ENV-08

Henderson–Hasselbalch pH

pH = pKa + log10([A−]/[HA]).

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Acid–baseHenderson 1908Hasselbalch

Governing equation

\mathrm{pH}=\mathrm{p}K_a+\log_{10}\dfrac{[A^-]}{[HA]}

where

\mathrm{p}K_a: pKa (—)
[A^-]: Conjugate base (mol/L)
[HA]: Acid (mol/L)
pH: pH (—)

Lecture brief

Historical brief

Streeter–Phelps (1925) oxygen sag, settling theory and Guldberg–Waage kinetics made water and air quality a rate problem. The lab computes sag, overflow and a snapshot of reactor mass balance. This sheet (ENV-08 — Henderson–Hasselbalch pH) is the form associated with Henderson 1908 · Hasselbalch. Working symbols: $\mathrm{p}K_a$ , $[A^-]$ , $[HA]$ $\rightarrow$ $pH$ . From Ka = [H+][A−]/[HA] one takes −log10 to obtain the buffer equation used in water chemistry and physiology.

Purpose

Purpose: compute

pH

from

\mathrm{p}K_a

[A^-]

[HA]

in Environmental via

\mathrm{pH}=\mathrm{p}K_a+\log_{10}\dfrac{[A^-]}{[HA]}

pH = pKa + log10([A−]/[HA]). Use it when a real environmental question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

\mathrm{p}K_a = 6.350\,\mathrm{—}

[A^-] = 0.010\,\mathrm{mol/L}

[HA] = 0.010\,\mathrm{mol/L}

, the governing relation

\mathrm{pH}=\mathrm{p}K_a+\log_{10}\dfrac{[A^-]}{[HA]}

yields

pH = 6.350\,\mathrm{—}

. Dilute aqueous monoprotic buffer, activity ≈ concentration. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

pKa \mathrm{p}K_a6.35 —Conjugate base [A^-]0.01 mol/LAcid [HA]0.01 mol/L

Outputs

pH pH6.350 —

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Narration of this film

Dilute aqueous monoprotic buffer, activity ≈ concentration.

From Ka = [H+][A−]/[HA] one takes −log10 to obtain the buffer equation used in water chemistry and physiology.

Reading speed

Watch on YouTube