Posted by: Mark Foreman | December 15, 2013

Titration I

Dear Student,

There appears to be no escape yet from equilibrium in aqueous media (maybe it is thermodynamically impossible to escape it), this in some ways is no bad thing as you need to learn these things in great depth to be good at physical chemistry.

Now imagine that we have 1 litre of 0.5 M acetic acid, what will the pH of this solution be ?

If we ignore the fact that the final acid concentration will be lower than the initial concentration, then we can use a very simple method.

KA = [H+][AcO]/[AcOH]

Then as [H+] = [AcO] rearrange to give

[H+]2 = KA[AcOH]

Then

[H+] = (KA[AcOH])½

Now if we mix the acetic acid solution with 500 ml of 0.5 M potassium hydroxide what will the pH be now.

We can attack the problem in a quick way or a more complex and better way.

Now the easy way is to use the buffer equation to deal with the problem.

pH = pKa + log10([AcO]/[AcOH])

For acetic acid pKa = 4.5

Now what is the pH of the mixture after 1000 ml of 0.5 M potassium hydroxide has been added.

The mixture is now 0.25 M potassium acetate

We have the equilibrium now of

AcO + H+ à AcOH

Which can be rewritten as

AcO + H2O à HO + AcOH

As

KW = KAKB

Then

pKW = pKA + pKB

So

pKW – pKA = pKB

Thus

pKB of acetate is 9.5

So KB = 3.16 x10-10

So if we assume that the final acetate concentration is equal to the initial then we can use

KB = [OH][AcOH]/[AcO]

Rearrange

[OH]2 = KB[AcO]

Next

[OH] = (KB[AcO])½

So for [OH] we get a value of (7.9 x 10-11) moles per litre

Which is 8.89 x 10-6 moles per litre

As

KW = [H+][HO]

Then we have a proton concentration of 1.125 x 10-9 moles per litre

This will give us a pH of 8.95

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