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-

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-

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-

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

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

Hi, what was the final answer for this question?

@sabsrus_ on instagram for 2020 papers:

AQA Chemistry

AQA Biology

and Edexcel Maths

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

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

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

(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

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

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.

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

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?

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

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)

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Last reply 6 days ago

Im confused about this chemistry question, why does it form these productsLast reply 6 days ago

Im confused about this chemistry question, why does it form these products