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Why proton concentration is divided by 10⁻⁷?


Unit of the equilibrium constant: contradiction of Bridgman's theorem?Chemical Equilibrium - Why do changes in pressure cause a shift in the ratio of products and reactants?Can you create a buffer with a strong acid?Why are weak acids weak?Which solution has the highest total molarity of ions and why?Ion concentration in acid and baseHow to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation?What is so special about positive hydrogen ions that they are used in pH calculation?What is the correct way of using the Law of Mass Action?Amount concentration versus number concentrationDoes the starting concentration in a isochoric container depend on the pressure?













7












$begingroup$


I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$. Why not just leave it at the antilog(-8.1) over $pu{1 M}$ like the other concentrations? I literally have see nothing else on the internet like this.



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    7












    $begingroup$


    I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$. Why not just leave it at the antilog(-8.1) over $pu{1 M}$ like the other concentrations? I literally have see nothing else on the internet like this.



    Scanned page










    share|improve this question









    New contributor




    user75312 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      7












      7








      7





      $begingroup$


      I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$. Why not just leave it at the antilog(-8.1) over $pu{1 M}$ like the other concentrations? I literally have see nothing else on the internet like this.



      Scanned page










      share|improve this question









      New contributor




      user75312 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$. Why not just leave it at the antilog(-8.1) over $pu{1 M}$ like the other concentrations? I literally have see nothing else on the internet like this.



      Scanned page







      physical-chemistry equilibrium ph concentration free-energy






      share|improve this question









      New contributor




      user75312 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      user75312 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 18 hours ago









      andselisk

      17.4k656117




      17.4k656117






      New contributor




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      Check out our Code of Conduct.









      asked 18 hours ago









      user75312user75312

      412




      412




      New contributor




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      Check out our Code of Conduct.





      New contributor





      user75312 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          14












          $begingroup$

          The textbook is precisely correct.
          The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
          In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
          One must be very fastidious with units when finding the equilibrium constant.
          For example, the reaction



          $$ce{aA + bB <=> cC + dD}$$



          equilibrium constant $K_c$ is exactly



          $$K_c = frac{([ce{C}]/c^⦵)^ccdot ([ce{D}]/c^⦵)^d}{([ce{A}]/c^⦵)^acdot ([ce{B}]/c^⦵)^b}$$



          For pure water in its standard state $c^⦵ = [ce{H+}] = pu{1e-7 M}$.
          It also correlates with so-called biological standard state of $mathrm{pH} = 7$.
          You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
          In this case those cannot be cancelled out, and must be written explicitly.





          In fact, your own textbook contains extensive explanation [1, p. 91]:




          For chemical reactions the standard state for solutes is defined as $pu{1 M}$; however, in living cells the concentration of $[ce{H+}]$ is roughly $10^{-7}~mathrm M$, much lower than the standard value of $pu{1 M}$.
          It is therefore appropriate to define the reference concentration of $ce{H+}$ in biochemical reactions relative to the $ce{H+}$ concentration found in the living state (i.e., $10^{-7}~mathrm M$), rather than the value $pu{1 M}$ defined by the chemical standard state.
          Recall that when a solute in a dilute solution has a concentration of $pu{1 M}$, the activity of that solute is unity.
          For the biochemical standard state we define the activity of $ce{H+}$ to be unity when $[ce{H+}] = 10^{-7}~mathrm M$.



          [...]




          1. The mass action expression $Q$ is unitless.
            We strip the units from each concentration term in $Q$ by dividing each by its
            proper standard concentration (e.g., $pu{1 M}$ for all solutes

            except $ce{H+}$; $10^{-7}~mathrm M$ for $ce{H+}$; $pu{1 bar}$ for gases, etc.).




          Refrences




          1. Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.






          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
            $endgroup$
            – Karsten Theis
            11 hours ago










          • $begingroup$
            Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
            $endgroup$
            – user75312
            3 hours ago










          • $begingroup$
            Nevermind, it is on page 59 of the second edition for those with the second edition.
            $endgroup$
            – user75312
            3 hours ago



















          4












          $begingroup$

          For an explanation you may want to inspect Recommendations for Biochemical Equilibrium Data 1, which states:




          Buffer and pH. If only a limited number of measurements are to be
          made, they should be carried out at pH = 7.0 and, if possible, also
          at a pH value at which the apparent equilibrium constant $K_c^prime$, has
          little or no dependence on pH. ($K_c^prime$ is defined in a later
          section.) If direct measurements at pH = 7.0 are not practicable, the
          calculated values for this pH should be reported.
          The procedure used
          in making these calculations must be carefully described. Care should
          be taken that the solution is adequately buffered so that the pH is
          well defined throughout the experiment. It is desirable to determine
          the effect of varying the nature and concentration of the buffer in
          order to identify buffer effects. Buffers that are known to interact
          with reactants (including macromolecules) or salts, such as phosphate
          or pyrophosphate in the presence of divalent metal ions, should be
          avoided.




          The highlighted portion means that reported values for biochemical reactions are (or should be) referenced to pH 7.0. The data in Table 3.6 refer to this biochemical standard state.



          Another question on the subject of equilibrium constants also addresses the importance of properly considering reference states.



          Reference



          1 The Journal of Biological Chemistry (1976), Vol. 261, No. 22, pp. 6859-6885.






          share|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            14












            $begingroup$

            The textbook is precisely correct.
            The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
            In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
            One must be very fastidious with units when finding the equilibrium constant.
            For example, the reaction



            $$ce{aA + bB <=> cC + dD}$$



            equilibrium constant $K_c$ is exactly



            $$K_c = frac{([ce{C}]/c^⦵)^ccdot ([ce{D}]/c^⦵)^d}{([ce{A}]/c^⦵)^acdot ([ce{B}]/c^⦵)^b}$$



            For pure water in its standard state $c^⦵ = [ce{H+}] = pu{1e-7 M}$.
            It also correlates with so-called biological standard state of $mathrm{pH} = 7$.
            You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
            In this case those cannot be cancelled out, and must be written explicitly.





            In fact, your own textbook contains extensive explanation [1, p. 91]:




            For chemical reactions the standard state for solutes is defined as $pu{1 M}$; however, in living cells the concentration of $[ce{H+}]$ is roughly $10^{-7}~mathrm M$, much lower than the standard value of $pu{1 M}$.
            It is therefore appropriate to define the reference concentration of $ce{H+}$ in biochemical reactions relative to the $ce{H+}$ concentration found in the living state (i.e., $10^{-7}~mathrm M$), rather than the value $pu{1 M}$ defined by the chemical standard state.
            Recall that when a solute in a dilute solution has a concentration of $pu{1 M}$, the activity of that solute is unity.
            For the biochemical standard state we define the activity of $ce{H+}$ to be unity when $[ce{H+}] = 10^{-7}~mathrm M$.



            [...]




            1. The mass action expression $Q$ is unitless.
              We strip the units from each concentration term in $Q$ by dividing each by its
              proper standard concentration (e.g., $pu{1 M}$ for all solutes

              except $ce{H+}$; $10^{-7}~mathrm M$ for $ce{H+}$; $pu{1 bar}$ for gases, etc.).




            Refrences




            1. Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.






            share|improve this answer











            $endgroup$









            • 3




              $begingroup$
              There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
              $endgroup$
              – Karsten Theis
              11 hours ago










            • $begingroup$
              Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
              $endgroup$
              – user75312
              3 hours ago










            • $begingroup$
              Nevermind, it is on page 59 of the second edition for those with the second edition.
              $endgroup$
              – user75312
              3 hours ago
















            14












            $begingroup$

            The textbook is precisely correct.
            The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
            In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
            One must be very fastidious with units when finding the equilibrium constant.
            For example, the reaction



            $$ce{aA + bB <=> cC + dD}$$



            equilibrium constant $K_c$ is exactly



            $$K_c = frac{([ce{C}]/c^⦵)^ccdot ([ce{D}]/c^⦵)^d}{([ce{A}]/c^⦵)^acdot ([ce{B}]/c^⦵)^b}$$



            For pure water in its standard state $c^⦵ = [ce{H+}] = pu{1e-7 M}$.
            It also correlates with so-called biological standard state of $mathrm{pH} = 7$.
            You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
            In this case those cannot be cancelled out, and must be written explicitly.





            In fact, your own textbook contains extensive explanation [1, p. 91]:




            For chemical reactions the standard state for solutes is defined as $pu{1 M}$; however, in living cells the concentration of $[ce{H+}]$ is roughly $10^{-7}~mathrm M$, much lower than the standard value of $pu{1 M}$.
            It is therefore appropriate to define the reference concentration of $ce{H+}$ in biochemical reactions relative to the $ce{H+}$ concentration found in the living state (i.e., $10^{-7}~mathrm M$), rather than the value $pu{1 M}$ defined by the chemical standard state.
            Recall that when a solute in a dilute solution has a concentration of $pu{1 M}$, the activity of that solute is unity.
            For the biochemical standard state we define the activity of $ce{H+}$ to be unity when $[ce{H+}] = 10^{-7}~mathrm M$.



            [...]




            1. The mass action expression $Q$ is unitless.
              We strip the units from each concentration term in $Q$ by dividing each by its
              proper standard concentration (e.g., $pu{1 M}$ for all solutes

              except $ce{H+}$; $10^{-7}~mathrm M$ for $ce{H+}$; $pu{1 bar}$ for gases, etc.).




            Refrences




            1. Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.






            share|improve this answer











            $endgroup$









            • 3




              $begingroup$
              There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
              $endgroup$
              – Karsten Theis
              11 hours ago










            • $begingroup$
              Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
              $endgroup$
              – user75312
              3 hours ago










            • $begingroup$
              Nevermind, it is on page 59 of the second edition for those with the second edition.
              $endgroup$
              – user75312
              3 hours ago














            14












            14








            14





            $begingroup$

            The textbook is precisely correct.
            The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
            In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
            One must be very fastidious with units when finding the equilibrium constant.
            For example, the reaction



            $$ce{aA + bB <=> cC + dD}$$



            equilibrium constant $K_c$ is exactly



            $$K_c = frac{([ce{C}]/c^⦵)^ccdot ([ce{D}]/c^⦵)^d}{([ce{A}]/c^⦵)^acdot ([ce{B}]/c^⦵)^b}$$



            For pure water in its standard state $c^⦵ = [ce{H+}] = pu{1e-7 M}$.
            It also correlates with so-called biological standard state of $mathrm{pH} = 7$.
            You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
            In this case those cannot be cancelled out, and must be written explicitly.





            In fact, your own textbook contains extensive explanation [1, p. 91]:




            For chemical reactions the standard state for solutes is defined as $pu{1 M}$; however, in living cells the concentration of $[ce{H+}]$ is roughly $10^{-7}~mathrm M$, much lower than the standard value of $pu{1 M}$.
            It is therefore appropriate to define the reference concentration of $ce{H+}$ in biochemical reactions relative to the $ce{H+}$ concentration found in the living state (i.e., $10^{-7}~mathrm M$), rather than the value $pu{1 M}$ defined by the chemical standard state.
            Recall that when a solute in a dilute solution has a concentration of $pu{1 M}$, the activity of that solute is unity.
            For the biochemical standard state we define the activity of $ce{H+}$ to be unity when $[ce{H+}] = 10^{-7}~mathrm M$.



            [...]




            1. The mass action expression $Q$ is unitless.
              We strip the units from each concentration term in $Q$ by dividing each by its
              proper standard concentration (e.g., $pu{1 M}$ for all solutes

              except $ce{H+}$; $10^{-7}~mathrm M$ for $ce{H+}$; $pu{1 bar}$ for gases, etc.).




            Refrences




            1. Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.






            share|improve this answer











            $endgroup$



            The textbook is precisely correct.
            The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
            In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
            One must be very fastidious with units when finding the equilibrium constant.
            For example, the reaction



            $$ce{aA + bB <=> cC + dD}$$



            equilibrium constant $K_c$ is exactly



            $$K_c = frac{([ce{C}]/c^⦵)^ccdot ([ce{D}]/c^⦵)^d}{([ce{A}]/c^⦵)^acdot ([ce{B}]/c^⦵)^b}$$



            For pure water in its standard state $c^⦵ = [ce{H+}] = pu{1e-7 M}$.
            It also correlates with so-called biological standard state of $mathrm{pH} = 7$.
            You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
            In this case those cannot be cancelled out, and must be written explicitly.





            In fact, your own textbook contains extensive explanation [1, p. 91]:




            For chemical reactions the standard state for solutes is defined as $pu{1 M}$; however, in living cells the concentration of $[ce{H+}]$ is roughly $10^{-7}~mathrm M$, much lower than the standard value of $pu{1 M}$.
            It is therefore appropriate to define the reference concentration of $ce{H+}$ in biochemical reactions relative to the $ce{H+}$ concentration found in the living state (i.e., $10^{-7}~mathrm M$), rather than the value $pu{1 M}$ defined by the chemical standard state.
            Recall that when a solute in a dilute solution has a concentration of $pu{1 M}$, the activity of that solute is unity.
            For the biochemical standard state we define the activity of $ce{H+}$ to be unity when $[ce{H+}] = 10^{-7}~mathrm M$.



            [...]




            1. The mass action expression $Q$ is unitless.
              We strip the units from each concentration term in $Q$ by dividing each by its
              proper standard concentration (e.g., $pu{1 M}$ for all solutes

              except $ce{H+}$; $10^{-7}~mathrm M$ for $ce{H+}$; $pu{1 bar}$ for gases, etc.).




            Refrences




            1. Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 8 hours ago

























            answered 18 hours ago









            andseliskandselisk

            17.4k656117




            17.4k656117








            • 3




              $begingroup$
              There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
              $endgroup$
              – Karsten Theis
              11 hours ago










            • $begingroup$
              Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
              $endgroup$
              – user75312
              3 hours ago










            • $begingroup$
              Nevermind, it is on page 59 of the second edition for those with the second edition.
              $endgroup$
              – user75312
              3 hours ago














            • 3




              $begingroup$
              There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
              $endgroup$
              – Karsten Theis
              11 hours ago










            • $begingroup$
              Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
              $endgroup$
              – user75312
              3 hours ago










            • $begingroup$
              Nevermind, it is on page 59 of the second edition for those with the second edition.
              $endgroup$
              – user75312
              3 hours ago








            3




            3




            $begingroup$
            There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
            $endgroup$
            – Karsten Theis
            11 hours ago




            $begingroup$
            There are two standard states for [H+]. In the absence of a prime after the $^circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.
            $endgroup$
            – Karsten Theis
            11 hours ago












            $begingroup$
            Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
            $endgroup$
            – user75312
            3 hours ago




            $begingroup$
            Thanks, I have the second edition and this explanation is not on this page 91. Is it from Chapter 3? If so I will find it.
            $endgroup$
            – user75312
            3 hours ago












            $begingroup$
            Nevermind, it is on page 59 of the second edition for those with the second edition.
            $endgroup$
            – user75312
            3 hours ago




            $begingroup$
            Nevermind, it is on page 59 of the second edition for those with the second edition.
            $endgroup$
            – user75312
            3 hours ago











            4












            $begingroup$

            For an explanation you may want to inspect Recommendations for Biochemical Equilibrium Data 1, which states:




            Buffer and pH. If only a limited number of measurements are to be
            made, they should be carried out at pH = 7.0 and, if possible, also
            at a pH value at which the apparent equilibrium constant $K_c^prime$, has
            little or no dependence on pH. ($K_c^prime$ is defined in a later
            section.) If direct measurements at pH = 7.0 are not practicable, the
            calculated values for this pH should be reported.
            The procedure used
            in making these calculations must be carefully described. Care should
            be taken that the solution is adequately buffered so that the pH is
            well defined throughout the experiment. It is desirable to determine
            the effect of varying the nature and concentration of the buffer in
            order to identify buffer effects. Buffers that are known to interact
            with reactants (including macromolecules) or salts, such as phosphate
            or pyrophosphate in the presence of divalent metal ions, should be
            avoided.




            The highlighted portion means that reported values for biochemical reactions are (or should be) referenced to pH 7.0. The data in Table 3.6 refer to this biochemical standard state.



            Another question on the subject of equilibrium constants also addresses the importance of properly considering reference states.



            Reference



            1 The Journal of Biological Chemistry (1976), Vol. 261, No. 22, pp. 6859-6885.






            share|improve this answer











            $endgroup$


















              4












              $begingroup$

              For an explanation you may want to inspect Recommendations for Biochemical Equilibrium Data 1, which states:




              Buffer and pH. If only a limited number of measurements are to be
              made, they should be carried out at pH = 7.0 and, if possible, also
              at a pH value at which the apparent equilibrium constant $K_c^prime$, has
              little or no dependence on pH. ($K_c^prime$ is defined in a later
              section.) If direct measurements at pH = 7.0 are not practicable, the
              calculated values for this pH should be reported.
              The procedure used
              in making these calculations must be carefully described. Care should
              be taken that the solution is adequately buffered so that the pH is
              well defined throughout the experiment. It is desirable to determine
              the effect of varying the nature and concentration of the buffer in
              order to identify buffer effects. Buffers that are known to interact
              with reactants (including macromolecules) or salts, such as phosphate
              or pyrophosphate in the presence of divalent metal ions, should be
              avoided.




              The highlighted portion means that reported values for biochemical reactions are (or should be) referenced to pH 7.0. The data in Table 3.6 refer to this biochemical standard state.



              Another question on the subject of equilibrium constants also addresses the importance of properly considering reference states.



              Reference



              1 The Journal of Biological Chemistry (1976), Vol. 261, No. 22, pp. 6859-6885.






              share|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                For an explanation you may want to inspect Recommendations for Biochemical Equilibrium Data 1, which states:




                Buffer and pH. If only a limited number of measurements are to be
                made, they should be carried out at pH = 7.0 and, if possible, also
                at a pH value at which the apparent equilibrium constant $K_c^prime$, has
                little or no dependence on pH. ($K_c^prime$ is defined in a later
                section.) If direct measurements at pH = 7.0 are not practicable, the
                calculated values for this pH should be reported.
                The procedure used
                in making these calculations must be carefully described. Care should
                be taken that the solution is adequately buffered so that the pH is
                well defined throughout the experiment. It is desirable to determine
                the effect of varying the nature and concentration of the buffer in
                order to identify buffer effects. Buffers that are known to interact
                with reactants (including macromolecules) or salts, such as phosphate
                or pyrophosphate in the presence of divalent metal ions, should be
                avoided.




                The highlighted portion means that reported values for biochemical reactions are (or should be) referenced to pH 7.0. The data in Table 3.6 refer to this biochemical standard state.



                Another question on the subject of equilibrium constants also addresses the importance of properly considering reference states.



                Reference



                1 The Journal of Biological Chemistry (1976), Vol. 261, No. 22, pp. 6859-6885.






                share|improve this answer











                $endgroup$



                For an explanation you may want to inspect Recommendations for Biochemical Equilibrium Data 1, which states:




                Buffer and pH. If only a limited number of measurements are to be
                made, they should be carried out at pH = 7.0 and, if possible, also
                at a pH value at which the apparent equilibrium constant $K_c^prime$, has
                little or no dependence on pH. ($K_c^prime$ is defined in a later
                section.) If direct measurements at pH = 7.0 are not practicable, the
                calculated values for this pH should be reported.
                The procedure used
                in making these calculations must be carefully described. Care should
                be taken that the solution is adequately buffered so that the pH is
                well defined throughout the experiment. It is desirable to determine
                the effect of varying the nature and concentration of the buffer in
                order to identify buffer effects. Buffers that are known to interact
                with reactants (including macromolecules) or salts, such as phosphate
                or pyrophosphate in the presence of divalent metal ions, should be
                avoided.




                The highlighted portion means that reported values for biochemical reactions are (or should be) referenced to pH 7.0. The data in Table 3.6 refer to this biochemical standard state.



                Another question on the subject of equilibrium constants also addresses the importance of properly considering reference states.



                Reference



                1 The Journal of Biological Chemistry (1976), Vol. 261, No. 22, pp. 6859-6885.







                share|improve this answer














                share|improve this answer



                share|improve this answer








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