Weak Acid Base pH

pH of a weak acid or weak base

Take a weak acid HA with dissociation constant K_a and analytical concentration C. If you ignore the self-ionization of water and lean on the approximation [H⁺] ≈ √(K_a·C), the pH works out to pH ≈ ½(pK_a − log C). The same reasoning for a weak base gives you the pOH instead, pOH ≈ ½(pK_b − log C), and then pH = 14 − pOH at 25 °C. A couple of examples. Acetic acid (pK_a = 4.76) at 0.1 mol/L comes out to pH ≈ ½(4.76 − log 0.1) = ½(4.76 + 1) = 2.88. Ammonia (pK_b = 4.75) at the same 0.1 mol/L gives pOH ≈ 2.88, so pH ≈ 11.12. Keep in mind this only holds when the degree of dissociation α stays small, usually below α < 5%.

Applications

It underpins ionic equilibrium coursework and shows up when you design buffer solutions through the Henderson–Hasselbalch equation. In biochemistry it explains amino acid side-chain protonation and how enzyme activity shifts with pH. You also reach for it to read titration curves of weak acids and bases, or just to estimate the pH of everyday samples like vinegar, household ammonia and carbonic-acid systems.

FAQ

When does the approximation fail? Very dilute acids, or cases where K_a just isn't small enough, push α up high enough that the simplification C − x ≈ C no longer holds. When that happens, solve the quadratic K_a = x²/(C − x) directly.

Why divide by 2? The square root is the reason. Since [H⁺] = √(K_a·C), taking −log of both sides gives pH = ½(pK_a + pC), and that's the same as ½(pK_a − log C).

What's pK_b for a weak base? It's just pK_b = −log K_b. And because a conjugate acid–base pair satisfies pK_a + pK_b = 14 at 25 °C, knowing one of the two hands you the other.