Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation , also called the buffer equation , describes the relationship between the pH value and the position of the equilibrium of an acid-base reaction between a medium-strength acid and its corresponding medium-strength base in dilute (≤ 1 mol / l) aqueous Solutions.

It goes back to Lawrence J. Henderson and Karl Albert Hasselbalch . Henderson developed his equation named after him in 1908. Hasselbalch was able to confirm the Henderson equation experimentally in human blood and rewrote the equation in 1916 in order to calculate the pH value instead of the hydrogen ion concentration. Mistakenly, the equation, often in literature, as Henderson Hassel bach called-equation.

This equation is used in particular when calculating the pH value of buffer solutions and describes part of the course of acid-base titration curves for medium-strength acids or medium-strength bases. It is derived from a general acid-base reaction:

${\displaystyle \mathrm {HA+H_{2}O\rightleftharpoons \ A^{-}+H_{3}O^{+}} }$;

here HA is a general acid and A ^- its corresponding base. The law of mass action can be formulated for this reaction equation using the acid constant of HA. After taking the logarithm, the value can be converted using simple transformations ${\displaystyle K_{\mathrm {S} }=K\cdot c(\mathrm {H_{2}O} )}$ ${\displaystyle \mathrm {p} K_{\mathrm {S} }}$

${\displaystyle \mathrm {p} K_{\mathrm {S} }=-\log _{10}\left(K_{\mathrm {S} }\cdot \mathrm {\frac {l}{mol}} \right)}$

and the pH

${\displaystyle \mathrm {pH} =-\log _{10}\left[c\left(\mathrm {{H}_{3}O^{+}} \right)\,\cdot \mathrm {\frac {l}{mol}} \right]}$

into the equation, which gives the Henderson-Hasselbalch equation. There are two equivalent versions of this, which can be converted into one another using a calculation rule that applies to logarithms:

• ${\displaystyle \mathrm {pH} =\mathrm {p} K_{\mathrm {S} }+\log _{10}{\frac {c\left(\mathrm {A^{-}} \right)}{c\left(\mathrm {HA} \right)}}}$
• ${\displaystyle \mathrm {pH} =\mathrm {p} K_{\mathrm {S} }-\log _{10}{\frac {c\left(\mathrm {HA} \right)}{c\left(\mathrm {A^{-}} \right)}}}$

Mathematical transformations to derive the Henderson-Hasselbalch equation

The following form of the law of mass action is suitable for deriving the Henderson-Hasselbalch equation from the above equation of a general acid-base reaction: ${\displaystyle K={\frac {c(\mathrm {{H}_{3}O^{+}} )\cdot c(\mathrm {A^{-}} )}{c(\mathrm {H_{2}O} )\cdot c(\mathrm {HA} )}};}$ here is the equilibrium constant . - Multiplying the equation by and inserting gives: ${\displaystyle K}$ ${\displaystyle c(\mathrm {H_{2}O} )\cdot \mathrm {\frac {l}{mol}} }$ ${\displaystyle K_{\mathrm {S} }=K\cdot c(\mathrm {H_{2}O} )}$ ${\displaystyle K_{\mathrm {S} }\cdot \mathrm {\frac {l}{mol}} ={\frac {c(\mathrm {{H}_{3}O^{+}} )\cdot c(\mathrm {A^{-}} )}{c(\mathrm {HA} )}}\cdot \mathrm {\frac {l}{mol}} ;}$ ${\displaystyle K_{\mathrm {S} }\cdot \mathrm {\frac {l}{mol}} =c(\mathrm {{H}_{3}O^{+}} )\cdot \mathrm {\frac {l}{mol}} \cdot {\frac {c(\mathrm {A^{-}} )}{c(\mathrm {HA} )}};}$ Logarithmizing to base 10, application of the calculation rule for the logarithm of products : ${\displaystyle \log _{10}\left(K_{S}\cdot \mathrm {\frac {l}{mol}} \right)=\log _{10}\left(c(\mathrm {{H}_{3}O^{+}} )\cdot \mathrm {\frac {l}{mol}} \right)+\log _{10}\left({\frac {c(\mathrm {A^{-}} )}{c(\mathrm {HA} )}}\right)}$ Subtract the left side and the left addend of the right side: ${\displaystyle -\log _{10}\left(c(\mathrm {{H}_{3}O^{+}} )\cdot \mathrm {\frac {l}{mol}} \right)=-\log _{10}\left(K_{S}\cdot \mathrm {\frac {l}{mol}} \right)+\log _{10}\left({\frac {c(\mathrm {A^{-}} )}{c(\mathrm {HA} )}}\right)}$ Application of the definitions of and given in the text : ${\displaystyle \mathrm {pH} }$ ${\displaystyle \mathrm {pK_{s}} }$ ${\displaystyle \mathrm {pH} =\mathrm {pK_{s}} +\log _{10}{\frac {c\left(\mathrm {A^{-}} \right)}{c\left(\mathrm {HA} \right)}};}$ with the calculation rule for the logarithm of quotients : ${\displaystyle +\log _{10}{\frac {c\left(\mathrm {A^{-}} \right)}{c\left(\mathrm {HA} \right)}}\quad =\log _{10}c\left(\mathrm {A^{-}} \right)-\log _{10}c\left(\mathrm {HA} \right)\quad =-[\log _{10}c\left(\mathrm {HA} \right)-\log _{10}c\left(\mathrm {A^{-}} \right)]\quad =-\log _{10}{\frac {c\left(\mathrm {HA} \right)}{c\left(\mathrm {A^{-}} \right)}}}$ used: ${\displaystyle \mathrm {pH} =\mathrm {pK_{s}} -\log _{10}{\frac {c\left(\mathrm {HA} \right)}{c\left(\mathrm {A^{-}} \right)}}.}$

In the buffer area of ​​the acid-base titration the ratio corresponds to the ratio , so that one can write: ${\displaystyle {\frac {c\left(\mathrm {A^{-}} \right)}{c\left(\mathrm {HA} \right)}}}$ ${\displaystyle {\frac {\tau }{1-\tau }}}$

${\displaystyle \mathrm {pH} =\mathrm {p} K_{\mathrm {S} }+\log _{10}{\frac {\tau }{1-\tau }}}$.

${\displaystyle \tau }$(the degree of titration ) is the ratio of the amount of substance (or the concentration) of the added standard solution to the amount of substance (or the concentration) of the substance to be determined.

In the range of or with greater dilution (less than 0.01 M), calculations with this formula become increasingly inaccurate, because then the low protolysis of HA or A ^- with the solvent used or the autoprotolysis of the water around the pH value 7 (e.g. for the phosphate buffer) would have to be included in the calculation of the concentration; if these concentrations are not taken into account, deviations of up to 0.4 pH units from the calculated value can occur. ${\displaystyle \tau \approx 0}$ ${\displaystyle \tau \approx 1}$

For an exact calculation of such pH values, the required equations are derived from the law of mass action for the components involved, with activities being to be calculated precisely not with concentrations, but with activities .

In order to theoretically derive the above-mentioned partial course of acid-base titration curves from the Henderson-Hasselbalch equation, its notation with the degree of titration can be understood as a function and examined using the means of the curve discussion . The defined arguments are in the open interval . This mathematical investigation gives: ${\displaystyle \tau }$ ${\displaystyle \mathrm {pH} =f(\tau )}$ ${\displaystyle f(\tau )}$ ${\displaystyle \tau }$ ${\displaystyle (0,1)}$

• ${\displaystyle f(\tau )}$everywhere is strictly increasing monotonously . (1)
• ${\displaystyle f(\tau )}$has exactly one turning point where the slope of the graph of is minimal, so the when changing from varies least. (2) ${\displaystyle W(\textstyle \tau ={\frac {1}{2}}|\mathrm {pH} =\mathrm {p} K_{\mathrm {S} })}$ ${\displaystyle f(\tau )}$ ${\displaystyle \mathrm {pH} }$ ${\displaystyle \tau }$
• ${\displaystyle f(\tau )}$is point symmetrical too . (3) ${\displaystyle W}$

The turning point turns out to be the half-equivalence point of the alkalimetric titration of a medium-strength acid. In a suitable environment of (i.e. in one in which the effects mentioned above for or can be neglected) the curve shape that can be theoretically derived from the Henderson-Hasselbalch equation agrees well with that found empirically. In particular, the statement (2) means that in the partial course of the titration considered, the buffering is strongest. ${\displaystyle W}$ ${\displaystyle W}$ ${\displaystyle \tau \approx 0}$ ${\displaystyle \tau \approx 1}$ ${\displaystyle \mathrm {pH} \approx \mathrm {p} K_{\mathrm {S} }}$

Proof of assertions (1), (2), (3) by means of the curve discussion

With the faster notation of the value (constant for a given titration) and the base conversion , in which the natural logarithm is, the representation follows: ${\displaystyle c=:\mathrm {p} K_{\mathrm {S} }}$ ${\displaystyle \mathrm {p} K_{\mathrm {S} }}$ ${\displaystyle \textstyle \log _{10}(x)={\frac {\ln(x)}{\ln(10)}}}$ ${\displaystyle \ln }$ ${\displaystyle \mathrm {pH} =\quad f(\tau )=\quad \mathrm {p} K_{\mathrm {S} }+\log _{10}({\frac {\tau }{1-\tau }})=\quad c+\log _{10}({\frac {\tau }{1-\tau }})=\quad c+{\frac {1}{\ln(10)}}\cdot \ln({\frac {\tau }{1-\tau }});}$ with the derivative rule for the constant function , derivative , the chain rule and the quotient rule for the inner function is: ${\displaystyle \textstyle \ln '(x)={\frac {1}{x}}}$ ${\displaystyle f'(\tau )=\quad (c+{\frac {1}{\ln(10)}}\cdot \ln({\frac {\tau }{1-\tau }}))'=\quad {\frac {1}{\ln(10)}}\cdot {\frac {1-\tau }{\tau }}\cdot {\frac {1-\tau +\tau }{(1-\tau )^{2}}}=\quad {\frac {1}{\ln(10)}}\cdot {\frac {1}{\tau (1-\tau )}}=\quad {\frac {1}{\ln(10)}}\cdot (\tau -\tau ^{2})^{-1};}$ that the functional term of is positive for all is sufficient for (1). ${\displaystyle {\frac {1}{\ln(10)}}\cdot {\frac {1}{\tau (1-\tau )}}}$ ${\displaystyle f'(\tau )}$ ${\displaystyle \tau \in (0,1)}$ For the proof of (2) it is sufficient to show that at has a global minimum (2 ') . - With the power rule and the chain rule : ${\displaystyle f'(\tau )}$ ${\displaystyle \textstyle \tau ={\frac {1}{2}}}$ ${\displaystyle f''(\tau )=\quad ({\frac {1}{\ln(10)}}\cdot (\tau -\tau ^{2})^{-1})'=\quad {\frac {1}{\ln(10)}}\cdot {\frac {-(1-2\tau )}{(\tau -\tau ^{2})^{2}}}=\quad {\frac {{\frac {1}{\ln(10)}}\cdot (2\tau -1)}{(\tau (1-\tau ))^{2}}};}$ the denominator function of is positive for all (of different and therefore), the numerator function is a strictly monotonically increasing linear function with exactly the zero . So it also has exactly the zero and is negative for smaller ones , positive for larger ones . That is sufficient for (2 '). ${\displaystyle f''(\tau )}$ ${\displaystyle \tau \in (0,1)}$ ${\displaystyle 0}$ ${\displaystyle \textstyle \tau ={\frac {1}{2}}}$ ${\displaystyle f''(\tau )}$ ${\displaystyle \textstyle \tau ={\frac {1}{2}}}$ ${\displaystyle \tau }$ ${\displaystyle \tau }$ The functional value of the turning point is ${\displaystyle \mathrm {pH} ({\frac {1}{2}})=\quad f({\frac {1}{2}})=\quad c+\log _{10}{\frac {\textstyle {\frac {1}{2}}}{\textstyle 1-{\frac {1}{2}}}}=\quad c+\log _{10}(1)=\quad c+0=\quad \mathrm {p} K_{\mathrm {S} }}$; The point symmetry at the point of inflection (claim (3)) follows from the transformation with the reciprocity law for logarithms ${\displaystyle f({\frac {1}{2}}-\tau _{0})-c=\quad c+\log _{10}{\frac {{\frac {1}{2}}-\tau _{0}}{1-({\frac {1}{2}}-\tau _{0})}}-c=\quad \log _{10}{\frac {{\frac {1}{2}}-\tau _{0}}{{\frac {1}{2}}+\tau _{0}}}=\quad -\log _{10}{\frac {{\frac {1}{2}}+\tau _{0}}{{\frac {1}{2}}-\tau _{0}}}=\quad c-(c+\log _{10}{\frac {{\frac {1}{2}}+\tau _{0}}{1-({\frac {1}{2}}+\tau _{0})}})=c-f({\frac {1}{2}}+\tau _{0})}$.

Individual evidence

1. ↑ Entry on Henderson – Hasselbalch equation . In: IUPAC Compendium of Chemical Terminology (the “Gold Book”) . doi : 10.1351 / goldbook.H02781 .
2. ↑ Hans P. Latscha, Uli Kazmaier, Helmut Klein: Chemistry for Pharmacists , p. 157 ff .; ISBN 978-3540427551 .

Web links

• The Henderson-Hasselbalch equation for indicators