pKa Value (Acid Constant)

Take the negative base-10 logarithm of the acid constant: pKa = −log Ka. Example acetic acid: Ka = 1.78×10⁻⁵ mol/L gives pKa = −log(1.78×10⁻⁵) = 4.75. Conversely you recover the acid constant via Ka = 10^(−pKa). The acid constant itself comes from the law of mass action of the protolysis HA + H₂O ⇌ H₃O⁺ + A⁻: Ka = [H₃O⁺]·[A⁻]/[HA], where the practically constant water concentration is already absorbed into Ka. The logarithm makes the unwieldy powers of ten comparable: instead of values between 10⁶ and 10⁻¹⁴ you work with a scale from about −6 to +14. One pKa unit of difference means a factor of 10 in acid strength.

The smaller the pKa, the stronger the acid, because the larger Ka is and the more completely the acid donates its proton to water. Rough orientation: very strong acids have negative pKa values (HCl ≈ −6, sulfuric acid first step ≈ −3), moderately strong ones lie between 0 and about 4 (phosphoric acid first step 2.1), weak ones above (acetic acid 4.75, carbonic acid 6.5, ammonium 9.25). Beware of the classic mix-up: a large pKa means a weak acid, not a strong one. For the conjugate base, pKb = 14 − pKa holds at 25 °C; a very weak acid therefore has a comparatively strong conjugate base.

For weak acids the approximation pH = ½(pKa − log c₀) holds, where c₀ is the initial acid concentration. Example: 0.10 mol/L acetic acid with pKa = 4.75 yields pH = ½(4.75 − log 0.1) = ½(4.75 + 1) = 2.88. The approximation assumes that only a small fraction of the acid protolyzes and that the self-ionization of water is negligible; that fits here, since the degree of protolysis is only about 1.3 %. For strong acids such as HCl the formula is wrong; there, because of complete protolysis, simply pH = −log c₀ applies. Borderline case of moderately strong acids: they require the quadratic equation from the law of mass action.

At the half-equivalence point of a titration exactly half of the weak acid is neutralized, so acid HA and conjugate base A⁻ are present in equal concentration. In the Henderson-Hasselbalch equation pH = pKa + log([A⁻]/[HA]) the logarithm then becomes log(1) = 0, leaving pH = pKa. This is doubly useful: experimentally you read off the pKa of an unknown acid simply from the titration curve halfway to the equivalence point. And conceptually this point marks the maximum buffering effect, because there the system best absorbs additions of acid or base. Exam tasks like to combine both aspects.

The pKa is a substance constant: it characterizes how readily a particular acid donates its proton and depends only on the acid and the temperature. The pH, by contrast, is a state variable of the solution: it describes the current hydronium concentration and changes with concentration, dilution or addition of other substances. The same acetic acid (pKa = 4.75) produces quite different pH values depending on concentration: 2.88 at 0.1 mol/L, 3.38 at 0.01 mol/L. The two quantities are linked via the law of mass action, most directly in the Henderson-Hasselbalch equation. Only in the special case of equal acid and base concentrations, for example at the half-equivalence point, do pH and pKa coincide.