# Confused with acid/base question

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#1
Please can someone help with the question attached
They've stated on the mark scheme that at half of the equivalence point [A-] = [HA]
and then stated pKA = pH (when is this true??)

Then 'or'
moles OH added = 1.875 × 10–3 = moles A = moles HA left
and i don't understand where any of this came from!

Any help would be greatly appreciated Thank you!
2
3 years ago
#2
What exam board is this?
0
3 years ago
#3
(Original post by jazz_xox_)
Please can someone help with the question attached
They've stated on the mark scheme that at half of the equivalence point [A-] = [HA]
and then stated pKA = pH (when is this true??)

Then 'or'
moles OH added = 1.875 × 10–3 = moles A = moles HA left
and i don't understand where any of this came from!

Any help would be greatly appreciated Thank you!
the Henderson equation is pH = pKa + log(salt/acid)

at half neutralisation, half the acid is neutralised so half the salt has formed - therefore the conc of acid and salt are = they therefore cancel out leaving pH = pKa

Have you left out the Ka?
1
3 years ago
#4
(Original post by jazz_xox_)
Please can someone help with the question attached
They've stated on the mark scheme that at half of the equivalence point [A-] = [HA]
and then stated pKA = pH (when is this true??)

Then 'or'
moles OH added = 1.875 × 10–3 = moles A = moles HA left
and i don't understand where any of this came from!

Any help would be greatly appreciated Thank you!
You are given the moles of the weak acid = molarity x volume = 0.15 x 0.025 = 3.75 x 10-3

So at the half equivalence point the moles of base reacted must be half of this = 1.875 x 10-3
0
#5
(Original post by Kyber Ninja)
the Henderson equation is pH = pKa + log(salt/acid)

at half neutralisation, half the acid is neutralised so half the salt has formed - therefore the conc of acid and salt are = they therefore cancel out leaving pH = pKa

Have you left out the Ka?
Thank you! And I've never learnt the Henderson equation and it's not in either of my text books- it's not stated on my spec but maybe it's assumed knowledge so thank you for that
In that equation are the salt and acid in moles?

So at half neutralisation the acid and salt will equal, meaning pH will always equal pKa? Can I also ask, what exactly is pKa- is it still a measurement of the dissociation of the acid?

Sorry for so many questions
0
#6
(Original post by Yme2day)
What exam board is this?
AQA, it seems it requires knowledge not specifically stated on the spec though - someone on this thread stated an equation which helps me answer the question but i'd never really come across it before?
0
#7
(Original post by charco)
You are given the moles of the weak acid = molarity x volume = 0.15 x 0.025 = 3.75 x 10-3

So at the half equivalence point the moles of base reacted must be half of this = 1.875 x 10-3
Could you explain what you would do after this using this method?
So have 1.275x10-3 moles of both the acid and base reacted here, is there another way to do this here mathematically rather than stating the Henderson equation? Thank you
0
3 years ago
#8
(Original post by jazz_xox_)
Could you explain what you would do after this using this method?
So have 1.275x10-3 moles of both the acid and base reacted here, is there another way to do this here mathematically rather than stating the Henderson equation? Thank you
yes

You can use the acid dissociation equation:

ka = [H+][A-]/[HA]

at the half equivalence point [A-] = [HA]

so

ka = [H+]
0
3 years ago
#9
(Original post by jazz_xox_)
Thank you! And I've never learnt the Henderson equation and it's not in either of my text books- it's not stated on my spec but maybe it's assumed knowledge so thank you for that
In that equation are the salt and acid in moles?

So at half neutralisation the acid and salt will equal, meaning pH will always equal pKa? Can I also ask, what exactly is pKa- is it still a measurement of the dissociation of the acid?

Sorry for so many questions
pKa is the logorithmic version Ka, which as you said is the dissociation constnat of an acid.
0
#10
Another question I've come across involving this is:

A mixture, formed by adding 50 cm3 of 1.00 mol dm–3 salicylic acid solution to 25 cm3 of 1.00 mol dm–3 sodium hydroxide solution, can be used to determine the pKa of salicylic acid. State one measurement that must be made for this mixture and explain how this measurement can be used to determine the pKa of salicylic acid.

The mark scheme simply states:
pH
[HA] = [A-] (half neutralised)
pH = pKA

So if you determine the pH of the solution, are they saying you then have to make it that it's half neutralised? I'm so confused I've never come across this before!
0
3 years ago
#11
(Original post by jazz_xox_)
AQA, it seems it requires knowledge not specifically stated on the spec though - someone on this thread stated an equation which helps me answer the question but i'd never really come across it before?
I mean I do OCR and have never come across the Henderson equation either :/
1
#12
(Original post by charco)
yes

You can use the acid dissociation equation:

ka = [H+][A-]/[HA]

at the half equivalence point [A-] = [HA]

so

ka = [H+]
Thank you SO much! would you be able to take a look at the other question I posted, still on this thread? It's not mathematical, just regarding the theory of pKa
0
#13
(Original post by Yme2day)
pKa is the logorithmic version Ka, which as you said is the dissociation constnat of an acid.
Got it, thank you so much Could you look at the question I've also posted on this thread- not sure exactly what it means but it's still regarding pKa and half-neutralisation. thanks again!
0
#14
(Original post by Yme2day)
I mean I do OCR and have never come across the Henderson equation either :/
I had 2 tutors to do this topic (both with chemistry phds) and never came across it
I guess it's rare in questions but sometimes they must expect you to know it
0
3 years ago
#15
(Original post by jazz_xox_)
Thank you! And I've never learnt the Henderson equation and it's not in either of my text books- it's not stated on my spec but maybe it's assumed knowledge so thank you for that
In that equation are the salt and acid in moles?

So at half neutralisation the acid and salt will equal, meaning pH will always equal pKa? Can I also ask, what exactly is pKa- is it still a measurement of the dissociation of the acid?

Sorry for so many questions
Sorry, no its not moles its the concentration of the salt and the conc of the acid

yes

yes, but the lower the pka, the stronger the acid, where as with Ka, the higher the stronger.

you can convert from pKa to Ka by doing 10^-pKa
1
3 years ago
#16
(Original post by Yme2day)
I mean I do OCR and have never come across the Henderson equation either :/
(Original post by jazz_xox_)
I had 2 tutors to do this topic (both with chemistry phds) and never came across it
I guess it's rare in questions but sometimes they must expect you to know it
This is strange

Is it on the old spec of your exam boards? It was on mine (WJEC)

You don't need to use the equation. Using the standard Ka expression is fine, it's just the mark scheme went with the simpler method. You would've got the same answer using your method

So don't bother with the HH equation if you don't want to - its just a simplification
0
#17
(Original post by Kyber Ninja)
This is strange

Is it on the old spec of your exam boards? It was on mine (WJEC)

You don't need to use the equation. Using the standard Ka expression is fine, it's just the mark scheme went with the simpler method. You would've got the same answer using your method

So don't bother with the HH equation if you don't want to - its just a simplification
Possibly old spec, the questions I'm doing are occassionally old spec so that might be the case.
It's something to think of I suppose in a half-neutralisation question I'm sure it'll jump out at me in future haha! Thanks again
0
3 years ago
#18
(Original post by jazz_xox_)
Another question I've come across involving this is:

A mixture, formed by adding 50 cm3 of 1.00 mol dm–3 salicylic acid solution to 25 cm3 of 1.00 mol dm–3 sodium hydroxide solution, can be used to determine the pKa of salicylic acid. State one measurement that must be made for this mixture and explain how this measurement can be used to determine the pKa of salicylic acid.

The mark scheme simply states:
pH
[HA] = [A-] (half neutralised)
pH = pKA

So if you determine the pH of the solution, are they saying you then have to make it that it's half neutralised? I'm so confused I've never come across this before!
If you measure the pH, you can get [H+] from it and plug it into the normal expression for ka and then convert to pKa

the MS is stupid, examiners are using positive marking - you can do all that the normal way like I've said
0
#19
(Original post by Kyber Ninja)
If you measure the pH, you can get [H+] from it and plug it into the normal expression for ka and then convert to pKa

the MS is stupid, examiners are using positive marking - you can do all that the normal way like I've said
Okay thank you! So in that case you could just do Ka = [H+]^2 / [HA], to find Ka and then to -log10Ka?
0
3 years ago
#20
(Original post by jazz_xox_)
Okay thank you! So in that case you could just do Ka = [H+]^2 / [HA], to find Ka and then to -log10Ka?
For just acids yes.

But for buffer questions, A- isn't the same as H+. You get A- concentration from the base after it's reacted with H+. I.e. the number of mol left if that makes sense
0
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