Concentration of a species outside buffer region (polyprotic acid titration)

Question

I'm wondering if you titrate a polyprotic acid, say, $\ce{H3PO4}$, with a strong base, and you successfully pass the first equivalence point and are maybe at the $\mathrm{pH}=\mathrm{pK}_{a2}$ point now, how would you find the concentration of $\ce{H3PO4}$?

Is there none left? I don't have a specific homework question, I'm just wondering.

Attempt at a solution:
I thought maybe you would find the equilibrium concentration of $\ce{H2PO4-}$, find the $\mathrm{pH}$, and use the $\mathrm{pH}$ to find $\ce{OH-}$. Then, you could write the equilibrium, $$\ce{H2PO4- + H2O <=> OH- + H3PO4},$$ and without making an ICE table, substitute your equilibrium values into the $\mathrm K_a$ expression.

Answer 1

When you have some solution of $\ce{H3PO4}$, or its various anions, then you'll always have some of all the species ($\ce{H3PO4}$, $\ce{H2PO4^-}$, $\ce{HPO4^{2-}}$ and $\ce{PO4^{3-}}$)present. The relative proportion changes with pH of course.

If you titrate $\ce{H3PO4}$ with a strong base like $\ce{NaOH}$ then there will be three equivalence points. If the $\ce{H3PO4}$ is 1.00 molar, then the first equivalence point will be when 0.33 moles of $\ce{NaOH}$ have been added. The second equivalence point will be when 0.66 moles of $\ce{NaOH}$ have been added. The third equivalence point will be when 1.00 moles of $\ce{NaOH}$ have been added. So the equivalence point is based on stoichiometry and not really the concentration of a particular species of ion.

Of course using the values for pKa1, pKa2, and pKa3 you can calculate the pH at which the three equivalence points take place.

The other point here is the "sharpness" of the equivalence points. If the pka values are close together there isn't a large pH change at the equivalence point. This means that it is hard to detect the precise equivalence point. So essentially you'd like a large pH change for a small volume of titrant change. If the pKa values are close together then you get the opposite, a relatively small pH change for a large volume of titrant.

Answer 2

At the one side, you can always calculate the percentages of the acid or it's ions in solution: $$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{a,j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{a,j}}$$ with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons.

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But on the other hand there are some special points during titrations. One of those points did you choose, which is where $\mathrm{pH} = \mathrm{pK}_{a,i}$.

For $\ce{H3PO4}$ there exist three different acidity constants, which describe the three possible dissociation steps. For the first step, the reaction is as you described: $$\ce{H3PO4 + H2O \xrightleftharpoons{k_{a,1}} H2PO4- + H3O+}$$

A little work on it gives you the Henderson-Hasselbalch-equation, which usually is good approximation to buffer regions: $$\mathrm{pH} = \mathrm{pK}_a - \lg\left(\frac{[HA]}{[A^-]}\right)$$

If you enter the same values for pH and pKa you have to solve $$-\lg(x)=0$$ which is true for $x=1$. So you know that you have a 1:1 ratio at this point.

This means for phosphoric acid:
At $\mathrm{pH=pK}_{a1}$, you've got exactly $50~\%~\ce{H3PO4}$ and $50~\%~\ce{H2PO4-}$ (blue, yellow). The same procedure can be applied to the other dissociation steps. Therefor at $\mathrm{pH=pK}_{a2}$ you've got $50~\%~\ce{H2PO4-}$ and $50~\%~\ce{HPO4^2-}$ (yellow, green) and at the last possible $\mathrm{pH=pK}_{a3}$ there are $50~\%~\ce{HPO4^2-}$ and $50~\%~\ce{PO4^3-}$ (green, red).

This can be applied to most acid-base dissociation steps but you have to pay attention for close successive $\mathrm{pK}_a$ values, as if the gap gets too small, the 1:1 ratio doesn't hold.