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    okay, my classnotes and revision guide say two different things and i don't know what to believe!!

    so aluminium oxide is amphoteric, and i know sometimes they ask you to write to equations to show it behaving as both a base and an acid. It's the base reactions that are causing me confusion.

    my notes:
    Al203 reacts as an acid like this
    Al203 + NaOH + 3H2O --> 2NaAl(OH)4

    then revision guide:
    Al203 reacts as an acid like this
    AL203 + 6OH- + 3H20 --> [Al(OH)6]3-

    EEEH??? I get the revision guide gives me ionic equations, but I do not understand that equation at all. Which one is right?? Or are they both right somehow??

    And could someone also clear up the acid/base reactions of Aluminium HYDROXIDE too for me? THANKS!!!
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    (Original post by cjoc28)
    It's the base* reactions that are causing me confusion.
    *that should say ACID not base!
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    (Original post by cjoc28)
    *that should say ACID not base!
    No coz Al2O3 acts as an acid with a base, i think both versions are correct, one is just the ionic one as Na is not used in the equation. Learn whichever u prefer cos there are loads on the mark scheme from the examiner to chose from. Hope this helps
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    (Original post by cjoc28)

    then revision guide:
    Al203 reacts as an acid like this
    AL203 + 6OH- + 3H20 --> [Al(OH)6]3-

    isnt it Al(OH)3 (s) + 3OH- --> [Al(OH)6]3- (aq)

    :confused:
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    search the site - i've posted a complete explanation about this before
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    (Original post by mackin boi)
    isnt it Al(OH)3 (s) + 3OH- --> [Al(OH)6]3- (aq)

    :confused:
    that equation's correct for aluminium hyrdroxide acting as an acid.

    i was confused about the aluminium oxide. it was just the (OH)4 and (OH)6 that were confusing me. i still don't get it thought, like the rest of the whole syllabus... i guess i'll just learn them and pray they don't ask too many questions on it.

    thanks for help
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    it was just the (OH)4 and (OH)6 that were confusing me.
    I think it is OH6 in the ionic equation because they are assuming that aluminium is forming an octahedral complex ion, and the OH- are acting as ligands, which is actually what happens in aqueous solution. However, when there are sodium ions present they will react with the complex aluminium ion to form a salt, hence the OH4.

    However, I am simply applying my limited chemistry knowledge, so this explanation could be completely wrong! Please can someone say if I have got completely the wrong end of the stick?!?!
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    Aluminium salts exist as the hexaaqua complex ion in solution [Al(H2O)6]3+

    If base is added (OH-) then this can abstract protons from the water ligands one at a time (depending on the concentration of the OH-)

    [Al(H2O)6]3+(aq) + OH- --> [Al(H2O)5(OH)]2+(aq) + H2O

    [Al(H2O)5(OH)]2+(aq) + OH- --> [Al(H2O)4(OH)2]+(aq) + H2O

    both of the above are soluble ions but when one more proton is abstracted...

    [Al(H2O)4(OH)2]+(aq) + OH- --> [Al(H2O)3(OH)3](ppt) + H2O

    the aluminium hydroxide appears as a white precipitate (the species is not charged and therefore cannot be further pulled into solution) This white ppt is sometimes written as though the water ligands did not exist as Al(OH)3 .... this makes no difference.

    Further addition of more OH- will abstract another H+ an the ppt redissolves

    [Al(H2O)3(OH)3](s) + OH- --> [Al(H2O)2(OH)4]-(aq) + H2O

    The species formed is the tetrahydroxy diaqua aluminate (III) ion (mouthful), usually just called the aluminate ion. The formula can be abbreviated to AlO2(-) by subtracting water formula units (strictly speaking inaccurate but WTF)

    OK?
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    follow the revision guide cos its got same equations as the marking schemes.
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    Aluminium salts exist as the hexaaqua complex ion in solution [Al(H2O)6]3+

    If base is added (OH-) then this can abstract protons from the water ligands one at a time (depending on the concentration of the OH-)

    [Al(H2O)6]3+(aq) + OH- --> [Al(H2O)5(OH)]2+(aq) + H2O

    [Al(H2O)5(OH)]2+(aq) + OH- --> [Al(H2O)4(OH)2]+(aq) + H2O

    both of the above are soluble ions but when one more proton is abstracted...

    [Al(H2O)4(OH)2]+(aq) + OH- --> [Al(H2O)3(OH)3](ppt) + H2O

    the aluminium hydroxide appears as a white precipitate (the species is not charged and therefore cannot be further pulled into solution) This white ppt is sometimes written as though the water ligands did not exist as Al(OH)3 .... this makes no difference.

    Further addition of more OH- will abstract another H+ an the ppt redissolves

    [Al(H2O)3(OH)3](s) + OH- --> [Al(H2O)2(OH)4]-(aq) + H2O

    The species formed is the tetrahydroxy diaqua aluminate (III) ion (mouthful), usually just called the aluminate ion. The formula can be abbreviated to AlO2(-) by subtracting water formula units (strictly speaking inaccurate but WTF)

    OK?
    I understand all the deprotination bit. Sooo, basically are you saying that the aluminium just gets completely deprotinated by the OH- ions to form [Al(OH-)6]3- when added to a base? And when sodium is present it actually forms Na+[AL(H2O)2(OH)4]- ?
 
 
 
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