Short Cut Method to Calculating pH of Polyprotic Acid?

Question

I am doing the following question:

Calculate the pH of a $4.00\ \mathrm{mol\ L^{-1}}$ solution of citric acid. $\mathrm{pK_{a1}} = 3.09~~~~~ \mathrm{pK_{a2}} = 4.75~~~~~ \mathrm{pK_{a3}} = 6.40$

Usually for polyprotic acids, we can assume that the second, third and other dissociations of $\ce{H+}$ ions are negligible when compared to the first dissociation. However I believe that this is only valid when the pKa values are separated by around 4. So for citric acid, this would be an invalid assumption as the pKa values are pretty close to each other.

Now, I can find the pH of the solution by doing the long way by first finding the amount of $\ce{H+}$ ions formed by the first dissociation, then using that to find the amount of $\ce{H+}$ ions formed by the second dissociation and so on.

However the room that they have given for this question is pretty small and has only has enough room for a couple of lines of working out and I definitely wouldn't be able to fit all my working out in that box.

So am I missing some short cut method to finding the pH in cases like this, or is the above assumption that I have outline at the start still correct for citric acid?

Answer 1

The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid.

You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps.

(For the sake of simplicity I write $c(\ce{H+})=x$ and $k_{a,n}=k_n$.)

Exact: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$ Neglecting $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$ Neglecting $k_2$ and $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{k_1}{x+k_1}$$

Solving those equations give the following pH values (with super unrealistic high precision):

• Exact: 1.2469291050971560
• w/o $k_3$: 1.2469291065371300
• w/o $k_2$ and $k_3$: 1.2470654318802320

As a pH can only be measured precisely within something like $\pm \left(10^{-2}\ldots10^{-3}\right)~\mathrm{pH}$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected.

So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days.

Answer 2

Since the solution will be acid in water, the $\mathrm{pH}$ will be below all the $\mathrm{p}K_\mathrm{a}$ values.

Given that the $\mathrm{pH}$ is below all the $\mathrm{p}K_\mathrm{a}$ values, an upper limit is that the second $\mathrm{p}K_\mathrm{a}$ will contribute $10^{(3.09-4.75)} = 0.022$ protons for each proton that the first $\mathrm{p}K_\mathrm{a}$ contributes.

The effect of these 2% more protons on $\mathrm{pH}$ would be about $\log (1.022) = 0.0009$ $\mathrm{pH}$ units.

In real life, the problems calculating $\mathrm{pH}$ from the information given in the question, would be:

1. $\pu{4 M}$ citric acid means 768 grams of citric acid in $\pu{1 L}$ of solution. Will that much really dissolve?

2. If that much did dissolve, and concentration of $\ce{H2O}$ is down to about $\pu{22 M}$ instead of the usual $\pu{55 M}$ , how do you account for that in the equilibrium calculations?

3. The 3 $\mathrm{p}K_\mathrm{a}$ values are calculated based upon extrapolating ionic strength to zero, what would these values be at $\pu{4 M}$ citric acid?

The $\mathrm{pH}$ of 50% citric acid is reported to be 0.6, and if you could make $\pu{4 M}$ citric acid, it would be still lower $\mathrm{pH}$. It doesn't matter if you use 1, 2, or 3 $\mathrm{p}K_\mathrm{a}$ values, you wouldn't calculate the right value (to say 0.1 $\mathrm{pH}$ units) for a solution that concentrated.