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
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-3 NaHSO4 is also x mol dm-3 HSO4-
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 3 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 3 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 year ago)
Original post by crazy_diamond
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
Thank you so much I was missing the taking away part and it had me wondering
Original post by crazy_diamond
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
but I have a question, does this only apply when there is a 1:1 ratio of the ionic products and the moles of the ionic products are the same?? pleaseee someone answer cause I'm loosing it. I'm going back on all kinds of this questions I've solved to check and it's getting really confusing
Original post by Emmaclems
but I have a question, does this only apply when there is a 1:1 ratio of the ionic products and the moles of the ionic products are the same?? pleaseee someone answer cause I'm loosing it. I'm going back on all kinds of this questions I've solved to check and it's getting really confusing
Hello Emmaclems!
A dissociation reaction is a type of reaction where a single compound breaks down into two or more simpler compounds.
The equilibrium constant of a dissociation reaction you how much a reaction likes to make products versus reactants. It's a fixed number, no matter how much compound you begin with.
For how things break apart, here's the formula:
K = [products]^n / [reactant]
[products] is how much product there is
[reactant] is how much starting material is there.
n is just a number from the balanced equation.
So, say hydrogen iodide (HI) splits into hydrogen (H^+) and iodide (I^-):
HI ==> H^+ + I^-
To get K, we need the amounts of everything when it's settled down. Say we have:
[HI] = 0.1 M
[H^+] = 0.05 M
[I^-] = 0.03 M
Then:
K = ([H^+] [I^-]) / [HI] = (0.05 M x 0.03 M) / 0.1 M = 0.015
That's K for this reaction. It doesn't change even if you start with way more HI.
Ratio: K = [H^+] [I^-] / [HI]
Like, up HI to 0.5 M, and K is still 0.015. Or cut H^+ down to 0.01 M, K is still 0.015.
It's special to each reaction and says where the balance is when things are steady. Changing the amounts doesn't change K itself.
Another one:
Okay, let's look at how iron(III) hydroxide, or Fe(OH)3, falls apart in water:
Fe(OH)3 ==> Fe^3+ + 3 OH^-
So, one Fe(OH)3 becomes one iron(III) ion (Fe^3+) and three hydroxide ions (OH^-).
To figure out the equilibrium constant (K) for this, we need the amounts of everything when the reaction's at equilibrium. Imagine we have:
[Fe(OH)3] = 0.1 M
[Fe^3+] = 0.05 M
[OH^-] = 0.15 M
The concentrations [Fe(OH)3] = 0.1 M, [Fe^3+] = 0.05 M, and [OH^-] = 0.15 M would be specifically chosen to represent a condition where the system is at equilibrium.
Here, the concentrations are picked to illustrate:
Relative value of [Fe(OH)3] is relatively high, which means that iron(III) hydroxide is abundant as precipitate.
Fe^3+ being low means that we have very few iron(III) ions.
This makes [OH^-] fairly high, since even a large concentration of hydroxide ions will still represent a small fraction of the total concentration of weak base.
In fact, by picking these concentrations, the example is probably meant to show how K is computed and how K itself is related to those values.
We can calculate K like this:
K = ([Fe^3+] [OH-]^3) / [Fe(OH)3] = (0.05 M x (0.15 M)^3) / 0.1 M = 0.0225
That K value (0.0225) is specific to this reaction and stays the same no matter how much stuff you start with.
For instance, even if we raise the Fe(OH)3 to 0.5 M, K is still 0.0225. Or, if we lower the Fe^3+ to 0.01 M, K remains at 0.0225.
That's because K is constant for the reaction itself. Changing how much of each substance you have doesn't change K; it's all about the balance between the stuff reacting and the stuff being made once equilibrium is reached.
The ratio is: K = [Fe^3+] [OH^-]^3 / [Fe(OH)3]
Hopefully, this helps you understand equilibrium constants a bit better!
Bye,
Sandro
(edited 2 weeks ago)
Original post by Emmaclems
but I have a question, does this only apply when there is a 1:1 ratio of the ionic products and the moles of the ionic products are the same?? pleaseee someone answer cause I'm loosing it. I'm going back on all kinds of this questions I've solved to check and it's getting really confusing
no it works for different ratios as well. you just multiply the change by the ratio so if there was a 1:2 ratio of idk HSO4^- and H^+ in this reaction then you would take away 0.019055 from the conc of HSO4^- like we did, but add 2 x 0.019055 to the conc of H^+. you just adapt this to any ratio but make sure you find the single change in conc first before you do.
also the explain above seems thorough so have a look at that
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