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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.