TargetLost
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#1
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#1
"Your factory has accidentally spilled 1078 kg of phosphoric acid into a lake, making it very acidic. How much solid caustic soda (NaOH) would you have to add to get the pH to 8.0?"

The answer is 840kg

I have no idea how to get this though - can anyone help?
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Pigster
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#2
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#2
(Original post by TargetLost)
"Your factory has accidentally spilled 1078 kg of phosphoric acid into a lake, making it very acidic. How much solid caustic soda (NaOH) would you have to add to get the pH to 8.0?"

The answer is 840kg

I have no idea how to get this though - can anyone help?
Oh come on, you must have some ideas and must have tried some calculations.
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GiveMeCoffee4
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#3
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#3
First find the number of hydrogen ions that have dissociated into the lake.
Then find the number of hydroxide ions to neutralise this, and find the mass of NaOH needed to produce this many hydroxide ions.
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Pigster
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#4
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(Original post by GiveMeCoffee4)
First find the number of hydrogen ions that have dissociated into the lake.
Given that H3PO4 is a weak acid with three different Ka values, it would be much simpler to work out how many mol of acid are added and how many mol of NaOH would be needed to neutralise them and then how many mol of NaOH would be required to go from pH 7 to pH 8. BUT since we don't know the volume of the lake, we cannot possibly know the last bit. Never mind the problem of the conjugate bases produced by neutralising the H3PO4 themselves being alkaline which would mean that at equivalence point, it would likely be quite a ways higher than pH 8, making this a hella difficult problem even if we had the lake volume, which we don't. This means that the person setting this Q hasn't really thought it through properly.
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TargetLost
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#5
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#5
(Original post by Pigster)
Given that H3PO4 is a weak acid with three different Ka values, it would be much simpler to work out how many mol of acid are added and how many mol of NaOH would be needed to neutralise them and then how many mol of NaOH would be required to go from pH 7 to pH 8. BUT since we don't know the volume of the lake, we cannot possibly know the last bit. Never mind the problem of the conjugate bases produced by neutralising the H3PO4 themselves being alkaline which would mean that at equivalence point, it would likely be quite a ways higher than pH 8, making this a hella difficult problem even if we had the lake volume, which we don't. This means that the person setting this Q hasn't really thought it through properly.
This is what I tried I think... but it gave me 1320kg of NaOH before even trying to find the amount from 7 to 8 pH
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Pigster
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#6
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#6
(Original post by TargetLost)
This is what I tried I think... but it gave me 1320kg of NaOH before even trying to find the amount from 7 to 8 pH
1320 kg is the correct mass of NaOH needed to reach the equivalence point.
Last edited by Pigster; 2 years ago
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TargetLost
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#7
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#7
(Original post by Pigster)
You missed my point, without knowing the volume of the lake it is an impossible ask.

If the lake were 1 cm3 it would require a whole lot less NaOH to raise the pH from 7 to 8 than would a 1x10^100 cm3 lake.
Yes I saw that, I was just saying that only to neutralise the acid requires 1320kg (I think?) of NaOH but the answer is just 840kg...

But how would you work it out using pK values?
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Pigster
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#8
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#8
(Original post by TargetLost)
Yes I saw that, I was just saying that only to neutralise the acid requires 1320kg (I think?) of NaOH but the answer is just 840kg...

But how would you work it out using pK values?
Do you know the volume of the lake?
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TargetLost
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#9
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#9
(Original post by Pigster)
Do you know the volume of the lake?
No
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Pigster
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#10
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#10
(Original post by TargetLost)
No
What level are you? I'm guessing A level.
Was this a Q you were just given and expected to be able to answer? i.e was this a Q you were expected to do research to be able to answer?
Were you given the Ka values? If it isn't (as I suspect) a research activity, if you weren't given the Ka values, then they can't be needed and therefore worrying about them is the wrong thing to do.

How do you know the answer is 840 kg?
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TargetLost
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#11
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(Original post by Pigster)
What level are you? I'm guessing A level.
Was this a Q you were just given and expected to be able to answer? i.e was this a Q you were expected to do research to be able to answer?
Were you given the Ka values? If it isn't (as I suspect) a research activity, if you weren't given the Ka values, then they can't be needed and therefore worrying about them is the wrong thing to do.

How do you know the answer is 840 kg?
Actually I just started a biochemistry degree, and from the question (I forgot to put in the OP) "pKs of phosphate = 2, 7 and 11"
However we haven't learnt much about acid dissociation and I'm not sure what I need to know for this question or in general

the answer is 840kg because it's from an online multiple choice question I was given
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Pigster
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#12
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#12
(Original post by TargetLost)
Actually I just started a biochemistry degree, and from the question (I forgot to put in the OP) "pKs of phosphate = 2, 7 and 11"
However we haven't learnt much about acid dissociation and I'm not sure what I need to know for this question or in general

the answer is 840kg because it's from an online multiple choice question I was given
Aha, now we're talking.

There are three reactions. The first is
H3PO4 + OH- -> H2PO4^- + H2O (this needs 440 kg of NaOH).
H2PO4^- is still pretty damn acidic, so you'll still be nowhere near the pH you need.

The second is:
H2PO4^- + OH- -> HPO4^2- + H2O
And this is the one where it gets interesting. Hopefully you did A level chem and this should ring a bell or two (if you didn't do A level chem, I'd start boning up as quick as you can, especially about buffers, which this is an example of)

Ka = [H+] [HPO4^2-] / [H2PO4^-]
Since pKa = 7, then Ka = 10^-pKa = 10^-7. Also the desired pH = 8, [H+] = 10^-pH = 10^-8
Rearranging the above to Ka / [H+] = [HPO4^2-] / [H2PO4^-] = 10 (since 10^-7 / 10^-8 = 10) i.e. you need 10 [HPO4^2-] compared to [H2PO4^-].
Now since the volume of the lake is the same for both, the volume concentration terms cancels out, so you therefore need 10x the amount of HPO4^2- compared to the amount of H2PO4^-.
Since it takes 440 kg to convert all of the H2PO4^- to HPO4^2- and we need a 10:1 ratio, we need to use 10/11th of the 440, i.e. 400.
440 + 400 = 840.
w00t

There probably is a much more elegant degree level way of doing it, but this works.
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TargetLost
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#13
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#13
(Original post by Pigster)
Aha, now we're talking.

There are three reactions. The first is
H3PO4 + OH- -> H2PO4^- + H2O (this needs 440 kg of NaOH).
H2PO4^- is still pretty damn acidic, so you'll still be nowhere near the pH you need.

The second is:
H2PO4^- + OH- -> HPO4^2- + H2O
And this is the one where it gets interesting. Hopefully you did A level chem and this should ring a bell or two (if you didn't do A level chem, I'd start boning up as quick as you can, especially about buffers, which this is an example of)

Ka = [H+] [HPO4^2-] / [H2PO4^-]
Since pKa = 7, then Ka = 10^-pKa = 10^-7. Also the desired pH = 8, [H+] = 10^-pH = 10^-8
Rearranging the above to Ka / [H+] = [HPO4^2-] / [H2PO4^-] = 10 (since 10^-7 / 10^-8 = 10) i.e. you need 10 [HPO4^2-] compared to [H2PO4^-].
Now since the volume of the lake is the same for both, the volume concentration terms cancels out, so you therefore need 10x the amount of HPO4^2- compared to the amount of H2PO4^-.
Since it takes 440 kg to convert all of the H2PO4^- to HPO4^2- and we need a 10:1 ratio, we need to use 10/11th of the 440, i.e. 400.
440 + 400 = 840.
w00t

There probably is a much more elegant degree level way of doing it, but this works.
Thank you very much!
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