evelync12
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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
Give your answer to three significant figures.
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-
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charco
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(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
Give your answer to three significant figures.
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
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evelync12
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(Original post by charco)
It's a typo. The 4 is missing
i still dont understand
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charco
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(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-
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Fiz_31
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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
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kewrkewrkwje
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Hi, what was the final answer for this question?
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Sabisaurus2
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@sabsrus_ on instagram for 2020 papers:
AQA Chemistry
AQA Biology
and Edexcel Maths
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lilyfrocks1
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Hiya. I can't find you on insta? Do you have the mark schemes for the 2021 papers ?
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deskochan
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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
Give your answer to three significant figures.
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
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Khalid Elmi
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(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
Give your answer to three significant figures.
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.
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deskochan
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(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
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14whollier
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#12
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(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
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sefth3
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(Original post by 14whollier)
I got the same answer bro
I got that as well
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