BUFFER SOLUTIONS

This page describes simple acidic and alkaline buffer solutionsand explains how they work.

What is a buffer solution?

Definition

A buffer solution is one which resists changes in pH when smallquantities of an acid or an alkali are added to it.

Acidic buffer solutions

An acidic buffer solution is simply one which has a pH less than 7.Acidic buffer solutions are commonly made from a weak acid andone of its salts - often a sodium salt.

A common example would be a mixture of ethanoic acid andsodium ethanoate in solution. In this case, if the solution containedequal molar concentrations of both the acid and the salt, it wouldhave a pH of 4.76. It wouldn't matter what the concentrations were,as long as they were the same.

You can change the pH of the buffer solution by changing the ratioof acid to salt, or by choosing a different acid and one of its salts.

Note: If you need to know about calculations involving buffersolutions, you may be interest in my chemistry calculationsbook.

Alkaline buffer solutions

An alkaline buffer solution has a pH greater than 7. Alkaline buffersolutions are commonly made from a weak base and one of itssalts.

A frequently used example is a mixture of ammonia solution andammonium chloride solution. If these were mixed in equal molarproportions, the solution would have a pH of 9.25. Again, it doesn'tmatter what concentrations you choose as long as they are thesame.

How do buffer solutions work?

A buffer solution has to contain things which will remove anyhydrogen ions or hydroxide ions that you might add to it - otherwisethe pH will change. Acidic and alkaline buffer solutions achieve thisin different ways.

Acidic buffer solutions

We'll take a mixture of ethanoic acid and sodium ethanoate astypical.

Ethanoic acid is a weak acid, and the position of this equilibriumwill be well to the left:

Adding sodium ethanoate to this adds lots of extra ethanoate ions.According to Le Chatelier's Principle, that will tip the position of theequilibrium even further to the left.

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Note: If you don't understand Le Chatelier's Principle, followthis link before you go any further, and make sure that youunderstand about the effect of changes of concentration onthe position of equilibrium.

Use the BACK button on your browser to return to this page.

The solution will therefore contain these important things:

lots of un-ionised ethanoic acid;

lots of ethanoate ions from the sodium ethanoate;

enough hydrogen ions to make the solution acidic.

Other things (like water and sodium ions) which are present aren'timportant to the argument.

Adding an acid to this buffer solution

The buffer solution must remove most of the new hydrogen ionsotherwise the pH would drop markedly.

Hydrogen ions combine with the ethanoate ions to make ethanoicacid. Although the reaction is reversible, since the ethanoic acid isa weak acid, most of the new hydrogen ions are removed in thisway.

Since most of the new hydrogen ions are removed, the pH won'tchange very much - but because of the equilibria involved, it will falla little bit.

Adding an alkali to this buffer solution

Alkaline solutions contain hydroxide ions and the buffer solutionremoves most of these.

This time the situation is a bit more complicated because thereare two processes which can remove hydroxide ions.

Removal by reacting with ethanoic acid

The most likely acidic substance which a hydroxide ion is going tocollide with is an ethanoic acid molecule. They will react to formethanoate ions and water.

Note: You might be surprised to find this written as a slightlyreversible reaction. Because ethanoic acid is a weak acid, itsconjugate base (the ethanoate ion) is fairly good at picking uphydrogen ions again to re-form the acid. It can get these fromthe water molecules. You may well find this reaction writtenas one-way, but to be fussy about it, it is actually reversible!

Because most of the new hydroxide ions are removed, the pHdoesn't increase very much.

Removal of the hydroxide ions by reacting with hydrogen ions

Remember that there are some hydrogen ions present from theionisation of the ethanoic acid.

Hydroxide ions can combine with these to make water. As soon asthis happens, the equilibrium tips to replace them. This keeps on

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happening until most of the hydroxide ions are removed.

Again, because you have equilibria involved, not all of thehydroxide ions are removed - just most of them. The water formedre-ionises to a very small extent to give a few hydrogen ions andhydroxide ions.

Alkaline buffer solutions

We'll take a mixture of ammonia and ammonium chloride solutionsas typical.

Ammonia is a weak base, and the position of this equilibrium willbe well to the left:

Adding ammonium chloride to this adds lots of extra ammoniumions. According to Le Chatelier's Principle, that will tip the positionof the equilibrium even further to the left.

The solution will therefore contain these important things:

lots of unreacted ammonia;

lots of ammonium ions from the ammonium chloride;

enough hydroxide ions to make the solution alkaline.

Other things (like water and chloride ions) which are present aren'timportant to the argument.

Adding an acid to this buffer solution

There are two processes which can remove the hydrogen ions thatyou are adding.

Removal by reacting with ammonia

The most likely basic substance which a hydrogen ion is going tocollide with is an ammonia molecule. They will react to formammonium ions.

Most, but not all, of the hydrogen ions will be removed. Theammonium ion is weakly acidic, and so some of the hydrogen ionswill be released again.

Removal of the hydrogen ions by reacting with hydroxide ions

Remember that there are some hydroxide ions present from thereaction between the ammonia and the water.

Hydrogen ions can combine with these hydroxide ions to makewater. As soon as this happens, the equilibrium tips to replace thehydroxide ions. This keeps on happening until most of thehydrogen ions are removed.

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Again, because you have equilibria involved, not all of thehydrogen ions are removed - just most of them.

Adding an alkali to this buffer solution

The hydroxide ions from the alkali are removed by a simplereaction with ammonium ions.

Because the ammonia formed is a weak base, it can react with thewater - and so the reaction is slightly reversible. That means that,again, most (but not all) of the the hydroxide ions are removed fromthe solution.

Calculations involving buffer solutions

This is only a brief introduction. There are more examples,including several variations, over 10 pages in my chemistrycalculations book.

Acidic buffer solutions

This is easier to see with a specific example. Remember that anacid buffer can be made from a weak acid and one of its salts.

Let's suppose that you had a buffer solution containing 0.10 moldm-3 of ethanoic acid and 0.20 mol dm-3 of sodium ethanoate.How do you calculate its pH?

In any solution containing a weak acid, there is an equilibriumbetween the un-ionised acid and its ions. So for ethanoic acid, youhave the equilibrium:

The presence of the ethanoate ions from the sodium ethanoate willhave moved the equilibrium to the left, but the equilibrium stillexists.

That means that you can write the equilibrium constant, Ka, for it:

Where you have done calculations using this equation previouslywith a weak acid, you will have assumed that the concentrations ofthe hydrogen ions and ethanoate ions were the same. Everymolecule of ethanoic acid that splits up gives one of each sort ofion.

That's no longer true for a buffer solution:

If the equilibrium has been pushed even further to the left, thenumber of ethanoate ions coming from the ethanoic acid will be

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completely negligible compared to those from the sodiumethanoate.

We therefore assume that the ethanoate ion concentration is thesame as the concentration of the sodium ethanoate - in this case,0.20 mol dm-3.

In a weak acid calculation, we normally assume that so little of theacid has ionised that the concentration of the acid at equilibrium isthe same as the concentration of the acid we used. That is evenmore true now that the equilibrium has been moved even further tothe left.

So the assumptions we make for a buffer solution are:

Now, if we know the value for Ka, we can calculate the hydrogenion concentration and therefore the pH.

Ka for ethanoic acid is 1.74 x 10-5 mol dm-3.

Remember that we want to calculate the pH of a buffer solutioncontaining 0.10 mol dm-3 of ethanoic acid and 0.20 mol dm-3 ofsodium ethanoate.

Then all you have to do is to find the pH using the expression pH = -log10 [H+]

You will still have the value for the hydrogen ion concentration onyour calculator, so press the log button and ignore the negativesign (to allow for the minus sign in the pH expression).

You should get an answer of 5.1 to two significant figures. Youcan't be more accurate than this, because your concentrationswere only given to two figures.

You could, of course, be asked to reverse this and calculate inwhat proportions you would have to mix ethanoic acid and sodiumethanoate to get a buffer solution of some desired pH. It is no moredifficult than the calculation we have just looked at.

Suppose you wanted a buffer with a pH of 4.46. If you un-log this tofind the hydrogen ion concentration you need, you will find it is 3.47x 10-5 mol dm-3.

Feed that into the Ka expression.

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All this means is that to get a solution of pH 4.46, the concentrationof the ethanoate ions (from the sodium ethanoate) in the solutionhas to be 0.5 times that of the concentration of the acid. All thatmatters is that ratio.

In other words, the concentration of the ethanoate has to be halfthat of the ethanoic acid.

One way of getting this, for example, would be to mix together 10cm3 of 1.0 mol dm-3 sodium ethanoate solution with 20 cm3 of 1.0mol dm-3 ethanoic acid. Or 10 cm3 of 1.0 mol dm-3 sodiumethanoate solution with 10 cm3 of 2.0 mol dm-3 ethanoic acid. Andthere are all sorts of other possibilities.

Note: If your maths isn't very good, these examples can looka bit scary, but in fact they aren't. Go through thecalculations line by line, and make sure that you can seeexactly what is happening in each line - where the numbersare coming from, and why they are where they are. Then goaway and practise similar questions.

If you are good at maths and can't see why anyone shouldthink this is difficult, then feel very fortunate. Most peoplearen't so lucky!

Alkaline buffer solutions

We are talking here about a mixture of a weak base and one of itssalts - for example, a solution containing ammonia and ammoniumchloride.

The modern, and easy, way of doing these calculations is to re-think them from the point of view of the ammonium ion rather thanof the ammonia solution. Once you have taken this slightly differentview-point, everything becomes much the same as before.

So how would you find the pH of a solution containing 0.100 moldm-3 of ammonia and 0.0500 mol dm-3 of ammonium chloride?

The mixture will contain lots of unreacted ammonia molecules andlots of ammonium ions as the essential ingredients.

The ammonium ions are weakly acidic, and this equilibrium is setup whenever they are in solution in water:

You can write a Ka expression for the ammonium ion, and makethe same sort of assumptions as we did in the previous case:

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The presence of the ammonia in the mixture forces the equilibriumfar to the left. That means that you can assume that the ammoniumion concentration is what you started off with in the ammoniumchloride, and that the ammonia concentration is all due to theadded ammonia solution.

The value for Ka for the ammonium ion is 5.62 x 10-10 mol dm-3.

Remember that we want to calculate the pH of a buffer solutioncontaining 0.100 mol dm-3 of ammonia and 0.0500 mol dm-3 ofammonium chloride.

Just put all these numbers in the Ka expression, and do the sum:

Questions to test your understanding

If this is the first set of questions you have done, please read theintroductory page before you start. You will need to use the BACK BUTTONon your browser to come back here afterwards.

questions on buffer solutions

answers

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© Jim Clark 2002 (last modified November 2013)

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