The Henderson-Hasselbalch equation

Result

The Henderson-Hasselbalch equation

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Statement

Let a closed aqueous solution at fixed temperature contain a single monoprotic weak acid \(\text{HA}\) and its conjugate base \(\text{A}^-\), introduced at formal (analytical) concentrations \(C_{\text{HA}}\) and \(C_{\text{A}}\) with a spectator cation of charge \(+1\). Write \(K_a\) for the acid dissociation constant appropriate to the concentration convention in use, and \(\text{p}K_a=-\log_{10}K_a\). Then mass balance and charge balance alone give the exact relation \(\text{pH}=\text{p}K_a+\log_{10}\dfrac{C_{\text{A}}+[\text{H}^+]-[\text{OH}^-]}{C_{\text{HA}}-[\text{H}^+]+[\text{OH}^-]}\), in which the numerator and denominator are the true equilibrium concentrations \([\text{A}^-]\) and \([\text{HA}]\). The familiar working form \(\text{pH}=\text{p}K_a+\log_{10}\big(C_{\text{A}}/C_{\text{HA}}\big)\) is the limit of this exact statement when \(\big|[\text{H}^+]-[\text{OH}^-]\big|\ll\min(C_{\text{A}},C_{\text{HA}})\); it is an approximation of controlled and computable error, not a definition, and the \(\text{p}K_a\) it requires is the apparent constant at the working ionic strength, not the tabulated thermodynamic value.

Why it matters

Almost every quantitative statement in this unit is a mass balance in disguise: mole-avogadro fixes how many particles a weighing corresponds to, concentration-dilution fixes how those particles are distributed through a volume, and limiting-reagent fixes what happens when two such counts meet in a stoichiometric ratio. The Henderson–Hasselbalch equation is what those bookkeeping rules become when the reaction in question is proton transfer and the reaction does not go to completion. It is the point at which stoichiometry and equilibrium have to be done simultaneously, and the point at which a student first meets the distinction — invisible in a limiting-reagent calculation, decisive here — between the amount of a species put into a flask and the amount present in it at equilibrium.

Its practical reach is enormous. Every enzyme assay, every chromatographic separation of ionisable analytes, every drug-formulation decision about oral absorption, and every clinical interpretation of an arterial blood gas is a Henderson–Hasselbalch calculation. Yet the equation is also the single most misused relation in undergraduate chemistry, precisely because the working form is so easy to write down that its two hidden approximations — that formal concentrations may replace equilibrium concentrations, and that concentrations may replace activities — are almost never stated. Both fail in exactly the situations where the answer matters most: dilute buffers, buffers far from \(\text{p}K_a\), and buffers at the \(0.15\ \text{mol\,L}^{-1}\) ionic strength of a physiological fluid, where the second approximation alone costs between \(0.1\) and \(0.5\) pH units. The purpose of this page is to derive the equation in a form that makes both errors explicit and computable.

Hypotheses

The system is closed: no component enters or leaves.Total \(\text{A}\)-containing material is conserved, which is what makes \(C_{\text{HA}}+C_{\text{A}}=[\text{HA}]+[\text{A}^-]\) available as a constraint. An open buffer — blood in contact with alveolar gas, or an uncapped carbonate solution — violates this: the amount of dissolved \(\text{CO}_2\) is fixed by Henry’s law from the gas-phase partial pressure and is replenished as fast as it is consumed, so the conserved quantity is not the total carbonate. The correct open-system form is derived in Problems, 5, and it behaves very differently.

Exactly one proton-transfer equilibrium is significant over the pH range considered.For a polyprotic acid this requires the neighbouring \(\text{p}K_a\) values to be far away. Carbonic acid, with \(\text{p}K_{a1}=6.35\) and \(\text{p}K_{a2}=10.33\) at \(25^\circ\text{C}\), satisfies this comfortably near pH \(6.5\); citric acid, with \(\text{p}K_a\) values \(3.13\), \(4.76\) and \(6.40\), does not, and at pH \(4.76\) all three of \(\text{H}_3\text{Cit}\), \(\text{H}_2\text{Cit}^-\) and \(\text{HCit}^{2-}\) are present at percent level (polyprotic-acids).

The \(K_a\) used matches the concentration convention used.Three distinct constants circulate under the same symbol: the thermodynamic \(K_a^\circ=a_{\text{H}^+}a_{\text{A}^-}/a_{\text{HA}}\), a pure number; the mixed or Brønsted constant \(K_a^{\text{mix}}=a_{\text{H}^+}[\text{A}^-]/[\text{HA}]\); and the stoichiometric constant \(K_a^{c}=[\text{H}^+][\text{A}^-]/[\text{HA}]\). Because a glass electrode reports \(-\log_{10}a_{\text{H}^+}\), it is the mixed constant that makes computed and measured pH agree. Substituting a tabulated \(K_a^\circ\) into the working form and comparing with a meter is a category error whose size is exactly \(\log_{10}(\gamma_{\text{A}^-}/\gamma_{\text{HA}})\).

Formal concentrations approximate equilibrium concentrations.This is the buffer approximation proper, and Step 8 shows it fails when \([\text{H}^+]-[\text{OH}^-]\) is not small against both \(C_{\text{A}}\) and \(C_{\text{HA}}\). Two regimes break it: a dilute buffer (millimolar or below), and a buffer whose pH is far from \(\text{p}K_a\), where one of the two denominators is small even though the total concentration is not. Problems, 2 quantifies a case in which the working form is wrong by \(0.38\) pH units.

The temperature is fixed and stated, and \(K_a\) is evaluated at it.\(\text{p}K_a\) is temperature-dependent through van-t-hoff-equation, \(d\,\text{p}K_a/dT=-\Delta H_a^\circ/(2.303RT^2)\). For carboxylic acids \(\Delta H_a^\circ\approx0\) and the drift is a few thousandths of a unit per kelvin; for amine and Tris buffers \(\Delta H_a^\circ\approx+47\ \text{kJ\,mol}^{-1}\) and \(\text{p}K_a\) falls by roughly \(0.028\) per kelvin, so a Tris buffer titrated to pH \(8.00\) on the bench at \(25^\circ\text{C}\) is near pH \(7.7\) once it reaches \(37^\circ\text{C}\). The pH of water itself moves too: \(\text{p}K_w=14.00\) at \(25^\circ\text{C}\) and \(13.62\) at \(37^\circ\text{C}\) (water-autoionisation-ph).

Proof

The derivation is deliberately done twice over: first exactly, from the three conservation statements that any aqueous solution must satisfy, and only then in the approximate working form, so that the discarded terms are visible and can be estimated. Every symbol is rearranged before a number appears anywhere.

\[ \text{HA}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{H}_3\text{O}^+(aq)+\text{A}^-(aq), \qquad K_a^\circ=\frac{a_{\text{H}^+}\,a_{\text{A}^-}}{a_{\text{HA}}} \]

The Brønsted–Lowry dissociation (bronsted-lowry) and its thermodynamic equilibrium constant, which by gibbs-equilibrium-constant is \(K_a^\circ=\exp(-\Delta G_a^\circ/RT)\) and is therefore a dimensionless number fixed by the standard states, not by the solution. Water’s activity is set to unity for a dilute aqueous solution and folded into \(K_a^\circ\). A

\[ a_i=\gamma_i\frac{[i]}{c^\circ} \quad\Longrightarrow\quad K_a^{\text{mix}}\equiv\frac{a_{\text{H}^+}[\text{A}^-]}{[\text{HA}]}=K_a^\circ\,\frac{\gamma_{\text{HA}}}{\gamma_{\text{A}^-}} \]

Introducing molar activity coefficients referred to \(c^\circ=1\ \text{mol\,L}^{-1}\) and collecting the two solute coefficients into a single factor; \(c^\circ\) cancels between the two solute activities, so \(K_a^{\text{mix}}\) is dimensionless like \(K_a^\circ\). Taking logarithms, \(\text{p}K_a^{\text{mix}}=\text{p}K_a^\circ+\log_{10}\dfrac{\gamma_{\text{A}^-}}{\gamma_{\text{HA}}}\). This is the constant that pairs with an electrode reading, and the whole ionic-strength dependence of a buffer sits in this one factor. B

\[ \log_{10}\gamma_i=-A z_i^{2}\left[\frac{\sqrt{I}}{1+\sqrt{I}}-0.3\,I\right], \qquad I=\tfrac12\sum_i c_i z_i^{2}, \qquad A=0.509\ (\text{water},\ 25^\circ\text{C}) \]

The Davies equation, the extended-Debye–Huckel form used for \(I\lesssim0.5\ \text{mol\,L}^{-1}\) (debye-huckel). Neutral \(\text{HA}\) has \(z=0\) and hence \(\gamma_{\text{HA}}\approx1\), so for a neutral acid \(\text{p}K_a^{\text{mix}}=\text{p}K_a^\circ-A\big[\sqrt{I}/(1+\sqrt{I})-0.3I\big]\); for an anionic acid such as \(\text{H}_2\text{PO}_4^-\) the bracket carries a factor \(z_{\text{A}}^2-z_{\text{HA}}^2=4-1=3\) instead, and the shift is three times larger. B

\[ C_{\text{HA}}+C_{\text{A}}=[\text{HA}]+[\text{A}^-] \]

Mass balance on the \(\text{A}\) fragment: every unit of \(\text{A}\) placed in the flask is present at equilibrium as one or the other of the two species, since the system is closed (Hypotheses) and no third \(\text{A}\)-containing species exists. This is the same accounting that mole-avogadro makes for unreactive matter, now applied across a reaction that has not gone to completion. A

\[ [\text{M}^+]+[\text{H}^+]=[\text{A}^-]+[\text{OH}^-], \qquad [\text{M}^+]=C_{\text{A}} \]

Electroneutrality. The buffer is prepared from \(C_{\text{HA}}\) of the free acid plus \(C_{\text{A}}\) of its salt \(\text{MA}\), so the spectator cation is present at exactly \(C_{\text{A}}\) and is inert. Preparing the same buffer instead by adding \(C_{\text{A}}\) of \(\text{MOH}\) to \(C_{\text{HA}}+C_{\text{A}}\) of \(\text{HA}\) gives an identical charge balance, which is why the two recipes are interchangeable. B

\[ [\text{A}^-]=C_{\text{A}}+[\text{H}^+]-[\text{OH}^-], \qquad [\text{HA}]=C_{\text{HA}}-[\text{H}^+]+[\text{OH}^-] \]

The first line is Step 5 rearranged; the second follows by substituting it into Step 4 and cancelling \(C_{\text{A}}\). No approximation has been made: these two identities are exact for any concentration whatever. The physical content is that the deviation of each equilibrium concentration from its formal value is the same quantity \(\delta\equiv[\text{H}^+]-[\text{OH}^-]\), with opposite signs — every proton that dissociates converts one \(\text{HA}\) into one \(\text{A}^-\). B

\[ a_{\text{H}^+}=K_a^{\text{mix}}\,\frac{[\text{HA}]}{[\text{A}^-]}=K_a^{\text{mix}}\,\frac{C_{\text{HA}}-\delta}{C_{\text{A}}+\delta} \quad\Longrightarrow\quad \text{pH}=\text{p}K_a^{\text{mix}}+\log_{10}\frac{C_{\text{A}}+\delta}{C_{\text{HA}}-\delta} \]

Step 2 rearranged for the hydrogen-ion activity, Step 6 substituted, then \(-\log_{10}\) applied to both sides using \(\text{pH}\equiv-\log_{10}a_{\text{H}^+}\) (the IUPAC definition) and \(-\log_{10}(x/y)=\log_{10}(y/x)\). This is the exact Henderson–Hasselbalch equation. It is still one equation in one unknown, since \(\delta\) depends on \(\text{pH}\) through \(\delta=[\text{H}^+]-K_w^{c}/[\text{H}^+]\) with \([\text{H}^+]=a_{\text{H}^+}/\gamma_{\text{H}^+}\); clearing the fractions turns it into a cubic in \([\text{H}^+]\). B

\[ \begin{aligned} \text{pH}&=\text{p}K_a^{\text{mix}}+\log_{10}\frac{C_{\text{A}}}{C_{\text{HA}}}+\log_{10}\!\left(1+\frac{\delta}{C_{\text{A}}}\right)-\log_{10}\!\left(1-\frac{\delta}{C_{\text{HA}}}\right)\\[2pt] &\approx\text{p}K_a^{\text{mix}}+\log_{10}\frac{C_{\text{A}}}{C_{\text{HA}}}+\frac{\delta}{\ln 10}\left(\frac{1}{C_{\text{A}}}+\frac{1}{C_{\text{HA}}}\right) \end{aligned} \]

Splitting the logarithm of Step 7 and expanding \(\log_{10}(1+x)=x/\ln10+O(x^2)\) for \(|x|\ll1\). The first line is exact; the second isolates the leading error of the working form as a single, computable term. Dropping it gives the equation in its usual shape, and the criterion for doing so is now explicit and two-sided: \(|\delta|\) must be small against \(C_{\text{A}}\) and against \(C_{\text{HA}}\), not merely against their sum. C

\[ \text{pH}=\text{p}K_a^{\text{mix}}+\log_{10}\frac{C_{\text{A}}}{C_{\text{HA}}} \qquad\text{when}\qquad \big|[\text{H}^+]-[\text{OH}^-]\big|\ll\min\!\big(C_{\text{A}},C_{\text{HA}}\big) \]

The working form, now carrying its own domain of validity rather than being asserted. Note that only the ratio appears, so amounts in moles may be substituted for concentrations provided both refer to the same volume — the standard shortcut when a strong acid or base has been added and the bookkeeping is easier in millimoles. A

\[ \beta\equiv\frac{dC_{\text{base added}}}{d\,\text{pH}}=\ln(10)\left([\text{H}^+]+[\text{OH}^-]+C_T\,\frac{K_a[\text{H}^+]}{\big(K_a+[\text{H}^+]\big)^2}\right),\qquad \beta_{\max}=\frac{\ln 10}{4}C_T\approx0.576\,C_T \]

The Van Slyke buffer capacity, obtained by writing \(C_{\text{base added}}\) from the charge balance of Step 5 as a function of \([\text{H}^+]\) and differentiating with respect to \(\text{pH}=-\log_{10}[\text{H}^+]\), using \(d[\text{H}^+]/d\,\text{pH}=-\ln(10)[\text{H}^+]\). Writing \(f_{\text{HA}}\) and \(f_{\text{A}^-}\) for the fractions of total buffer \(C_T=C_{\text{HA}}+C_{\text{A}}\) in each form, the third term is \(\ln(10)\,C_T f_{\text{HA}}f_{\text{A}^-}\), which is maximal at \(f_{\text{HA}}=f_{\text{A}^-}=\tfrac12\), i.e. at \(\text{pH}=\text{p}K_a\). This is the quantitative statement of what a buffer is, and it is invisible in the working form. C

Result

\[ \text{pH}=\text{p}K_a^{\text{mix}}+\log_{10}\frac{[\text{A}^-]}{[\text{HA}]} =\text{p}K_a^{\text{mix}}+\log_{10}\frac{C_{\text{A}}+\delta}{C_{\text{HA}}-\delta} \;\xrightarrow{\ \ |\delta|\,\ll\,\min(C_{\text{A}},C_{\text{HA}})\ \ }\; \text{p}K_a^{\text{mix}}+\log_{10}\frac{C_{\text{A}}}{C_{\text{HA}}} \]

Reading. A buffer’s pH is its acid’s \(\text{p}K_a\) shifted by the base-ten logarithm of the base-to-acid ratio. The middle expression, with \(\delta=[\text{H}^+]-[\text{OH}^-]\), is exact; the right-hand limit is the working form. Every decade of ratio moves the pH by exactly one unit, so a buffer spanning \(\text{p}K_a\pm1\) uses ratios between \(1{:}10\) and \(10{:}1\).

Units check. \(C_{\text{A}}\), \(C_{\text{HA}}\) and \(\delta\) all carry \(\text{mol\,L}^{-1}\), so the argument of the logarithm is dimensionless as it must be; \(\text{p}K_a\) and pH are dimensionless by construction. Because only the ratio enters, the working form is invariant under any change of concentration unit and under multiplication of both concentrations by a common factor — the formal statement that ideal dilution does not change buffer pH.

Scope. Closed system, one dominant equilibrium, and a \(\text{p}K_a\) evaluated at the working ionic strength and temperature. The working form additionally requires \(|\delta|\ll\min(C_{\text{A}},C_{\text{HA}})\); Step 8 gives the leading correction, \(\Delta\text{pH}\approx\delta\big(C_{\text{A}}^{-1}+C_{\text{HA}}^{-1}\big)/\ln10\), when it does not hold.

Corollaries & converses

Half-equivalence. Halfway to the equivalence point of a weak-acid titration, exactly half the acid has been converted, \(C_{\text{A}}=C_{\text{HA}}\), and \(\text{pH}=\text{p}K_a^{\text{mix}}\). Reading \(\text{p}K_a\) off the midpoint of a titration curve (titration-curves) is the standard laboratory determination, and it delivers the mixed constant at whatever ionic strength the titration ran at.
The decade rule. \(\text{pH}-\text{p}K_a=\log_{10}(C_{\text{A}}/C_{\text{HA}})\) means one pH unit per factor of ten. At \(\text{p}K_a\pm2\) the minor species is \(1\%\) of the total, which is why the useful buffering window is conventionally \(\text{p}K_a\pm1\).
Buffer capacity peaks at \(\text{p}K_a\). From Step 10, \(\beta=\ln(10)C_Tf_{\text{HA}}f_{\text{A}^-}\) is maximal at equal fractions, with \(\beta_{\max}\approx0.576\,C_T\), and has fallen to \(0.33\,\beta_{\max}\) one pH unit away. Capacity scales linearly with total buffer concentration but falls only quadratically — and hence flatly — with position — a strong argument for choosing an acid whose \(\text{p}K_a\) is near the target rather than over-concentrating a poorly matched one.
Conjugate-base form. For a base \(\text{B}\) with conjugate acid \(\text{BH}^+\), \(\text{pOH}=\text{p}K_b+\log_{10}([\text{BH}^+]/[\text{B}])\) is the same statement, converted by \(\text{p}K_a+\text{p}K_b=\text{p}K_w\) (water-autoionisation-ph). Working in \(\text{p}K_a\) throughout avoids the sign errors this alternative invites.
Converse: buffer design. Given a target pH and a chosen acid, \(C_{\text{A}}/C_{\text{HA}}=10^{\,\text{pH}-\text{p}K_a^{\text{mix}}}\) fixes the ratio and the desired capacity fixes \(C_T\); the two together fix both concentrations uniquely, since \(C_{\text{A}}=C_Tr/(1+r)\) and \(C_{\text{HA}}=C_T/(1+r)\). Worked example 2 carries this out.
Polyprotic reduction. For a polyprotic acid whose successive \(\text{p}K_a\) values are separated by more than about three units, the working form applies to each conjugate pair independently within its own window; phosphate is used as three practically separate buffers near pH \(2.1\), \(7.2\) and \(12.3\) for exactly this reason (polyprotic-acids).
Partition and absorption. Combining the Result with a partition coefficient gives the pH-partition hypothesis: the fraction of a weak acid in the neutral, membrane-permeable form is \(f_{\text{HA}}=\big(1+10^{\,\text{pH}-\text{p}K_a}\big)^{-1}\), which is why a \(\text{p}K_a\ 3.5\) drug is absorbed from the stomach and a \(\text{p}K_a\ 9\) base is not (chromatography-partition uses the same expression for retention).

Fails without

Formal concentrations dropped for equilibrium concentrations — the dilute-buffer regime: a buffer \(1.00\times10^{-3}\ \text{mol\,L}^{-1}\) in both a \(\text{p}K_a=3.00\) acid and its salt has \(\delta\) comparable to the concentrations themselves. Solving Step 7 exactly gives \([\text{H}^+]=4.14\times10^{-4}\ \text{mol\,L}^{-1}\) and \(\text{pH}=3.38\), while the working form insists on \(3.00\). The error, \(0.38\) units, is a factor of \(2.4\) in hydrogen-ion concentration (Problems, 2).
The same failure at fixed total concentration but extreme ratio: a buffer with \(C_T=0.10\ \text{mol\,L}^{-1}\) but \(C_{\text{A}}/C_{\text{HA}}=10^{4}\) has \(C_{\text{HA}}\approx10^{-5}\ \text{mol\,L}^{-1}\), so for an acid of \(\text{p}K_a\approx6\) the predicted pH is \(10\) and \([\text{OH}^-]=10^{-4}\ \text{mol\,L}^{-1}\) is ten times \(C_{\text{HA}}\) itself and the working form breaks even though the buffer is nominally concentrated. This is why the criterion in Step 9 is stated against \(\min(C_{\text{A}},C_{\text{HA}})\) rather than against \(C_T\).
Activity coefficients ignored at real ionic strength: the thermodynamic \(\text{p}K_{a2}^\circ=7.20\) of phosphoric acid becomes \(\text{p}K_a^{\text{mix}}=6.84\) at \(I=0.150\ \text{mol\,L}^{-1}\) by Step 3, because the charge-squared difference between \(\text{HPO}_4^{2-}\) and \(\text{H}_2\text{PO}_4^-\) is \(3\). A phosphate buffer mixed from the tabulated value lands at pH \(7.04\) instead of the intended \(7.40\) (Worked example 2) — an error of \(0.36\) units, which in a physiological experiment is the difference between normal and frank acidosis.
Closure dropped — the open system: for the \(\text{CO}_2/\text{HCO}_3^-\) pair in contact with a gas reservoir, the denominator is not a conserved formal concentration but \(\alpha\,p_{\text{CO}_2}\) fixed by Henry’s law. The correct relation is \(\text{pH}=\text{p}K_{a1}'+\log_{10}\big[\text{HCO}_3^-\big]/(\alpha p_{\text{CO}_2})\), and the system’s buffer capacity against added acid is many times the closed-system value because the \(\text{CO}_2\) generated is exhaled rather than accumulating (Problems, 5).
A second equilibrium intrudes: if \(\text{A}^-\) also precipitates (solubility-product) or complexes a metal ion present in the solution, mass balance in Step 4 acquires a further term and both the pH and the capacity are wrong. Carbonate buffers in hard water are the standard laboratory example: \(\text{CaCO}_3\) deposition silently removes the basic component.

Common errors

“\(\text{pH}=\text{p}K_a+\log([\text{HA}]/[\text{A}^-])\).” The ratio is inverted; the base goes on top. The sign check that never fails: adding more conjugate base must raise the pH, so \([\text{A}^-]\) must sit in the numerator.
“Look up \(\text{p}K_a\) and use it.” Tables give \(\text{p}K_a^\circ\), extrapolated to infinite dilution. At any working ionic strength the constant you need is \(\text{p}K_a^{\text{mix}}\), which differs by \(\log_{10}(\gamma_{\text{A}^-}/\gamma_{\text{HA}})\) — about \(-0.11\) for a neutral acid at \(I=0.10\), about \(-0.36\) for a \(-1/-2\) pair at \(I=0.15\), and positive for a cationic acid such as \(\text{NH}_4^+\), where the charged species is the acid.
“Dilution does not change buffer pH.” True only in the ideal limit. Diluting tenfold lowers \(I\) and therefore shifts \(\text{p}K_a^{\text{mix}}\) back towards \(\text{p}K_a^\circ\), and it simultaneously enlarges \(\delta/C\), so the neglected term in Step 8 grows. Both effects are small at \(0.1\ \text{mol\,L}^{-1}\) and neither is at \(10^{-4}\ \text{mol\,L}^{-1}\).
Using the initial ratio after adding strong acid or base. A strong acid converts \(\text{A}^-\) to \(\text{HA}\) mole for mole; the ratio to substitute is the one after that stoichiometric conversion, which is a limiting-reagent calculation done first, with the equilibrium calculation done second on its output.
“A buffer holds pH constant.” It holds \(d\,\text{pH}/dC\) small, and only while both components remain. Once one is consumed the solution is a plain strong-acid or weak-acid solution and the pH moves freely. Capacity is a finite, computable number (Step 10), not a property that is present or absent.
Applying the equation to a strong acid. For \(\text{HCl}\) there is no undissociated \(\text{HA}\) at equilibrium, \(C_{\text{HA}}-\delta\to0\), and the logarithm diverges. The correct treatment is the charge balance of Step 5 on its own.
Reporting more digits than \(\text{p}K_a\) supports. A \(\text{p}K_a\) known to \(\pm0.02\) cannot give a pH to three decimals, however many the ratio carries; and the Davies equation itself is good to only about \(\pm0.02\) in \(\log_{10}\gamma\) at \(I=0.1\).

Discussion

Lawrence Joseph Henderson, a physiologist working on the buffering of blood, published the underlying relation in non-logarithmic form in 1908; Karl Albert Hasselbalch recast it in the logarithmic pH notation Søren Sørensen had introduced in 1909, and it is that 1917 recasting that gives the equation its modern shape and its double name. The order matters historically: the equation entered chemistry from physiology, as a tool for a specific open buffer system in blood, and only afterwards became a general-purpose bench formula. Some of the confusion around it — particularly the habit of quoting an apparent \(\text{p}K_{a1}'\) of \(6.1\) for carbonic acid, which is neither the thermodynamic value nor even a constant for the same reaction — is a direct inheritance of that origin.

What is worth being clear about is that the equation contains no chemistry that the law of mass action does not already contain. It is an algebraic rearrangement, and Steps 1 to 7 make no approximation at all. All of its content, and all of its failure modes, lie in the substitutions made afterwards: which \(K_a\) is used, and whether formal concentrations may stand in for equilibrium ones. This is why it is simultaneously trivial and treacherous. A student who derives it once from mass balance and charge balance, as above, will never again wonder whether it “applies” to a given problem — the exact form always applies, and the only question is whether the two correction terms are small.

The activity question deserves one further turn of the screw. The pH that a glass electrode reports is not \(-\log_{10}a_{\text{H}^+}\) in any operationally exact sense, because single-ion activities are not thermodynamically measurable: no experiment can separate \(a_{\text{H}^+}\) from the activity of its counter-ion. The IUPAC operational scale defines pH by comparison with a series of primary standard buffers whose assigned values themselves rest on the Bates–Guggenheim convention for the chloride activity coefficient, and it carries an unknown residual liquid-junction potential of order a few millivolts when the sample and the standard differ in ionic strength or in medium. The practical consequence is that a pH computed with a mixed constant and a pH read from a meter agree to about \(\pm0.02\) in dilute aqueous solution and can diverge by \(0.1\) or more in concentrated brine, in a mixed solvent, or in a suspension. None of this is a defect of the Henderson–Hasselbalch equation; it is a limit on what the symbol pH means, and it sets the floor on how precisely any buffer calculation can be verified.

Common misconceptions. That a buffer resists pH change symmetrically: at \(\text{pH}=\text{p}K_a-1\) there is ten times more acid than base, so the solution absorbs added base ten times more readily than added acid, and \(\beta\) as defined in Step 10 is a single number only because it is a derivative at a point. That buffer capacity depends on the ratio: it depends chiefly on \(C_T\), with the ratio setting only the factor \(f_{\text{HA}}f_{\text{A}^-}\le\tfrac14\). And that a buffer’s job is to hold a particular pH: in a titration or an enzymatic assay its job is to hold pH against a known load, which is a statement about \(\beta\) and \(C_T\), not about \(\text{p}K_a\) alone.

Worked examples

Example 1. One litre of solution is prepared \(0.100\ \text{mol\,L}^{-1}\) in acetic acid and \(0.100\ \text{mol\,L}^{-1}\) in sodium acetate at \(25^\circ\text{C}\). Find its pH; find the pH after \(10.0\ \text{mmol}\) of \(\text{HCl}\) is added with negligible volume change; verify the buffer approximation against the exact form of Step 7; and check the result against the buffer capacity of Step 10. Standard values used: \(\text{p}K_a^\circ(\text{CH}_3\text{COOH})=4.756\) at \(25^\circ\text{C}\), Davies constant \(A=0.509\) for water at \(25^\circ\text{C}\), \(K_w=1.0\times10^{-14}\).

\[ I=\tfrac12\big[(0.100)(+1)^2+(0.100)(-1)^2\big]=0.100\ \text{mol\,L}^{-1} \]

Only the sodium acetate contributes: undissociated acetic acid is neutral, and \([\text{H}^+]\) is of order \(10^{-5}\ \text{mol\,L}^{-1}\), four orders below the salt. Ionic strength must be computed before \(\text{p}K_a\) can be fixed, so this is the first step, not an afterthought. A

\[ \log_{10}\gamma_{\pm}=-0.509\left[\frac{0.31623}{1.31623}-0.3(0.100)\right]=-0.509\,(0.24025-0.03000)=-0.1070 \]

Davies (Step 3 of the Proof) with \(z=1\); \(\sqrt{0.100}=0.31623\). Hence \(\gamma_{\text{CH}_3\text{COO}^-}=10^{-0.1070}=0.782\), while \(\gamma_{\text{CH}_3\text{COOH}}\approx1\) because the species is uncharged. B

\[ \text{p}K_a^{\text{mix}}=\text{p}K_a^\circ+\log_{10}\frac{\gamma_{\text{A}^-}}{\gamma_{\text{HA}}}=4.756+(-0.107)=4.649 \]

Step 2 of the Proof. The buffer is a tenth of a unit more acidic than the tabulated \(\text{p}K_a\) would suggest, and this shift is fully ten times the precision to which an ordinary pH meter reads. B

\[ \text{pH}=4.649+\log_{10}\frac{0.100}{0.100}=4.649+0=4.65 \]

The working form of Step 9 at unit ratio: the buffer sits exactly at its apparent \(\text{p}K_a\), which is also the point of maximum capacity. A

\[ \begin{aligned} \text{CH}_3\text{COO}^-+\text{H}^+&\longrightarrow\text{CH}_3\text{COOH} \quad(\text{quantitative})\\ C_{\text{A}}&=0.100-0.010=0.090\ \text{mol\,L}^{-1},\qquad C_{\text{HA}}=0.100+0.010=0.110\ \text{mol\,L}^{-1} \end{aligned} \]

The strong acid is consumed completely by the conjugate base — a limiting-reagent step done in full before any equilibrium arithmetic (Common errors). Ionic strength is unchanged: \(\text{Na}^+\) is still \(0.100\), and the \(0.010\) of \(\text{Cl}^-\) replaces exactly the \(0.010\) of acetate consumed, so \(I=\tfrac12(0.100+0.010+0.090)=0.100\ \text{mol\,L}^{-1}\) and \(\text{p}K_a^{\text{mix}}\) does not move. B

\[ \text{pH}=4.649+\log_{10}\frac{0.090}{0.110}=4.649-0.0872=4.56 \]

The same working form on the post-neutralisation amounts. For comparison, the same \(10.0\ \text{mmol}\) of \(\text{HCl}\) in a litre of pure water gives \(\text{pH}=-\log_{10}(0.0100)=2.00\), a fall of five units from neutrality; the buffer confines the change to \(0.087\). A

\[ [\text{H}^+]=\frac{a_{\text{H}^+}}{\gamma_{\text{H}^+}}=\frac{10^{-4.562}}{0.782}=3.51\times10^{-5}\ \text{mol\,L}^{-1},\qquad [\text{OH}^-]\approx3\times10^{-10}\ \text{mol\,L}^{-1} \]

Converting the activity back to a concentration so that \(\delta=[\text{H}^+]-[\text{OH}^-]=3.51\times10^{-5}\ \text{mol\,L}^{-1}\) can be evaluated. The hydroxide term is five orders smaller and is discarded. C

\[ \Delta\text{pH}\approx\frac{\delta}{\ln 10}\left(\frac{1}{C_{\text{A}}}+\frac{1}{C_{\text{HA}}}\right)=\frac{3.51\times10^{-5}}{2.3026}\big(11.11+9.09\big)=+3.1\times10^{-4} \]

The neglected term of Step 8 of the Proof. The exact pH is \(4.5621\) against the approximate \(4.5618\): three ten-thousandths of a unit, far below both the electrode’s resolution and the uncertainty in \(\text{p}K_a^\circ\). The buffer approximation is vindicated here — and Problems, 2 shows a case where it is not. C

\[ \beta=\ln(10)\,C_T\,f_{\text{HA}}f_{\text{A}^-}=2.3026\,(0.200)(0.5)(0.5)=0.115\ \text{mol\,L}^{-1}\,\text{pH}^{-1} \quad\Longrightarrow\quad \Delta\text{pH}\approx-\frac{0.0100}{0.115}=-0.087 \]

Step 10 of the Proof, evaluated at the initial state where \(C_T=0.200\ \text{mol\,L}^{-1}\) and the fractions are equal. The linearised prediction \(-0.0869\) reproduces the exact \(-0.0872\) to within \(0.0003\) because the load is small enough that \(\beta\) has barely changed over the interval — a consistency check on both routes. B

\[ \text{pH}=4.65 \;\xrightarrow{\ \ +10.0\ \text{mmol HCl}\ \ }\; \text{pH}=4.56, \qquad \beta=0.115\ \text{mol\,L}^{-1}\,\text{pH}^{-1}, \qquad \text{p}K_a^{\text{mix}}=4.65 \]

Reading. An equimolar acetate buffer sits at its apparent \(\text{p}K_a\), not at the tabulated \(4.76\); the \(0.11\)-unit ionic-strength shift is the largest single correction in the calculation, some three hundred times larger than the buffer-approximation error it is usually confused with.

Scope. \(25^\circ\text{C}\), \(I=0.100\ \text{mol\,L}^{-1}\), Davies activity coefficients (good to about \(\pm0.02\) in \(\log_{10}\gamma\) here), and volume change on adding the \(\text{HCl}\) neglected.

Example 2. Design one litre of phosphate buffer at pH \(7.400\) and \(25^\circ\text{C}\) with total phosphate \(C_T=0.0500\ \text{mol\,L}^{-1}\) and total ionic strength adjusted to \(I=0.150\ \text{mol\,L}^{-1}\) with sodium chloride. Give the masses of \(\text{NaH}_2\text{PO}_4\), \(\text{Na}_2\text{HPO}_4\) and \(\text{NaCl}\) required, the buffer capacity obtained, and the pH that would have resulted from using the tabulated \(\text{p}K_a\). Standard values used: \(\text{p}K_{a2}^\circ(\text{H}_3\text{PO}_4)=7.198\) at \(25^\circ\text{C}\); \(A=0.509\); molar masses from the standard atomic weights \(\text{Na}\ 22.99\), \(\text{H}\ 1.008\), \(\text{P}\ 30.97\), \(\text{O}\ 16.00\), \(\text{Cl}\ 35.45\), giving \(M(\text{NaH}_2\text{PO}_4)=119.98\), \(M(\text{Na}_2\text{HPO}_4)=141.96\), \(M(\text{NaCl})=58.44\ \text{g\,mol}^{-1}\).

\[ \text{H}_2\text{PO}_4^-\rightleftharpoons\text{H}^++\text{HPO}_4^{2-}, \qquad z_{\text{HA}}=-1,\quad z_{\text{A}}=-2,\quad z_{\text{A}}^2-z_{\text{HA}}^2=3 \]

Identifying the conjugate pair and, crucially, its charges. Unlike acetic acid, here both members are charged, and it is the difference of the squared charges that drives the ionic-strength shift (Step 3 of the Proof). B

\[ \left[\frac{\sqrt{I}}{1+\sqrt{I}}-0.3I\right]_{I=0.150}=\frac{0.38730}{1.38730}-0.04500=0.27916-0.04500=0.23416 \]

The Davies bracket at the target ionic strength, evaluated once and reused for both ions. Note the design proceeds at fixed \(I\): the recipe is engineered to land on \(I=0.150\), so no iteration is needed. A

\[ \text{p}K_a^{\text{mix}}=\text{p}K_a^\circ-A\big(z_{\text{A}}^2-z_{\text{HA}}^2\big)(0.23416)=7.198-(0.509)(3)(0.23416)=7.198-0.358=6.840 \]

Substituting \(\log_{10}(\gamma_{\text{A}}/\gamma_{\text{HA}})=-A(z_{\text{A}}^2-z_{\text{HA}}^2)\times(\text{bracket})\) into Step 2 of the Proof. The shift is \(0.36\) units, three times the acetate case of Example 1 and far too large to ignore. This number, \(6.84\), is why physiological phosphate buffers are always tabulated near \(6.8\) rather than \(7.2\). B

\[ r\equiv\frac{C_{\text{A}}}{C_{\text{HA}}}=10^{\,\text{pH}-\text{p}K_a^{\text{mix}}}=10^{\,7.400-6.840}=10^{0.560}=3.631 \]

The converse form of the Result, solved for the ratio. Comfortably inside the \(1{:}10\) to \(10{:}1\) window, so the buffer will be effective in both directions. A

\[ C_{\text{A}}=C_T\frac{r}{1+r}=0.0500\left(\frac{3.631}{4.631}\right)=0.03920,\qquad C_{\text{HA}}=\frac{C_T}{1+r}=0.01080\ \text{mol\,L}^{-1} \]

Ratio and total together determine both concentrations; the algebra is rearranged in symbols first, then evaluated. A

\[ \begin{aligned} [\text{Na}^+]&=C_{\text{HA}}+2C_{\text{A}}=0.01080+0.07840=0.08920\ \text{mol\,L}^{-1}\\ I_{\text{phos}}&=\tfrac12\big[(0.08920)(1)+(0.01080)(1)+(0.03920)(4)\big]=\tfrac12(0.25681)=0.12841\ \text{mol\,L}^{-1} \end{aligned} \]

The ionic strength the phosphate salts themselves deliver. The doubly-charged \(\text{HPO}_4^{2-}\) contributes four times its concentration, so it alone supplies \(0.0784\) of the total — more than the sodium. B

\[ c_{\text{NaCl}}=I_{\text{target}}-I_{\text{phos}}=0.1500-0.1284=0.0216\ \text{mol\,L}^{-1} \]

A 1:1 electrolyte contributes exactly its own concentration to \(I\), so the make-up is a simple difference. Had \(I_{\text{phos}}\) exceeded the target — which it would at \(C_T=0.100\ \text{mol\,L}^{-1}\), where it reaches \(0.257\) — the specification would be infeasible and either \(C_T\) or \(I\) would have to be renegotiated. B

\[ \begin{aligned} m(\text{NaH}_2\text{PO}_4)&=(0.01080)(119.98)=1.296\ \text{g}\\ m(\text{Na}_2\text{HPO}_4)&=(0.03920)(141.96)=5.565\ \text{g}\\ m(\text{NaCl})&=(0.02160)(58.44)=1.262\ \text{g} \end{aligned} \]

Amount times molar mass, per litre of final solution — the mole-avogadro conversion that turns an equilibrium specification into something that can be weighed out. A

\[ \beta=\ln(10)\,C_Tf_{\text{HA}}f_{\text{A}}=2.3026\,(0.0500)(0.2160)(0.7840)=0.0195\ \text{mol\,L}^{-1}\,\text{pH}^{-1} \]

Step 10 of the Proof. Because the buffer sits \(0.56\) units above its \(\text{p}K_a\), the fraction product \(f_{\text{HA}}f_{\text{A}}=0.169\) is well below its maximum of \(0.25\), so the capacity is only \(68\%\) of what the same total phosphate would give at pH \(6.84\). B

\[ r_{\text{naive}}=10^{\,7.400-7.198}=1.593 \;\Longrightarrow\; \frac{C_{\text{A}}}{C_{\text{HA}}}=\frac{0.03071}{0.01929} \;\Longrightarrow\; \text{pH}_{\text{actual}}=6.840+\log_{10}(1.593)=7.042 \]

The counterfactual: mixing to the tabulated \(\text{p}K_a^\circ\) produces a solution whose real pH, governed by the mixed constant of Step 3, is \(7.04\). The error is \(0.36\) units in the direction of acidity — the whole of the ionic-strength shift, transmitted unchanged into the product. B

\[ 1.296\ \text{g NaH}_2\text{PO}_4+5.565\ \text{g Na}_2\text{HPO}_4+1.262\ \text{g NaCl}\ \text{per litre} \;\Longrightarrow\; \text{pH}=7.400,\ I=0.150,\ \beta=0.0195\ \text{mol\,L}^{-1}\text{pH}^{-1} \]

Reading. A complete buffer recipe follows from three numbers — target pH, total buffer concentration, target ionic strength — once \(\text{p}K_a\) is corrected from \(7.198\) to \(6.840\) for the medium. Using the uncorrected constant would have delivered pH \(7.04\).

Scope. \(25^\circ\text{C}\); Davies coefficients at \(I=0.150\ \text{mol\,L}^{-1}\), near the upper end of that equation’s reliable range, so the computed pH carries perhaps \(\pm0.03\); the buffer approximation is excellent here since \(\delta\sim10^{-7}\ \text{mol\,L}^{-1}\) against concentrations of \(10^{-2}\ \text{mol\,L}^{-1}\).

Problems

A buffer contains \(0.200\ \text{mol\,L}^{-1}\) \(\text{NH}_3\) and \(0.300\ \text{mol\,L}^{-1}\) \(\text{NH}_4\text{Cl}\) in \(0.500\ \text{L}\) at \(25^\circ\text{C}\); \(\text{p}K_a^\circ(\text{NH}_4^+)=9.25\). (a) Find the pH using the tabulated constant. (b) Find the pH after adding \(10.0\ \text{mmol}\) of solid \(\text{NaOH}\). (c) Correct both answers for ionic strength and explain why the correction has the opposite sign to the acetate case of Worked example 1.
Solution
(a) Amounts: \(n(\text{NH}_3)=(0.200)(0.500)=0.100\ \text{mol}\), \(n(\text{NH}_4^+)=(0.300)(0.500)=0.150\ \text{mol}\). Since only the ratio enters (Step 9), moles may be used directly: \(\text{pH}=9.25+\log_{10}(0.100/0.150)=9.25-0.176=9.07\).
(b) Hydroxide converts the acid to the base, mole for mole: \(n(\text{NH}_3)=0.110\), \(n(\text{NH}_4^+)=0.140\ \text{mol}\). Then \(\text{pH}=9.25+\log_{10}(0.110/0.140)=9.25-0.105=9.14\). The load has moved the pH by \(+0.07\); the same \(10.0\ \text{mmol}\) of \(\text{NaOH}\) in \(0.500\ \text{L}\) of water would give \([\text{OH}^-]=0.0200\ \text{mol\,L}^{-1}\), pOH \(1.70\), pH \(12.30\).
(c) The ionic strength comes from \(\text{NH}_4\text{Cl}\): \(I=\tfrac12[(0.300)(1)+(0.300)(1)]=0.300\ \text{mol\,L}^{-1}\) (the added \(\text{Na}^+\) and the \(\text{NH}_4^+\) it consumes leave \(I\) unchanged). Davies bracket: \(\sqrt{0.300}=0.54772\), \(0.54772/1.54772=0.35389\), minus \(0.3(0.300)=0.0900\), giving \(0.26389\). Here the acid \(\text{NH}_4^+\) is charged and the base \(\text{NH}_3\) is neutral, so \(z_{\text{A}}^2-z_{\text{HA}}^2=0-1=-1\) and \(\text{p}K_a^{\text{mix}}=9.25+(0.509)(0.26389)=9.25+0.134=9.38\).
Corrected: (a) \(9.38-0.176=9.21\); (b) \(9.38-0.105=9.28\). The shift is \(+0.13\) rather than \(-0.11\) because the sign of \(\log_{10}(\gamma_{\text{A}}/\gamma_{\text{HA}})\) follows the difference of squared charges: stabilising the charged species by its ionic atmosphere favours dissociation when the product is charged (acetic acid, shift negative) and disfavours it when the reactant is charged (ammonium, shift positive). Note also that \(I=0.300\ \text{mol\,L}^{-1}\) is near the limit of the Davies equation, so \(\pm0.03\) is a fair uncertainty on the correction.
A solution is \(1.00\times10^{-3}\ \text{mol\,L}^{-1}\) in a weak acid with stoichiometric \(\text{p}K_a=3.00\) and \(1.00\times10^{-3}\ \text{mol\,L}^{-1}\) in its sodium salt. Compute the pH from the working form and then exactly from Step 7, and state the error. Work throughout on a concentration basis (activity coefficients are within \(0.02\) of unity at this ionic strength).
Solution
Working form. The ratio is \(1\), so \(\text{pH}=\text{p}K_a=3.00\).
Exact. Write \(h=[\text{H}^+]\) and neglect \([\text{OH}^-]\sim10^{-11}\ \text{mol\,L}^{-1}\), so \(\delta=h\). Step 6 gives \([\text{A}^-]=C+h\) and \([\text{HA}]=C-h\) with \(C=1.00\times10^{-3}\). Substituting into \(K_a=h[\text{A}^-]/[\text{HA}]\) and rearranging in symbols first:
\(K_a(C-h)=h(C+h)\ \Longrightarrow\ h^2+(C+K_a)h-K_aC=0\ \Longrightarrow\ h=\dfrac{-(C+K_a)+\sqrt{(C+K_a)^2+4K_aC}}{2}\).
With \(K_a=1.00\times10^{-3}\) and \(C=1.00\times10^{-3}\), \(C+K_a=2.00\times10^{-3}\) and \((C+K_a)^2+4K_aC=4.00\times10^{-6}+4.00\times10^{-6}=8.00\times10^{-6}\), whose square root is \(2.828\times10^{-3}\). Hence \(h=(-2.000+2.828)\times10^{-3}/2=4.14\times10^{-4}\ \text{mol\,L}^{-1}\), and \(\text{pH}=-\log_{10}(4.14\times10^{-4})=3.38\).
Check against Step 7. \([\text{A}^-]=1.000+0.414=1.414\times10^{-3}\), \([\text{HA}]=1.000-0.414=0.586\times10^{-3}\), ratio \(2.414\); \(\text{pH}=3.00+\log_{10}(2.414)=3.00+0.383=3.383\). Consistent.
Error. \(0.38\) pH units, a factor of \(2.4\) in \([\text{H}^+]\). The cause is visible in Step 8: \(\delta/C=0.41\), nowhere near small. Two features conspire — the buffer is dilute, and \(K_a\) is large enough that the acid is genuinely strong at this dilution. The rule of thumb that follows is that the working form needs \(C\gtrsim100K_a\) as well as \(C\gg[\text{H}^+]\).
Compute the apparent \(\text{p}K_a\) of acetic acid at \(I=0.500\ \text{mol\,L}^{-1}\) on both the mixed and the stoichiometric conventions, taking \(\text{p}K_a^\circ=4.756\) and \(A=0.509\). Which one should be used to predict a glass-electrode reading, and which to predict a species concentration?
Solution
Davies bracket. \(\sqrt{0.500}=0.70711\); \(0.70711/1.70711=0.41421\); \(0.3(0.500)=0.1500\); bracket \(=0.26421\). For a singly charged ion, \(\log_{10}\gamma_{\pm}=-(0.509)(0.26421)=-0.1345\), so \(\gamma_{\pm}=0.734\).
Mixed constant. \(\text{p}K_a^{\text{mix}}=\text{p}K_a^\circ+\log_{10}(\gamma_{\text{A}^-}/\gamma_{\text{HA}})=4.756-0.134=4.622\), since \(\gamma_{\text{HA}}\approx1\) for the neutral acid.
Stoichiometric constant. \(K_a^c=[\text{H}^+][\text{A}^-]/[\text{HA}]=K_a^\circ\gamma_{\text{HA}}/(\gamma_{\text{H}^+}\gamma_{\text{A}^-})\), so \(\text{p}K_a^{c}=\text{p}K_a^\circ+\log_{10}(\gamma_{\text{H}^+}\gamma_{\text{A}^-})=4.756-2(0.134)=4.487\).
Which is which. A glass electrode responds to hydrogen-ion activity, so a predicted meter reading must come from \(\text{p}K_a^{\text{mix}}=4.62\) paired with concentration ratios. A calculation of how much \(\text{CH}_3\text{COO}^-\) is actually present — for a kinetics rate law, a solubility balance, or a mass balance — needs concentrations on both sides and therefore \(\text{p}K_a^{c}=4.49\), paired with \([\text{H}^+]\) rather than \(a_{\text{H}^+}\). Mixing the two conventions is the single commonest source of a systematic \(0.13\)-unit discrepancy in this system.
Caveat: \(I=0.500\ \text{mol\,L}^{-1}\) is at the outer limit of the Davies equation, which was fitted to data up to about this ionic strength; the numbers above are good to perhaps \(\pm0.05\).
One litre of acetate buffer is \(0.100\ \text{mol\,L}^{-1}\) in acetic acid and \(0.100\ \text{mol\,L}^{-1}\) in sodium acetate, with \(\text{p}K_a^{\text{mix}}=4.649\) as established in Worked example 1. How many millilitres of \(1.00\ \text{mol\,L}^{-1}\) \(\text{HCl}\) can be added before the pH falls to \(4.00\)? Compare with the estimate obtained by treating the buffer capacity as constant, and explain the discrepancy. Neglect dilution.
Solution
Exact route. At the endpoint, \(\log_{10}(C_{\text{A}}/C_{\text{HA}})=4.000-4.649=-0.649\), so \(r=10^{-0.649}=0.2244\). With \(C_T=0.200\ \text{mol\,L}^{-1}\) conserved (strong acid interconverts the two forms but does not destroy total acetate),
\(C_{\text{A}}=C_T\dfrac{r}{1+r}=0.200\left(\dfrac{0.2244}{1.2244}\right)=0.03665\ \text{mol\,L}^{-1}\).
Each mole of \(\text{HCl}\) consumes one mole of acetate, so \(n(\text{HCl})=0.1000-0.03665=0.0633\ \text{mol}\), requiring \(V=0.0633/1.00=63.3\ \text{mL}\) of \(1.00\ \text{mol\,L}^{-1}\) acid.
Constant-capacity estimate. From Worked example 1, \(\beta=0.115\ \text{mol\,L}^{-1}\,\text{pH}^{-1}\) at the starting point. Treating it as constant, \(n\approx\beta\,|\Delta\text{pH}|=(0.115)(0.649)=0.0747\ \text{mol}\), i.e. \(74.7\ \text{mL}\).
Discrepancy. The linear estimate overshoots by \(18\%\). The reason is that \(\beta\) is a derivative evaluated at \(\text{pH}=\text{p}K_a\), which is precisely where it is maximal; as acid is added the buffer moves off its optimum and \(\beta\) falls. At the endpoint \(f_{\text{A}}=0.183\) and \(f_{\text{HA}}=0.817\), giving \(\beta=2.3026(0.200)(0.183)(0.817)=0.0688\ \text{mol\,L}^{-1}\,\text{pH}^{-1}\), only \(60\%\) of the initial value. The exact answer is the integral \(\int\beta\,d\,\text{pH}\) over the interval, which necessarily lies between \(0.0688(0.649)=0.0447\) and \(0.115(0.649)=0.0747\ \text{mol}\); the true \(0.0633\ \text{mol}\) does.
Two neglected effects. Adding \(63.3\ \text{mL}\) dilutes the litre by \(6\%\), lowering \(I\) from \(0.100\) to about \(0.094\ \text{mol\,L}^{-1}\) and raising \(\text{p}K_a^{\text{mix}}\) by \(0.002\) — negligible. Ionic strength is otherwise unchanged, since each \(\text{Cl}^-\) added replaces an acetate consumed.
Derive the open-system form of the equation for the \(\text{CO}_2/\text{HCO}_3^-\) pair in blood, and use it. Take the physiological constants \(\text{p}K_{a1}'=6.10\) and Henry solubility \(\alpha=0.0301\ \text{mmol\,L}^{-1}\,\text{mmHg}^{-1}\), both referred to \(37^\circ\text{C}\) and \(I=0.15\ \text{mol\,L}^{-1}\). (a) Show that arterial blood at \(\text{pH}=7.40\) with \(p_{\text{CO}_2}=40\ \text{mmHg}\) has \([\text{HCO}_3^-]=24\ \text{mmol\,L}^{-1}\). (b) A patient presents with \([\text{HCO}_3^-]=12\ \text{mmol\,L}^{-1}\) and \(p_{\text{CO}_2}=26\ \text{mmHg}\); find the pH. (c) Find the \(p_{\text{CO}_2}\) that would restore pH \(7.40\) at that bicarbonate, and comment. (d) Explain why this buffer is far more effective than a closed buffer of the same \(\text{p}K_a\) and concentration, even though \(\text{p}K_{a1}'=6.10\) is \(1.3\) units from the target pH.
Solution
Derivation. The relevant equilibrium is \(\text{CO}_2(aq)+\text{H}_2\text{O}\rightleftharpoons\text{H}^++\text{HCO}_3^-\), whose composite constant \(K_{a1}'\) lumps the hydration step with the dissociation of \(\text{H}_2\text{CO}_3\). Closure (Hypotheses) fails: \([\text{CO}_2(aq)]\) is not set by mass balance but by Henry’s law against the alveolar gas, \([\text{CO}_2(aq)]=\alpha p_{\text{CO}_2}\). Substituting this in place of \(C_{\text{HA}}\) in Step 7 and dropping \(\delta\) (utterly negligible at \(10^{-7}\) against millimolar concentrations):
\(\text{pH}=\text{p}K_{a1}'+\log_{10}\dfrac{[\text{HCO}_3^-]}{\alpha\,p_{\text{CO}_2}}\).
(a) Rearranged for the unknown: \([\text{HCO}_3^-]=\alpha p_{\text{CO}_2}\,10^{\,\text{pH}-\text{p}K_{a1}'}=(0.0301)(40)\,10^{1.30}=(1.204)(19.95)=24.0\ \text{mmol\,L}^{-1}\), the standard reference value.
(b) \(\alpha p_{\text{CO}_2}=(0.0301)(26)=0.7826\ \text{mmol\,L}^{-1}\); ratio \(=12/0.7826=15.33\); \(\log_{10}(15.33)=1.186\); \(\text{pH}=6.10+1.186=7.29\). A metabolic acidosis (bicarbonate halved) that is partly compensated: the low \(p_{\text{CO}_2}\) shows the patient is hyperventilating, which has held the pH at \(7.29\) instead of the \(7.10\) that \(p_{\text{CO}_2}=40\ \text{mmHg}\) would have given — check: \(6.10+\log_{10}(12/1.204)=6.10+0.999=7.10\).
(c) Setting \(\text{pH}=7.40\): \(12/(0.0301\,p_{\text{CO}_2})=10^{1.30}=19.95\), so \(0.0301\,p_{\text{CO}_2}=0.6015\) and \(p_{\text{CO}_2}=20.0\ \text{mmHg}\). Full normalisation would require driving \(p_{\text{CO}_2}\) from the already low \(26\ \text{mmHg}\) down to \(20\ \text{mmHg}\) — half the normal \(40\ \text{mmHg}\), and a further \(23\%\) below what the patient is already sustaining, i.e. near-maximal hyperventilation held indefinitely. Respiratory compensation is in practice never complete, and the empirical clinical expectation for a metabolic acidosis (Winter’s formula, \(p_{\text{CO}_2}\approx1.5[\text{HCO}_3^-]+8\)) predicts \(1.5(12)+8=26\ \text{mmHg}\) — exactly what this patient shows, so the compensation is appropriate and the residual acidaemia is expected.
(d) Three reasons, all traceable to the open denominator. First, the closed-system capacity formula of Step 10 does not apply: because \(\alpha p_{\text{CO}_2}\) is held constant by ventilation, added acid converts \(\text{HCO}_3^-\) to \(\text{CO}_2\) which is then exhaled, so the denominator does not rise as the numerator falls. Differentiating the open-system relation at fixed \(p_{\text{CO}_2}\) gives \(\beta_{\text{open}}=\ln(10)[\text{HCO}_3^-]=2.303(24.0)=55\ \text{mmol\,L}^{-1}\,\text{pH}^{-1}\), against \(\beta_{\text{closed}}=\ln(10)C_Tf_{\text{HA}}f_{\text{A}}=2.303(25.2)(0.0455)=2.6\) for the same solution sealed — a factor \(C_T/[\text{CO}_2]=1+10^{\,\text{pH}-\text{p}K_{a1}'}\approx21\). Only at \(\text{pH}=\text{p}K_a\), where the two terms would move equally, does that factor fall to \(2\). Second, \(p_{\text{CO}_2}\) is itself actively regulated, so the system is not merely open but feedback-controlled, and the effective capacity is set by the gain of that loop rather than by \(C_T\). Third, \(\beta\) for the closed form at \(1.3\) units from \(\text{p}K_a\) would be only \(\ln(10)C_Tf_{\text{HA}}f_{\text{A}}\) with \(f_{\text{HA}}f_{\text{A}}=0.045\), about \(18\%\) of maximum — which is exactly why a closed bicarbonate buffer at pH \(7.4\) would be a poor choice, and why the physiological system works only because it is open.