The Student Room Group

AQA A-level Chemistry 7405 - Paper 3 - 19th June 2019 unofficial mark scheme

Scroll to see replies

Why are you overcomplicating a simple 6 marker?
Original post by sam23478
For the Ka question I'm not convinced you guys have the right answer. Usually we're supposed to assume concentration of HX is constant whereas some of it will react to form H+/X-. It usually changes so little that the concentration is the same to 3sf. However in this case this was not a valid assumption as HSO4- was actually quite a strong acid and the actual concentration was not the same as the original concentration to 1sf even. Not only that but if you take this into account usually when finding ph this is quite tough but in this case it was very straightforward as you knew the concentration of H+ already.
Could you have done colorimetry for the HCl sodium thiosulfate experiment?
Original post by zixd
I don’t think it was -0.8V it was positive cause 0.8+0.33=1.13






Agreed. E cell value has to be positive
Cause I think that that is the right thing to do. Also I wasn't overcomplicating it, since its all in a fixed volume to find the new [HSO4-] you just subtract the [H] from the original one. So much of that HSO4- is going to dissociate that I don't see how it can be right not to take it into account.
Original post by lmaooome
Why are you overcomplicating a simple 6 marker?
Reply 84
Original post by CeliaaaWuuu
isn't the % uncertainty =0.005/0.685 something


I got 1.47 too
Reply 85
Uncertainty was in two values so you had to x answer by 2
Reply 86
Original post by sam23478
Cause I think that that is the right thing to do. Also I wasn't overcomplicating it, since its all in a fixed volume to find the new [HSO4-] you just subtract the [H] from the original one. So much of that HSO4- is going to dissociate that I don't see how it can be right not to take it into account.


One of the assumptions for weak acids is that it doesn’t dissociate much. So HSO4- initial = HSO4- equilibrium
Original post by Jdjdjj
I got 1.47 too


But doesn’t 0.005/0.685 * 100 (%) get you like 0.7%?
Reply 88
Yeah I did that
Original post by mpatel12
It was once used to measure mass of container and once to measure mass of container and solid. Uncertainty will be ×2
Reply 89
Yh that’s right but multiply by 2
Original post by science_geeks
But doesn’t 0.005/0.685 * 100 (%) get you like 0.7%?
Original post by Jdjdjj
Yh that’s right but multiply by 2


Why would you need to multiply by 2? Feel like I’ve done something wrong here...
Everyone saying salt bridge for the question asking what is A. They asked for the APPARATUS, not what it is. So it's a U - TUBE. the next question you could say use it as a salt bridge
Gb predictions? 246 was A* in 2017 and 2018 was 241.
'A weak acid is an acid chemical which does not dissociate (split into ions) completely in water solution. This means it does not give all its hydrogen ions into the water. Weak acids typically have a pH between 3 and 6.' There is nothing about assuming [HX] being constant in that definition. The fact that the PH was like 1.4 or something showed it wasn't a typical weak acid and that assumption is just not valid. Usually if you work it out the concentrations are the same to 3sf but in this case they weren't to 1sf.
Original post by achiu
One of the assumptions for weak acids is that it doesn’t dissociate much. So HSO4- initial = HSO4- equilibrium
I thought the same as you but thought that since it was a 6 marker and not worth more marks (as it wouldn't be in an A2 question anyway) I just assumed that [HA] remained constant, so did mass/volume. I think you're right though - [HA] would decrease so you would expect a larger Ka value. The real standard Ka value for HSO4^- is apparently 1.2*10^-2moldm^-3 which is larger than 7*10^-3mol^dm^-3 that was calculated so this matches up with the previous statement
'A weak acid is an acid chemical which does not dissociate (split into ions) completely in water solution. This means it does not give all its hydrogen ions into the water. Weak acids typically have a pH between 3 and 6.' There is nothing about assuming [HX] being constant in that definition. The fact that the PH was like 1.4 or something showed it wasn't a typical weak acid and that assumption is just not valid. Usually if you work it out the concentrations are the same to 3sf but in this case they weren't to 1sf.
It’s a salt bridge
Original post by tsr unofficial
Everyone saying salt bridge for the question asking what is A. They asked for the APPARATUS, not what it is. So it's a U - TUBE. the next question you could say use it as a salt bridge
It could have easily been a strip of filter paper soaked in KNO3 solution. The diagram didn't specify, so it's safe to assume that putting 'salt bridge' is probably okay
Original post by tsr unofficial
Everyone saying salt bridge for the question asking what is A. They asked for the APPARATUS, not what it is. So it's a U - TUBE. the next question you could say use it as a salt bridge
Thanks! Yeah I got 1.2*10^-2.
Original post by pRex
I thought the same as you but thought that since it was a 6 marker and not worth more marks (as it wouldn't be in an A2 question anyway) I just assumed that [HA] remained constant, so did mass/volume. I think you're right though - [HA] would decrease so you would expect a larger Ka value. The real standard Ka value for HSO4^- is apparently 1.2*10^-2moldm^-3 which is larger than 7*10^-3mol^dm^-3 that was calculated so this matches up with the previous statement
Walk us through how u did that
Original post by sam23478
Thanks! Yeah I got 1.2*10^-2.
Hoping they'll credit both answers though. It all depends on the assumptions you make, how weak or strong the acid is etc. Don't think they've ever posed a question (even in the previous AQA spec, not just this one) where they specifically expect you to deduct [H+] from initial [HA] for the final [HA], even though it's technically the right thing to do (with really weak acids, it doesn't make much of a difference so the simplification is okay to use)
Original post by sam23478
Thanks! Yeah I got 1.2*10^-2.

Quick Reply

Latest

Trending

Trending