Skip to main content
Biology LibreTexts

2.3: Buffering against pH Changes in Biological Systems

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Search Fundamentals of Biochemistry


    As one way to ensure homeostasis, the pH is maintained between 7.35 and 7.45 in humans. (Much lower pH values, around 4.5, are found in the lysosome). Lower pH values are associated with metabolic and respiratory acidosis while higher pH values are characteristic of metabolic and respiratory alkalosis. pH is maintained by buffering systems that consist of a weak acid and base. If you understand the Henderson-Hasselbalch equation from the previous section, buffer systems become easy to understand.

    p H=p K_a+\log \frac{\left[A^{-}\right]}{H A}

    At the inflection point of the curve, pH = pKa and the system is most resistant to changes in pH on the addition of either acid or base. At this pH, [HA]=[A-].

    If a bit of a strong acid is added, it would react with the strongest base in the solution, which would be the conjugate base of the weak acid:

    HCl + A- --> HA + Cl-

    The reaction goes from a strong acid, HCl, to the weak acid, HA. Its concentration would increase a bit but since it's a weak acid, it will only ionize to a small extent. The [HA] in the Henderson-Hasselbalch equations increases a bit but not enough to change the pH significantly. If the same amount of HCl were added to pure water, it would react completely to form an equal amount of H3O+ which would significantly alter the pH of pure water (7.0).

    If a bit of a strong base is added, it would react with the strongest acid in the solution which would be HA:

    HA + OH- --> H2O + A-

    The reaction goes from a strong base to the weak acid A-. Its concentration would increase a bit but since it's a weak base, it won't affect the pH significantly. The [A-] in the Henderson-Hasselbalch equations increases a bit but not enough to change the pH significantly. If the same amount of NaOH were added to pure water, it would react to make the solution basic and significantly alter the pH of pure water (7.0).

    To review, buffer solutions contain a weak acid and its conjugate base. They have maximal buffering capacity at a pH = pKa of the weak acid. In general, a buffered solution is best able to withstand a change in pH only with + 1 pH unit from the pKa.

    Biological Buffering Agents

    The most relevant systems for biology are the carbonic acid/carbonate buffering system, which controls blood pH and cells and the phosphate buffering system. Proteins, which have many weak acid and base functional groups, can also act as buffering agents.

    Carbonic acid/carbonate buffering system: At first glance, the reaction of carbonic acid can be written as follows:

    H2CO3 (aq) + H2O(l) ßà H3O+(aq) + HCO3-(aq) pKa = 3.6

    However, this system is a bit more complex since we must consider CO2 (g) solubility and reactivity as well. The overall chemical reactions look like this, where H2CO3 is the weak oxyacid, carbonic acid and HCO3-(aq) is the weak conjugate based, bicarbonate (or hydrogen carbonate). The [CO2(aq)] >> [H2CO3 (aq)]

    Rx 1: CO2 (g) ßà CO2(aq) + H2O (l) ßà H2CO3 (aq) + H2O(l) ßà H3O+(aq) + HCO3-(aq)

    The respiratory system can quickly adjust pH simply by increasing the exhalation of CO2. The kidneys can respond in a slower fashion to remove H3O+ and retain HCO3-. The carbonic acid/bicarbonate buffering system can help us understand how shifting equilibria caused by excessive CO2 released from rapid deep breathing or decreased CO2 release associated with pulmonary disease or shallow rapid breathing can lead to respiratory alkalosis and acidosis, respectively.

    • Respiratory alkalosis can be caused by “hyperventilation” - breathing rapidly. This would lead to breathing out too much CO2, shifting the above equilibrium to the left, consuming H3O+, and increasing pH, making the blood more alkaline. You could breathe into a bag to increase your CO2 levels.
    • Respiratory acidosis is caused by increased CO2, which can occur when the lungs aren’t working well, and you can’t get rid of CO2 you produce during respiration Respiratory acidosis can happen with asthma, pneumonia, lung disease or anything that decreases respiration rate.

    Inhaling CO2 can lead to panic. This makes sense as it would mimic suffocation which is lethal to humans. A suffocation response follows. High CO2 would drive the equilibrium to the right, leading to H3O+ production. An acid-sensing ion channel-1a (ASIC1a) in the amygdala has been discovered which appears to mediate the effect. Panic attacks are sometimes associated with hyperventilation which leads to alkalosis, not acidosis. Less noted is that when some people panic, they take short shallow breaths (in a way almost stopping their breath). This would lead to a buildup of CO2 since it wouldn’t be released in exhalation. The acid channel in the amygdala would be activated and the panic response ensues.

    A Dilemma?

    How can carbonic acid with a pKa of 3.6 act as a buffer component at pH 7.5?

    An astute student might have picked up this conundrum.

    The solution to this problem involves looking at the full set of reactions for the components of the buffer system.

    Here is the complete set of reactions again:

    CO2 (g) ßà CO2(aq) + H2O (l) ßà H2CO3 (aq) + H2O(l) ßà H3O+(aq) + HCO3-(aq)

    Let's simplify it since there would be no free "gas bubbles" in blood, so CO2 (g) = CO2(aq):

    CO2(aq) + H2O (l) ßà H2CO3 (aq) + H2O(l) ßà H3O+(aq) + HCO3-(aq)

    H2CO3 (aq) participates in two different reactions.

    Rightwards from H2CO3 (aq) :

    H2CO3 (aq) + H2O(l) ßà H3O+(aq) + HCO3-(aq)

    Using the simplified equation with H+ gives

    K_a=\frac{\left[H^{+}\right]\left[\mathrm{HCO}_3^{-}\right]}{\mathrm{H}_2 \mathrm{CO}_3}


    \left[\mathrm{H}_2 \mathrm{CO}_3\right]=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_3^{-}\right]}{K_a}

    Leftwards from H2CO3 (aq) :

    H2CO3 (aq) ßàCO2(aq) + H2O (l)

    K_2=\frac{\left[\mathrm{CO}_2\right]}{\mathrm{H}_2 \mathrm{CO}_3}


    \left[\mathrm{H}_2 \mathrm{CO}_3\right]=\frac{\left[\mathrm{CO}_2\right]}{K_a}

    Setting 2.3.3 and 2.3.5 equal to each other gives:

    \left[\mathrm{H}_2 \mathrm{CO}_3\right]=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_3^{-}\right]}{K_a}=\frac{\left[\mathrm{CO}_2\right]}{K_2}

    Solving for [H+] gives:

    \left[H^{+}\right]=\frac{\left[\mathrm{CO}_2\right]\left(K_a\right)}{\left[H C O_3^{-}\right]\left(K_2\right)}

    Now take the -log of each side to produce an equation similar to the Henderson-Hasselbalch equation.

    &-\log \left[H^{+}\right]=-\log \left(\frac{\left[\mathrm{CO}_2\right]}{\left[\mathrm{HCO}_3^{-}\right.}\right)-\log \left(\frac{K_a}{K_2}\right) \\
    &p H=p K_{a E F F E C T I V E}-\log \left(\frac{\left[\mathrm{CO}_2\right]}{\left[\mathrm{HCO}_3^{-}\right]}\right)


    K_{a E F F E C T I V E}=\frac{K_a}{K_2}

    This Henderson-Hasselbalch-like equation shows the pH is determined by the ratio \(K_a/K_2\) ratio. pKa EFFECTIVE = 6.3. This gives a ratio of \(CO_2/HCO_3^{-}\) of 0.08 = 8/100. There is effectively 12-13 x as much HCO3-(aq) as CO2, making the system primed to react with acid produced metabolically. Yet a second conundrum exists. The pH of the blood (7.4) is outside of the optimal range for a buffer system (in this case + 1 pH unit from the pKa which is 6.3 in this case). Again, the system is primed to react with acid as it would move the pH close to the optimal buffering pH of 6.3. Other biological systems also must be involved in maintaining pH.

    Phosphate buffering system: Phosphates, in the form of dihydrogen (H2PO4-) and monohydrogen phosphate (HPO42-) are also present in the blood. Given the pKa of HPO42-, why is PO43- not present to any significant degree? Since the concentration of phosphates are low in blood, this system is a minor player in blood.

    Proteins: Proteins are found in all cellular and extracellular fluids and they contain weak acids as buffer components. Proteins contain two amino acids, aspartic acid, and glutamic acid) that contain carboxylic acid side chains. Each comprises about 6% of the proteins. In blood, hemoglobin is the most abundant protein by far. Its role in buffering and in O2 and CO2 will be discussed in a subsequent chapter.

    Making Buffers in the Lab

    When studying biomolecules like proteins and nucleic acids in the lab, the pH of the solution is usually maintained under physiological conditions. These macromolecules are either dissolved in or diluted into a buffer solution. Sometimes it's important to study their properties and activities as a function of pH. A wide variety of buffer systems have been developed for the lab study of these molecules. The dihydrogen (H2PO4-)/monohydrogen phosphate (HPO42-) pair are commonly used as the pKa of H2PO4- is 7.21, which makes it physiologically relevant. Care must be taken in selecting buffer systems as some of them might bind calcium ions. The pKa of some weak acids vary significantly with temperature as well. Some common biological buffers are listed below.


    (at 25°C)

    MES 6.10
    Bis-Tris 6.50
    ACES 6.78
    PIPES 6.76
    MOPSO 6.90
    MOPS 7.20
    HEPES 7.48
    Tris 8.06
    Tricine 8.05
    Gly-Gly 8.20
    Bicine 8.26
    TAPS 8.40
    AMPSO 9.00
    CAPS 10.40

    There are 3 general ways to make a buffered solution:

    1. Make us separate equal concentration solutions of both the weak acid (for example Na(H2PO4) and its conjugate base (for example Na2(HPO4). Use the Henderson-Hasselbalch equation to calculate how much of each should be added to give the correct [A-]/[HA] ratio (in the case [HPO42-]/[H2PO4-1]) to give the correct pH.
    2. Use a pH meter and monitor the pH when adding both solutions together until the desired pH is reached.
    3. Make a solution of one of the components (weak acid or its conjugate base) and bring to near its correct volume for the desired molarity. Monitor the pH as you add a concentrated solution of either HCl or NaOH to get the desired pH. Then bring the solution to the correct volume in a volumetric flask. With this method, you will be adding counter ions (Cl- with HCl and Na+ with NaOH) which you may not want in your buffer solution. Often it is not a problem.

    This page titled 2.3: Buffering against pH Changes in Biological Systems is shared under a not declared license and was authored, remixed, and/or curated by Henry Jakubowski and Patricia Flatt.