Titrations are often recorded on graphs calledtitration curves, which generally contain the volume of thetitrant as theindependent variable and thepH of the solution as thedependent variable (because it changes depending on the composition of the two solutions).^[1]

Theequivalence point on the graph is where all of the starting solution (usually anacid) has been neutralized by the titrant (usually abase). It can be calculated precisely by finding thesecond derivative of thetitration curve and computing thepoints of inflection (where the graph changesconcavity); however, in most cases, simplevisual inspection of the curve will suffice. In the curve given to the right, both equivalence points are visible, after roughly 15 and 30mL ofNaOH solution has been titrated into theoxalic acid solution. To calculate the logarithmicacid dissociation constant (pK_a), 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, thendisodium 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_a.

In weakmonoprotic acids, 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, theHenderson-Hasselbalch equation can be solved in this manner:

${\displaystyle \mathrm{p}\mathrm{H}=\mathrm{p}{K}_{\mathrm{a}}+\log \left(\frac{[\text{base}]}{[\text{acid}]}\right)}$

${\displaystyle \mathrm{p}\mathrm{H}=\mathrm{p}{K}_{\mathrm{a}}+\log (1)}$

${\displaystyle \mathrm{p}\mathrm{H}=\mathrm{p}{K}_{\mathrm{a}}}$

Therefore, one can easily find the pK_a 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 constantK_a = 10^-pKa would be approximately 1.78×10⁻⁵ from visual inspection (the actualK_a2 is 1.7×10⁻⁵)

Forpolyprotic 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_a 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 asphosphoric acid).

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