An equal volume mixture of which two solutions will produce a buffer solution?

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Because most of the new hydroxide ions are removed, the pH 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 soon as this happens, the equilibrium tips to replace them. This keeps on happening until most of the hydroxide ions are removed.

Again, because you have equilibria involved, not all of the hydroxide ions are removed - just most of them. The water formed re-ionises to a very small extent to give a few hydrogen ions and hydroxide ions.

Alkaline buffer solutions

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

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

Adding ammonium chloride to this adds lots of extra ammonium ions. According to Le Chatelier's Principle, that will tip the position of 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't important to the argument.

Adding an acid to this buffer solution

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

Removal by reacting with ammonia

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

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

Removal of the hydrogen ions by reacting with hydroxide ions

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

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

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

Adding an alkali to this buffer solution

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

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

Calculations involving buffer solutions

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

Acidic buffer solutions

This is easier to see with a specific example. Remember that an acid 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 mol dm-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 equilibrium between the un-ionised acid and its ions. So for ethanoic acid, you have the equilibrium:

The presence of the ethanoate ions from the sodium ethanoate will have moved the equilibrium to the left, but the equilibrium still exists.

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

Where you have done calculations using this equation previously with a weak acid, you will have assumed that the concentrations of the hydrogen ions and ethanoate ions were the same. Every molecule of ethanoic acid that splits up gives one 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, the number of ethanoate ions coming from the ethanoic acid will be completely negligible compared to those from the sodium ethanoate.

We therefore assume that the ethanoate ion concentration is the same 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 the acid has ionised that the concentration of the acid at equilibrium is the same as the concentration of the acid we used. That is even more true now that the equilibrium has been moved even further to the left.

So the assumptions we make for a buffer solution are:

Now, if we know the value for Ka, we can calculate the hydrogen ion 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 solution containing 0.10 mol dm-3 of ethanoic acid and 0.20 mol dm-3 of sodium 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 on your calculator, so press the log button and ignore the negative 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 concentrations were only given to two figures.

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