Weak Bases, K_b and the K_a-K_b Relationship

Ammonia has no hydroxide in its formula, yet its solution is basic because the nitrogen lone pair pulls a proton off water, leaving OH⁻ behind. The equilibrium constant for that process is K_b.

For a neutral base B, write B + H₂O ⇌ BH⁺ + OH⁻, so K_b = [BH⁺][OH⁻]/[B] in the dilute model. For an anionic base such as acetate, write CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; the charge pattern changes but the same base-accepts-a-proton reasoning applies. Water is omitted as solvent, not because it is chemically uninvolved.

Build the ICE table with x as the amount of base that reacts. Under the small-x conditions x = sqrt(K_b x C). The trap is stopping there: x is [OH⁻], so take pOH = -log x and then pH = 14.00 - pOH.

Starting with C of base alone and neglecting water's background ions, K_b = x²/(C − x), where x = [OH⁻]. Try x ≈ √(KbC), test x/C, and retain the denominator if that test fails. With 0.250 M NH₃ and K_b = 1.80 × 10⁻⁵, x ≈ 0.00212 M and x/C ≈ 0.85%, validating the shortcut. Then pOH ≈ 2.67 and pH ≈ 11.33 at 25 °C using guard digits. Taking −log x and reporting it as pH would describe an acidic solution inconsistent with a base.

Multiply the K_a expression of HA by the K_b expression of A- and every concentration except the water ions cancels, leaving K_w. So K_a x K_b = 1.0 × 10⁻¹⁴ and pK_a + pK_b = 14.00 at 25 C.

Add HA + H₂O ⇌ H₃O⁺ + A⁻ to A⁻ + H₂O ⇌ HA + OH⁻. Cancelling HA and A⁻ gives water autoionization, so multiplying their constants gives K_a(HA)K_b(A⁻) = K_w. The two constants must belong to conjugates at the same temperature. K_a for acetic acid cannot be paired with K_b for ammonia. At other temperatures use that temperature's K_w and pK_w rather than automatically 10⁻¹⁴ and 14.

Amines are ammonia with organic groups replacing hydrogens, and common simple amines are weak bases in water: methylamine has K_b = 4.4 × 10⁻⁴, somewhat stronger than ammonia at 1.8 × 10⁻⁵. Anions of weak acids, such as CN⁻, F⁻ and CH₃COO⁻, are also weak bases, while anions of strong acids are not.

An amine's available nitrogen lone pair can accept H⁺. Alkyl substitution often increases basicity relative to ammonia, but solvent stabilization and steric effects also affect the trend. In an amide, the nitrogen lone pair is delocalized toward a carbonyl, greatly reducing ordinary nitrogen basicity. Amines and amides therefore should not be assigned the same strength merely because both contain N. Use a supplied K_b or conjugate-acid pK_a when a numerical ranking is needed.