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Edexcel Chem 4 15/06/11

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Original post by Kirbester

Original post by Kirbester
That's okay, just know the two types I suppose, you'll be fine :smile:


Thanks. :wink:
So, how's revision?
Reply 21
Avatar for Kirbester
Kirbester
Original post by Vampire-Love4ever
Thanks. :wink:
So, how's revision?


Good, but I do need to learn quite a bit! Unit 4 is mostly application of knowledge though, so it should be okay :smile: you?
Original post by Kirbester

Original post by Kirbester
Good, but I do need to learn quite a bit! Unit 4 is mostly application of knowledge though, so it should be okay :smile: you?


Not so good. :frown: I feel like everything is evaporating.
Reading the textbook takes so much time and I need to do the pastpapers as well. I am really scared. :|
Reply 23
Avatar for Rosie2039
Rosie2039
Help me with this question please :smile:
10. This question concerns the titration of a solution of sodium hydroxide with a solution of
hydrochloric acid. As the titration proceeds the pH of the mixture changes.
(a) What was the pH when 24.95 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11

(b) What was the pH when 25.05 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11
Reply 24
Avatar for Rosie2039
Rosie2039
Original post by Vampire-Love4ever
Could anyone please explain this? :smile: (The ratio thing) :s-smilie:


See it basically uses the (n+1) formula, n is the number of neighboring protons and when u put the number in the formula u get the peak get it?
like in the first column its zero ya? in left side? for that u use the formula (0+1) = 1 <--- that's the no. of peak(s)
Hey does anybody know if we have to learn calculations for partial coefficient between two immiscible solvents?:smile:
(edited 12 years ago)
Original post by Rosie2039

Original post by Rosie2039
See it basically uses the (n+1) formula, n is the number of neighboring protons and when u put the number in the formula u get the peak get it?
like in the first column its zero ya? in left side? for that u use the formula (0+1) = 1 &lt;--- that's the no. of peak(s)


But, I don't get the ratio thing. :/
Reply 27
Avatar for Rosie2039
Rosie2039
Original post by Vampire-Love4ever
But, I don't get the ratio thing. :/


wat ratio thing? can u type it here? no wait
i think its the number of peaks not the ratio!
Reply 28
Avatar for mairori
mairori
How are preparations for unit 4 chemistry?
Original post by mairori
How are preparations for unit 4 chemistry?


Hey do you or anybody have practice questions for chemistry?? like a compilation of questions?? or anything .. would really help :smile: ...
peopleeeeee just one day?
What should we study, i mean , book is huge:\

well approx how much time is it going to take to read the unit 4 from book?
Original post by Vampire-Love4ever
Could anyone please explain this? :smile: (The ratio thing) :s-smilie:


Heyyy this not related to ratio as far as I think
Its just the peaks and stuff :\
But yeaa even I need to get clear with this topic, like the spectroscopy and all.

Kc Kp is fun anyways :biggrin:
Original post by Rosie2039
Help me with this question please :smile:
10. This question concerns the titration of a solution of sodium hydroxide with a solution of
hydrochloric acid. As the titration proceeds the pH of the mixture changes.
(a) What was the pH when 24.95 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11

(b) What was the pH when 25.05 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11


This question is crazy crazy hard, it took me aaagggeessss but basically for :

10a. Once you've added the two solutions in effect there is 0.05 cm3 of acid left un-neutralised, so do V/1000 x [] = no of moles
0.05/1000 x 1 = 5x10-5 moles
then all together you now have 49.95 cm3 of solution left so to find the [H+] and therefore pH you do V/1000 x [] = no of moles, again to get 49.95/1000 x [] = 5x10-5
(the 5x10-5 is from the previous answer) and on rearranging it [] = 1/999 (or 1.001x10-3) which you then -log10 to get a pH of 2.9996 (ie 3) so the answer is A. 3

b.This one i'm yet to work out because it has something to do with Kw I assume haha

but I hope the first one helped!
Reply 33
Avatar for cie6868
cie6868
Original post by Rosie2039
Help me with this question please :smile:
10. This question concerns the titration of a solution of sodium hydroxide with a solution of
hydrochloric acid. As the titration proceeds the pH of the mixture changes.
(a) What was the pH when 24.95 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11

(b) What was the pH when 25.05 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11


I just thought that (a) must be 3 because there is more acid than alkali. And (b) is 11 because there is more alkali than acid. I dunno if that's correct thinking, but it did work. Also consider that both solutions are 1M and both dissociate fully.
Reply 34
Avatar for aaronwu
aaronwu
Original post by Rosie2039
Help me with this question please :smile:
10. This question concerns the titration of a solution of sodium hydroxide with a solution of
hydrochloric acid. As the titration proceeds the pH of the mixture changes.
(a) What was the pH when 24.95 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11

(b) What was the pH when 25.05 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to
25 cm3 of 1.00 mol dm–3 HCl(aq)?
A 3
B 6
C 8
D 11



Okay, as the answer for (a) has been given already. Here's how to do (b) : use the same method as u did to find the answer to (a). You'd get 5x10^-5. which is the excess NaOH in the mixture. Then divide by 50.05/1000( total volume ) to get 9.99x10^-4 mol dm-3.

Then you -log 9.99x10^-4 to get 3 which is the pOH. As you know, pOH + pH = 14. So all you have to do is 14 - 3 = 11. Thus the answer is D. Hope i'm being clear enough. :/
i am tired and so not ready for chem unit 4 :\:angry:
i think everyone is sleeping :redface:
Reply 37
Avatar for Rosie2039
Rosie2039
Original post by hannahschmitty
This question is crazy crazy hard, it took me aaagggeessss but basically for :

10a. Once you've added the two solutions in effect there is 0.05 cm3 of acid left un-neutralised, so do V/1000 x [] = no of moles
0.05/1000 x 1 = 5x10-5 moles
then all together you now have 49.95 cm3 of solution left so to find the [H+] and therefore pH you do V/1000 x [] = no of moles, again to get 49.95/1000 x [] = 5x10-5
(the 5x10-5 is from the previous answer) and on rearranging it [] = 1/999 (or 1.001x10-3) which you then -log10 to get a pH of 2.9996 (ie 3) so the answer is A. 3

b.This one i'm yet to work out because it has something to do with Kw I assume haha

but I hope the first one helped!



Original post by cie6868
I just thought that (a) must be 3 because there is more acid than alkali. And (b) is 11 because there is more alkali than acid. I dunno if that's correct thinking, but it did work. Also consider that both solutions are 1M and both dissociate fully.



Original post by aaronwu
Okay, as the answer for (a) has been given already. Here's how to do (b) : use the same method as u did to find the answer to (a). You'd get 5x10^-5. which is the excess NaOH in the mixture. Then divide by 50.05/1000( total volume ) to get 9.99x10^-4 mol dm-3.

Then you -log 9.99x10^-4 to get 3 which is the pOH. As you know, pOH + pH = 14. So all you have to do is 14 - 3 = 11. Thus the answer is D. Hope i'm being clear enough. :/


Thank u all so much!

The calculations seem kinda hard though and its just one mark question... :s-smilie:
(edited 12 years ago)
Reply 38
Avatar for Rizam
Rizam
I'm hating acid-base equilibria .. :/ mainly the buffer stuff :/
can anyone explain how to get the "half equivalence point" ... i have this guide which has questions after the explainations n stuff, and umm i solved one question that confused me alot !! so help? anyone?

P.S havent touched spectroscopy yet! x[
Original post by Rizam
I'm hating acid-base equilibria .. :/ mainly the buffer stuff :/
can anyone explain how to get the "half equivalence point" ... i have this guide which has questions after the explainations n stuff, and umm i solved one question that confused me alot !! so help? anyone?

P.S havent touched spectroscopy yet! x[


HEYYYY I'll explain you this tmrw. I like this stuff. :smile:

I just have to polish up the concepts so I'll get them done till tmrw inshaAllah. :smile:

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