# Phosphoric Acid Titration

I physics graduate student, but I teach an interdisciplinary lab for the university which incorporates biology, physics, and chemistry.

We recently added a weak-acid titration lab to the course and I got a result which surprised me.

The procedure was to neutralize 100 mL of 0.010 M Phosphoric acid by adding 0.10 M NaOH.

We have 0.001 mole of phosphoric acid in solution, but this acid is triprotic which means that a maximum of 0.003 moles of hydronium can be released into the solution. My understanding of the equillibrium reactions was that as I lowered the ambient hydronium ion concentration more protons would be released by the acid.

So my expectation was that I would need to add 30 mL of the NaOH solution to reach a pH of 7. In practice it only took between 12-15 mL.

I talked this over with a Chemist I work with and she wasn't sure why this discrepancy happened either. One possiblity is that the person who prepared the acid screwed up, but we don't have more stock for me to test that theory.

So my question is do these results sound normal or did I have the right idea? If I didn't have the right idea then what am I missing theoretically speaking?

• It's a long story; read about the $\mathrm{pK_a}$ and acid-base equilibrium, otherwise it would not make sense to you. The result is more or less OK. – Ivan Neretin Oct 10 '15 at 2:05
• You are right in calculating that you need 30 mL of NaOH for complete reaction with H3PO4. However, phosphoric acid is a weak acid and NaOH is a strong base. The pH at the end-point will therefore be greater than 7, and the volume of NaOH you need to reach pH 7 will be less than 30 mL. – orthocresol Oct 10 '15 at 10:14

You might want to read about the pH of a neutralized solution. But in case you don't, one of my answers here on chem.se covers the calculation of titration curves with and without increasing volume. It is quite easy to create a titration curve using an adapted (see exact solution for triprotic acids here) version of equation 8.

$$V(B)=\frac{V(H_3A)~\left(\frac{k_w}{x}-x+C(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)}$$

I did that for you, so you can see, why it is not wrong, that you only needed 12-15 ml to reach pH 7 but also that that pH 7 is not what you searched for.

What you probably actually wanted, was to reach one of the equivalence points at 10 or 20 ml. You will also not have a chance to reach the third EP in an aqueous solution, which is why 30 ml is not a good goal to go for.

• Thank you for the comprehensive answer. This is exactly what I was looking for. – Spencer Feb 16 '16 at 14:54
• Why is it not possible to reach the equivalence point in aqueous solution? Is it because the third neutralisation reaction with $\ce {NaOH}$ does not go to completion? – Tan Yong Boon Oct 16 at 9:46
• I think it is best to ask this as a separate question. – pH13 - Yet another Philipp Oct 17 at 15:28

Phosphoric acid (which is WEAK ACID) releases it's protons in 3 different stages. It has 3 pKa values for 3 dissociations. So it is impossible to get a 1:1 reaction , reactant condition when titrating.