of the volume of 0.10 mol dm–3 sodium
hydroxide needed for complete neutralisation.

HA + NaOH --------> NaA + H₂O

So first of all the moles of Acid HA.
n = C X V / 1000
n = (0.05 X 27) / 1000
n = 1.35X10^-3 moles

As the volume added was a 1/3 of the require volume to neutralise, the amount of moles in the volume actually added will be the moles required / 3.

So...
actual n = (1.35X10^-3) / 3
actual n = 4.5X10^-4

V added = (n X 1000) / C where C is the concentration of sodium hydroxide

V added = ( 4.5X10^-4 X 1000) / 0.1
V added = 4.5 cm³

5))ii)
Here you use a lot of the values you calculated above ^^^
moles of HA added to the solution = 1.35X10^-3 moles
moles of A^- added to the solution = 4.5X10^-4 moles

For part 2 I have 0.2 less than the mark scheme so going to have a quick check of the ol' solutions before I post this.


And for 1)c) if you draw out the structure you'll find the there is in fact an oxygen between the CH₃ and the H bonded to the oxygen. This there is no splitting.

Thanks for this!

For all those calculations, I still can't believe question 5.f.i was only one mark. Seems pretty harsh to be honest!

Why is there no splitting? wouldn't the CH3 have to split from the O atom?

[TiTo20]

(Original post by TheStudent.)
Thanks for this!

For all those calculations, I still can't believe question 5.f.i was only one mark. Seems pretty harsh to be honest!

Why is there no splitting? wouldn't the CH3 have to split from the O atom?

Only adjacent hydrogens next to the CH3 cause splitting, because oxygen isn't hydrogen, it doesn't split anything =]


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

[TheStudent.]

(Original post by TiTo20)
Only adjacent hydrogens next to the CH3 cause splitting, because oxygen isn't hydrogen, it doesn't split anything =]


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

We were never taught that but anyways thank youuuuuu! it all makes sense now lol.

Good luck for next wednesday guys!

[kishenp]

Does anyone have the Jan 2012 F335 and F334 paper and mark scheme ?
Thanks alot

[sunshinesmile10]

(Original post by kishenp)
Does anyone have the Jan 2012 F335 and F334 paper and mark scheme ?
Thanks alot

someones posted it above...read the thread first :P :P

[racing_stripes]

Does anyone have the marksheme for June 2012 salters a2 f335 just done it today and am very anxious!