Download REACTIONS OF HEXAAQUA METAL IONS WITH HYDROXIDE IONS

REACTIONS OF HEXAAQUA METAL IONS WITH HYDROXIDE IONS
The general case
Although there are only minor differences, for simplicity we will look at 2+ ions and 3+ ions
separately.
Adding hydroxide ions to 2+ hexaaqua ions
These have the form [M(H2O)6]2+. Their acidity is shown in the reaction of the hexaaqua ions with
water molecules from the solution:
They are acting as acids by donating hydrogen ions to water molecules in the solution.
Because of the confusing presence of water from two different sources (the ligands and the
solution), it is easier to simplify this:
Disturbing this equilibrium by adding hydroxide ions - stage 1
What happens if you add hydroxide ions to this equilibrium? There are two possible reactions.
Reaction of hydroxide ions with the hydroxonium ions (hydrogen ions)
According to Le Chatelier's Principle, the position of equilibrium will move to the right, producing
more of the new complex ion.
Reaction of hydroxide ions with the hexaaqua ion
Statistically, there is far more chance of a hydroxide ion hitting a hexaaqua metal ion than of hitting
a hydrogen ion. There are far more hexaaqua ions present.
If that happens, you get exactly the same new complex ion formed as above.
Notice that this isn't a ligand exchange reaction. The hydroxide ion has removed a hydrogen ion
from one of the ligand water molecules. The reaction has also become virtually one-way.
The second stage of the reaction
Whichever of the above reactions happens, you end up with [M(H2O)5(OH)]+ ions in solution. These
are also acidic, and can lose hydrogen ions from another of the water ligands.
Taking the easier version of the equilibrium:
Adding hydroxide ions again tips the equilibrium to the right - either by reacting with the hydrogen
ions, or by reacting directly with the complex on the left-hand side.
When this happens, the new complex formed no longer has a charge - we describe it as a "neutral
complex". In all the cases we are looking at, this neutral complex is insoluble in water - and so a
precipitate is formed.
This precipitate is often written without including the remaining water ligands. In other words we
write it as M(OH)2. A precipitate of the metal hydroxide has been formed.
Summarising what has happened so far
Going further
There is no logical reason why the removal of hydrogen ions from the complex
should stop at this point. Further hydrogen ions can be removed by hydroxide ions to produce
anionic complexes - complexes carrying negative charges. Whether this actually happens in the
test tube to any extent varies from metal to metal.
In fact, if you do this using sodium hydroxide solution of the usual concentrations, most of the 2+
ions that you will meet at this level don't go beyond the precipitate. The only one you are likely to
come across is the zinc case - and that has a complication. The final ion is [Zn(OH)4]2- - a
tetrahedral ion which has lost the remaining 2 water ligands.
Adding hydroxide ions to 3+ hexaaqua ions
The argument here is exactly as before - the only difference is the number of hydrogen ions which
have to be removed from the original hexaaqua complex to produce the neutral complex.
Going beyond the neutal complex is also rather more common with 3+ than with 2+ ions, and may
go as far as having a hydrogen ion removed from each of the six water molecules. This is
summarised in the same sort of flow scheme as before:
All you need to do is to understand that one hydrogen ion gets removed at a time. When you have a
neutral complex, it will form a precipitate. That is equally true of the 2+ or 3+ cases. Just be careful with
the charges on the complexes. Remember that for every hydrogen ion you remove, you will lose a
positive charge (or will gain one more negative charge).
Looking at the ions of specific metals
In each case the formula of the precipitate will be given as if it were the simple neutral complex. In
fact, these "hydroxide" precipitates sometimes rearrange by losing water from combinations of the
attached OH groups. This produces oxides closely associated with the lost water. These changes
are beyond the scope of this site.
2+ ions
hexaaquacobalt(II)
hexaaquacopper(II)
hexaaquairon(II)
Iron is very easily oxidised under alkaline conditions. Oxygen in the air oxidises the iron(II)
hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of
the precipitate comes from the same effect.
hexaaquamanganese(II)
I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is
virtually colourless. The pale brown precipitate is oxidised to darker brown manganese(III) oxide in
contact with oxygen from the air.
hexaaquanickel(II)
hexaaquazinc
You start and finish with colourless solutions, producing a white precipitate on the way.
3+ ions
hexaaquaaluminium
This looks exactly the same in the test tube as the corresponding zinc reaction above - but beware
the different formulae of the precipitate and the final solution.
hexaaquachromium(III)
hexaaquairon(III)
Reversing the process
If adding hydroxide ions removes hydrogen ions from the hexaaqua complex one at a time, it
doesn't seem unreasonable that you could put them back again by adding an acid. That's just what
happens!
We'll look in detail at what happens in the chromium(III) case, but exactly the same principle
applies to all the other examples we've looked at - whether for 2+ or 3+ ions. As long as you
understand what is happening, you can work out the details if you need to.
Adding acid to hexahydroxochromate(III) ions
These are the ions formed at the end of the sequence in which you add hydroxide ions to a solution
containing hexaaquachromium(III) ions. Their formula is [Cr(OH)6]3-.
A reminder of the colour changes when you add sodium hydroxide solution to a solution containing
hexaaquachromium(III) ions:
If you add an acid (dilute sulphuric acid, for example), the hydrogen ions get put back on one at a
time.
You already know the colours of the significant stages (the beginning, the end, and the neutral
complex). It isn't a separate bit of learning!
You can apply this to any case. If you know the colours as you remove hydrogen ions, you
automatically know them as you put the hydrogen ions back on again.
It also doesn't matter where you start from either - whether, for example, you add acid to an ionic
complex like [Cr(OH)6]3-, or a neutral one like [Fe(H2O)4(OH)2].
You will know that the [Fe(H2O)4(OH)2] is a dirty green precipitate. When you add the hydrogen
ions back to it, it will revert to the very pale green solution of the [Fe(H2O)6]2+ ion. None of this is a
new bit of learning - you just have to re-arrange what you already know!
(Note: In fact, in the chromium case, it is actually slightly more difficult than this, because the final
solution tends to end up green rather than the usual strange blue-grey-violet of the
hexaaquachromium(III) ion.
This happens whenever you produce hexaaquachromium(III) ions in a test tube. The problem is that
other ions from the solution (chloride or sulphate) replace one or more water molecules in the complex
and give green colours. This involves a Ligand exchange reaction.)
Amphoteric hydroxides
An amphoteric substance has both acidic and basic properties. In other words, it will react with both
bases and acids. Some of the metal hydroxides we've been looking at are doing exactly that.
Chromium(III) hydroxide as an amphoteric hydroxide
"Chromium(III) hydroxide" is a simple way of naming the neutral complex [Cr(H2O)3(OH)3]. You
have seen that it reacts with bases (hydroxide ions) to give [Cr(OH) 6]3-. It also reacts with acids
(hydrogen ions) to give [Cr(H2O)6]3+.
This is a good example of amphoteric behaviour.
Other examples of amphoteric hydroxides are zinc hydroxide and aluminium hydroxide.
Copper(II) hydroxide as a basic oxide
Quite a lot of metal hydroxides won't react any further with hydroxide ions if you use sodium
hydroxide solution at the sort of concentrations normally used in the lab. That means that they don't
have any significant acidic nature.
"Copper(II) hydroxide" is what we would normally call the neutral complex [Cu(H 2O)4(OH)2]. This
doesn't dissolve in sodium hydroxide solution at any concentration normally used in the lab. It
doesn't show any acidic character.
On the other hand, it will react with acids - replacing the lost hydrogen ions on the water ligands.
Because it is accepting hydrogen ions, it is acting as a base.
Hydroxides like this (which react with acids, but not bases) are not amphoteric - they are just
simple bases.
REACTIONS OF HEXAAQUA METAL IONS WITH AMMONIA SOLUTION
Reactions of the hexaaqua ions with ammonia solution are complicated by the fact that the
ammonia can have two quite different functions. It can act as a base (in the Bronsted-Lowry
sense), but it is also a possible ligand which can replace water molecules around the central metal
ion. When it acts as a ligand, it is acting as a Lewis base.
We need to look at these two functions separately.
Ammonia acting as a (Bronsted-Lowry) base
The general case
This is what happens when you only add small amounts of dilute ammonia solution to any of the
hexaaqua ions. The ligand effect only happens with an excess of ammonia or with concentrated
ammonia - and with some metals you don't even see it then.
We'll talk through what happens if you add a small amount of dilute ammonia solution to a solution
containing a 2+ hexaaqua ion.
These have the formula [M(H2O)6]2+, and they are acidic. Their acidity is shown in the reaction of
the hexaaqua ions with water molecules from the solution:
They are acting as acids by donating hydrogen ions to water molecules in the solution.
Because of the confusing presence of water from two different sources (the ligands and the
solution), it is easier to simplify this:
Adding ammonia solution to this equilibrium - stage 1
There are two possible reactions.
Reaction of ammonia with the hydroxonium ions (hydrogen ions)
Ammonia will react with these to produce ammonium ions.
According to Le Chatelier's Principle, the position of equilibrium will move to the right, producing
more of the new complex ion.
Reaction of ammonia with the hexaaqua ion
Statistically, there is far more chance of an ammonia molecule hitting a hexaaqua metal ion than of
hitting a hydrogen ion. There are far more hexaaqua ions present.
If that happens, you get exactly the same new complex ion formed as above.
Notice that this is still a reversible change (unlike the corresponding change when you add
hydroxide ions). Ammonia is only a weak base.
The second stage of the reaction
Whichever of the above reactions happens, you end up with [M(H2O)5(OH)]+ ions in solution. These
are also acidic, and can lose hydrogen ions from another of the water ligands.
Taking the easier version of the equilibrium:
Adding ammonia again tips the equilibrium to the right - either by reacting with the hydrogen ions,
or by reacting directly with the complex on the left-hand side. When this happens, the new
complex formed no longer has a charge - this is a "neutral complex". It is insoluble in water - and so
a precipitate is formed. This precipitate is often written without including the remaining water
ligands. In other words we write it as M(OH)2. A precipitate of the metal hydroxide has been
formed.
Summarising what has happened
You can also usefully write the complete change as an overall equilibrium reaction. This will be
important for later on.
If you did the same reaction with a 3+ ion, the only difference is that you would have to remove a
total of 3 hydrogen ions in order to get to the neutral complex. That would give the overall equation:
Looking at the ions of specific metals
Remember that we are concentrating for the moment on the ammonia acting as a base - in other
words, on the formation of hydroxide precipitates when you add small amounts of ammonia
solution to solutions containing hexaaqua metal ions.
The diagrams, however, will show the complete change so I don't have to repeat them later on.
Ignore the cases where the precipitate dissolves in excess ammonia for the moment
2+ ions
hexaaquacobalt(II)
Note: The final solution rapidly darkens in air to a deep red-brown. This is due to oxidation from
hexaamminecobalt(II) to hexaamminecobalt(III) ions.
hexaaquacopper(II)
hexaaquairon(II)
Iron is very easily oxidised under alkaline conditions. Oxygen in the air oxidises the iron(II)
hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of
the precipitate comes from the same effect. This is NOT a ligand exchange reaction.
hexaaquamanganese(II)
I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is
virtually colourless. The pale brown precipitate is oxidised to darker brown manganese(III) oxide in
contact with oxygen from the air. Again, this isn't a ligand exchange reaction.
hexaaquanickel(II)
hexaaquazinc
You start and finish with colourless solutions, producing a white precipitate on the way.
3+ ions
hexaaquaaluminium
Starting from a colourless solution, you get a white precipitate.
hexaaquachromium(III)
hexaaquairon(III)
Summary of the effect of adding small amounts of ammonia solution
In each case you get a precipitate of the neutral complex - the metal hydroxide. Apart from minor
differences in the exact shade of colour you get, these are almost all exactly the same as the
precipitates you get when you add a little sodium hydroxide solution to the solutions of the
hexaaqua ions. The only real difference lies in the colour of the cobalt precipitate.
Ammonia acting as a ligand
The ligand exchange reaction
In some cases, ammonia replaces water around the central metal ion to give another soluble
complex. This is known as a ligand exchange reaction, and involves an equilibrium such as this
one:
The formation of this new soluble complex causes the precipitate to dissolve.
The ammonia attaches to the central metal ion using the lone pair of electrons on the nitrogen
atom. Because it is a lone pair donor, it is acting as a Lewis base.
Explaining why the precipitate dissolves
Almost all text books leave the argument at this
point, assuming that it is obvious why the formation of the complex causes some precipitates to
dissolve. It isn't! If you want to know the quite complicated reasons, read on . . .
We'll take the copper case as typical of any of them.
There are two equilibria involved in this. The first is the one in which ammonia is acting as a base
and producing the precipitate:
The other one is the ligand exchange reaction:
Notice that the hexaaqua ion appears in both of these. There is now an interaction between the two
equilibria:
Looking at it like this is helpful in explaining why some precipitates dissolve in excess ammonia
while others don't. It depends on the positions of the equilibria.
To get the precipitate to dissolve, you obviously need the ligand exchange equilibrium to lie well to
the right, but you need the acid-base equilibrium to be easy to pull back to the left.
REACTIONS OF HEXAAQUA METAL IONS WITH CARBONATE IONS
The general cases
There is a difference in the reactions depending on whether the metal at the centre of the
hexaaqua ion carries a 2+ or a 3+ charge. We need to look at the two cases separately.
3+ ions reacting with carbonate ions
The 3+ hexaaqua ions are sufficiently acidic to react with carbonate ions to release carbon dioxide
gas. You also get a precipitate of the metal hydroxide.
This is how it happens . . .
The acidity of the 3+ hexaaqua ions
These have the formula [M(H2O)6]3+, and they are fairly acidic. They react with water molecules
from the solution:
They are acting as acids by donating hydrogen ions to water molecules in the solution.
Because of the confusing presence of water from two different sources (the ligands and the
solution), it is easier to simplify this:
Carbonate ions acting as a base
Carbonate ions combine with hydrogen ions in two stages - first to make hydrogencarbonate ions,
and then to give carbon dioxide and water.
Reacting carbonate ions with the 3+ hexaaqua ions
Provided the proportions are right, the 3+ hexaaqua ions are sufficiently acidic for the reactions to
go all the way to carbon dioxide. There are two possible reactions. Reaction of carbonate ions
with the hydroxonium ions (hydrogen ions)
According to Le Chatelier's Principle, as the hydrogen ions are removed, the position of equilibrium
will move to the right, producing more of the new complex ion.
Reaction of carbonate ions with the hexaaqua ion
Statistically, there is far more chance of a carbonate ion hitting a hexaaqua metal ion than of hitting
a hydrogen ion. It removes a hydrogen ion directly from the hexaaqua complex.
If that happens, you get exactly the same new complex ion formed as above.
But it doesn't stop there. Carbonate ions keep on removing hydrogen ions from the complex until
you end up with a neutral complex. You see that as a precipitate of the metal hydroxide.
You can also usefully write the complete change as an overall reaction.
2+ ions reacting with carbonate ions
The 2+ hexaaqua ions aren't strongly acidic enough to release carbon dioxide from carbonates. In
these cases, you still get a precipitate - but it is a precipitate of what is loosely described as the
"metal carbonate".
Looking at the ions of specific metals
We'll look at the reactions of three 3+ ions and three 2+ ions. The important ones are the 3+ ions.
3+ ions
hexaaquaaluminium
Starting from a colourless solution, you get a white precipitate - but with bubbles of gas as well. The
precipitate is identical to the one you get if you add small amounts of either sodium hydroxide or
ammonia solutions to a solution of the hexaaquaaluminium ions.
hexaaquachromium(III)
Again, the precipitate is just the same as if you had added small amounts of either sodium
hydroxide or ammonia solution.
hexaaquairon(III)
. . . and again, exactly the same precipitate as if you had added any other base.
Summary
In each case you get a precipitate of the neutral complex - the metal hydroxide. This is exactly the
same precipitate that you get if you add small amounts of either sodium hydroxide solution or
ammonia solution to solutions of these ions. Bubbles of carbon dioxide are also given off.
Note: You may miss this carbon dioxide if your proportions are wrong. Unless there is an excess of the
acidic hexaaqua ion, you may get hydrogencarbonate ions formed in solution instead of carbon dioxide.
2+ ions
hexaaquacobalt(II)
No gas this time - just a precipitate of "cobalt(II) carbonate".
hexaaquacopper(II)
Again, there isn't any carbon dioxide - just a precipitate of the "copper(II) carbonate".
hexaaquairon(II)
You get a precipitate of the "iron(II) carbonate", but no carbon dioxide.
Summary
Hexaaqua ions with a 2+ charge aren't sufficiently acidic to liberate carbon dioxide from carbonate
ions. Instead you get a precipitate which you can think of as being the metal carbonate.
The general cases
There is a difference in the reactions depending on whether the metal at the centre of the
hexaaqua ion carries a 2+ or a 3+ charge. We need to look at the two cases separately.
3+ ions reacting with carbonate ions
The 3+ hexaaqua ions are sufficiently acidic to react with carbonate ions to release carbon dioxide
gas. You also get a precipitate of the metal hydroxide.
This is how it happens . . .
The acidity of the 3+ hexaaqua ions
These have the formula [M(H2O)6]3+, and they are fairly acidic. They react with water molecules
from the solution:
They are acting as acids by donating hydrogen ions to water molecules in the solution.
Because of the confusing presence of water from two different sources (the ligands and the
solution), it is easier to simplify this:
Carbonate ions acting as a base
Carbonate ions combine with hydrogen ions in two stages - first to make hydrogencarbonate ions,
and then to give carbon dioxide and water.
Reacting carbonate ions with the 3+ hexaaqua ions
Provided the proportions are right, the 3+ hexaaqua ions are sufficiently acidic for the reactions to
go all the way to carbon dioxide. There are two possible reactions.
Reaction of carbonate ions with the hydroxonium ions (hydrogen ions)
According to Le Chatelier's Principle, as the hydrogen ions are removed, the position of equilibrium
will move to the right, producing more of the new complex ion.
Reaction of carbonate ions with the hexaaqua ion
Statistically, there is far more chance of a carbonate ion hitting a hexaaqua metal ion than of hitting
a hydrogen ion. It removes a hydrogen ion directly from the hexaaqua complex.
If that happens, you get exactly the same new complex ion formed as above but it doesn't stop
there. Carbonate ions keep on removing hydrogen ions from the complex until you end up with a
neutral complex. You see that as a precipitate of the metal hydroxide.
You can also usefully write the complete change as an overall reaction.
2+ ions reacting with carbonate ions
The 2+ hexaaqua ions aren't strongly acidic enough to release carbon dioxide from carbonates. In
these cases, you still get a precipitate - but it is a precipitate of what is loosely described as the
"metal carbonate".
REACTIONS OF HEXAAQUA METAL IONS WITH CARBONATE IONS
Looking at the ions of specific metals
We'll look at the reactions of three 3+ ions and three 2+ ions. The important ones are the 3+ ions.
3+ ions hexaaquaaluminium
Starting from a colourless solution, you get a white precipitate - but with bubbles of gas as well. The
precipitate is identical to the one you get if you add small amounts of either sodium hydroxide or
ammonia solutions to a solution of the hexaaquaaluminium ions
hexaaquachromium(III)
Again, the precipitate is just the same as if you had added small amounts of either sodium
hydroxide or ammonia solution.
hexaaquairon(III)
. . . and again, exactly the same precipitate as if you had added any other base.
Summary In each case you get a precipitate of the neutral complex - the metal hydroxide. This is
exactly the same precipitate that you get if you add small amounts of either sodium hydroxide
solution or ammonia solution to solutions of these ions. Bubbles of carbon dioxide are also given
off.
Note: You may miss this carbon dioxide if your proportions are wrong. Unless there is an excess of the
acidic hexaaqua ion, you may get hydrogencarbonate ions formed in solution instead of carbon dioxide.
2+ ions
hexaaquacobalt(II)
No gas this time - just a precipitate of "cobalt(II) carbonate".
hexaaquacopper(II)
Again, there isn't any carbon dioxide - just a precipitate of the "copper(II) carbonate".
hexaaquairon(II)
You get a precipitate of the "iron(II) carbonate", but no carbon dioxide.
Summary
Hexaaqua ions with a 2+ charge aren't sufficiently acidic to liberate carbon dioxide from carbonate
ions. Instead you get a precipitate which you can think of as being the metal carbonate.