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baking soda and vinegar


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#1 herpguy

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Posted 24 November 2005 - 03:43 AM

Why is there a chemical reaction when baking soda and vinegar are mixed together? I've always thought that the acid in vinegar breaks down each compound into individual atoms. the oxygen in the water from vinegar joins with the carbon from baking soda making carbon dioxide. Other oxygen atoms remain with hydrogen keeping some water. The salt from from the baking soda is dissolved into the water making acidic salt water. Am i right? If I'm not please tell me what really happens. Thanks.
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#2 Tetrahedrite

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Posted 24 November 2005 - 04:29 AM

Why is there a chemical reaction when baking soda and vinegar are mixed together? I've always thought that the acid in vinegar breaks down each compound into individual atoms. the oxygen in the water from vinegar joins with the carbon from baking soda making carbon dioxide. Other oxygen atoms remain with hydrogen keeping some water. The salt from from the baking soda is dissolved into the water making acidic salt water. Am i right? If I'm not please tell me what really happens. Thanks.

This is an acid base neutralisation reaction. The baking soda (aka sodium bicarbonate, NaHCO3) is the base and the vinegar (aka acetic acid, CH3COOH) is obviously the acid. The two react to form a salt, water, and carbon dioxide:

NaHCO3 + CH3COOH ----> CH3COO-Na+ + H2O + CO2

The resulting solution will in fact be basic rather than acidic as the CH3COO- ion is a medium strength base.
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#3 budullewraagh

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Posted 24 November 2005 - 02:33 PM

the reaction occurs via a decarboxylation mechanism
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#4 Yggdrasil

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Posted 25 November 2005 - 01:38 AM

In the reaction between baking soda (sodium bicarbonate) and vinegar (acetic acid), the acetic acid is not being decarboxylated. Rather, it's just a result of the equilibrium between carbonic acid and carbon dioxide:

HCO3- + H+ <--> H2CO3 <--> CO2 + H2O

Since the acetic acid supplies extra protons and carbon dioxide gets removed from the system (since it is a gas), reactions are pushed to favor the right side of the equilibrium.
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#5 budullewraagh

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Posted 25 November 2005 - 03:53 AM

the acetic acid isnt being decarboxylated, but the bicarbonate is
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#6 vrus

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Posted 25 November 2005 - 10:01 AM

I never thought of it that deeply. To me it was just ye average
metal + metal carbonate reaction yielding metal salt, CO2 and water !

I've heard the word decarboxyation before, but what does it mean ? Can someone please explain ?
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#7 AL

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Posted 25 November 2005 - 10:21 AM

NaHCO3 + CH3COOH ----> CH3COO-Na+ + H2O + CO2

The resulting solution will in fact be basic rather than acidic as the CH3COO- ion is a medium strength base.

It depends on initial concentration too. If you had 8 M acetic acid and tossed in 0.0001 moles of sodium bicarbonate, the solution would still be acidic. If not, change the numbers in my example until it's so.;)
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#8 Yggdrasil

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Posted 25 November 2005 - 08:34 PM

I've heard the word decarboxyation before, but what does it mean ? Can someone please explain ?


The way I've heard it used, decarboxylation refers to a reaction in which a carboxylate group (R-COO[sup-[/sup]) or carboxylic acid (R-COOH) functions as a leaving group, forming carbon dioxide. Since carboylates and carboxylic acids are organic functional groups, the term decarboxylation suggests a removal of carbon dioxide from an organic compound, which is why I corrected budullewraagh's post. I have not heard the term used in the contect of an inorganic acid such as carbonic acid, so I'm not sure if it is accurate to call this reaction a decarboxylation.
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#9 akcapr

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Posted 26 November 2005 - 10:15 PM

Id say no, because the COOH isnt removed as CO2 but it goes to a H+ and a R-COO-. So it doesnt turn into CO2 so i guess its not a decarboxylation.
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#10 Tetrahedrite

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Posted 28 November 2005 - 03:24 AM

It depends on initial concentration too. If you had 8 M acetic acid and tossed in 0.0001 moles of sodium bicarbonate, the solution would still be acidic. If not, change the numbers in my example until it's so.;)

This is quite true, I was assuming an approximate 1:1 molar ratio though
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