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# Urgent Chemistry help! Due tomorrow! watch

1. I have to write a plan for a Chemistry practical due tomorrow!
We have to figure out the concentration of H2SO4
The concentration is roughly 10M and we can choose the amount that we use.
We also have 0.1M concentration of NaOH, we can also choose how much of the NaOH that we use. We also have a solid lump of MgCo3.
Using 2 formulas that I have figured out:

H2SO4 + 2NaOH = Na2SO4 + H2O and
MgCO3 + H2SO4 = MgSO4 +H2O +CO2

To work out the concentration, ie molarity of the H2SO4 I would need the number of moles and the volume of the H2SO4 according to the formula N=CV.

We are told to use titration and gas collection. So I would be using the titration to figure out the number of moles, and the gas collection to figure out the volume. (Not sure but this is what I have worked out)

Now once I have my NaOH in a conical flask, and my acid (which i will have dilued to 0.1M) in a burette, using the Phenolphthalein (sp) indicator, I'll do the titration. At the end of this I will get a number which is the amount of acid that was used to neutralize the NaOH. Now I don't exactly know what to do next!!!
From this I should be able to figure out the nuber of moles of H2SO4 used to neutralize the NaOH. I know this would be something to do with the the 1:2 ratio in the formula, and figuring out the number of moles from the N=CV thing, but I can't see how to do this! Any help would be much appreciated!! Also, I really don't have much idea what to do for the gas collection part, I'm asusming somehow the amount of CO2 collected and the fact that the ratio is 1:1 and that there are 2400dm3 to 1 mole of gas must have something to do with something.....
Thanks a lot
Gem
2. When you do the titration, you'll have a certain volume of NaOH which you'll know (depends whether you are putting alkali or acid in the burrette). As the concentration is 0.1M:

moles of NaOH=0.1xvolume in dm3

Moles of H2SO4=NaOH/2 (as the stoichiometric (sp) ratio is 1:2)

You'll know the volume of H2SO4 that you've added, so then just use:

Concentration=moles/volume
3. So from this titration alone, I should be able to work out the concentration of the H2SO4???? I'm not questioning you, it's just that we are told that we have to do a titration and gas meaurement for this, surely they wouldn't ask us to do 2 experiments if we could work out the concentration from one simple titration??
4. Yeah, I don't understand that either, but you can definately get the concentration by a titration alone. I've never done gas collection before, so couldn't help you with that I'm afraid. Which board are you with?
5. I'm with the OCR board. It just seems a bit easy for an A level practical don't you think??
6. (Original post by Gem12318)
I'm with the OCR board. It just seems a bit easy for an A level practical don't you think??
Yeah, I know what you mean. Perhaps there's something on the net. Sorry I can't help you with the gas collection bit, but I'm sure there is someone on here who can. Perhaps try searching the other threads on this subsection for "gas collection".

I've just looked, and I think you need to collect the gas in a gas syringe. From this, you can use either pV=nRT or the 24000cm3 (although I think if you used pV=nRT, they'd be more impressed as this is more accurate if you're not dealing with standard temperature).

Then, repeat the process as before (except this time the ratio is 1:1, so of course the number of moles of H2SO4 is equal to the number of moles of CO2).
7. So I'd collect the CO2 gas, and would measure the amount that went into the gas syringe before it stopped making the gas?? Would it stop?? How do I know how much H2SO4 to use for this? I don't really understand, because if I used varying amounts of H2SO4 to react with the MgCO3 it would produce different amount of gas surely??
8. (Original post by Gem12318)
So I'd collect the CO2 gas, and would measure the amount that went into the gas syringe before it stopped making the gas?? Would it stop?? How do I know how much H2SO4 to use for this? I don't really understand, because if I used varying amounts of H2SO4 to react with the MgCO3 it would produce different amount of gas surely??
If you add an excess of the MgCO3, then all the H2SO4 should react with it. If you stick the MgCO3 in then immediately slam on the bung with the tube in so the gas can be collected into the syringe, then you shouldn't lose any. Have a look here:
http://www.wpbschoolhouse.btinternet...reparation.htm
9. Yes, but what I'm saying is, that the gas syringe only holds 100dm3 of gas, so how would I know how much H2SO4 to use, because it may produce more gas than this, then I wouldn't be able to record it all.
10. (Original post by Gem12318)
Yes, but what I'm saying is, that the gas syringe only holds 100dm3 of gas, so how would I know how much H2SO4 to use, because it may produce more gas than this, then I wouldn't be able to record it all.
Well, it says that the concentration is roughly 10M, so dilute it using distilled water then you can approximate how many moles of H2SO4 you've got in the volumetric flask/whatever you're reacting in so you have a decent amount of solution but not a whole lto of gas evolving. (So make the concentration lower so the number of moles is much lower. eg, if you diluted down to roughly 0.25M and used 0.01dm (10cm^3), you'd only evolve 60cm^3 of gas). Dilute!
11. So, if I was to say, dilute the acid to roughly 0.1M, how much would I have to use to produce a decent amount of gas ie 60dm3 plus. I'm sorry I'm sure these are really easy calculations my brain just seems to have died on me this evening! Way too much work and revision etc etc! I really do appreciate your responses though!
12. (Original post by Gem12318)
So, if I was to say, dilute the acid to roughly 0.1M, how much would I have to use to produce a decent amount of gas ie 60dm3 plus. I'm sorry I'm sure these are really easy calculations my brain just seems to have died on me this evening! Way too much work and revision etc etc! I really do appreciate your responses though!
If you diluted to 0.1M, then the volume of gas=volume of solution*conc*24 (in dm3)

So volume of solution=volume of gas/(24*conc)
(and I think you mean 0.6dm3 or 60cm3?)
So volume=0.6/(24*0.1)
=0.6/2.4
=0.25dm3
13. I thought it was 1000 dm3 to a cm3?? So wouldn't 60cm3 be 0.006dm3??
So if I needed 0.25dm3 of H2SO4 to make 60cm3 of gas that's 25cm3 of 0.1M acid? or 250?? Ahh, think I need sleep! lol, I'm not stupid really....
14. (Original post by Gem12318)
I thought it was 1000 dm3 to a cm3?? So wouldn't 60cm3 be 0.006dm3??
So if I needed 0.25dm3 of H2SO4 to make 60cm3 of gas that's 25cm3 of 0.1M acid? or 250?? Ahh, think I need sleep! lol, I'm not stupid really....
It's the other way around. 1000 cm3 make a dm3. I'm sure you're not stupid. (I've been in the tired place before). And yess, I've made a mistake. If you want 60cm3, it's 0.06dm3:

So volume of acid=0.06/2.4
=0.025dm3
=25cm3

Sorry! And good luck for tomorrow.
15. Thank you sooooooooo much, you've saved me from a certain failure!!! I have to ask? What do you do? Are you a student or ex student or something?
16. Oh ignore that, I just read your profile! Lol, You're 2 years younger than me and 100 times as smart!
Gemma
17. Sorry to be bombarding with posts, but with the titration, do you think that 1.0M of H2SO4 would be too strong, and that it would immediately neutralize the Base?? This could be a problem as it might mean that just one or 2 drops would be needed, or perhaps less and I wouldn't be able to read it accurately. Do you think 0.1M would be more suitable?? Or would that be too weak? Any ideas?
18. (Original post by Gem12318)
Sorry to be bombarding with posts, but with the titration, do you think that 1.0M of H2SO4 would be too strong, and that it would immediately neutralize the Base?? This could be a problem as it might mean that just one or 2 drops would be needed, or perhaps less and I wouldn't be able to read it accurately. Do you think 0.1M would be more suitable?? Or would that be too weak? Any ideas?
Approximately 0.1M concentration would be fine, as you'd need about half as much acid as alkali. The only difficulty would be in preparing such a solution, as you'll need a large volumetric flask to allow a good quantity of the concentrated acid to be added so the limitations of the equipment are reduced (you need a flask with 100 times the volume of the acid you are adding if you are diluting to 0.1M).
19. I was gonna use a 250cm3 flask. So a 1.0M acid solution would be too strong??? It is plausable, that using the 250 volumetric flask, I could add 2.5 cm3 of acid and then fill the rest up with water... That would make a 0.1M solution would it not?
20. (Original post by Gem12318)
I was gonna use a 250cm3 flask. So a 1.0M acid solution would be too strong??? It is plausable, that using the 250 volumetric flask, I could add 2.5 cm3 of acid and then fill the rest up with water... That would make a 0.1M solution would it not?
Yeah, correct. Or, you could put in 5cm3 and have a 0.2M solution, which wouldn't be too bad really, as you'd have a decent amount of acid to put in the volumetric flask and a decent amount in the titration.

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