Mod 6 pH calculation from NSG 2022 Q34 (7 marks) (1 Viewer)

[SadCeliac]

Hi all,

I completely messed up this pH calculation - can someone please help me figure out where I went wrong???
https://thsconline.github.io/s/v/1828/North%20Sydney%20Girls%202022%20w.%20sol <-- Question 34


My solution

Their solution



Thanks!

 

[SadCeliac]

Their solution
View attachment 39147
View attachment 39148

In their solution, why have they found moles of benzoic acid first??

 

[HSCya1234567]

I think they've calculated the moles of Benzoic acid first to find the limiting reagent because as you can see in the reaction they've given, although NaOH and Benzoic acid react in a 1 to 1 ratio, there are excess Benzoic acid moles, which means whilst the rest will be neutralised, the pH will be calculated from the H+ ions leftover that were not reacted with the NaOH, whereas I think you've just straightup assumed that all of the H+ in the reaction is present after the system has reached equilibrium?

Not sure why you have found OH- or how you got the value that you say it is?


Not sure if any of that made sense sorry!

 

[SadCeliac]

I think they've calculated the moles of Benzoic acid first to find the limiting reagent because as you can see in the reaction they've given, although NaOH and Benzoic acid react in a 1 to 1 ratio, there are excess Benzoic acid moles, which means whilst the rest will be neutralised, the pH will be calculated from the H+ ions leftover that were not reacted with the NaOH, whereas I think you've just straightup assumed that all of the H+ in the reaction is present after the system has reached equilibrium?

Not sure why you have found OH- or how you got the value that you say it is?


Not sure if any of that made sense sorry!

That... kinda makes sense?

I'm pretty sure when I did my calculations I thought that I had to take into account the weak acid equilibrium, and from there I tried finding H+ ion moles. But when I did that I got more OH- moles than H+ moles, so I found OH- XS concentration.... but they seem to have done it differently, using the Ka much later in the calculation - why??

 

[HSCya1234567]

You dont have to take into account the weak acid equilibrium because it doesn't really matter for the neutralisation reaction. Take acetic acid and NaOH for example. When they react, you wouldn't write the equation with a reversible arrow because it isn't really reversible. This is because the OH ions keep reacting with the H+ ions from the weak acid which DOES EFFECT the equilibrium of the weak acid, and by lcp, it will favour the forward dissociation reaction to keep producing H+ ions since they are removed by the OH ions. Hence the overall reaction is not really reversible. This is the same reason why the titration of equal concs of NaOH with HCl requires the same volume as Acetic acid and NaOH.


So now, given that terribly lengthy and confusing explanation, you don't have to take into account the weak acid just as yet because it' still reaction normally with the NaOH, thus meaning you use limiting reagent to find excess H+ ions over. Since those excess H+ ions are not reacting with any OH ions, then an equilibrium system forms and you use the Ka

 

[SadCeliac]

You dont have to take into account the weak acid equilibrium because it doesn't really matter for the neutralisation reaction. Take acetic acid and NaOH for example. When they react, you wouldn't write the equation with a reversible arrow because it isn't really reversible. This is because the OH ions keep reacting with the H+ ions from the weak acid which DOES EFFECT the equilibrium of the weak acid, and by lcp, it will favour the forward dissociation reaction to keep producing H+ ions since they are removed by the OH ions. Hence the overall reaction is not really reversible. This is the same reason that the titration of equal concs of NaOH with HCl requires the same volume as Acetic acid and NaOH.


So now, given that terribly lengthy and confusing explanation, you don't have to take into account the weak acid just as yet because it' still reaction normally with the NaOH, thus meaning you use limiting reagent to find excess H+ ions over. Since those excess H+ ions are not reacting with any OH ions, then an equilibrium system forms and you use the Ka

Oh I see... so I find the H+ ions in XS then use Ka??

Can you also explain why they then use the ICE table after finding XS H+? I think I'm just not following their solution

Thanks

 

[HSCya1234567]

because now the H+ ions aren't reacting with anything because all the OH- have been reacted so the equation is not shifted to the right anymore, the benzoic acid is left in equilibrium and you have to use the ICE table to find how much H+ has really dissociated from the weak acid

hope your trial goes well!

 

[SadCeliac]

because now the H+ ions aren't reacting with anything because all the OH- have been reacted so the equation is not shifted to the right anymore, the benzoic acid is left in equilibrium and you have to use the ICE table to find how much H+ has really dissociated from the weak acid

Oh okay thanks

hope your trial goes well!

That sarcasm goes so hard

 

[HSCya1234567]

no stop i mean it lol

 

[SadCeliac]

no stop i mean it lol

lol ik

thank you, it's on friday - plan is to just spam papers until then

 

[HSCya1234567]

same mine's on friday, currently doing the ruse 2020 paper

 

[SadCeliac]

same mine's on friday, currently doing the ruse 2020 paper

oh I did that the other day (think I got like 75% I keep dropping marks on all the multis)

 

[HSCya1234567]

so valid, ive been reading the soluns as i go along and its not looking good....

 

[SadCeliac]

so valid, ive been reading the soluns as i go along and its not looking good....

rip :/