chem -alevels Watch

studybloomer
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A chemical company has a waste tank of volume 25 000dm3. The tank is full of phosphoric acid (H3PO4) solution formed by adding some unwanted phosphorus (v) oxide to water in the tank.

A 25.0cm3 sample of this solution required 21.2cm3 of 0.500 moldm-3 sodium hydroxide solution for complete reaction.

Calculate the mass in kg of phosphorus(v) oxide that must have been added to the water in the waste tank.

The answer is 251kg, anyone care to explain please the equations to use?

I've seen another person equations they used but , it doesn't fit with ms , but they do get the same answer
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CheeseIsVeg
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(Original post by studybloomer)
A chemical company has a waste tank of volume 25 000dm3. The tank is full of phosphoric acid (H3PO4) solution formed by adding some unwanted phosphorus (v) oxide to water in the tank.

A 25.0cm3 sample of this solution required 21.2cm3 of 0.500 moldm-3 sodium hydroxide solution for complete reaction.

Calculate the mass in kg of phosphorus(v) oxide that must have been added to the water in the waste tank.

The answer is 251kg, anyone care to explain please the equations to use?

I've seen another person equations they used but , it doesn't fit with ms , but they do get the same answer
Hi there!
I've just done it, quite a nice little problem you have there!

now I've used the following equations:

H20 P2O5 --> H3PO4

H3PO4 NaOH --> Na3PO4 H2O

[ Neither are balanced in order to leave this as an exercise to you. ]

that last one is a bit tricky (or at least I thought it might be) I think different ones may be around, however I think this is the most simple!

I have worked it out and got the correct answer, if you're still having issues I can get back and help you

hope this helps,
cheese
(this was fun )
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Last edited by CheeseIsVeg; 4 weeks ago
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studybloomer
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HI, thanks for your help. I am working through it and i don't seem to understand why we need 2 separate equations? Also, should i change my moles from the 25 cm3 to 25000 dm3? How?

Thanksss

Okay, so I managed to convert and work out the new moles. I am stuck on why we need to 2 separate equations. Also, I worked out the moles of H3PO4, which was 3530 mol. (for the 2nd equation)
How do I now work out the mole ratio of that to P205? The fact that there are two equations, is throwing me off.

3H20 + P2O5 --> 2H3PO4

H3PO4 + 3NaOH --> Na3PO4+ 3 H2O
Last edited by studybloomer; 4 weeks ago
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CheeseIsVeg
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(Original post by studybloomer)
HI, thanks for your help. I am working through it and i don't seem to understand why we need 2 separate equations? Also, should i change my moles from the 25 cm3 to 25000 dm3? How?

Thanksss

Okay, so I managed to convert and work out the new moles. I am stuck on why we need to 2 separate equations. Also, I worked out the moles of H3PO4, which was 3530 mol. (for the 2nd equation)
How do I now work out the mole ratio of that to P205? The fact that there are two equations, is throwing me off.

3H20 + P2O5 --> 2H3PO4

H3PO4 + 3NaOH --> Na3PO4+ 3 H2O
Hey - no problem!

You nailed that balancing task :yy:

So you need two equations to relate to the data you are given in the question.

The first equation describes the solution in the big tank
The second equation describes the reaction that the data for NaOH comes from (it's a titration - used to work out the concentration of the phosphoric acid in the sample).

Your n for H3PO4 are correct! Nice job.
Your next step is to be to relate that to your solution in the tank.
HINT - that second equation and that you balanced it!!!

Hope this helps!
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studybloomer
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Thanks for explaining in detail and being encouraging. Please, may I ask: how am I meant to know this is a titration style question? Are there are clues?

I'm confused on finding the mole ratio between the H3P04 and P205 between the two equations.

From the second equation, I have 1 mole, for the first equation, I have 1:2 molar ratio between P205 AND H3P04.
Would I just divide the 3530 mol by 2?
This gives: 1765 mol.
To work out mass: mr x mol
The mr is 142 and multiplying it with 1765 gives: 250630 grams and converting this = 251 kg!

At last I got it! Wouldn't have been possible without your help! Thanks again, but would you mind explaining, how I am meant to know its a titration style question and know that I should not just add the h20 and p205 and NaOH in to one equation?

(Original post by CheeseIsVeg)
Hey - no problem!

You nailed that balancing task :yy:

So you need two equations to relate to the data you are given in the question.

The first equation describes the solution in the big tank
The second equation describes the reaction that the data for NaOH comes from (it's a titration - used to work out the concentration of the phosphoric acid in the sample).

Your n for H3PO4 are correct! Nice job.
Your next step is to be to relate that to your solution in the tank.
HINT - that second equation and that you balanced it!!!

Hope this helps!
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CheeseIsVeg
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(Original post by studybloomer)
Thanks for explaining in detail and being encouraging. Please, may I ask: how am I meant to know this is a titration style question? Are there are clues?

I'm confused on finding the mole ratio between the H3P04 and P205 between the two equations.

From the second equation, I have 1 mole, for the first equation, I have 1:2 molar ratio between P205 AND H3P04.
Would I just divide the 3530 mol by 2?
This gives: 1765 mol.
To work out mass: mr x mol
The mr is 142 and multiplying it with 1765 gives: 250630 grams and converting this = 251 kg!

At last I got it! Wouldn't have been possible without your help! Thanks again, but would you mind explaining, how I am meant to know its a titration style question and know that I should not just add the h20 and p205 and NaOH in to one equation?
I would say the major clue for the titration calculation is the data you were given.

You have a base volume and concentration (NaOH) and then the resulting volume of acid used to neutralise it.
For me it's an obvious indicator the equation n=cv has to be used and it's data from a titration

So glad you got it - you didn't even need my help :lol:

The two parts to the question are separate.

Ie: You cannot use the NaOH + the P2O5 without difficulty unless you split it into 2 different separate equations.
How else would you then calculate the mass of P2O5 and then get the molecular ratios correct for both parts of the equation?

The part I thought was hard, was knowing what the result of the H3PO4 + NaOH equation was. I had to Google to double check mine was right !

Have a great evening, if you need a hand with any other chemistry questions, feel free to tag me - I'll be making a big revision thread on the 30th
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