Henderson–Hasselbalch equation

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In chemistry, the Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity (using pKa, the negative log of the acid dissociation constant) in biological and chemical systems. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid-base reactions (it is widely used to calculate the isoelectric point of proteins).

The equation is given by:

\textrm{pH} = \textrm{pK}_{a}+ \log_{10} \left ( \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )

Here, [HA] is the molar concentration of the undissociated weak acid, [A⁻] is the molar concentration (molarity, M) of this acid's conjugate base and \textrm{pK}_{a} is -\log_{10} (K_{a}) where K_{a} is the acid dissociation constant, that is:

\textrm{pK}_{a} = - \log_{10} (K_{a}) = - \log_{10} \left ( \frac{[\mbox{H}_{3}\mbox{O}^+][\mbox{A}^-]}{[\mbox{HA}]} \right ) for the non-specific Brønsted acid-base reaction: \mbox{HA} + \mbox{H}_{2}\mbox{O} \rightleftharpoons \mbox{A}^- + \mbox{H}_{3}\mbox{O}^+

In these equations, \mbox{A}^- denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed.

For bases[edit]

For the standard base equation:[1]

\textrm{B} + \textrm{H}^{+} \rightleftharpoons \textrm{BH}^{+}

A second form of the equation, known as the Heylman Equation, expressed in terms of K_{b} where K_{b} is the base dissociation constant: \textrm{pK}_{b} = - \log_{10} (K_{b}) = - \log_{10} \left ( \frac{[\mbox{O}\mbox{H}^-][\mbox{HA}]}{[\mbox{A}^-]} \right )

In analogy to the above equations, the following equation is valid:

\textrm{pOH} = \textrm{pK}_{b}+ \log_{10}  \left ( \frac{[\textrm{BH}^+]}{[\textrm{B}]} \right )

Where BH+ denotes the conjugate acid of the corresponding base B. Using the properties of these terms at 25 degrees Celsius one can synthesise an equation for pH of basic solutions in terms of pKa and pH:

\textrm{pH} = \textrm{pK}_{a} + \log_{10} \left(\frac{[\textrm{B}]}{[\textrm{BH}^{+}]}\right)


The Henderson–Hasselbalch equation is derived from the acid dissociation constant equation by the following steps:[2]

K_a = \frac{[\textrm{H}^+][\textrm{A}^-]} {[\textrm{HA}]}

Taking the log, to base ten, of both sides gives:

\log_{10}K_a = \log_{10} \left ( \frac{[\textrm{H}^+][\textrm{A}^-]}{[\textrm{HA}]} \right )

Then, using the properties of logarithms:

\log_{10}K_a = \log_{10}[\textrm{H}^+] + \log_{10} \left ( \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )

Identifying the left-hand side of this equation as -pKa and the \log_{10} [\textrm{H}^{+}] as -pH:

-\textrm{pK}_a = -\textrm{pH} + \log_{10} \left ( \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )

Adding pH and pKa to both sides:

\textrm{pH} = \textrm{pK}_a + \log_{10} \left ( \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )

The ratio [A^-]/[HA] is unitless, and as such, other ratios with other units may be used. For example, the mole ratio of the components, n_{A^-}/n_{HA} or the fractional concentrations \alpha_{A^-}/\alpha_{HA} where \alpha_{A^-}+\alpha_{HA}=1 will yield the same answer. Sometimes these other units are more convenient to use.


Lawrence Joseph Henderson wrote an equation, in 1908, describing the use of carbonic acid as a buffer solution. Karl Albert Hasselbalch later re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation [1]. Hasselbalch was using the formula to study metabolic acidosis.


There are some significant approximations implicit in the Henderson–Hasselbalch equation. The most significant is the assumption that the concentration of the acid and its conjugate base at equilibrium will remain the same as the formal concentration. This neglects the dissociation of the acid and the binding of H+ to the base. The dissociation of water and relative water concentration itself is neglected as well. These approximations will fail when dealing with relatively strong acids or bases (pKa more than a couple units away from 7), dilute or very concentrated solutions (less than 1 mM or greater than 1M), or heavily skewed acid/base ratios (more than 100 to 1). In high buffer dilutions, where the concentration of protons arising from water become equally or more prevalent than the buffer species themselves (at pH 7, this means buffer component concentrations of <10−5 M formally, but practically much higher), the pKa of the 'buffer' system will tend towards neutrality.

Estimating blood pH[edit]

The Henderson–Hasselbalch equation can be applied to relate the pH of blood to constituents of the bicarbonate buffering system:[3]

 \textrm{pH} = \textrm{pK}_{a~\mathrm{H}_2\mathrm{CO}_3}+ \log_{10}  \left ( \frac{[\textrm{HCO}_3^-]}{[\textrm{H}_2\textrm{CO}_3]} \right )

, where:

This is useful in arterial blood gas, but these usually state PaCO2, that is, the partial pressure of carbon dioxide, rather than H2CO3. However, these are related by the equation:[3]

 [\textrm{H}_2\textrm{CO}_3] = k_{\rm H~CO_2}\, \times PaCO_2

, where:

  • [H2CO3] is the concentration of carbonic acid in the blood
  • kH CO2 is the Henry's law constant for the solubility of carbon dioxide in blood. kH CO2 is approximately 0.03 mmol/(L-torr)
  • PaCO2 is the partial pressure of carbon dioxide in the blood

Taken together, the following equation can be used to relate the pH of blood to the concentration of bicarbonate and the partial pressure of carbon dioxide:[3]

 \textrm{pH} = 6.1 + \log_{10}  \left ( \frac{[\textrm{HCO}_3^-]}{0.03 \times PaCO_2} \right )

, where:

  • pH is the acidity in the blood
  • [HCO3] is the concentration of bicarbonate in the blood
  • PaCO2 is the partial pressure of carbon dioxide in the arterial blood

See also[edit]


  1. ^ Larsen, D. "Henderson-Hasselbalch Approximation". Chemwiki. University of California. Retrieved 27 March 2014. 
  2. ^ Henderson Hasselbalch Equation: Derivation of pKa and pKb
  3. ^ a b c page 556, section "Estimating plasma pH" in: Bray, John J. (1999). Lecture notes on human physiolog. Malden, Mass.: Blackwell Science. ISBN 978-0-86542-775-4. 

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