Editing Titration curve (section)

{{Short description|Graph in acid-base chemistry}}
[[Image:Oxalic acid titration grid.png|thumb|A typical titration curve of a diprotic acid, [[oxalic acid]], titrated with a strong base, [[sodium hydroxide]]. Both equivalence points are visible.]]

[[Titration]]s are often recorded on graphs called '''titration curves''', which generally contain the volume of the [[titrant]] as the [[independent variable]] and the [[pH]] of the solution as the [[dependent variable]] (because it changes depending on the composition of the two solutions).<ref>
{{cite book
 |last=Skoog |first=D.A
 |author2=West, D.M.|author3= Holler, J.F.|author4= Crouch, S.R.
 |title=Fundamentals of Analytical Chemistry
 |publisher=Thomson Brooks/Cole
 |year=2004
 |edition=8th
 |isbn=0-03-035523-0
}} Section 14C: Titration curves for weak acis
</ref> 

The [[equivalence point]] on the graph is where all of the starting solution (usually an [[acid]]) has been neutralized by the titrant (usually a [[Base (chemistry)|base]]). It can be calculated precisely by finding the [[second derivative]] of the '''titration curve''' and computing the [[point of inflection|points of inflection]] (where the graph changes [[concave function|concavity]]); however, in most cases, simple [[visual inspection]] of the curve will suffice. In the curve given to the right, both equivalence points are visible, after roughly 15 and 30 [[milliliter|mL]] of [[Sodium hydroxide|NaOH solution]] has been titrated into the [[oxalic acid]] solution. To calculate the logarithmic [[acid dissociation constant]] (pK<sub>a</sub>), one must find the volume at the half-equivalence point, that is where half the amount of titrant has been added to form the next compound (here, sodium hydrogen oxalate, then [[disodium oxalate]]). Halfway between each equivalence point, at 7.5 mL and 22.5 mL, the pH observed was about 1.5 and 4, giving the pK<sub>a</sub>.

In weak [[monoprotic acid]]s, the point halfway between the beginning of the curve (before any titrant has been added) and the equivalence point is significant: at that point, the concentrations of the two species (the acid and conjugate base) are equal. Therefore, the [[Henderson-Hasselbalch equation]] can be solved in this manner:
:<math>\mathrm{pH} = \mathrm pK_{\mathrm a} + \log \left( \frac{[\mbox{base}]}{[\mbox{acid}]} \right)</math>
:<math>\mathrm{pH} = \mathrm pK_{\mathrm a} + \log(1)\,</math>
:<math>\mathrm{pH} = \mathrm pK_{\mathrm a} \,</math>

Therefore, one can easily find the pK<sub>a</sub> of the weak monoprotic acid by finding the pH of the point halfway between the beginning of the curve and the equivalence point, and solving the simplified equation. In the case of the sample curve, the acid dissociation constant ''K<sub>a</sub>'' = 10<sup>-pKa</sup> would be approximately 1.78×10<sup>−5</sup> from visual inspection (the actual ''K''<sub>a2</sub> is 1.7×10<sup>−5</sup>)

For [[polyprotic]] acids, calculating the acid dissociation constants is only marginally more difficult: the first acid dissociation constant can be calculated the same way as it would be calculated in a monoprotic acid. The pK<sub>a</sub> of the second acid dissociation constant, however, is the pH at the point halfway between the first equivalence point and the second equivalence point (and so on for acids that release more than two protons, such as [[phosphoric acid]]).