Henderson–Hasselbalch Equation (Weak Base Buffer)

\mathrm{pOH} = \mathrm{p}K_b + \log_{10}\!\frac{[\mathrm{BH^+}]}{[\mathrm{B}]}

Worked example: Ammonia buffer 0.200 M NH4+ / 0.100 M NH3 → pOH 5.0510 — press Try an example to run it live, then adjust anything.

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Henderson–Hasselbalch Equation (Weak Base Buffer) explained

Run Henderson and Hasselbalch's argument on the base side and everything mirrors: pOH is set by the base's pKb plus the logarithm of the conjugate-acid-to-base ratio. An ammonia buffer (pKb 4.75) holding 0.200 M NH₄⁺ against 0.100 M NH₃ has pOH = 4.75 + log₁₀(2) = 5.05, so at 25 °C its pH is 14 − 5.05 = 8.95 — comfortably alkaline, which is why ammonia/ammonium buffers are the standard choice for EDTA titrations of calcium and magnesium hardness.

The one thing to keep straight is which species goes on top. In the acid form the base is the numerator; here the conjugate acid is, because adding more BH⁺ makes the solution less basic. Get it upside down and your answer is wrong by twice the log term. Many chemists sidestep the whole issue by converting pKb to pKa (pKa = 14 − pKb at 25 °C) and using the acid form throughout — for ammonia that gives pKa(NH₄⁺) = 9.25, and 9.25 + log₁₀(0.100/0.200) = 8.95, the same answer by a different road.

Henderson–Hasselbalch Equation (Weak Base Buffer)

\mathrm{pOH} = \mathrm{p}K_b + \log_{10}\!\frac{[\mathrm{BH^+}]}{[\mathrm{B}]}

Where

$\mathrm{pOH}$ = pOH of the buffer
$\mathrm{p}K_b$ = pKb of the weak base
$[\mathrm{BH^+}]$ = Conjugate acid concentration (M)
$[\mathrm{B}]$ = Weak base concentration (M)

Missing one of these? Work it out first, then come back

pOH of the buffer — pH and pOH Relation, pOH from Hydroxide Ion Concentration
Conjugate acid concentration — Henderson–Hasselbalch Equation (Weak Acid Buffer), pH of a Weak Acid from Ka
Weak base concentration — Henderson–Hasselbalch Equation (Weak Acid Buffer), pH of a Weak Base from Kb

Part of this formula set

The buffer equations

4 formulasHenderson–Hasselbalch in both its acid and base forms, plus the pKa and pKb conversions that feed them — buffer pH from a ratio.