# chemistry aqa a-level question help

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#1
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-
1
1 year ago
#2
(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
0
#3
(Original post by charco)
It's a typo. The 4 is missing
i still dont understand
0
1 year ago
#4
(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-
1
1 year ago
#5
Hi, can you please explain how to get the answer to this question.
can you please post the markscheme?
Last edited by Fiz_31; 1 year ago
5
1 week ago
#6
Hi, what was the final answer for this question?
1
1 week ago
#7
@sabsrus_ on instagram for 2020 papers:
AQA Chemistry
AQA Biology
and Edexcel Maths
0
4 days ago
#8
Hiya. I can't find you on insta? Do you have the mark schemes for the 2021 papers ?
0
4 days ago
#9
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
Last edited by deskochan; 4 days ago
0
4 days ago
#10
(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.
0
4 days ago
#11
(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.
Last edited by deskochan; 4 days ago
0
4 days ago
#12
(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
1
2 days ago
#13
(Original post by 14whollier)
I got the same answer bro
I got that as well
0
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