# chemistry aqa a-level question help

A solution that contains 605 mg of NaHSO4 in 100 cm3 of solution has a pH of 1.72 Calculate the value of K for the hydrogensulfate ion (HSO −) that is behaving as a weak acid
weak acid.
State the units of Ka

i dont understand the mark scheme:
[H+] = 10–1.72 (= 0.0191 (mol dm–3))
amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol))
initial [NaHSO ] = [HSO –] = M2 x 10 (= 5.04 x 10–2 (mol dm–3))
why is NaHSO = HSO-
Original post by evelync12
A solution that contains 605 mg of NaHSO4 in 100 cm3 of solution has a pH of 1.72 Calculate the value of K for the hydrogensulfate ion (HSO −) that is behaving as a weak acid
weak acid.
State the units of Ka

i dont understand the mark scheme:
[H+] = 10–1.72 (= 0.0191 (mol dm–3))
amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol))
initial [NaHSO ] = [HSO –] = M2 x 10 (= 5.04 x 10–2 (mol dm–3))
why is NaHSO = HSO-

It's a typo. The 4 is missing
Original post by charco
It's a typo. The 4 is missing

i still dont understand
Original post by evelync12
i still dont understand

Ionic compounds dissociate 100% into ions in solution.
Hence x mol dm-3 NaHSO4 is also x mol dm-3 HSO4-
Hi, can you please explain how to get the answer to this question.
can you please post the markscheme?
(edited 3 years ago)
Hi, what was the final answer for this question?
@sabsrus_ on instagram for 2020 papers:
AQA Chemistry
AQA Biology
and Edexcel Maths
Hiya. I can't find you on insta? Do you have the mark schemes for the 2021 papers ?
A solution that contains 605 mg of NaHSO4 in 100 cm3 of solution has a pH of 1.72 Calculate the value of K for the hydrogensulfate ion (HSO −) that is behaving as a weak acid
weak acid.
State the units of Ka

[H+] = –log 1.72 (= 0.0191 (mol dm–3))
amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol))
initial [NaHSO4 ] = [HSO4 –] = 5.04 x 10–2 mol dm–3

HSO4- --> H+ + SO4 2-

5.04 x 10–2 - 0.0191 0.0191 0.0191

Ka = [H+][SO42-]/[HSO4-] , then answer will be produced
(edited 2 years ago)
Original post by deskochan
A solution that contains 605 mg of NaHSO4 in 100 cm3 of solution has a pH of 1.72 Calculate the value of K for the hydrogensulfate ion (HSO −) that is behaving as a weak acid
weak acid.
State the units of Ka

[H+] = –log 1.72 (= 0.0191 (mol dm–3))
amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol))
initial [NaHSO4 ] = [HSO4 –] = 5.04 x 10–2 mol dm–3

HSO4- --> H+ + SO4 2-

5.04 x 10–2 - 0.0191 0.0191 0.0191

Ka = [H+][SO42-]/[HSO4-] , then answer will be produced

Did you get the answer to be 7.21 x10-3 moldm-3, I assumed [H+]=[SO4(2-)] since its a weak acid.
Original post by Khalid Elmi
Did you get the answer to be 7.21 x10-3 moldm-3, I assumed [H+]=[SO4(2-)] since its a weak acid.

I get 0.0166
I do not have the model answer and just according to the concept to develop.
(edited 2 years ago)
Original post by Khalid Elmi
Did you get the answer to be 7.21 x10-3 moldm-3, I assumed [H+]=[SO4(2-)] since its a weak acid.

I got the same answer bro
Original post by 14whollier
I got the same answer bro

I got that as well
amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol))
initial [NaHSO4 ] = [HSO4 –] = 5.04 x 10–2 mol dm–3

why did u multiply the 5.04x10^-3 with 10?
Original post by ayeshaam2
amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol))
initial [NaHSO4 ] = [HSO4 –] = 5.04 x 10–2 mol dm–3

why did u multiply the 5.04x10^-3 with 10?

concentration = moles / volume
concentration = 5.04x10^-3 / 0.1
the 0.1 is the volume 100cm^3 converted to dm^3
since HSO4^- H^+ + SO4^2- is in an equilibrium
we know that [H+] is 0.19055moldm^-3 so there has been an increase from the initial moles (conc.?) of 0 to 0.19055 and the same happens to SO4^2- since they are in a 1:1 ratio
as the conc of the products increase the conc of the reactant HSO4^- decreases by the same amount

HSO4^- H^+ + SO4^2-
intial conc : 0.050375 , 0 , 0
change in conc : -0.019055 , +0.019055 , +0.019055
final conc : 0.03132 , 0.019055 , 0.019055

you then use that in the Ka equation Ka = [H+] [SO4^2-] / [HSO4^2-] giving us Ka = 0.0116 units mol dm^-3
(edited 1 month ago)