What are acid-base titrations?
When to a given amount of a solution of an acid is added an equivalent amount of a solution of a base, the equivalence point is reached, i.e., the number of equivalents of acid is equal to the number of equivalents of base.
In the case of a neutralization reaction between a strong acid and a strong base the equivalence point corresponds to pH=7, as for the reaction between a weak acid and a weak base, with equal dissociation constants.
In any other case the equivalence pH is determined by the hydrolysis.
Titration of a solution is the exact determination of its concentration, and we call titrated solution a solution whose concentration is known precisely.
An acid-base titration (also called acid-base or neutralization volumetry) consists of determining the concentration of a solution of an acid by means of a titrated solution of a base or vice versa.
At the equivalence point it must be fulfilled:
NAH·VAH = NB·VA
Let’s take a look at the possible cases that arise when performing an appraisal:
Titration of a strong acid with a strong base
If we observe the obtained values that are represented in the attached table and we make a graphical representation of them, we can observe that the pH increases very slowly at the beginning of the titration and suffers an abrupt increase in the proximity of the equivalence point so that, after this, the pH continues increasing more and more slowly. The non-symmetry of the curve is due to the effect of the solution.
Titration of a weak acid with a strong base
Initial pH: this is the pH of the acid since there is no base yet:
AH <–> A– + H+
Ka = [A–]·[H+]/[AH]
Considering that [A–] = [H+] and that [AH] = Ca –[H+]
Ka = [H+]2/(Ca-[H+])
Since it is a weak acid, the concentration of protons will be negligible [H+] compared to the initial concentration of the acid Ca.
Ka = [H+]2/Ca [H+] ≡ KACa pH = ½·pKA – ½·log Ca
pH during neutralization:
AH + BOH -> AB + H2O
Ka = [A–]·[H+]/[AH]
In this particular case [A–]≠[H+] and [AH]≈[Ca] weak acid, and [A–]≈[Cs] salt concentration
[H+] = Ka·[AH]/[A–] pH = pKa + log(Cs/Ca)
pH at the time of neutralization (equivalence point). The acid and base have neutralized and the salt is formed.
AB + H2O ≡ A– + B+ (anion hydrolysis)
A– + H2O ≡ AH + OH–
Kh = Kw/Ka = [AH]·[OH–]/[A–] where [AH]·= [OH–]
Kw/Ka = [OH–]2·[H+]2/[A–]·[H+]2 = Kw2/[A–]·[H+]2
[H+] ≡ (Kw·Ka/[A–]) ≡ (Kw·Ka/Cs) [A–] ≡ [Cs]
pH = 7 + ½·pKa + ½·log Cs
- titration of a strong acid with a weak base
Initial pH:
BOH ≡ B+ + OH–
Kb = [B+]·[OH–]/[BOH]
Considering [B+] = [OH–] and [BOH] = Cb –[OH–]
Ka = [OH–]2/(Cb-[OH–])
Since it is a weak base, the hydroxyl concentration [OH–] will be negligible compared to the initial concentration of the base Cb.
Kb = [OH–]2/Cb [OH–] ≡ KbCb pOH = ½·pKb – ½·log Cb
pH = 14 – ½·pKb + ½·log Cb
pH during neutralization:
BOH + AH ≡ AB + H2O
Kb = [B+]·[OH–]/[BOH]
In this particular case [BOH] ≡ [Cb] weak base, and [B+] ≡ [Cs] salt concentration
pOH = pKb + log(Cs/Cb) pH = 14 – pKb – log(Cs/Cb)
pH at the time of neutralization (equivalence point). Acid and base have been neutralized and the salt is formed.
AB + H2O ≡ A– + B+ (cation hydrolysis)
B+ + H2O ≡ BOH + H+
Kh = Kw/Kb = [BOH]·[H+]/[B+] where [BOH]·= [H+]
Kw/Kb = [H+]2/[B+] [H+] ≡ (Kw·[B+]/Kb) [B+] ≡ [Cs]
pH = 7 – ½·pKb – ½·log Cs
Summary
Summarizing, one can generalize by saying:
- In a strong acid-strong base neutralization, the pH at the equivalence point does not depend on the concentrations and has a value of 7.
- In a weak acid-strong base neutralization, the pH at the equivalence point depends on the concentrations and will be alkaline.
- In a strong acid-weak base neutralization, the pH at the equivalence point depends on the concentrations and will be acid.
