Experiment Overview

Although we use the term "polyprotic acid titrations", we usually don't titrate acids, but their salts. And for polyprotic acid this already gives a lot of options (for example for phosphates we can titrate \(M_3PO_4, M_2HPO_4, MH_2PO_4\))

Theory

From a practical point of view, these titrations work just like any other titration. However, theory is the most important. Suppose you're titrating a sample of tripotassium phosphate (\(K_3PO_4\)) with \(HCl\). What reaction takes place? There are at least 3 possible answers:

\(\ce{K_3PO_4 + HCl \rightarrow K_2HPO_4 + KCl}\)
\(\ce{K_3PO_4 + 2HCl \rightarrow KH_2PO_4 + 2KCl}\)
\(\ce{K_3PO_4 + 3HCl \rightarrow H_3PO_4 + 3KCl}\)

So? Which one is it? Turns out that it depends on one thing a lot of people overlook - the indicator.

In an acid base system (like the phosphate one), at every pH, we can calculate the fraction of each of the species (see the exact calculation below - in the IJSO you'll usually be given the diagram or only be asked to make qualitative reasoning). The diagram below shows the fractions for the phosphate system:

Now, looking once again at the titration of \(\ce{K_3PO_4}\) with \(\ce{HCl}\), if for an indicator you use:

thymolphtalein (which changes its color at pH = 10) - at pH = 10, from the graph, you see that almost all phosphate is found as \(HPO_4^{2-}\), so reaction 1 is the one to use.
bromocresol green (which changes its color at pH = 4.5) - at pH = 4.5, from the graph, you see that almost all phospahte is found as \(H_2PO_4^-\), so reaction 2 is the correct one to use.

Another frequent example is the titration of sodium carbonate. If you prepare sodium hydroxide and leave it uncovered for a long time, it reacts with the \(CO_2\) from air, to produce sodium carbonate. Then, if you titrate in the presence of phenolphtalein, the reactions are:

\(\ce{NaOH + HCl \rightarrow NaCl + H_2O}\)
\(\ce{Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl}\)

However, if you titrate in the presence of methyl orange, which changes colors at a lower (more acidic) pH, the reactions are:

\(\ce{NaOH + HCl \rightarrow NaCl + H_2O}\)
\(\ce{Na_2CO_3 + 2HCl \rightarrow H_2O + CO_2 + 2NaCl}\)

Using the two titrations, you can find the concentrations of both the hydroxide and the carbonate.

Apparatus required

Burette
Erlenmeyer flask
Pipette

Chemicals required

The sample you're titrating
A titrant (usually 0.1M hydrochloric acid)
Distilled water for eventual dilutions
One or more indicators - very important

Method

Generally, the experimental procedure coincides with that of any other titration, but some titrations have their particularities.

One such example is the titration of carbonates, in the presence of methyl orange. The redness of the indicator (the color it has in acidic medium) is supposed to indicate excess HCl. However, some of the \(\ce{CO_2}\) remains dissolved as \(\ce{H_2CO_3}\) which unwantedly lowers the pH. To get rid of this, we bring the solution to a boil, in order to decompose the carbonic acid and force it out of solution, after which, if the color reverts back from red to yellow or orange, we continue the titration.

For some indicators, it is possible to do both titrations at once. For the example above, you can add phenolphtalein that turns pink and titrate until it's colorless (see the reactions above). After that, you add methyl orange, and only convert the already existing bicarbonate into \(\ce{CO_2}\).

Extra: finding the exact fractions yourself

We will use a general triprotic acid \(H_3X\) as an example. We can write the acidity constants:

\(\ce{H_3X \ce{<=>} H_2X^- + H^+}\), for which \(k_{a1} = \frac{[\ce{H_2X^-}][\ce{H^+}]}{[\ce{H_3X}]}\)
\(\ce{H_2X^- \ce{<=>} HX^{2-} + H^+}\), for which \(k_{a2} = \frac{[\ce{HX^{2-}}][\ce{H^+}]}{[\ce{H_2X^-}]}\)
\(\ce{HX^{2-} \ce{<=>} X^{3-} + H^+}\), for which \(k_{a3} = \frac{[\ce{HX^{2-}}][\ce{H^+]}}{[\ce{X^{3-}}]}\)

Then we have the following concentrations:

\([\ce{H_2X^-}] = \frac{k_{a1}}{[\ce{H^+}]}[\ce{H_3X}]\)
\([HX^{2-}] = \frac{k_{a2}}{[H^+]}[H_2X^-] = \frac{k_{a1}k_{a2}}{[H^+]^2}[H_3X]\)
\([\ce{X^{3-}}] = \frac{k_a3}{[H^+]}[HX^{2-}] = \frac{k_{a1}k_{a2}k_{a3}}{[H^+]^3}[H_3X]\)

If the total concentration of the acid species is c, we have \([H_3X] + [H_2X^-] + [HX^{2-}] + [X^{3-}] = c\), which, after substituting, gives:

\[[H_3X]\bigg{(}1 + \frac{k_{a1}}{[H^+]} + \frac{k_{a1}k_{a2}}{[H^+]^2} + \frac{k_{a1}k_{a2}k_{a3}}{[H^+]^3}\bigg{)}=c\]

\[[H_3X] = \frac{[H^+]^3}{[H^+]^3 + k_{a1}[H^+]^2 + k_{a1}k_{a2}[H^+] + k_{a1}k_{a2}k_{a3}}c\]

Then, using the equation above and the equations for the other 3 concentrations, we have: