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Topic: Potassium bicarbonate model Help  (Read 13167 times)

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Offline jwlaurie

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Potassium bicarbonate model Help
« on: February 13, 2008, 11:52:22 AM »
is this correct ?

         HO-C-O-K
               |
               O
               |
         HO-C-O-K

 

Offline Arkcon

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Re: Potassium bicarbonate model Help
« Reply #1 on: February 13, 2008, 12:01:11 PM »
No, for starters, that model has the wrong amounts of atoms, compared to the formula.  Potassium bicarbonate doesn't contain 2 carbonates, or 2 potassium ions, the bi is an ancient, obsolete chemical term meaning "one more hydrogen than."  The correct chemical name is potassium hydrogen carbonate.
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Offline jwlaurie

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Re: Potassium bicarbonate model Help
« Reply #2 on: February 13, 2008, 12:20:38 PM »
I am a firefighter taking fire chemistry and I need to make the model for Potassium bicarbonate can you help me and show me how it would look ?

Offline Borek

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Re: Potassium bicarbonate model Help
« Reply #3 on: February 13, 2008, 12:29:38 PM »
bi is an ancient, obsolete chemical term meaning "one more hydrogen than."

AFAIR bi refers to the stoichiometry of neutralization when using diprotic acids - it means "compound made when you use twice as much acid as necessary".
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Offline Borek

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Re: Potassium bicarbonate model Help
« Reply #4 on: February 13, 2008, 12:31:16 PM »
I am a firefighter taking fire chemistry and I need to make the model for Potassium bicarbonate can you help me and show me how it would look ?

First of all - please check the correct formula, you will find it (for example) in wikipedia.
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Offline Arkcon

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Re: Potassium bicarbonate model Help
« Reply #5 on: February 13, 2008, 12:31:38 PM »
OK, here's the {wikipedia} page for sodium bicarbonate, just replace the Na with a K
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Offline jwlaurie

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Re: Potassium bicarbonate model Help
« Reply #6 on: February 13, 2008, 12:57:23 PM »
it would look like this ?

       K+ -O   OH
              \  /
               C         
               ||
               O

Offline Arkcon

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Re: Potassium bicarbonate model Help
« Reply #7 on: February 13, 2008, 12:59:16 PM »
bi is an ancient, obsolete chemical term meaning "one more hydrogen than."

AFAIR bi refers to the stoichiometry of neutralization when using diprotic acids - it means "compound made when you use twice as much acid as necessary".

Really, so, how is that done, you've got NaOH, and instead of neutralizing it with HCO3 to get NaCO3, you use H2CO3 (the one that actually exists, heh) to get NaHCO3?  Naw, I'm not getting it.  Sodium Bicarbonate, sodium bisulfite, sodium phosphate dibasic -- they only barely taught these to me when I was in high school, they were so sure everyone was going to only use the IUPAC name, forever.
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Offline Arkcon

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Re: Potassium bicarbonate model Help
« Reply #8 on: February 13, 2008, 01:00:59 PM »
it would look like this ?

       K+ -O   OH
              \  /
               C         
               ||
               O


Yeah, that'll work.  If you're bored, poke around the web a bit to find bond angles, it's kinda interesting.
« Last Edit: February 13, 2008, 06:47:55 PM by Arkcon »
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Offline jwlaurie

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Re: Potassium bicarbonate model Help
« Reply #9 on: February 13, 2008, 01:05:28 PM »
Thank you for the help

Offline Borek

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Re: Potassium bicarbonate model Help
« Reply #10 on: February 13, 2008, 01:40:59 PM »
you've got NaOH, and instead of neutralizing it with HCO3 to get NaCO3, you use H2CO3 (the one that actually exists, heh) to get NaHCO3?

Are we talking chemistry here? HCO3? NaCO3? Time to wake up ;)

Stoichiometry - 1 mole of diprotic acid per 2 moles of monoprotic base:

2NaOH + H2CO3 -> Na2CO3 + 2H2O

Product: carbonate.

Now, double amount of acid - ie 2 moles of diprotic acid per 2 moles of base, or simply 1:1:

NaOH + H2CO3 -> NaHCO3 + H2O

Product: bicarbonate.

Its not that hard ;)
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Offline jwlaurie

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Re: Potassium bicarbonate model Help
« Reply #11 on: February 13, 2008, 01:47:56 PM »
you've got NaOH, and instead of neutralizing it with HCO3 to get NaCO3, you use H2CO3 (the one that actually exists, heh) to get NaHCO3?

Are we talking chemistry here? HCO3? NaCO3? Time to wake up ;)

Stoichiometry - 1 mole of diprotic acid per 2 moles of monoprotic base:

2NaOH + H2CO3 -> Na2CO3 + 2H2O

Product: carbonate.

Now, double amount of acid - ie 2 moles of diprotic acid per 2 moles of base, or simply 1:1:

NaOH + H2CO3 -> NaHCO3 + H2O

Product: bicarbonate.

Its not that hard ;)



so is the above modle wrong ?

Offline Arkcon

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Re: Potassium bicarbonate model Help
« Reply #12 on: February 13, 2008, 01:56:03 PM »

Are we talking chemistry here? HCO3? NaCO3? Time to wake up ;)

Stoichiometry - 1 mole of diprotic acid per 2 moles of monoprotic base:

2NaOH + H2CO3 -> Na2CO3 + 2H2O

Product: carbonate.

Now, double amount of acid - ie 2 moles of diprotic acid per 2 moles of base, or simply 1:1:

NaOH + H2CO3 -> NaHCO3 + H2O

Product: bicarbonate.

Its not that hard ;)

Ok, so you mix NaOH and H2CO3, and you get, which one, carbonate or bicarbonate?  Or some of one and the other, maybe, with great excess of H2CO3, more bicarbonate than carbonate, is that it?

so is the above modle wrong ?

No it's fine, I'm just trying to learn from Borek about the bi- prefix, and what it means.
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Offline Borek

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Re: Potassium bicarbonate model Help
« Reply #13 on: February 13, 2008, 02:32:47 PM »
Ok, so you mix NaOH and H2CO3, and you get, which one, carbonate or bicarbonate?  Or some of one and the other, maybe, with great excess of H2CO3, more bicarbonate than carbonate, is that it?

You will probably get mixture, unless you will mix exact ratio of acid and base.

Quote
I'm just trying to learn from Borek about the bi- prefix, and what it means.

Note: meaning of the prefix is of no practical importance, I believe it is just echo of some early chemistry. Imagine one of those early chemists not knowing anything that we take for granted today before we even touch the beaker. He finds out that mixing 4 mass units (be it grams, ounces or anything else) of sodium hydroxide with 10 mass units of sulfuric acid he gets solution that - once dried out - gives nice crystals and nothing is left - neither base nor acid. That's great! But now he does the same using 4 mass units of base and 20 mass units of the same acid - and he gets crystallic substance as well... And once again nothing is left! Oops, how do I tell which one is which one? Oh well, I have used twice the amount of sulfuric acid, it is doubled, let's call it BIsulfate...

Note2: NaOH and H2SO4 just because I happen to remember their molar masses precisely enough :) And it is not exactly 4:10, but close enough. And I am assuming pure sulfruic acid, not some solution. Don't be to picky and you should get the idea behind :)
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