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Why does the half-equivalence point of HCl in the case of titrating strong base into HCl not yield the pKa?
If I have 20 mL of 0.1 M HCl and I titrate it with 0.1 M NaOH, the equivalence point should be about at pH = 7, and 20 mL of titrant.
The pH of 0.1 M HCl is about -log(0.1) = 1, so the pH at the half-equivalence point has to be between 1 and 7. If this was a weak acid being titrated with a strong base then this point would yield the pKa.
However, for HCl the half-equivalence point doesn't yield the pKa. Lots of literature suggests that the pKa of HCl is less than 0.
Is this determined by titration somehow?
1 Answer
- 冷眼旁觀Lv 71 year ago
Consider that a weak acid HA is titrated with a strong base like NaOH.
HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l)
At the half equivalence point, half of the HA is neutralized to give A⁻ ions. Consider the dissociation of HA.
HA(aq) + H₂O(aq) ⇌ H₃O⁺(aq) + A⁻(aq) …… Ka
Because of very small Ka and the common ion effect in the presence of A⁻ ions, the dissociation of HA is to a very small and affect insignificantly [HA] and [A⁻].
Therefore, at the half equivalence point, [HA] = [A⁻]
According to Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa
HCl is a strong acid, which is completely dissociates in aqueous solution to give H₃O⁺.
The titration of HCl against a strong base like NaOH is actually the neutralization between H₃O⁺ and OH⁻ ions. At the half equivalence point, half of the H₃O⁺ ions are neutralized but half is left unreacted. At this point,
pH = -log[H₃O⁺]
where [H₃O⁺] = (moles of H₃O⁺ left unreacted) / (Volume of the solution in L)
Obviously, pH is independent of the pKa of HCl. Therefore, there is no way to find pKa by the half equivalence point of the titration against a strong acid (like HCl) against a strong base (like NaOH).
The pKa of strong base is negative. For example, In 0.1 M solution of a strong acid HY, 95% of the acid dissociates to Y⁻ and H₃O⁺ ions, and 5% of the acid is left.
HY(aq) + H₂O(aq) ⇌ Y⁻(aq) + H₃O⁺(aq) …… Ka
pKa = -log([Y⁻][H₃O⁺]/[HY]) = -log(95%²/5%) = -1.25
To determine Ka for a strong acid, refer to the following webpage:
