Now sometimes we want to calculate the ”proton” concentration in an aqueous solution of an acid from the initial concentration of the acid [HA]_{initial} and the dissociation constant (K_{A}) of the acid.

Now when the initial concentration is high and the dissociation constant is small then we can take a shortcut, here we will also ignore the autoioniziation of the water.

As

K_{A} = [H^{+}][A^{–}]/[HA]

Then

K_{A} = [H^{+}]^{2}/[HA]

And then

[H+] = (K_{A}[HA])^{½}

Here we assume that [HA]_{initial} = [HA]_{final}

While this is OK for things like a 1 M solution of acetic acid it will fall down for many other things. One problem is that it predicts more than 100 % ioniziation of the acid. Here is a graph showing the bad prediction of how much of the acid will be ionised.

Now a slightly more rigorous method is to use the following mass balance equation

[HA]_{initial} = [HA]_{final} + [A^{–}]

And the charge balance equation

[H^{+}] = [OH^{–}] + [A^{–}]

Together with the buffer equation

-log_{10}[H^{+}] = pH = pKa + log_{10}([A^{–}]/[HA])

We will still ignore the autoionization of the water, so we assume that

[H^{+}] = [A^{–}]

Now we need to make what I call a brute force attack on the problem, this is one which is quite simple in terms of maths but is one in which close to every possible combination is tried. It is like going through all the combinations which are possible on a safe. Given time you will get lucky and open the safe.

Now

[HA]_{final} = K_{A} / [H^{+}]^{2}

We can get [A^{–}] from the buffer equation, and we can then calculate for a series of proton concentrations the values of [A^{–}] and [HA]_{final}. As the sum of [A^{–}] and [HA]_{final} is [HA]_{initial} then we can work backwards and make a graph of pH vs [HA]_{initial}. Here is a graph of pH against the original concentration of the acid.

The good thing about the new model is that it does not predict more than 100 % ionization of the acid, well that is a good step forward !

So now we have a better model for the pH range around 4, but as the pH rises further and further this new model will become worse and worse.

So we should go back to our model and improve it further.

If

[H^{+}] = [OH^{–}] + [A^{–}] and K_{w} = [H^{+}][OH^{–}]

Then

[A^{–}] = [H^{+}] – (K_{W}/[H^{+}])

We can use

[H^{+}]^{3} + K_{A}[H^{+}]^{2} = (K_{W} + K_{A}[HA]_{initial}) [H^{+}] + K_{A}K_{W}

As we know K_{A} and K_{W} we can now create a table of values for (K_{W} + K_{A}[HA]_{initial}) for different proton concentrations.

From this we can get the [HA]_{initial} concentrations for a range of different proton concentrations.

From this we can get [A^{–}] and [HA]_{final} to finish off our problem. Here is the graph of pH vs original acid concentration.

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