Chemguide: Core Chemistry 14 - 16


Testing for positive ions

This page looks at tests for a number of positive ions in solution using sodium hydroxide solution and ammonia solution.

There is a wide variation between what various syllabuses might want you to know about this, and it is is essential that you find out what level of detail your examiners want. The more you have to learn, the more confusing it can get.

If you are doing a UK-based syllabus, you can find links to the Exam Boards' websites where you can download a copy of your syllabus and other useful stuff on the about this part of Chemguide page.


An easy bit - testing for ammonium ions

This is easy because it is unique - you can't confuse it with anything else.

You can test for ammonium ions either in a solid or in solution by adding sodium hydroxide solution and warming gently.

If the compound contains ammonium ions, you will get ammonia gas produced which you can test with damp red litmus paper which it turns blue.


Note:  The video shows some very poor practice by heating a solution in a test tube using a very hot Bunsen flame. That is asking for trouble because there is a real risk of hot solution spurting out of the tube. Liquids should never be heated in a test tube with anything other than a small blue flame with no blue cone in it.


This also works perfectly well if you add the sodium hydroxide solution to some solid containing ammonium ions and warm that mixture.

The video gives an example equation:

NH4Cl(aq) + NaOH(aq)   NaCl(aq) + NH3(g) + H2O(l)

If you had ammonium sulfate, then the equation would be

(NH4)2SO4(aq) + 2NaOH(aq)   Na2SO4(aq) + 2NH3(g) + 2H2O(l)

In fact, it is easier to write the ionic equation, which is the same for all cases - just choosing the (aq) or (s) as appropriate.

NH4+(aq or s) + OH-(aq)   NH3(g) + H2O(l)


Testing for iron(II), iron(III) and copper(II) ions

These are easy because they give different, easily recognisable coloured precipitates.

These tests for metal ions are always done in solution and adding either sodium hydroxide solution or ammonia solution.

You start with a solution of your compound and then add a few drops of sodium hydroxide solution or ammonia solution. Note what happens.

Then go on to add an excess of the solution you are adding. Again, note what happens.

Testing these ions using sodium hydroxide solution

The video shows what happens when you add sodium hydroxide solution to these three ions (and also to Al3+ ions - ignore that for now.)

The Fe2+ precipitate often starts a lighter green than that, but darkens rapidly.

That colour is shown better in this very short video using, I suspect, more dilute solutions.

Results

Solution
contains
few drops NaOH(aq)excess NaOH(aq)
Fe2+dirty green pptno change
Fe3+orange-brown pptno change
Cu2+pale blue pptno change

In each case, you get a precipitate of the metal hydroxide.

It is much simpler to write ionic equations for these reactions.

Fe2+(aq) + 2OH-(aq)   Fe(OH)2(s)

Fe3+(aq) + 3OH-(aq)   Fe(OH)3(s)

Cu2+(aq) + 2OH-(aq)   Cu(OH)2(s)

You should, of course, be able to write full equations if you were asked, but they are more time-consuming, with more chance of making mistakes.

For example:

FeCl3(aq) + 3NaOH(aq)   Fe(OH)3(s) + 3NaCl(aq)


Testing these ions using ammonia solution

Ammonia reacts with water to some extent to form ammonium hydroxide.

NH3(g) + H2O(l)    NH4+(aq) + OH-(aq)

The presence of the hydroxide ions means that you might expect ammonia solution to react with metal ions in the same way as sodium hydroxide solution.

It does, but there are occasionally differences.

Solution
contains
few drops NH3(aq)excess NH3(aq)
Fe2+dirty green pptno change
Fe3+orange-brown pptno change
Cu2+pale blue pptvery dark blue solution formed

In each case the formation of the precipitates involves exactly the same chemistry as with sodium hydroxide solution. All that matters is the presence of the hydroxide ions, not where they came from.

The formation of the precipitates is shown by the same ionic equations as before.

Fe2+(aq) + 2OH-(aq)   Fe(OH)2(s)

Fe3+(aq) + 3OH-(aq)   Fe(OH)3(s)

Cu2+(aq) + 2OH-(aq)   Cu(OH)2(s)

The next video shows the formation of the dark blue solution with copper(II) ions.

What is happening here is that the excess ammonia solution reacts with the copper(II) ions to give the deep blue soluble ion [Cu(NH3)4]2+ where ammonia molecules have attached themselves to the copper ion.


Note:  Strictly speaking this should be the [Cu(NH3)4(H2O)2]2+ ion. At this level, all you may be expected to know is its deep blue colour.

Check to see if this reaction is on your syllabus. If it is, look at past exam papers and mark schemes to find out exactly what your examiners expect you to say about it.




Testing for magnesium and calcium ions

Testing these ions using sodium hydroxide solution

Solution
contains
few drops NaOH(aq)excess NaOH(aq)
Mg2+white pptno change
Ca2+white pptno change

In each case, the white precipitate is of the metal hydroxide.

Mg2+(aq) + 2OH-(aq)   Mg(OH)2(s)

Ca2+(aq) + 2OH-(aq)   Ca(OH)2(s)

Testing these ions using ammonia solution

Solution
contains
few drops NH3(aq)excess NH3(aq)
Mg2+white pptno change
Ca2+no ppt (or only
very slight ppt)
no change

The next video shows these two reactions. If you look very closely at the ammonia one, you will see a slight trace of a precipitate.

The problem in the calcium case is that calcium hydroxide is slightly soluble - we use its dilute solution as lime water.

Ammonia solution contains far fewer hydroxide ions than sodium hydroxide solution of the same concentration. So you don't form enough calcium hydroxide to form a reasonable precipitate.

You can use this difference to distinguish between magnesium ions and calcium ions.

  • Magnesium ions give a white precipitate insoluble in excess in both sodium hydroxide or ammonia solution.

  • Calcium ions give a white precipitate insoluble in excess sodium hydroxide solution, but no precipitate or a very slight precipitate in ammonia solution.


Testing for aluminium and zinc ions

This is where things start to get more than a bit confusing!

Testing these ions using sodium hydroxide solution

Solution
contains
few drops NaOH(aq)excess NaOH(aq)
Al3+white pptppt dissolves in excess to give colourless solution
Zn2+white pptppt dissolves in excess to give colourless solution

So both of these give a white precipitate of the metal hydroxide and they both dissolve in excess sodium hydroxide to give a colourless solution.

The precipitates are easy:

Al3+(aq) + 3OH-(aq)   Al(OH)3(s)

Zn2+(aq) + 2OH-(aq)   Zn(OH)2(s)

They dissolve in excess sodium hydroxide solution to produce soluble complex ions in which hydroxide groups are bound to the metal ion.

  • In the zinc case the final solution contains [Zn(OH)4]2- ions.

  • In the aluminium case the final solution contains [Al(OH)4]- ions.

Testing these ions using ammonia solution

Solution
contains
few drops NH3(aq)excess NH3(aq)
Al3+white pptno change
Zn2+white pptppt dissolves in excess to give colourless solution

Again, the precipitates are the metal hydroxides. Remember that ammonia solution contains some hydroxide ions.

So the ionic equations for the formation of the precipitates are the same as before.

This time, only the zinc hydroxide precipitate redissolves - for the same reason that the copper(II) hydroxide that we have already looked at dissolves in excess ammonia solution.

The solution formed contains the soluble complex ion [Zn(NH3)4]2+.


There are two useful short pieces of video showing what happens in these reactions. They look at the aluminium ion reactions first, followed by the zinc ion reactions.

First the aluminium ion reactions . . .

. . . and then the zinc ion reactions.


And finally . . .

There is probably no other page in this part of Chemguide where it is so important to find out exactly what your examiners want you to know. You need to know what the syllabus says, and what sort of questions your examiners are going to ask you.

If you are really lucky, you may only need to know about the metals which form coloured precipitates - and you might not even need to know about the reactions with ammonia solution.

The further you need to work down this page, the more it just becomes a case of tediously learning stuff. You need to avoid that if you possibly can!

It is quite likely that you won't need to know the formulae of the complex ions that are formed in some cases, but you can't rely on that. You have to check your syllabus and exam papers (and mark schemes as well) very carefully to try to save you unnecessary effort.


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