pH of a Weak Base from Kb

A weak base does not fully accept protons any more than a weak acid fully donates them. Ammonia in water sets up NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, and the equilibrium sits far to the left: K_b for ammonia is 1.8 × 10⁻⁵, so in a 0.10 M solution only about 1.3% of it has reacted. Write the equilibrium expression, assume the amount reacted is small compared with the formal concentration, and it collapses to [OH⁻] = √(K_b·C).

Worked: household ammonia diluted to 0.100 M. K_b·C = 1.8 × 10⁻⁵ × 0.100 = 1.8 × 10⁻⁶, whose square root is 1.34 × 10⁻³ M hydroxide. That is a pOH of 2.87 — and then the step people drop — pH = 14 − 2.87 = 11.13. Reporting 2.87 as the pH turns a cleaning product into something close to lemon juice, and it is the single most common error on this calculation. A base calculation naturally produces pOH; the conversion is not optional.

The approximation being made is that [B] at equilibrium is still C, which is only true when very little of the base has reacted. The usual rule of thumb is that C/K_b should be at least 100, or equivalently that less than about 5% has ionised. Below that, solve the full quadratic x² + K_b·x − K_b·C = 0 instead; the square-root shortcut will read too basic, because it credits the solution with base that has already been used up.

Handbooks more often tabulate pK_b than K_b, and many tabulate only the conjugate acid's pK_a — for ammonium, 9.25. The bridge is pK_a + pK_b = 14 at 25 °C, the same 14 as everywhere else on the pH scale and with the same temperature caveat attached to it. And the whole calculation assumes a monoprotic base with no other base in the flask; carbonate, phosphate and the amines with two nitrogens each need their stepwise constants treated in turn.