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Index/Physical chemistry/CHM-09

CHM-09

Henderson–Hasselbalch

pH = pKa + log10([A−]/[HA]). Buffer equation.

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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−]∈[1×10−6,2][A^-] \in \left[1\times 10^{-6}, 2\right]

where

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

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CHM-09 · gauge
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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−]∈[1×10−6,2][A^-] \in \left[1\times 10^{-6}, 2\right]

[A−]=1×10−6[A^-] = 1\times 10^{-6}pH=−0.24pH = -0.24

②

Domain of the variable — blue

[A−]min⁡=1×10−6,[A−]max⁡=2[A^-]_{\min} = 1\times 10^{-6},\quad [A^-]_{\max} = 2

[A−]∈[1×10−6,2][A^-] \in \left[1\times 10^{-6}, 2\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−]∈[1×10−6,2][A^-] \in \left[1\times 10^{-6}, 2\right]

[A−]=1×10−6[A^-] = 1\times 10^{-6}pH=−0.24pH = -0.24

④

Rate of change — amber

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

[A−]∈[1×10−6,2][A^-] \in \left[1\times 10^{-6}, 2\right]

[A−]=0.013[A^-] = 0.013pH′=174.9pH^{\prime} = 174.9

Lecture brief

Historical brief

Ideal-gas law, van ’t Hoff, Nernst, Michaelis–Menten and Clausius–Clapeyron are physical chemistry’s working equations of equilibrium and rate. The lab is pressure, potential and kinetics. This sheet (CHM-09 — Henderson–Hasselbalch) 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. Best buffering when [A−] ≈ [HA].

Purpose

Purpose: compute pHpH from pKa\mathrm{p}K_a, [A−][A^-], [HA][HA] in Physical chemistry via pH=pKa+log⁡10[A−][HA]\mathrm{pH}=\mathrm{p}K_a+\log_{10}\dfrac{[A^-]}{[HA]} pH = pKa + log10([A−]/[HA]). Buffer equation. Use it when a real physical chemistry question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given pKa=4.760 —\mathrm{p}K_a = 4.760\,\mathrm{—}, [A−]=0.100 mol/L[A^-] = 0.100\,\mathrm{mol/L}, [HA]=0.100 mol/L[HA] = 0.100\,\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=4.760 —pH = 4.760\,\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 pH4.760 —
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CHM-09 · gauge
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Narration of this film

Dilute aqueous monoprotic buffer, activity ≈ concentration.

From Ka = [H+][A−]/[HA] one takes −log10. Best buffering when [A−] ≈ [HA].

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