Buffer Intensity and Buffer Capacity

Buffers and Alkalimetric Titration

A buffer is a mixture of an acid and a base. Let’s consider a diprotic acid H2A to which a monoacidic base BOH (with B+ = Na+, K+, or NH4+) is added:

(1) H2A + n BOH  =  H2-n Bn A + n H2O

Here n acts as a stoichiometric coefficient that embodies the ratio of added amount of strong base, CB, to the total amount of the diprotic acid CT:

(2)   n = CB /CT     or     CB = n CT

Both CB and CT are given in molar units (mol/L) while n is a dimensionless number (unit-free). Moreover, the amount of strong base is just the total alkalinity, Alk = CB. Thus we have:

(3)   n = Alk /CT     or     Alk = n CT

The variation of n (or CB) in the reaction formula (1) by adding a strong base is called alkalimetric titration.

Note: The reason for our focus on diprotic acids is that it comprises carbonic acid H2CO3 as the most prominent representative (i.e. A-2 = CO3-2).

Notation. To keep the notation straight we abbreviate the concentration of H+ simply by x. Its relationship with pH is then given by:1

(4)   x = [H+] = 10-pH pH = –log [H+] = –log x

In addition, we introduce the “alkalinity of pure water” by

(5)   w  =  [OH-] – [H+]  =  Kw /x – x

where Kw = 10-14 is the equilibrium constant of the self-ionization of water.

Starting Point: Basic Set of Equations

The set of mathematical equations to describe the alkalimetric titration is exactly the same as we already introduced for diprotic acids and their conjugate ampholytes and bases:2

(6a) K1 =  {H+} {HA-} / {H2A} (1st diss. step)
(6b) K2 =  {H+} {A-2} / {HA-} (2nd diss. step)
(6c) Kw =  {H+} {OH-} (self-ionization)
(6d) CT =  [H2A] + [HA-] + [A-2] (mass balance)
(6e) 0 =  [H+] + n [H2A] + (n-1) [HA-] + (n-2) [A-2] – [OH-] (proton balance)

The only difference to the previous approach is that n is now a smooth variable of any value (rather than merely a fixed integer n = 0, 1, 2).

For the special case of n = 0, the set reduces to the description of the pure diprotic-acid system, where [H+] (or pH) is exactly determined by the amount of CT, and vice versa. Now, with the new variable n the system gets an additional degree of freedom.

The set of equations given above can be cast into a polynomial of 4th order in x (= 10-pH):

(7)   x4 + {K1 + nCT} x3 + {K1K2 + (n–1)CT K1 – Kw} x2 + K1 {(n–2)CT K2 – Kw} x – K1K2Kw  =  0

This quartic equation predicts the pH for any given pair of CT and n.

Once the pH (or x) is calculated, the concentrations of the three dissolved species, H2 A, HA-, and A-2, can be expressed by the ionization fractions ai as follows

(8)   [H2-i A-i]  =  CT ai (x) for   i = 0, 1, 2

The whole set of 6, on the one side, and 7 together with 8, on the other side, contain exactly the same information. In the text below, we provide additional closed-form equations which do the same work (but save us from solving a quartic equation).

Set of Equations based on Alkalinity

Until now the basic set of equations was built on both CT and n. Now, our aim is to re-establish the whole framework on CT and alkalinity (instead n). For this purpose we plug 6d into 6e which yields

(9a)   n CT  =  ([OH-] – [H+]) + [HA-] + 2 [A-2]

The right-hand side is just the definition of total alkalinity (also known as M alkalinity in carbonate systems),

(9b) alkalinity: Alk  ≡  ([OH-] – [H+]) + [HA-] + 2 [A-2]

Comparing both equations it reconfirms 3: Alk = n CT.

Set of Equations. The new set of equations is obtained by replacing the proton balance in 6e with the alkalinity in 9b:

(10a) K1 =  {H+} {HA-} / {H2A} (1st diss. step)
(10b) K2 =  {H+} {A-2} / {HA-} (2nd diss. step)
(10c) Kw =  {H+} {OH-} (self-ionization)
(10d) CT =  [H2A] + [HA-] + [A-2] (mass balance)
(10e) Alk =  ([OH-] – [H+]) + [HA-] + 2 [A-2] (alkalinity)

In this way, the polynomial in 7 becomes:

(11)   x4 + {K1 + Alk} x3 + {K1K2 + (Alk – CT) K1 – Kw} x2 + K1 {(Alk – 2CT) K2 – Kw} x – K1K2Kw  =  0

The concentrations of the three dissolved species remain unaffected by this change and are given by exactly the same relationship as in 8 above.

Ionization Fractions (a0, a1, a2)

From a mathematical point of view, the ionization fractions are nothing else than a clever way to combine 6a and (6b). They are given by

(12a)   a0   =   [ 1 + K1/x + K1K2/x2 ]-1 = (x/K1) a1
(12b)   a1   =   [ x/K1 + 1 + K2/x ]-1    
(12c)   a2   =   [ x2/(K1K2) + x/K2 + 1 ]-1 = (K2/x) a1

They are solely functions of x (or pH); the only other ingredients are two equilibrium constants, K1 and K2. Their pH dependence is displayed here. Other nice features are:

(13a) a0 + a1 + a2 = 1 (mass balance)

and the fact that each ai is imprisoned between 0 and 1, i.e. it neither become negative nor greater than 1. Strictly speaking, the functions will come very close to the boundaries, but will never actually reach the values 0 and 1:

(13b) 0 < ai < 1 for   i = 0, 1, 2

Asymptotic values. For two extreme conditions (opposite ends of pH scale) we can deduce the asymptotic values:

(14a) strong acidic: pH → 0 (or [H+] → ∞): a0 = 1 a1 = 0 a2 = 0
(14b) strong basic: pH → 14 (or [H+] → 0): a0 = 0 a1 = 0 a2 = 1

In addition, there is an exceptional point characterized by:

(14c)   x = (K1K2)1/2   a0 = a2

In the further treatment, the three ionization fractions (a0, a1, and a2) will become the building blocks of almost all relevant quantities. In this respect, don’t confuse the smooth variable n with the index i for three integers i = 0, 1, 2. The latter are dedicated to the three ionization fractions ai and to the three dissolved species, H2A, HA-, and A-2.

Exact Description by Closed-Form Equations

The set of five mathematical equations in 6a to (6e) (or, alternatively, in 10a to (10e)) contains all information about the diprotic-acid system when it is attacked by a strong base (alkalimetric titration). In particular, the actual equilibrium state (i.e. the concentrations of the dissolved species H2A, HA-, and A-2) is completely controlled by two parameters chosen from the triple (CT, n, pH) or (CT, Alk, pH). Once you know two of them, the third is automatically determined:

(16a)   pH (CT,n) = positive root of 7 [i] = CT ai
(16b)   pH (CT,Alk) = positive root of 11 [i] = CT ai
(17a)   n (CT,pH) = a1 + 2a2 + w/CT [i] = CT ai
(17b)   Alk (CT,pH) = CT (a1 + 2a2) + w [i] = CT ai
(18a)   CT (n,pH) = w/(n – a1 – 2a2) [i] = [w/(n – a1 – 2a2)] ai
(18b)   CT (Alk,pH) = (Alk – w)/(a1 + 2a2) [i] = [(Alk – w)/(a1 + 2a2)] ai

where pH is expressed by x, and where the concentrations of the three dissolved species are abbreviated by

(19)   [i]  ≡  [H2-i A-i] for   i = 0, 1, 2

To sum up: All formulas presented here are different encodings of one and the same thing, namely the set of equations (6a) to (6e).

It’s quite instructive to exhibit the non-linearity of all these equations graphically. The diagrams below display all possible combinations of one dependent and two independent variables (taken from the triplet CT, n, and pH) for the carbonate system:

dependent and independent variables taken from the triplet (CT, n, and pH) of the carbonate buffer system

The same can be done when the ‘stoichiometric coefficient’ n is replaced by the Alkalinity (Alk = n CT):

dependent and independent variables taken from the triplet (CT, Alkalinity, and pH) of carbonate buffer system

Other Examples. Application of 17a to a pure CO2 system provides the (blue) titration curves in the diagram below. On the other hand, for the special case of n = 0, 1, 2, 18a was used to plot the pH dependence of three equivalence points of the carbonate system.

Buffer Intensity vs Buffer Capacity

In literature, both buffer intensity and buffer capacity are commonly used as synonyms. For our purpose, however, we make the following difference:

(20a) buffer intensity: β  = dn / d pH [unitless]
(20b) buffer capacity: βc = dCB / d pH [mol/L] or [eq/L]

where CB = nCT. Both quantities differ by their physical units only. Both the buffer intensity and capacity express the inverse change in pH per amount of strong base. Thus, the pH where β reaches its maximum represents the optimal buffer range (bounded by pH±1).

The starting point is 17a:

(21)   n (CT,x)  = 

The buffer intensity is then obtained by derivation of this equation with respect to pH:


or, more explicitly,


Some hints how to perform this derivation are given in the Appendix.

Example: The buffer intensity of a pure CO2 system is displayed as green curve in the diagram below.

Derivative of the Buffer Intensity

The derivative of the buffer intensity is given by:


with the abbreviation

(24)   f(x) = (x/K1 – K2/x) (x/K1 + K2/x – 1 + 8 K2/K1)

Example: The derivative of the buffer intensity of a pure CO2 system is displayed as red curve in the diagram below. The zeros of dβ/d(pH) are marked by blue circles.

Example Calculation for the Carbonate System

Given is a pure CO2 system with CT = 100 mM, 10 mM, and 1 mM H2CO3.3 Three quantities are plotted as function of pH:

•  n the (normalized) amount of base added4 (as blue curve)
•  β = dn/d(pH) the buffer intensity (as green curve)
•  dβ/d(pH) derivative of the buffer intensity (as red curve)

All three quantities as per definition are unit-less. The small blue dots mark the zeros of dβ/d(pH).

aqion - buffer intensities of carbonate system

Since the (blue) titration curve is an ever-increasing function, its derivation, i.e. the buffer intensity β, is always positive (see green curve). This is in full accord with Le Châtelier’s principle that every solution resists pH changes.

Appendix – Derivatives with Respect to pH

The first and second derivatives of x = [H+] with respect to pH are

(A1)   dx/d(pH)  =  (–ln 10) x and   d2x/d(pH)2  =  (–ln 10)2 x

This result can be used to differentiate any given function f(x) with respect to pH (by application of the chain rule):


Hence, for the “pure-water alkalinity” introduced in 5, w(x) = Kw/x – x, we get the amazing result:

(A3a)   dw/d(pH) =   ln 10 {Kw/x + x} =   ln 10 {w + 2x}
(A3b)   d2w/d(pH)2 =   (ln 10)2 {w + 2x – 2x} =   (ln 10)2 w

All derivatives of even degree remain w unchanged.

Ionization Fractions. The three ionization fractions, introduced in 12a to (12c), obey the following relations (where the sum runs over i from i=0 to 2):

(A4a)   Σi ai =   a0 + a1 + a2 =   1
(A4b)   Σi i ai =   0⋅a0 + 1⋅a1 + 2⋅a2 =   a1 + 2a2

The first derivative of the ionization fractions is given by

(A5)   dai /d(pH)  =  ln 10 {i – (a1 + 2a2)} for   i=0, 1, 2

Applying it to the two sums in A4a and (A4b) yields

(A6a)   (d/d pH) Σi ai =   ln 10 {(a1 + 2a2) – (a1 + 2a2)}   =   0
(A6b)   (d/d pH) Σi i ai =   ln 10 {(a1 + 4a2) – (a1 + 2a2)2}
      =   ln 10 {(a2 + a0) – (a2 – a0)2}

The first relation, which yields zero, is obvious because it represents the derivation of a constant, namely d1/d(pH) = 0. In the last line, A4a was applied. It enters the buffer intensity in 22a.

Remarks & Footnotes

  1. To be correct, pH is based on the activity of H+: pH = –log {H+}.

  2. Here and in the following we assume that all activities (expressed by curly braces) are replaced by molar concentrations (expressed by square brackets). This is legitimate for dilute systems or by switching to conditional equilibrium constants.

  3. The two equilibrium constants of the carbonic acid system are: K1 = 10-6.35 and K2 = 10-10.33.

  4. The formula allows also negative values of n. This mimics the withdrawal of the strong base from the solution or the addition of a monoprotic acid (e.g. HCl).

[last modified: 2016-12-08]