Overview Activity Chem Equilibrium Acids Carbonate System Mineral Phases

Buffer Intensity and Buffer Capacity

pH Buffers and Alkalimetric Titration

A pH buffer is a mixture of a weak 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

(1)	H2A + n BOH  =  BnH2-n A + n H2O

Example: For H₂CO₃ and NaOH we obtain – in the special case of n = 0, 1, and 2 – pure solutions of H₂CO₃, NaHCO₃ and Na₂CO₃ (see Equivalence Points).

The parameter 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 = CB /CT     or     CB = n CT

Both C_B and C_T are given in molar units (mol/L) while n is dimensionless (unitless).

The entity B_nH_2-nA in reaction (1) does not survive in water; it dissociates into several aqueous species – as indicated in the right picture.

As we will see below, the amount of strong base is just the total alkalinity, Alk = C_B. Thus we have:

(3)

n = Alk /CT     or     Alk = n CT

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 H₂CO₃ as the most prominent representative (i.e. A^-2 = CO₃^-2).

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

(4)

x ≡ {H+} = 10-pH

⇔

pH = – lg {H+} = – lg x

The (molar) concentrations of the acid species [H₂A], [HA^-] and [A^-2] are abbreviated as

(5)

[j]  =  [H2-j A-j]

with   j = 0, 1, 2

In addition, we introduce the “alkalinity of pure water” by²

(6)

w  =  [OH-] – [H+]  ≈  Kw /x – x

where K_w = 10^-14 is the equilibrium constant of autoprotolysis (self-ionization of water). As we will see later, w represents the contribution of self-ionization to the alkalinity. For pure water – defined by x = (K_w)^1/2 = 10^-7, i.e. pH = 7, – this contribution vanishes: w = 0.

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 bases:³

(1.1a)		K1	=  {H+} {HA-} / {H2A}	(1st diss. step)
(1.1b)		K2	=  {H+} {A-2} / {HA-}	(2nd diss. step)
(1.1c)		Kw	=  {H+} {OH-}	(self-ionization)
(1.1d)		CT	=  [H2A] + [HA-] + [A-2]	(mass balance)
(1.1e)		n CT	=  [HA-] + 2 [A-2] + [OH-] – [H+]	(charge 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 determined by the amount of C_T, and vice versa. Now, with the new variable n the system gets an additional degree of freedom.

Alkalinity.  The right-hand side of 1.1e is just the definition of the total alkalinity (also known as M alkalinity in carbonate systems),

(1.2)

Alk  ≡  ([OH-] – [H+]) + [HA-] + 2 [A-2]

This justifies our statement, which we originally made in 3:

(1.3)

Alk  =  CB  =  n CT

Ionization Fractions (a₀, a₁, a₂)

Our equation system consists of 5 equations. Everything concerning the acid itself is contained in a subset of three equations: (1.1a), (1.1b), and (1.1d). The information inherent in this subsystem can be expressed by ionization fractions (instead of the three acid species [j]):

(2.1)

aj  =  [j] / CT

with   j = 0, 1, 2

These ‘normalized concentrations’, which indicate the degree of dissociation, are completely independent of the amount of acid C_T. Dividing 1.1d by C_T yields:

(2.2)

1  =  a0 + a1 + a2

(mass balance)

From a mathematical point of view, the ionization fractions are nothing else than a clever way to combine 1.1a and (1.1b) using 2.2:

(2.3a)		a0   =   [ 1 + K1/x + K1K2/x2 ]-1
(2.3b)		a1   =   [ x/K1 + 1 + K2/x ]-1
(2.3c)		a2   =   [ x2/(K1K2) + x/K2 + 1 ]-1

They are solely functions of x (or pH); the only other ingredients are two equilibrium constants, K₁ and K₂. Their pH dependence is shown here. Each a_j 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 their boundaries, but never really reach 0 and 1:

(2.4)

0 < aj < 1

for   j = 0, 1, 2

There are three exceptional points of pH (or x):⁴

(2.5a)		pH = pK1	x = K1	⇔	a1 = a0 ≈ ½	a2 ≈ 0
(2.5b)		pH = ½ (pK1+pK2)	x = (K1K2)1/2	⇔	a0 = a2 ≈ 0	a1 ≈ 1
(2.5c)		pH = pK2	x = K2	⇔	a1 = a2 ≈ ½	a0 ≈ 0

In what follows, the three ionization fractions a₀, a₁, and a₂ will become the building blocks of almost all relevant acid-base quantities. In this respect, don’t confuse the smooth variable n with the index j for the three integers j = 0, 1, 2. The latter are dedicated to the three ionization fractions a_j and the three dissolved species, H₂A, HA^-, and A^-2.

Analytical Solution of the System of Equations

The set of 5 equations in (1.1) can be cast into a single analytical formula:

(3.1a)

\(C_T(x) \, = \, \dfrac{x - K_w/x} {a_1 + 2a_2 - n}\)

Here, the pH dependence (in form of x = 10^-pH) is also in a₁ and a₂. To make it evident, you must insert (2.3b) and (2.3c) into 3.1a:

(3.1b)

\(C_T(x) \, = \, \left(x-\dfrac{K_w}{x}\right) \ \left(\dfrac{1+2K_2/x} {x/K_1 + 1 + K_2/x}-n\right)^{-1}\)

Both formulas calculate C_T for a given pH (or x) and n. Often, however, one is interested in the “inverse task”: to calculate the pH for a given C_T and/or n. For this purpose, you need to rearrange 2.1b to isolate x. But that’s impossible. The only thing you can achieve is a polynomial of 4^th order in x:

(3.2)

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

This quartic equation (quartic!, not quadratic)⁵ predicts the pH for any given pair of C_T and n. Replacing nC_T by C_B in 3.2, you get an alternative form:

(3.3)

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

Once the pH value (or x) is known, the concentrations of the acid species, H₂A, HA^-, and A^-2, can be calculated via the ionization fractions a_j as follows:

(3.4)

[H2-j A-j]  =  CT aj (x)

for   j = 0, 1, 2

Exact Relationships between pH, C_T, and n (or Alk = nC_T)

The set of five mathematical equations in (1.1a) to (1.1e) contains all information about the “diprotic-acid plus strong base” system (alkalimetric titration). In particular, the actual equilibrium state (i.e. the concentrations of the acid species H₂A, HA^-, and A^-2) is 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:

(4.1a)		pH (CT,n)	= –lg x1  with  x1 = positive root of 3.2	[j] = CT aj
(4.1b)		pH (CT,Alk)	= –lg x1  with  x1 = positive root of 3.3	[j] = CT aj

(4.2a)		n (CT,pH)	= a1 + 2a2 + w/CT		[j] = CT aj
(4.2b)		Alk (CT,pH)	= CT (a1 + 2a2) + w		[j] = CT aj
(4.3a)		CT (n,pH)	= w/(n – a1 – 2a2)		[j] = [w/(n – a1 – 2a2)] aj
(4.3b)		CT (Alk,pH)	= (Alk – w)/(a1 + 2a2)		[j] = [(Alk – w)/(a1 + 2a2)] aj

with w(x) defined in 6. Here [j] stands for the molar concentration of the three acid species.

Note: All formulas presented here are different encodings of one and the same thing, namely the set of equations (1.1a) to (1.1e).

Cross Plots.  It’s quite instructive to exhibit the non-linearity of these relationships 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:

The diagrams can be redrawn for the case when n is replaced by the alkalinity (Alk = n C_T):

Other Examples.  Application of 4.2a 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, 4.3a was used to plot the pH dependence of three equivalence points.

Buffer Capacity

Alkalinity is known to be a buffer capacity (and therefore it is what we have called C_B). Thus, the titration curves based on 4.2b display the buffer capacity as a function of pH. In this sense, n(pH) = C_B/C_T can be understood as the “normalized” or unitless buffer capacity. From the above formulas we get:

(5.1a)		buffer capacity (in mol/L):		CB	=	(a1 + 2a2) CT + w
(5.1b)		normalized buffer capacity:		n	=	(a1 + 2a2) + w/CT

Each individual term in these “titration formulas” is pH-dependent:  a₁(x), a₂(x), C_T(x), and w(x). In particular, 5.1b can be written as:

(5.2)

n (CT,x)  =  \(a_1 + 2a_2 + \dfrac{w}{C_T} \ =\ \dfrac{1+2K_2/x} {x/K_1 + 1 + K_2/x} + \dfrac{K_w/x - x}{C_T}\)

Buffer Intensity

The buffer intensity is the 1^st derivative of the buffer capacity with respect to pH. Again, we distinguish between two types of buffer intensities:

(6.1a)		buffer intensity (in mol/L):	βc	=	dCB / d pH  =  β CT
(6.1b)		normalized buffer intensity:	β	=	dn / d pH

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 1^st pH-derivative of 5.2 yields (see Appendix):

(6.2a)

β  =  ln 10 {(a2 + a0) – (a2 – a0)2} + ln 10 (w+2x) /CT

or, more explicitly,

(6.2b)

\(\beta \ = \ ln\,10 \, \left\{\dfrac{x/K_1+4K_2/K_1+K_2/x}{(x/K_1+1+K_2/x)^2} \, +\ \dfrac{x+K_w/x}{C_T} \right\}\)

Equation (6.2b), that contains only positive summands, tells us 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 = K1	maximum	⇒	optimal buffer range
x = (K1K2)1/2	minimum
x = K2	maximum	⇒	optimal buffer range (small CT 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 first-order derivative of β is zero.

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

Buffer Capacity. The buffer capacity is just the integral of the buffer intensity:

(6.3)

buffer capacity = \(\int_a^b \beta_c \, dpH\)

Derivative of the Buffer Intensity

The derivative of the buffer intensity is given by:

(7.1)

\(\dfrac{d\beta}{d\,pH} \ = \ (ln\,10)^2 \, \left\{\dfrac{f(x)}{(x/K_1+1+K_2/x)^3} \, +\ \dfrac{K_w/x - x}{C_T} \right\}\)

with the abbreviation

(7.2)

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 C_T = 100 mM, 10 mM, and 1 mM H₂CO₃.⁶ Three quantities are plotted as function of pH:

•  n(pH)	(normalized) amount of strong base added7	(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 ever-increasing function, its derivative, i.e. the buffer intensity β, is always positive (see green curve). This is in full accord with Le&nsp;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):

(A2)

\(\dfrac{d\,f(x)}{d\,pH} \ = \ \left(\dfrac{dx}{d\,pH}\right) \left(\dfrac{d\,f(x)}{dx}\right) \ = \ (-ln\,10)\, x \ \dfrac{d\,f(x)}{dx}\)

Hence, for w(x) = K_w/x – x, introduced in 6, 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 higher degree repeat this pattern:

(A3c)

\(\dfrac{d^{k}w}{d\,(pH)^k} \ = \ (ln 10)^k \, \left\{\begin{matrix}\ w& &for & k\ even & \\ \ w+2x && for & k\ odd & \end{matrix}\right.\)

Ionization Fractions. The three ionization fractions, introduced in 2.3a to (2.3c), obey the following relations (where the sum runs from i = 0 to 2):

(A4a)		Y0  ≡  Σi ai	=  a0 + a1 + a2	=  1
(A4b)		Y1  ≡  Σi i ai	=  0⋅a0 + 1⋅a1 + 2⋅a2	=  a1 + 2 a2
(A4c)		Yk  ≡  Σi ik ai	=  0⋅a0 + 1⋅a1 + 2k⋅a2	=  a1 + 2k a2	(for all k > 0)

Quantities so defined are also called kth moments. Here in our case, Y₀ represents the mass balance, Y₁ enters the alkalinity (as its main part), and Y₂ together with Y₃ will be used for the buffer intensity and its derivative. All Y_k (excluding Y₀) are positive numbers, living in the range 0 < Y_k < 2^k.

The first derivative of the ionization fractions is given by

(A5)

dai /d(pH)  =  ln 10 (i – a1 – 2a2) ai  =  ln 10 (i – Y1) ai

for i = 0, 1, 2

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

(A6a)		(d/d pH) Σi ai	=   ln 10 Σi (i – Y1) ai	=   ln 10 (Y1 – Y1)  =  0
(A6b)		(d/d pH) Σi i ai	=   ln 10 Σi i (i – Y1) ai	=   ln 10 (Y2 – Y12)

The first relation, which delivers zero, is obvious because it represents the derivation of a constant, namely d1/d(pH) = 0. In general we have

(A7)

d Yk /d pH   =   ln 10 Σi ik (i – Y1) ai   =   ln 10 (Yk+1 – Y1Yk)

The second derivative of Y₁ is given by

(A8a)		\(\dfrac{d^2\,Y_1}{d(pH)^2}\)	=   \((ln\,10) \,\dfrac{d}{d\, pH}\, (Y_2-Y_1^2)\)
			=   (ln 10)2 (Y3 – Y1Y2 – 2Y1 (Y2 – Y12))
			=   (ln 10)2 (Y3 – 3Y1Y2 + 2Y13)

and the third derivative of Y₁ is

(A8b)

\(\dfrac{d^3\,Y_1}{d(pH)^3}\)

=   \((ln\,10)^2 \,\dfrac{d}{d\, pH}\, (Y_3-3Y_1Y_2+2Y_1^3)\)

=   (ln 10)3 {(Y4–Y1Y3) – 3Y2 (Y2–Y12) – 3Y1 (Y3–Y1Y2) + 6Y12 (Y2–Y12)}

Buffer Intensity & Co. The above equations can now be employed to derive the buffer intensity and its derivative introduced in 6.2 and 7.1. In particular, we have:

(A9a)		n	=	(Y1 + w/CT)
(A9b)		β	=   d /d pH	(Y1 + w/CT)
(A9c)		dβ/d pH	=   d2/d (pH)2	(Y1 + w/CT)

which yields

(A10a)		n	=   (Y1 + w/CT)
(A10b)		β	=   (ln 10)  {(Y2 – Y12) + (w+2x) /CT}
(A10c)		dβ/d pH	=   (ln 10)2 {(Y3 – 3Y1Y2 + 2Y13) + w/CT}

If desired, the Y’s can be replaced by the a’s. For example, the central piece of A10b is given by:

(A11)

Y2 – Y12   =   (a1 + 4a2) – (a1 + 2a2)2   =   (a2 + a0) – (a2 – a0)2

Here, in the last relation A4a was applied.

Remarks & Footnotes