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OfflineRob1983
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Registered: 06/26/15
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Last seen: 8 years, 7 months
Need help with some chemistry
    #21860910 - 06/26/15 05:01 PM (8 years, 7 months ago)

Hi there, I'm new to the board. I have a question for any chemistry nerds out there. There's a tek from this site that's fairly well known, and I'm not actually doing it, but I'm confused about one part.

So the part I'm confused about goes like this:

"240 ml of acetone were dried with anhydrous potassium carbonate and an equimolar quantity of HCL (6.58 ml) were added to it"

I take this to mean that FIRST the acetone was dried with potassium carbonate, the PC is then filtered out, and an equimolar quantity of HCL was added to it.

The HCL used in the tek is 37% w/w. But by my calculations, an equimolar quantity of HCL for 240 ml of acetone should be way more then 6.58 ml.

Acetone has a molar mass of 58.08 g/mol and HCL has 36.46 g/mol. Accoording to my calculations, an equimolar quantity should be almost equal.

Here's my math, maybe someone can tell me where I'm wrong:

100 ml of acetone weighs 78 g (0.78 g/ml). 78 g of acetone is 1.34 mols. So 1.34 mols of HCL should be 48.58 g. in a 37% solution, that should be about 130 ml added to the 100 ml of acetone.

What am I missing?


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OfflineMarkostheGnostic
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Re: Need help with some chemistry [Re: Rob1983]
    #21879976 - 06/30/15 08:14 PM (8 years, 6 months ago)

Awww, #1 Post. I'm honored to respond .:wink:  Check with the Chemistry & Pharmacology forum. There are some professional chemist/biochemists there who have been most helpful to me in the past. I do not want to put anyone out there, but that's where you should go.


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γνῶθι σαὐτόν - Gnothi Seauton - Know Thyself


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Offlinetbaggin
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Registered: 06/05/17
Posts: 2
Last seen: 6 years, 6 months
Re: Need help with some chemistry [Re: Rob1983]
    #24497876 - 07/21/17 02:17 AM (6 years, 6 months ago)

hey ! I'm on the same homework than you ... but my problem is I need to know the quantity of potassium carbonate to use to dry this 240ml of acetone before adding the 6,58ml of HCL acid ... Would you help me , since this is the only thing I don't understand in this situation ...

thank you .
Re: Need help with some chemistry
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OfflineKryptos
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Re: Need help with some chemistry [Re: MarkostheGnostic]
    #24498935 - 07/21/17 02:47 PM (6 years, 6 months ago)

Quote:

MarkostheGnostic said:
Awww, #1 Post. I'm honored to respond .:wink:  Check with the Chemistry & Pharmacology forum. There are some professional chemist/biochemists there who have been most helpful to me in the past. I do not want to put anyone out there, but that's where you should go.




C&P forum requires 10 posts to access. It's why my first ten posts on this site were basically bumps.

OP:

I am not certain as to the specifics of the problem...from what I understand, you are correct, you would need a lot more than just 6 ml of HCl, especially if using 37%.

However, another way of reading this problem is from the standpoint of titration. As K2CO3 is basic, they might be adding HCl to neutralize the leftover K2CO3. However, this would be counterproductive as far as drying goes, because you would end up generating KCl, H2O, and CO2 according to the following reaction:

2HCl + K2CO3 --> 2KCl + H2O + CO2

I hope this helps? Could you be more specific with the problem/tek? A direct quote would help...

I guess I'll also add these:

K2CO3 + H2O --> KOH + KHCO3

KOH + HCL --> KCl + H2O

That's the mechanism by which potassium carbonate absorbs water (I'm pretty sure), and this is what leads me to think that you are titrating the KOH byproduct by adding HCl. If it is a genchem problem, as mentioned by the third post, the question is likely one where you are required to calculate the original water content of the acetone, given that it took 6ml of 37% HCl to neutralize the KOH.


Edited by Kryptos (07/21/17 03:07 PM)


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Offlineilrb2
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Registered: 06/04/17
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Last seen: 6 years, 4 months
Re: Need help with some chemistry [Re: Kryptos]
    #24525487 - 08/02/17 05:44 PM (6 years, 5 months ago)

I think Kryptos is right, the quantity of HCl you need to add should be equimolar to the potassium carbonate you are using to dry, not to the acetone. What exactly are you trying to do with this? Because assuming you have ultra-dry acetone, adding HCl wouldn't do anything to it besides generate weird mixtures of products. I don't recall my orgo super well but I think you'd just get some aldol reactions, followed by acid-catalyzed dehydrations that would also just generate water again.


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OfflineSpanishfly
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Re: Need help with some chemistry [Re: Kryptos]
    #24655966 - 09/24/17 12:27 PM (6 years, 4 months ago)

Quote:

Kryptos said:
Quote:

MarkostheGnostic said:
Awww, #1 Post. I'm honored to respond .:wink:  Check with the Chemistry & Pharmacology forum. There are some professional chemist/biochemists there who have been most helpful to me in the past. I do not want to put anyone out there, but that's where you should go.




C&P forum requires 10 posts to access. It's why my first ten posts on this site were basically bumps.

OP:

I am not certain as to the specifics of the problem...from what I understand, you are correct, you would need a lot more than just 6 ml of HCl, especially if using 37%.

However, another way of reading this problem is from the standpoint of titration. As K2CO3 is basic, they might be adding HCl to neutralize the leftover K2CO3. However, this would be counterproductive as far as drying goes, because you would end up generating KCl, H2O, and CO2 according to the following reaction:

2HCl + K2CO3 --> 2KCl + H2O + CO2

Please  get the annotation right - K2CO3

I hope this helps? Could you be more specific with the problem/tek? A direct quote would help...

I guess I'll also add these:

K2CO3 + H2O --> KOH + KHCO3

KOH + HCL --> KCl + H2O

That's the mechanism by which potassium carbonate absorbs water (I'm pretty sure), and this is what leads me to think that you are titrating the KOH byproduct by adding HCl. If it is a genchem problem, as mentioned by the third post, the question is likely one where you are required to calculate the original water content of the acetone, given that it took 6ml of 37% HCl to neutralize the KOH.




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I am currently BANNED from using Private Messages - so can anyone who wants to contact me do it via my Journal thread.  Link is https://www.shroomery.org/forums/showflat.php/Number/23831115

Maybe some mod or whatever might think this has now been long enough.


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OfflineMartianman420
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Registered: 01/11/17
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Re: Need help with some chemistry [Re: Spanishfly]
    #24719616 - 10/18/17 02:30 PM (6 years, 3 months ago)

just keep responding and then go lol


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Thinking is not an option


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OfflineZenmaster6
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Registered: 11/04/17
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Re: Need help with some chemistry [Re: Rob1983]
    #25086126 - 03/23/18 09:39 PM (5 years, 10 months ago)

looks right to me.... I'm way to baked to do chemistry right now. May come back to it later. Worked at a pharmacy took AP chem and passed with an A+ and I'm getting a minor in Chemistry currently. Could be of some help if I wasn't so fucking high. Uhg I hate being this high


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Grandma: Were those really Portobello mushrooms!?!

 


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