I was doing a question that asks the molarity of a 10 mL of tartaric acid that is titrated to phenolphthalein endpoint with 20 mL of 1.0 M NaOH.

I worked it out by equating the molarity of the NaOH times its volume to that one of the tartaric acid and got a value of 2.0 M but as tartaric acid is a diprotic acid does it affect this calculation or does the above method work with diprotic acids too?


1 Answer 1


Tartaric acid (2,3-dihydroxybutanedioic acid) is a diprotic acid; i.e. it is capable of donating two $\ce{H+}$ ions per molecule. Thus, $1\ \mathrm{mol}$ of tartaric acid can donate $2\ \mathrm{mol}$ of $\ce{H+}$ ions. Therefore, in order to neutralize $1\ \mathrm{mol}$ of tartaric acid, you need to add $2\ \mathrm{mol}$ of $\ce{NaOH}$.

The logarithmic acid dissociation constants of naturally occurring tartaric acid are $\mathrm pK_{\mathrm a1}=2.98$ and $\mathrm pK_{\mathrm a2}=4.34$; i.e. the second acid group of tartaric acid is roughly only as strong as acetic acid. Therefore, the $\mathrm{pH}$ at the second equivalence point is slightly alkaline (about $\mathrm{pH}\approx9).$ However, since you are using phenolphthalein, which changes from colourless to pink in the $\mathrm{pH}$ range of $8.2{-}10.0$, you are observing the correct equivalence point.

  • $\begingroup$ Note to OP - You have to be careful about pKa's... Phosphoric acid has 3 protons, but only two would be titrated with NaOH and phenolphthalein. $\text{pKa}_3$ for phosphoric acid is 12.319. $\endgroup$
    – MaxW
    Commented Jan 17, 2017 at 22:45

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