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    Hi, could anyone explain what partial and complete neutralization reaction are? There is a question in the chemistry ocr textbook which asks to write equations for the partial and complete neutralization reaction for carbonic acid with sodium hydroxide. There are two answers; one forms na2co3 and the other one forms NaHco3. Can anyone explain? Thanks
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    Salts form by removing H+ and replacing with a metal ion (or ammonium ions).

    If there are two H+s that can be removed, e.g. in H2CO3, then the acid can be partially neutralised by replacing just one H+, i.e. NaHCO3 or fully neutralised i.e. Na2CO3.
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    (Original post by Pigster)
    Salts form by removing H+ and replacing with a metal ion (or ammonium ions).

    If there are two H+s that can be removed, e.g. in H2CO3, then the acid can be partially neutralised by replacing just one H+, i.e. NaHCO3 or fully neutralised i.e. Na2CO3.
    Oh that makes sense! So would the half equation for the partial neutralization be H+ + OH- ----> h2o and one for complete neutralization is 2H+ +2OH- --> 2H2O? so the partial neutralization part, does it mean that the salt formed is acid salt because there is still one hydrogen ion remaining which acts as an acid? Thank you, you are the only one helping!
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    I wouldn't call it a half-equation, they are for redox equations... it is more like an ionic equation. Regardless, you're correct.

    It is an acid salt.

    If you want more people to help, I'd recommend posting in the chemistry forum, rather than the chemistry exams forum, i.e. one level up.

    http://www.thestudentroom.co.uk/forumdisplay.php?f=130
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    (Original post by Pigster)
    I wouldn't call it a half-equation, they are for redox equations... it is more like an ionic equation. Regardless, you're correct.

    It is an acid salt.

    If you want more people to help, I'd recommend posting in the chemistry forum, rather than the chemistry exams forum, i.e. one level up.

    http://www.thestudentroom.co.uk/forumdisplay.php?f=130
    so is my ionic equation right as well? I will do next time, thanks again.
 
 
 
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