Buffer Intensity and Buffer Capacity
Buffers and Alkalimetric Titration
A buffer is a mixture of a weak acid and a base. Let’s consider a diprotic acid H_{2}A to which a monoacidic base BOH (with B^{+} = Na^{+}, K^{+}, or NH_{4}^{+}) is added:^{1}
(1)  H_{2}A + n BOH = B_{n}H_{2n }A + n H_{2}O 
Example: For H_{2}CO_{3} and NaOH we obtain – in the special case of n = 0, 1, and 2 – pure solutions of H_{2}CO_{3}, NaHCO_{3} and Na_{2}CO_{3} (see Equivalence Points).
In general, n acts as a stoichiometric coefficient that embodies the ratio of added amount of strong base, C_{B}, to the total amount of the diprotic acid C_{T}:
(2)  n = C_{B }/C_{T} or C_{B} = n C_{T} 
Both C_{B} and C_{T} are given in molar units (mol/L) while n is a dimensionless number (unitfree).
The entity B_{n}H_{2n }A in reaction (1) does not survive in water; it dissociates into several aqueous species – as shown in the right scheme.
As we will see below, the amount of strong base is just the total alkalinity, Alk = C_{B}. Thus we have:
(3)  n = Alk /C_{T} or Alk = n C_{T} 
The variation of n (or C_{B}) in 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 H_{2}CO_{3} as the most prominent representative (i.e. A^{2} = CO_{3}^{2}).
Notation. To keep the notation straight we abbreviate the concentration of H^{+} simply by x. Its relationship with pH is given by:^{2}
(4)  x = [H^{+}] = 10^{pH}  ⇔  pH = –log [H^{+}] = –log x 
In addition, we introduce the “alkalinity of pure water” by
(5)  w = [OH^{}] – [H^{+}] = K_{w }/x – x 
where K_{w} = 10^{14} is the equilibrium constant of autoprotolysis (selfionization of water). As we’ll see later, w represents the (additive) contribution of selfionization to the alkalinity of the buffer system. For pure water – defined by x = (K_{w})^{1/2}, or pH = 7 – this contribution vanishes exactly: w = 0.
StartingPoint: 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:^{3}
(6a)  K_{1}  = {H^{+}} {HA^{}} / {H_{2}A}  (1^{st} diss. step) 
(6b)  K_{2}  = {H^{+}} {A^{2}} / {HA^{}}  (2^{nd} diss. step) 
(6c)  K_{w}  = {H^{+}} {OH^{}}  (selfionization) 
(6d)  C_{T}  = [H_{2}A] + [HA^{}] + [A^{2}]  (mass balance) 
(6e)  0  = [H^{+}] + n [H_{2}A] + (n1) [HA^{}] + (n2) [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 diproticacid system, where [H^{+}] (or pH) is exactly determined by the amount of C_{T}, 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 4^{th} order in x (= 10^{pH}):
(7)  x^{4} + {K_{1} + nC_{T}} x^{3} + {K_{1}K_{2} + (n–1)C_{T }K_{1} – K_{w}} x^{2} + K_{1} {(n–2)C_{T }K_{2} – K_{w}} x – K_{1}K_{2}K_{w} = 0 
This quartic equation (quartic!, not quadratic) predicts the pH for any given pair of C_{T} and n.
Once the pH value (or x) is calculated, the concentrations of the three dissolved species, H_{2 }A, HA^{}, and A^{2}, can be expressed by the ionization fractions a_{i} as follows
(8)  [H_{2i }A^{i}] = C_{T} a_{i }(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. Below we provide additional closedform equations which will 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 C_{T} and n. Now, our aim is to reestablish the whole framework on C_{T} and alkalinity (instead n). For this purpose we plug 6d into 6e which yields
(9a)  n C_{T} = ([OH^{}] – [H^{+}]) + [HA^{}] + 2 [A^{2}] 
The righthand 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 C_{T}.
Set of Equations. The new set of equations is obtained by replacing the proton balance in 6e with the alkalinity definition in 9b:
(10a)  K_{1}  = {H^{+}} {HA^{}} / {H_{2}A}  (1^{st} diss. step) 
(10b)  K_{2}  = {H^{+}} {A^{2}} / {HA^{}}  (2^{nd} diss. step) 
(10c)  K_{w}  = {H^{+}} {OH^{}}  (selfionization) 
(10d)  C_{T}  = [H_{2}A] + [HA^{}] + [A^{2}]  (mass balance) 
(10e)  Alk  = ([OH^{}] – [H^{+}]) + [HA^{}] + 2 [A^{2}]  (alkalinity) 
In this way, the polynomial in 7 becomes:
(11)  x^{4} + {K_{1} + Alk} x^{3} + {K_{1}K_{2} + (Alk – C_{T}) K_{1} – K_{w}} x^{2} + K_{1} {(Alk – 2C_{T}) K_{2} – K_{w}} x – K_{1}K_{2}K_{w} = 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 (a_{0}, a_{1}, a_{2})
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)  a_{0} = [ 1 + K_{1}/x + K_{1}K_{2}/x^{2} ]^{1}  =  (x/K_{1}) a_{1}  
(12b)  a_{1} = [ x/K_{1} + 1 + K_{2}/x ]^{1}  
(12c)  a_{2} = [ x^{2}/(K_{1}K_{2}) + x/K_{2} + 1 ]^{1}  =  (K_{2}/x) a_{1} 
They are solely functions of x (or pH); the only other ingredients are two equilibrium constants, K_{1} and K_{2}. Their pH dependence is displayed here. Other nice features are:
(13a)  a_{0} + a_{1} + a_{2} = 1  (mass balance) 
and the fact that each a_{i} 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 < a_{i} < 1  for i = 0, 1, 2 
There are three exceptional points of pH (or x):^{4}
(14a)  pH = pK_{1}  x = K_{1}  ⇔  a_{1} = a_{0} ≈ ½  a_{2} ≈ 0  
(14b)  pH = ½ (pK_{1}+pK_{2})  x = (K_{1}K_{2})^{1/2}  ⇔  a_{0} = a_{2} ≈ 0  a_{1} ≈ 1  
(14c)  pH = pK_{2}  x = K_{2}  ⇔  a_{1} = a_{2} ≈ ½  a_{0} ≈ 0 
Asymptotic values. For two extreme conditions (opposite ends of pH scale) we can deduce the asymptotic values:
(14d)  strong acidic:  pH → 0  (or x → ∞):  a_{0} = 1  a_{1} = 0  a_{2} = 0 
(14e)  strong basic:  pH → 14  (or x → 0):  a_{0} = 0  a_{1} = 0  a_{2} = 1 
In the further treatment, the three ionization fractions (a_{0}, a_{1}, and a_{2}) 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 a_{i} and to the three dissolved species, H_{2}A, HA^{}, and A^{2}.
Exact Relationships between pH, C_{T}, and n (or Alk = nC_{T})
The set of five mathematical equations in 6a to (6e) (or, alternatively, in 10a to (10e)) contains all information about the diproticacid 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 H_{2}A, HA^{}, and A^{2}) is completely controlled by two parameters chosen from the triple (C_{T}, n, pH) or (C_{T}, Alk, pH). Once you know two of them, the third is automatically determined:
(16a)  pH (C_{T},n)  = –log x_{1} with x_{1} = positive root of 7  [i] = C_{T} a_{i}  
(16b)  pH (C_{T},Alk)  = –log x_{1} with x_{1} = positive root of 11  [i] = C_{T} a_{i} 
(17a)  n (C_{T},pH)  = a_{1} + 2a_{2} + w/C_{T}  [i] = C_{T} a_{i}  
(17b)  Alk (C_{T},pH)  = C_{T }(a_{1} + 2a_{2}) + w  [i] = C_{T} a_{i}  
(18a)  C_{T }(n,pH)  = w/(n – a_{1} – 2a_{2})  [i] = [w/(n – a_{1} – 2a_{2})] a_{i}  
(18b)  C_{T }(Alk,pH)  = (Alk – w)/(a_{1} + 2a_{2})  [i] = [(Alk – w)/(a_{1} + 2a_{2})] a_{i} 
where pH is expressed by x, and where the concentrations of the three dissolved species are abbreviated by
(19)  [i] ≡ [H_{2i }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).
Cross Plots. It’s quite instructive to exhibit the nonlinearity of all these equations graphically. The diagrams below display all possible combinations of one dependent and two independent variables (taken from the triplet C_{T}, n, and pH) for the carbonate system:
All diagrams can be redrawn for the case when the ‘stoichiometric coefficient’ n is replaced by the Alkalinity (Alk = n C_{T}):
Other Examples. Application of 17a to a pure CO_{2} 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
Definition. The (normalized) buffer intensity is the derivative of the ‘titration curve’ n(pH) with respect to pH. We distinguish between two types of buffer intensities:
(20a)  normalized:  β_{ } = dn / d pH  [unitless] 
(20b)  nonnormalized:  β_{c} = dC_{B} / d pH = β C_{T}  [mol/L] 
where C_{B} = n^{ }C_{T}. Both quantities differ by their physical units only.
The steeper the slope of a titration curve the higher is the buffer intensity β = dn/d(pH), i.e. the higher is the system’s resistance to pH changes (caused by a strong base). Thus, the pH where β reaches its maximum represents the optimal buffer range (bounded by pH_{max }±1) – see example below:
Formulas. The starting point is the titration curve described by 17a:
(21)  n (C_{T},x) = 
The buffer intensity is then obtained by derivation of this equation with respect to pH (see Appendix):
(22a)  β  = ln 10 {(a_{2} + a_{0}) – (a_{2} – a_{0})^{2}} + ln 10 (w+2x) /C_{T} 
or, more explicitly,
(22b) 
Equation (22b), that contains positive summands only, exhibits the fact that β is always positive (it never becomes negative or zero).
Local Extrema. The local maxima (and minima) of the buffer intensity are at the following points:
x = K_{1}  maximum  ⇒  optimal buffer range 
x = (K_{1}K_{2})^{1/2}  minimum  
x = K_{2}  maximum  ⇒  optimal buffer range (small C_{T} excluded) 
Please note that this is not strictly exact, but it’s a very good approximation. Mathematically, local extrema of β are found where the firstorder derivative of β is zero.
Example: The buffer intensity of a pure CO_{2} system is displayed as green curve in the diagram below.
Buffer Capacity. The buffer capacity is just the integral of the buffer intensity:
(23)  buffer capacity = 
Derivative of the Buffer Intensity
The derivative of the buffer intensity is given by:
(24) 
with the abbreviation
(25)  f_{ }(x) = (x/K_{1} – K_{2}/x) (x/K_{1} + K_{2}/x – 1 + 8 K_{2}/K_{1}) 
Example: The derivative of the buffer intensity of a pure CO_{2} 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 CO_{2} system with C_{T} = 100 mM, 10 mM, and 1 mM H_{2}CO_{3}.^{5} Three quantities are plotted as function of pH:
• n(pH)  (normalized) amount of base added^{6}  (as blue curve) 
• β = dn/dpH  (normalized) buffer intensity  (as green curve) 
• dβ/dpH  derivative of the buffer intensity  (as red curve) 
All three quantities as per definition are unitless. The small blue dots mark the zeros of dβ/d(pH).
Since the (blue) titration curve is an everincreasing 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  d^{2}x/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):
(A2) 
Hence, for w(x) = K_{w}/x – x, introduced in 5, we get the amazing result:
(A3a)  dw/d(pH)  = ln 10 (K_{w}/x + x)  = ln 10 (w + 2x)  
(A3b)  d^{2}w/d(pH)^{2}  = (ln 10)^{2} (w + 2x – 2x)  = (ln 10)^{2} w 
All derivatives of higher degree repeat this pattern:
(A3c) 
Ionization Fractions. The three ionization fractions, introduced in 12a to (12c), obey the following relations (where the sum runs from i = 0 to 2):
(A4a)  Y_{0} ≡ Σ_{i} a_{i}  = a_{0} + a_{1} + a_{2}  = 1  
(A4b)  Y_{1} ≡ Σ_{i} i a_{i}  = 0⋅a_{0} + 1⋅a_{1} + 2⋅a_{2}  = a_{1} + 2 a_{2}  
(A4c)  Y_{k} ≡ Σ_{i} i^{k} a_{i}  = 0⋅a_{0} + 1⋅a_{1} + 2^{k}⋅a_{2}  = a_{1} + 2^{k }a_{2}  (for all k > 0) 
Quantities so defined are also called kth moments. Here in our case, Y_{0} represents the mass balance, Y_{1} enters the alkalinity (as its main part), and Y_{2} together with Y_{3} will be used for the buffer intensity and its derivative. All Y_{k} (excluding Y_{0}) are positive numbers, living in the range 0 < Y_{k} < 2^{k}.
The first derivative of the ionization fractions is given by
(A5)  da_{i }/d(pH) = ln 10 (i – a_{1} – 2a_{2}) a_{i} = ln 10 (i – Y_{1}) a_{i}  for i = 0, 1, 2 
Applying it to the two sums in A4a and (A4b) yields
(A6a)  (d/d pH) Σ_{i} a_{i}  = ln 10 Σ_{i} (i – Y_{1}) a_{i}  = ln 10 (Y_{1} – Y_{1}) = 0  
(A6b)  (d/d pH) Σ_{i} i a_{i}  = ln 10 Σ_{i} i (i – Y_{1}) a_{i}  = ln 10 (Y_{2} – Y_{1}^{2}) 
The first relation, which delivers zero, is obvious because it represents the derivation of a constant, namely d^{ }1/d(pH) = 0. In general we have
(A7)  d Y_{k }/d pH = ln 10 Σ_{i} i^{k} (i – Y_{1}) a_{i} = ln 10 (Y_{k+1} – Y_{1}Y_{k}) 
The second derivative of Y_{1} is given by
(A8a)  =  
= (ln 10)^{2} (Y_{3} – Y_{1}Y_{2} – 2Y_{1} (Y_{2} – Y_{1}^{2}))  
= (ln 10)^{2} (Y_{3} – 3Y_{1}Y_{2} + 2Y_{1}^{3}) 
and the third derivative of Y_{1} is
(A8b)  = 
= (ln 10)^{3} {(Y_{4}–Y_{1}Y_{3}) – 3Y_{2} (Y_{2}–Y_{1}^{2}) – 3Y_{1} (Y_{3}–Y_{1}Y_{2}) + 6Y_{1}^{2 }(Y_{2}–Y_{1}^{2})} 
Buffer Intensity & Co. The above equations can now be employed to derive the buffer intensity and its derivative introduced in 22 and 24. In particular, we have:
(A9a)  n  =  (Y_{1} + w/C_{T})  
(A9b)  β  = d /d pH  (Y_{1} + w/C_{T})  
(A9c)  dβ/d pH  = d^{2}/d (pH)^{2}  (Y_{1} + w/C_{T}) 
which yields
(A10a)  n  = (Y_{1} + w/C_{T})  
(A10b)  β  = (ln 10) {(Y_{2} – Y_{1}^{2}) + (w+2x) /C_{T}}  
(A10c)  dβ/d pH  = (ln 10)^{2} {(Y_{3} – 3Y_{1}Y_{2} + 2Y_{1}^{3}) + w/C_{T}} 
If desired, the Y’s can be replaced by the a’s. For example, the central piece of A10b is given by:
(A11)  Y_{2} – Y_{1}^{2} = (a_{1} + 4a_{2}) – (a_{1} + 2a_{2})^{2} = (a_{2} + a_{0}) – (a_{2} – a_{0})^{2} 
Here, in the last relation A4a was applied.
Remarks & Footnotes

This topic is also discussed in the PowerPoint presentation. ↩

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

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. ↩

The equilibrium constant K can also be expressed by pK = –log K. ↩

The two equilibrium constants of the carbonic acid are: K_{1} = 10^{6.35} and K_{2} = 10^{10.33} (see also here). ↩

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). ↩