About titration curve (quick)

(1) To obtain Ka by measurement of pH of a knownconcentration weak acid

Consider the dissociation of a weak acid HA in a dilute aqueous solution :
HA(aq) ⇌ H⁺(aq) + A⁻(aq) .. Ka

Ka = [H⁺][A⁻]/[HA]

Since HA is a weak acid, it can be assumed that thedissociation of HA is negligible and thus it can be assumed that [HA] ≈ [HA]o
where [HA]o is the initial concentration of HA before dissociation.
Neglect the self dissociation of water, [H⁺] ≈[A⁻]
The equilibrium expression can thus be written as :
Ka = [H⁺]²/[HA]o

Take -log :
-log(Ka) = -log[H⁺]² + log[HA]o
-log(Ka) = 2(-log[H⁺]) + log[HA]o
pKa = 2pH + log[HA]o

According to the above expression, pK­a can be found by measuring thepH of a known concentration (i.e.[HA]o)of the weak acid HA. Hence, Ka can be found from pKa.


(2) To obtain Ka by measurementof pH at half equivalence point

Consider the following equilibrium :
HA(aq) ⇌ H⁺(aq) + A⁻(aq) .. Ka

Ka = [H⁺][A⁻]/[HA]
Taking -log :
-log(Ka) = -log[H⁺] +log([A⁻]/[HA])
pKa = pH + log([A⁻]/[HA])...... (*)

At the half equivalence point, half of HA is neutralized to give A⁻ bythe alkali added.
Since HA is a weak acid and the reaction mixture is a buffer solution, due tothe common ion effect, the dissociation of HA is negligible and thus [HA] ≈ [A⁻]
At the half equivalence point, (*) can thus be written as :
pKa = pH

According to the above expression, pK­a can be found by measuring thepH at the half equivalencepoint. Hence, Ka can be found from pKa.


(3) To compare(1) and (2)

In both (1) and (2), it is assumed that the dissociation of the acid isnegligible. The solution in (2) is abuffer solution, while that in (1) is only a dilute solution of a weak acid. Dueto the common ion effect, the degree of dissociation of a weak acid in thebuffer solution is much smaller than that in a solution of the acid. Therefore, the assumption is closer to thereality in (2) than in (1), and thus Ka value obtained from (2) ismuch closer to the true Ka value.