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ENV-08

Henderson–Hasselbalch pH

pH = pKa + log10([A−]/[HA]).

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Acid–baseHenderson 1908Hasselbalch

Governing equation

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

[A−]∈[1×10−6,1][A^-] \in \left[1\times 10^{-6}, 1\right]

where

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

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ENV-08 · phase
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①

Governing law — red

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

[A−]∈[1×10−6,1][A^-] \in \left[1\times 10^{-6}, 1\right]

[A−]=1×10−6[A^-] = 1\times 10^{-6}pH=2.35pH = 2.35

②

Domain of the variable — blue

[A−]min⁡=1×10−6,[A−]max⁡=1[A^-]_{\min} = 1\times 10^{-6},\quad [A^-]_{\max} = 1

[A−]∈[1×10−6,1][A^-] \in \left[1\times 10^{-6}, 1\right]

[A−]=1×10−6[A^-] = 1\times 10^{-6}[A−]=1×10−6[A^-] = 1\times 10^{-6}

③

Response curve — green

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

[A−]∈[1×10−6,1][A^-] \in \left[1\times 10^{-6}, 1\right]

[A−]=1×10−6[A^-] = 1\times 10^{-6}pH=2.35pH = 2.35

④

Rate of change — amber

dpHd[A−]\frac{\mathrm{d}pH}{\mathrm{d}[A^-]}

[A−]∈[1×10−6,1][A^-] \in \left[1\times 10^{-6}, 1\right]

[A−]=0.006[A^-] = 0.006pH′=325.9pH^{\prime} = 325.9

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: pKa\mathrm{p}K_a, [A−][A^-], [HA][HA] →\rightarrow pHpH. From Ka = [H+][A−]/[HA] one takes −log10 to obtain the buffer equation used in water chemistry and physiology.

Purpose

Purpose: compute pHpH from pKa\mathrm{p}K_a, [A−][A^-], [HA][HA] in Environmental via pH=pKa+log⁡10[A−][HA]\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 pKa=6.350 —\mathrm{p}K_a = 6.350\,\mathrm{—}, [A−]=0.010 mol/L[A^-] = 0.010\,\mathrm{mol/L}, [HA]=0.010 mol/L[HA] = 0.010\,\mathrm{mol/L}, the governing relation pH=pKa+log⁡10[A−][HA]\mathrm{pH}=\mathrm{p}K_a+\log_{10}\dfrac{[A^-]}{[HA]} yields pH=6.350 —pH = 6.350\,\mathrm{—}. Dilute aqueous monoprotic buffer, activity ≈ concentration. Move a slider: the numbers are this situation, not a canned story.

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Outputs

  • pH pH6.350 —
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ENV-08 · phase
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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