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Topic: Proving an acid is dibasic  (Read 89729 times)

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Offline HDW

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Re: Proving an acid is dibasic
« Reply #15 on: March 28, 2008, 05:30:28 AM »
I'm doing the same thing as you.  For the gas collection, I've been thinking of doing an electrolysis in which hydrogen would be collected at the cathode (+ve ions gain electrons).  Can't see how it would prove it though :/

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Re: Proving an acid is dibasic
« Reply #16 on: March 28, 2008, 06:04:50 AM »
Electrolysis won't work - you will electrolyse water, not H+ from the acid.

As for equivalents.

Imagine you have a sample of acid of known mass.

Equivalent is formally defined as amount of substance that reacts with 1 mole of electrons. This definition is a little bit to narrow IMHO. Druing neutralization there are no electrons involved (well, they are, but it is hardly reaction WITH electrons). But - if you will react H+ with Zn - you will see that exactly 1 mole of H+ reacts with 1 mole of electrons. In the case of acids 1 equivalent is 1 acidic proton.

When you titrate it with base, you know how many acid equivalents have been neutralized. You don't know how many moles of acid it was. This gives you information about mass of equivalent, not about molar mass.

There are other methods of determining molar mass - freezing point depression, boiling point depression, volume of gas (PV=nRT) to name a few. Unfortunately, they are not easy to use in the case of sulfuric acid, but we are talking about general approach.

So, what you can do is to determine mass of equivalent by titration, then molar mass by other means. If they are identical - you know there is one acidic proton per molecule. If they differ - molar mass should be a multiply of equivalent, molar/equivalent mass ratio gives you number of protons.
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Offline merkl

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Re: Proving an acid is dibasic
« Reply #17 on: March 31, 2008, 07:40:38 PM »
That equivalent method seems so much more definite. When I get back to school, I'll ask my teachers advice on it *as I dont know if we are expected to use that method or not*, then post back.

Offline willowe

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Re: Proving an acid is dibasic
« Reply #18 on: April 02, 2008, 09:22:18 AM »
hi eragon 11
i appear to be doing the same thing ...
im not sure what to do for the titration so if you could possibly repeat it in layman's terms.

However, in terms of the gas collection i was going to react both a monoprotic and a diprotic acid with magnesium carbonate MGCO3, recording the amount of CO2 given off, as the diprotic h2so4 would give off twice as much CO2
Am i going horribly wrong?

Offline chocoholic4lyf

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Re: Proving an acid is dibasic
« Reply #19 on: April 03, 2008, 10:24:52 AM »
can someone run through the method and calculations for the titration and gas collection experiment please? i have no idea and need help !!!
« Last Edit: April 03, 2008, 10:40:19 AM by chocoholic4lyf »

weiguxp

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Re: Proving an acid is dibasic
« Reply #20 on: April 03, 2008, 08:38:46 PM »
ever tried titrating it and looking at the pH curve? it should have 2 equivalence points

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Re: Proving an acid is dibasic
« Reply #21 on: April 04, 2008, 03:00:06 AM »
ever tried titrating it and looking at the pH curve? it should have 2 equivalence points

Not necesarilly. See plot.
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Offline Jack Joseph

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Re: Proving an acid is dibasic
« Reply #22 on: April 04, 2008, 06:51:23 AM »
However, in terms of the gas collection i was going to react both a monoprotic and a diprotic acid with magnesium carbonate MGCO3, recording the amount of CO2 given off, as the diprotic h2so4 would give off twice as much CO2
Am i going horribly wrong?

Err yes, dibasic is how many h+ ions released. Keep it simply Do: Metal + Acid -> Salt + Hydrogen. a dibasic acid will produce twice as much h2 as a monobasic                 

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Re: Proving an acid is dibasic
« Reply #23 on: April 04, 2008, 06:59:42 AM »
However, in terms of the gas collection i was going to react both a monoprotic and a diprotic acid with magnesium carbonate MGCO3, recording the amount of CO2 given off, as the diprotic h2so4 would give off twice as much CO2
Am i going horribly wrong?

Err yes, dibasic is how many h+ ions released. Keep it simply Do: Metal + Acid -> Salt + Hydrogen. a dibasic acid will produce twice as much h2 as a monobasic

Willowe asks about CO2, you answer about H2. Sometimes it helps to read the question before answering ;)
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Offline merkl

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Re: Proving an acid is dibasic
« Reply #24 on: April 04, 2008, 05:54:03 PM »
I see a few british people have had their AS chemistry plans through...

Ahh, refreshing. So glad Ive done it now.

Offline ladybird

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Re: Proving an acid is dibasic
« Reply #25 on: April 08, 2008, 03:56:32 PM »
so your second method, involving calculating how much base is needed to neutralise HCL and sulphuric acid before comparing the amount needed is the right way to go? that kind of makes sense to me but am quite unsure about the calculations invloved afterwards.

the gas collection...that just needs an acid and a metal reacting, and then the amount of gas collected and moles calculated blah blah?

sorry i'm a bit simple, and ive left all this way too late.

Offline Lil-lindz

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Re: Proving an acid is dibasic
« Reply #26 on: April 09, 2008, 02:52:51 PM »
Hey just got the plans for the planning stage of my AS chemistry coursework and i was just wondering if anyone could suggest a suitable indicator for use in the titration against NaOH, and also the colour change that you should expect to see.
Also in the gas collection, are you simply comaparing the amount of gas collected for a monoprotic and diprotic acid (obviously the diprotic will release twice as much) but im unsure of any calculations that could be done afterwards. I was thinking about weighing the metal before the reaction and working out how many mols of gas will be realeased for one mole on monoprotic and one mole of diprotic acid, but that essentially just shows what your experiment showed and so there doesn't seem much point

Also for the titration, my understanding is that it is used for finding concentrations of unknown soloutions, but since you should know the concentration of both soloutions (which i assume should be the same) how can calculations show sulphuric acid is diprotic??

Wow thats a mouthful, but if someone could give me some hints on that i would be really grateful!!

Offline jef749

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Re: Proving an acid is dibasic
« Reply #27 on: April 10, 2008, 10:47:13 AM »
For the gas collection, how would you go about proving how much sulphuric acid you need and how much metal (in my case iron) you need? the metal needs to be in excess, but how do i know how much that is? any help is greatly appreciated.

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Re: Proving an acid is dibasic
« Reply #28 on: April 11, 2008, 09:25:48 AM »
Hey Eragon Cud U Pls Explain The Format Of Dis Plannin 2 Me
Cos I Dnt Undastand It & Cud U Pls Tel Me Which Acid Acts As d Standard Solution?
Fanxs

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Offline danielle21_711

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Re: Proving an acid is dibasic
« Reply #29 on: April 14, 2008, 09:41:04 AM »
is this for the chemistry plan? I'm totally lost as well... I don't know where to start really.

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