buffer action and concepts involved

8/12/2019 Buffer Action and Concepts involved

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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 small

quantities 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 than7. Acidic buffer solutions are commonly made from a weak acidand one 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 solutioncontained equal molar concentrations of both the acid and the

salt, it would have a pH of 4.76. It wouldn't matter what theconcentrations were, as long as they were the same.

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

Note: If you need to know about calculations involvingbuffer solutions, you may be interest in mychemistrycalculations book.

Alkaline buffer solutions

An alkaline buffer solution has a pH greater than 7. Alkalinebuffer solutions are commonly made from a weak base and oneof its salts.
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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, itdoesn't matter what concentrations you choose as long as theyare the same.

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 -otherwise the pH will change. Acidic and alkaline buffersolutions achieve this in different ways.


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 ethanoateions. According to Le Chatelier's Principle, that will tip the

position of the equilibrium even further to the left.

Note: If you don't understandLe 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
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aren't important to the argument.

Adding an ac id to th is bu f fer solut ion

The buffer solution must remove most of the new hydrogen ions

otherwise the pH would drop markedly.

Hydrogen ions combine with the ethanoate ions to makeethanoic acid. Although the reaction is reversible, since theethanoic acid is a weak acid, most of the new hydrogen ions areremoved in this way.

Since most of the new hydrogen ions are removed, the pH won'tchange very much - but because of the equilibria involved,

it willfall a little bit.

Add ing an alkali to this buffer solut ion

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

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

Removal by reacting with ethanoic acid

The most likely acidic substance which a hydroxide ion is goingto collide with is an ethanoic acid molecule. They will react toform ethanoate ions and water.

Note: You might be surprised to find this written as aslightly reversible reaction. Because ethanoic acid is a weakacid, its conjugate base (the ethanoate ion) is fairly good atpicking up hydrogen ions again to re-form the acid. It can get

these from the water molecules. You may well find thisreaction written as one-way, but to be fussy about it, it isactually reversible!

Because most of the new hydroxide ions are removed, the pH


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doesn't increase very much.

Removal of the hydroxide ions by reacting with hydrogen ions

Remember that there are some hydrogen ions present from the

ionisation of the ethanoic acid.

Hydroxide ions can combine with these to make water. As soonas this happens, the equilibrium tips to replace them. This keepson happening until most of the hydroxide ions are removed.

Again, because you have equilibria involved, not allof thehydroxide ions are removed - just most of them. The waterformed re-ionises to a very small extent to give a few hydrogenions and hydroxide ions.


We'll take a mixture of ammonia and ammonium chloridesolutions as 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 theposition of the equilibrium even further to the left.


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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 presentaren't important to the argument.

Adding an ac id to th is bu f fer solut ion

There are twoprocesses which can remove the hydrogen ionsthat you are adding.

Removal by reacting with ammonia

The most likely basic substance which a hydrogen ion is goingto collide 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 hydrogenions will 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 replacethe hydroxide ions. This keeps on happening until most of thehydrogen ions are removed.


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

Add ing an alkali to this buffer solut ion

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 withthe water - and so the reaction is slightly reversible. That meansthat, again, most (but not all) of the the hydroxide ions areremoved from the solution.

Calculations involving buffer solutions

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


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-3of ethanoic acid and 0.20 mol dm-3of 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,
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you have the equilibrium:

The presence of the ethanoate ions from the sodium ethanoate

will have moved the equilibrium to the left, but the equilibriumstill exists.

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

Where you have done calculations using this equationpreviously with a weak acid, you will have assumed that the

concentrations of the hydrogen ions and ethanoate ions werethe same. Every molecule of ethanoic acid that splits up givesone of each sort of ion.

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 becompletely 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 thiscase, 0.20 mol dm-3.

In a weak acid calculation, we normally assume that so little of

the acid has ionised that the concentration of the acid atequilibrium is the same as the concentration of the acid weused. That is even more true now that the equilibrium has beenmoved even further to the left.

So the assumptions we make for a buffer solution are:


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

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

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

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

+]

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

You should get an answer of 5.1 to two significant figures. You

can'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 andsodium ethanoate to get a buffer solution of some desired pH. It


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is no more difficult than the calculation we have just looked at.

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

Feed that into the Kaexpression.

All this means is that to get a solution of pH 4.46, theconcentration of the ethanoate ions (from the sodium ethanoate)in the solution has to be 0.5 times that of the concentration ofthe acid. All that matters 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

10 cm3

of 1.0 mol dm-3

sodium ethanoate solution with 20 cm3

of1.0 mol dm-3ethanoic acid. Or 10 cm3of 1.0 mol dm-3sodiumethanoate solution with 10 cm3of 2.0 mol dm-3ethanoic acid.And there are all sorts of other possibilities.

Note: If your maths isn't very good, these examples canlook a 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!


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We are talking here about a mixture of a weak base and one ofits salts - for example, a solution containing ammonia and

ammonium chloride.

The modern, and easy, way of doing these calculations is to re-think them from the point of view of the ammonium ion ratherthan of the ammonia solution. Once you have taken this slightlydifferent view-point, everything becomes much the same asbefore.

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

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

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

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

The presence of the ammonia in the mixture forces theequilibrium far to the left. That means that you can assume thatthe ammonium ion concentration is what you started off with inthe ammonium chloride, and that the ammonia concentration isall due to the added ammonia solution.


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

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

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


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HNO3+ CH3COONa -------> NaNO3+ CH3COOH

Here we have a mixture of salt of a strong acid and strong base and a weak acid on the

product side.

So do you mean CH3COOH and CH3COONa (at equilibrium) are responsible for buffer here?

What about NaNO3and HNO3? Do they play any role? Give reason for your answer.

Borek #4

Oct26-10, 07:56 AM

Admin

P: 21,842

Buffer Solution

Quote by Abdul Quadeer

1) A solution of weak acid and its salt with a strong base.

Agreed - that will be mixture of acetic acid and sodium acetate.

2) A solution of strong base and its salt with a strong acid.

Like mixture of NaON and NaCl? It is not a classic buffer. Relatively

concentrated solutions of strong bases or strong acids do resist pH changes, so

in a way they are similar to buffers, but they don't require presence of salts.

3) A solution of a salt of weak acid and a weak base.

Like solution of ammonium acetate? Buffering capacity close to zero, much

lower than in both earlier cases.

How can they act as a buffer?

What happens if you mix - say - 2 moles of sodium acetate, with 1 mole ofnitric acid?

Edit: I see you were on the right track already. Think about composition of the

solution after the reaction. Does it contain nitric acid?
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Abdul Quadeer #5

Oct26-10, 08:15 AM

P: 1,395Quote by Borek

What happens if you mix - say - 2 moles of sodium acetate, with 1 mole of

nitric acid?

Edit: I see you were on the right track already. Think about composition of

the solution after the reaction. Does it contain nitric acid?

HNO3would be completely consumed and 1 mole of sodium acetate would

remain as a reactant.

1 mole each of NaNO3and CH3COOH are present in product side.

So here CH3COOH and CH3COONa form a buffer solution.But we still have 1 mole of NaNO3. Will it not affect the buffer?

espen180 #6

Oct26-10, 01:25 PM

P: 835 What kind of acid is nitric acid? Will this kind of acid make a buffer?

Borek #7

Oct26-10, 01:49 PM

Admin

P: 21,842

Quote by Abdul Quadeer

But we still have 1 mole of NaNO3. Will it not affect the buffer?

No, it just changes ionic strength of the solution.

Abdul Quadeer #8

Oct26-10, 02:24 PM
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P: 1,395 Thanks Mr. PH

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