Aluminium oxide amphoteric
I know that aluminium oxide reacts with both acids and alkalis, however im confused at when it acts acidic e.g.
Aluminium oxide + sodium hydroxide + water --> NaAl(OH)4
This may be a stupid question but why is the water on the left hand side of the equation? Shouldnt an acid + base give salt + water
Aluminium oxide + sodium hydroxide + water --> NaAl(OH)4
This may be a stupid question but why is the water on the left hand side of the equation? Shouldnt an acid + base give salt + water
Original post by mushed
I know that aluminium oxide reacts with both acids and alkalis, however im confused at when it acts acidic e.g.
Aluminium oxide + sodium hydroxide + water --> NaAl(OH)4
This may be a stupid question but why is the water on the left hand side of the equation? Shouldnt an acid + base give salt + water
Aluminium oxide + sodium hydroxide + water --> NaAl(OH)4
This may be a stupid question but why is the water on the left hand side of the equation? Shouldnt an acid + base give salt + water
The water is there to make everything balance.
Usually a salt and water are formed, but not always, e.g. ammonia + sulfuric acid.
I think you'll agree that this reaction is rather unusual in its own right, a GCSE student would think Al2O3 were a metal oxide and think this reaction is base + base + water -> base (since a metal hydroxide is a base).
Original post by Pigster
The water is there to make everything balance.
Usually a salt and water are formed, but not always, e.g. ammonia + sulfuric acid.
I think you'll agree that this reaction is rather unusual in its own right, a GCSE student would think Al2O3 were a metal oxide and think this reaction is base + base + water -> base (since a metal hydroxide is a base).
Usually a salt and water are formed, but not always, e.g. ammonia + sulfuric acid.
I think you'll agree that this reaction is rather unusual in its own right, a GCSE student would think Al2O3 were a metal oxide and think this reaction is base + base + water -> base (since a metal hydroxide is a base).
can you expand on this please?
Original post by miac0328
can you expand on this please?
Happy to assist, which bit/s do you need help on?
Original post by miac0328
Thanks for replying.
i just dont understand why it balances it out and why the water needs to be on the left, why cant it be on the right like it normally is? and if the water is on the right, would we get a different product and not NaAl(OH)4?
i just dont understand why it balances it out and why the water needs to be on the left, why cant it be on the right like it normally is? and if the water is on the right, would we get a different product and not NaAl(OH)4?
NaAl(OH)4 exists as two ions: Na^+ and Al(OH)4^-
Al(OH)4^- exists as an Al3+ covalently bonded to 4x OH- ions in a tetrahedral arrangement just like CH4.
Al(OH)4^- might seem rather unusual, but it is stable and it does form, but how?
If you try to balance the equation, you'd start with:
Al2O3 + NaOH -> NaAl(OH)4
Now clearly, since there are 2x Al on the LHS, you'll quickly get to:
Al2O3 + 2NaOH -> 2NaAl(OH)4
But hopefully, you'd notice that you're 6x H and 3x O short on the LHS, hence you need 3H2O to make it all balance:
Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4
Al2O3 is acting as an base as it is accepting H+ from water; which is therefore acting as an acid (H2O <-> H+ + OH-). The NaOH isn't actually doing anything apart from providing the alkaline conditions that allow Al(OH)4^- to be stable.
Original post by Pigster
NaAl(OH)4 exists as two ions: Na^+ and Al(OH)4^-
Al(OH)4^- exists as an Al3+ covalently bonded to 4x OH- ions in a tetrahedral arrangement just like CH4.
Al(OH)4^- might seem rather unusual, but it is stable and it does form, but how?
If you try to balance the equation, you'd start with:
Al2O3 + NaOH -> NaAl(OH)4
Now clearly, since there are 2x Al on the LHS, you'll quickly get to:
Al2O3 + 2NaOH -> 2NaAl(OH)4
But hopefully, you'd notice that you're 6x H and 3x O short on the LHS, hence you need 3H2O to make it all balance:
Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4
Al2O3 is acting as an base as it is accepting H+ from water; which is therefore acting as an acid (H2O <-> H+ + OH-). The NaOH isn't actually doing anything apart from providing the alkaline conditions that allow Al(OH)4^- to be stable.
Al(OH)4^- exists as an Al3+ covalently bonded to 4x OH- ions in a tetrahedral arrangement just like CH4.
Al(OH)4^- might seem rather unusual, but it is stable and it does form, but how?
If you try to balance the equation, you'd start with:
Al2O3 + NaOH -> NaAl(OH)4
Now clearly, since there are 2x Al on the LHS, you'll quickly get to:
Al2O3 + 2NaOH -> 2NaAl(OH)4
But hopefully, you'd notice that you're 6x H and 3x O short on the LHS, hence you need 3H2O to make it all balance:
Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4
Al2O3 is acting as an base as it is accepting H+ from water; which is therefore acting as an acid (H2O <-> H+ + OH-). The NaOH isn't actually doing anything apart from providing the alkaline conditions that allow Al(OH)4^- to be stable.
AFAIK the aluminate ion is octahedral, with two water ligands and four hydroxyl ligands.
Certainly the aluminium ion is octahedral with six water ligands.
Addition of base to aluminium salts serially replaces the hydrogen ions from the water ligands:
[Al(H2O)6]3+ + OH- ==>> [Al(H2O)5OH]2+ + H2O
[Al(H2O)5OH]2+ + OH- ==>> [Al(H2O)4(OH)2]+ + H2O
[Al(H2O)4(OH)2]+ + OH- ==>> [Al(H2O)3(OH)3] + H2O
At this point the precipitation of aluminium hydroxide appears as it is not charged. On further addition of base it redissolves:
[Al(H2O)3(OH)3] + OH- ==>> [Al(H2O)2(OH)4]- + H2O
Original post by charco
AFAIK the aluminate ion is octahedral, with two water ligands and four hydroxyl ligands.
Certainly the aluminium ion is octahedral with six water ligands.
Addition of base to aluminium salts serially replaces the hydrogen ions from the water ligands:
[Al(H2O)6]3+ + OH- ==>> [Al(H2O)5OH]2+ + H2O
[Al(H2O)5OH]2+ + OH- ==>> [Al(H2O)4(OH)2]+ + H2O
[Al(H2O)4(OH)2]+ + OH- ==>> [Al(H2O)3(OH)3] + H2O
At this point the precipitation of aluminium hydroxide appears as it is not charged. On further addition of base it redissolves:
[Al(H2O)3(OH)3] + OH- ==>> [Al(H2O)2(OH)4]- + H2O
Certainly the aluminium ion is octahedral with six water ligands.
Addition of base to aluminium salts serially replaces the hydrogen ions from the water ligands:
[Al(H2O)6]3+ + OH- ==>> [Al(H2O)5OH]2+ + H2O
[Al(H2O)5OH]2+ + OH- ==>> [Al(H2O)4(OH)2]+ + H2O
[Al(H2O)4(OH)2]+ + OH- ==>> [Al(H2O)3(OH)3] + H2O
At this point the precipitation of aluminium hydroxide appears as it is not charged. On further addition of base it redissolves:
[Al(H2O)3(OH)3] + OH- ==>> [Al(H2O)2(OH)4]- + H2O
I probably should have checked, rather than guessing
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