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Topic: NaOH + H3C6H5O7 -----> Na3 C6H5O7 => Sodium citrate  (Read 40902 times)

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Dufault

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NaOH + H3C6H5O7 -----> Na3 C6H5O7 => Sodium citrate
« on: November 03, 2005, 12:55:30 PM »
Hi, I have a lab due tommorow and I am unsure as to how to balance this equation because Im not sure if just one hydrogen comes off or all of them. I think its the case where its all of them but Im having troubles getting it to balance. Here is the equation..

__NaOH + __H3C6H507 ----> Na_________ +__H20

Any help you could offer would be great. Thanks!
« Last Edit: November 03, 2005, 05:02:14 PM by Mitch »

Offline Alberto_Kravina

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Re:Need *delete me* balancing citric acid for a lab
« Reply #1 on: November 03, 2005, 01:24:04 PM »
 ACID + BASE ----------> SALT + WATER, you said that all hydrogen ions are replaced with sodium

3 NaOH + H3C6H5O7 -----> Na3 C6H5O7 => Sodium citrate

In this case all H+ react with the base to form water.

Dufault

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Re:Need *delete me* balancing citric acid for a lab
« Reply #2 on: November 03, 2005, 01:47:30 PM »
Hey, thanks for the help, so would this be the final balance equation then?

3NaOH + H3C6H5O7--------->Na3C6H5O7 + 3 H20

Can you quickly explain to me why the Na in the products has a subscript of 3, is this because of the charge on it or what? thanks.

Offline jdurg

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Re:Need *delete me* balancing citric acid for a lab
« Reply #3 on: November 03, 2005, 02:34:22 PM »
Hey, thanks for the help, so would this be the final balance equation then?

3NaOH + H3C6H5O7--------->Na3C6H5O7 + 3 H20

Can you quickly explain to me why the Na in the products has a subscript of 3, is this because of the charge on it or what? thanks.

Yup.  Sodium ions have the same charge as a hydrogen ion, so in your formula for citric acid, you are replacing the H at the front with Na.  Since you are replacing 3 of the hydrogen ions, you need 3 sodium ions to balance this out.

A good thing to remember is that whenever you react a weak acid with a strong base, ALL of the acidic hydrogen ions will come off if there is enough base present.  So in this case, the three acidic hydrogens of citric acid come off and are replaced with the counter ion of the base (in this case, Na+).    
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Dufault

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Re:Need *delete me* balancing citric acid for a lab
« Reply #4 on: November 03, 2005, 03:44:26 PM »
Thanks alot, I would of been here forever trying to figure it out. It was for a titration lab. Thanks for explaining it, I new that they all had to come off I just didnt know how the Na fit into the equation.

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