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Equation for Citric acid reacting with Sodium hydrogen-carbonate

  1. Sep 22, 2010 #1
    Hi, Is this the correct equation for Citric acid reacting with Sodium hydrogen-carbonate.

    C6H8O7 + 3NaHCO3 -> 3CO2 + 3Na+1 + C6H5O73-


    Is there supposed to be H2O present also??
     
  2. jcsd
  3. Sep 23, 2010 #2
    Re: Stoichiometry

    Hey Sirsh

    So you balanced your equation almost correctly for both stoichiometric equivalents and charge, great! Except this is indeed the wrong product formation.

    One thing to notice is that your main carbon molecule (citric acid) never gains an oxygen while your bicarbonate magically loses one on its way to CO2. This doesn't happen. Indeed it is probable that water is formed because you are forming CO2 from the bicarbonate by losing a proton and an oxygen.You'd have to check the mechanism on that one. If I were doing this problem I'd say though that the bicarbonate attacks the citric acid on one of the carbonyl bonds to form a dihydroxy intermediate that then evolves CO2. One of the hydroxyls then would be protonated by acid lying around in solution and the carbonyl oxygen would kick off the water molecule. This would happen on all three carboxylic acid groups on the CA thus 3 H20 would be formed.

    Just a guess though, I'm not sure what you're products are but just going off gen/org chem knowledge I'd say:

    C6H807 + 3NaHCO3 -> 3CO2 + 3H20 + 3Na(+1) + C6H8O7(3-)

    Review the acid-catalyzed esterification mechanism in your ochem text. Instead of a carboxy acid and an alcohol just pretend the alcohol is bicarbonate (HO-R' = HO-CO2). Also, instead of going all the way through with the reaction, just stop at the part where the carbonyl oxygen has a negative charge on it and that's where the ionic bond from sodium will form.

    Lastly, double check your hydrogen count on both sides, it seems you lost 3 on your way to products. Balancing hydrogens is the last step of balancing equations but don't forget it!

    Hope that helps!
    theLHR
     
  4. Sep 23, 2010 #3

    Borek

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    Staff: Mentor

    Re: Stoichiometry

    I think you are overcomplicating things, this is a simple inorganic chemistry. In low pH HCO3- is converted to carbonic acid, carbonic acid decomposes to water and carbon dioxide.

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  5. Sep 23, 2010 #4
    Re: Stoichiometry

    Hey Borek,

    If that what you're saying is true, then could you please show me by writing a balanced equation so I can see where your charges are going? Charges must be conserved on either side of the equation so if it is just water solvating the ions then you end up +3 on the product side which cannot happen. Also, if it was as simple as inorganic chemistry, why write the citric acid in? Confused!

    Thanks!
    theLHR

    P.S. Sirsh I seem to have lost a proton too, oops! Still trying to figure this out, lets see what Borek says.
     
    Last edited: Sep 23, 2010
  6. Sep 23, 2010 #5

    Borek

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    Staff: Mentor

    Re: Stoichiometry

    Your reaction equation is almost correct, you just have citrate anion wrong (correct charge, wrong number of H).

    Citric acid just acidifies the solution. However, it is a weak acid, so it is usually written in undissociated form before the reaction. As bicarbonate is a base (albeit weak), it neutralizes the acid, hence it becomes an anion after the reaction.

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  7. Sep 23, 2010 #6
    Re: Stoichiometry

    Nice! Yea, that definitely sounds more right than what I was doing. Thanks Borek.

    theLHR
     
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