## Acids and Bases

An acid HA is a proton donor; it produces H+ ions (or H3O+) when dissolved in water:1

 (1) HA  =  H+ + A- (2) HA + H2O  =  H3O+ + A-

In the following we prefer the short notation of 1, but keep in mind that H+ ions do not exist in a free state; they are extremely reactive and form hydronium ions H3O+.

In contrast, a base BOH (such like NaOH, KOH or NH4OH) is a proton acceptor:

 (3) BOH + H+  =  H2O + B+ (or   OH- + H+ = H2O)

According to Brønsted, protons are transferred between conjugate acid-base pairs, which becomes evident when 1 is added to 3:

(4a)   HA + BOH = H2O + A- + B+
(4b)   acid + base = conjugate acid + conjugate base
(of base BOH)   (of acid HA)

[Note: The cation B+ itself acts like a spectator which can be cancelled on both sides of the reaction formula. In fact, B+ doesn’t play any role in the further treatment.]

One very special case is the self-dissociation of water:

 (5) H2O + H2O  =  H3O+ + OH-

Here, water acts as both acid and base. Such substances are called ampholytes.

Acidity Constants Ka (and pKa)

The equilibrium constant of reaction (1) is called ‘acid dissociation constant’ or

 (6) acidity constant: $K_a = \dfrac{\{H^+\}\{A^-\}}{\{HA\}}$

The value of Ka resembles the strength of the acid (large numbers are strong acids, small numbers are weak acids). In practice, it is more convenient to use the (base-10) logarithmic form of 6:

 (7) log Ka  =  log {H+} + log {A-} – log {HA}

The negative decadic logarithm of the acidity constant is abbreviated by pKa:

 (8) pKa = – log Ka

Thus, using the definition of pH as pH = –log {H+}, 7 converts to

 (9) pKa  =  pH – log {A-} + log {HA}

The pKa value allows a classification into strong and weak acids: The smaller the pKa the stronger the acid – quite the opposite to a Ka-based ranking. In literature, several types of such classification exist. How this is handled in aqion is described here.

Polyprotic Acids

Acids can donate one, two or more protons (H+). Typical examples are:

Monoprotic acid (HA) Diprotic acid (H2A) Triprotic acid (H3A)
HCl H2CO3 H3PO4
HNO3 H2SO4 H3AsO4
HI H2CrO4 H3BO3
HF H2SeO4 citric acid
formic acid oxalic acid
acetic acid, …

A monoprotic acid is characterized by a single acidity constant K1 (= Ka), a diprotic acid by two acidity constants (K1, K2), and a triprotic acid by three acidity constants (K1, K2, K3):

 1st dissociation step: H3A = H+ + H2A- K1 2nd dissociation step: H2A- = H+ + HA-2 K2 3rd dissociation step: HA-2 = H+ + A-3 K3

The three reaction steps of a triprotic acid can also be written as:

 H3A  =  H+ + H2A- K1 H3A  =  2H+ + HA-2 K1K2 H3A  =  3H+ + A-3 K1K2K3

with equilibrium constants constructed from K1, K2, and K3. In fact, this procedure can be extended to any N-protic acid with N dissociation steps.

Ranking. Protons are released sequentially one after the other, with the first proton being the fastest and most easily lost, then the second, and then the third (which is most strongly bound). This yields the following ranking of acidity constants of a polyprotic acid:2

 (10) K1 > K2 > K3 or pK1 < pK2 < pK3

For example, phosphoric acid has pK1 = 2.147, pK2 = 7.207, and pK3 = 12.346.

In the following we focus on diprotic acids (including carbonic acid as a case of exceptional interest). An alternative description, based on the tableaux method,3 is presented as PowerPoint.

Diprotic Acids

When a diprotic acid H2A is added to pure water the equilibrium system is characterized by five dissolved species: H+, OH-, H2A, HA-, and A-2.

Thus, five equations are required for a rigor mathematical description:

 (11a) K1 =  {H+} {HA-} / {H2A} (1st diss. step) (11b) K2 =  {H+} {A-2} / {HA-} (2nd diss. step) (11c) Kw =  {H+} {OH-} (self-ionization of water) (11d) CT =  [H2A] + [HA-] + [A-2] (mass balance) (11e) 0 =  [H+] – [HA-] – 2 [A-2] – [OH-] (charge balance)

The first three equations are mass-action laws; the last two equations represent the mass balance and the charge balance. While the mass-action laws are based on activities (denoted by curly braces), the mass-balance and charge-balance equations rely on molar concentrations (denoted by square brackets).

An exact solution in closed form (i.e. analytical formula) is only possible, when the activities in the first three equations are replaced by molar concentrations. This is valid either in (very) dilute systems or by switching to conditional equilibrium constants cK. In the following we assume that this has been done (without explicitly introducing the notation cK).

Ionization Fractions. The simplest description of the species distribution as a function of pH can be obtained just from a subset of three equations: 11a, (11b) and (11d). From the first two equations one gets (with the abbreviation x = [H+] = 10-pH):

 (12) [H2A] = (x/K1) [HA-]     and     [A-2] = (K2/x) [HA-]

Entering it into 11d yields

 (13) CT  =  (x/K1 + 1 + K2/x) [HA-]

This allow us to write for the three dissolved species the following simple formulas:

 (14) [H2A] = CT a0 [HA-] = CT a1 [A-2] = CT a2

based on three ionization fractions:

 (15a) a0   =   [ 1 + K1/x + K1K2/x2 ]-1 = (x/K1) a1 (15b) a1   =   [ x/K1 + 1 + K2/x ]-1 (15c) a2   =   [ x2/(K1K2) + x/K2 + 1 ]-1 = (K2/x) a1

It’s easy to check that all three coefficients add up to 1:

 (16) a0 + a1 + a2 = 1 (mass balance)

Due to their elegance and simplicity, diagrams of ionization fractions appear in almost any textbook on hydrochemistry. Below is an example for the carbonic acid system (with pK1 = 6.35, pK2 = 10.33):

The formulas in 14 which predict the pH dependence of the three dissolved species can also be put into a more compact form:

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

Be careful though, these equations suggest that the amount of each species is determined by both pH (or x) and CT. But this is not true. CT cannot be chosen freely; its value is completely determined by pH too, but until now, we ignored charge balance and the self-ionization of water, i.e. 11e and (11c). Thus, for very small concentrations when CT < 10-7 M the simple ionization-fractions approach fails.

Exact Analytical Solution. The problem of the simple ionization-fractions approach is that CT is not a free parameter. This problem is solved by incorporating charge balance and the self-ionization of water. Setting 11d into 11e and using 17, together with the abbreviation yi = [H2-i A-i], yields:

 0 =   x – y1 – 2 y2 – Kw/x =   x – Kw/x – (y1 + 2 y2) =   x – Kw/x – CT (a1 + 2 a2)

This gives us the exact relation between the total amount of acid and pH (= –log x):

 (18) $C_T(x) \ =\ \dfrac{x-K_w/x}{a_1 + 2a_2} \ =\ \left(x-\dfrac{K_w}{x}\right) \ \dfrac{K_2/x + 1 + x/K_1} {1 + 2K_2/x}$

In fact, 18 represents the analytical solution of the complete set of five equations defined in 11. Instead of CT(x) we can “reverse” this closed-form expression into the inverse function x(CT). In doing so we obtain a polynomial of 4th order in x, i.e. a quartic equation:

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

To recapitulate: Either 18 or 19 or the set of five equations defined in 11 represent one and the same thing: the complete mathematical description of a diprotic acid. Surely, calculating CT for a given x (or pH) by 18 is much easier than to solve a 4th order equation to get x (or pH) for a given value of CT. [To provide an example, the equations above were applied for the description of the closed and open CO2 systems.]

Based on 18 we are now able to replace the approximations in 17 by an exact formula for the three dissolved species:

 (20) [H2-i A-i]  =  $\left( \dfrac{x-K_w/x}{a_1 + 2a_2} \right) a_i$ for   i = 0, 1, 2

Diprotic Acids including Ampholytes and Conjugate Bases

Any diprotic acid is tight-knit with its conjugate base(s), H2A ⇔ BHA ⇔ B2A, where B refers to a monoacidic base (B+ = Na+, K+, or NH4+). For example: the triple H2CO3, NaHCO3, and Na2CO3 represent such an acid-ampholyte-base system.

Let’s denote the (stoichiometric) number of B+ by n, then we get the compact notation:

 (21) BnH2-n A (or  H2-n A-n) with n = 0 for acid (H2A) n = 1 for ampholyte (BHA) n = 2 for base (B2A)

The set of equations to describe such a system is

 (22a) K1 =  {H+} {HA-} / {H2A} (1st diss. step) (22b) K2 =  {H+} {A-2} / {HA-} (2nd diss. step) (22c) Kw =  {H+} {OH-} (self-ionization) (22d) CT =  [H2A] + [HA-] + [A-2] (mass balance) (22e) 0 =  [H+] + n [H2A] + (n-1) [HA-] + (n-2) [A-2] – [OH-] (proton balance)

It differs from the set of equations (11) only by one equation, namely the last equation, where “charge balance” is replaced by the more general concept of proton balance. For n=0, and only for this case, proton balance and charge balance coincide.

Remarkably enough, the last equation (22e) is the sole equation that explicitly depends on n. The other four equations are independent on the type of reactant we add to water (acid, ampholyte or base). In particular, the ionization fractions derived in 15a to (15c) for the diprotic acid, H2A, are independent of n; they remain exactly the same in our general approach.

What is so great about the set of equations (11) is that it also represents the mathematical heart of buffer systems.

Exact Relation between pH and CT

The extended system of diprotic acids, that includes the conjugate ampholytes and bases, is defined in the set of 22. These five equations can be converted into a single closed-form expression, quite similar as it was done above in 18:

 (23) CT(n,x)  =  $\dfrac{x-K_w/x}{a_1 + 2a_2 - n} \, =\, \left(x-\dfrac{K_w}{x}\right)\, \left(\dfrac{1+2K_2/x} {x/K_1 + 1 + K_2/x} - n\right)^{-1}$

For n=0 it falls back to 18. Again, the function CT(n,x) can be converted into the inverse function x(n,CT). In doing so we obtain a polynomial of 4th order in x, i.e. a quartic equation:

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

In fact, each formula, 23 and 24, comprise three equations in compact form: one for an acid (n=0) – already presented in 18 and 19, one for an ampholyte (n=1), and one for a base (n=2).

Based on 23 we get – in place of 20 – the generalized expression for the three dissolved species:

 (25) [H2-i A-i]  =  $\left( \dfrac{x-K_w/x}{a_1 + 2a_2 - n} \right) a_i$ for   i = 0, 1, 2

Plots. The diagram below displays CT as a function of pH. The solid lines represent 23 for n = 0, 1, and 2. The dots are exact results calculated with aqion (or PhreeqC), where activity corrections are considered. [Activity corrections are relevant for high concentrations of Na2CO3.]

Irrespective of the type of acid, the first factor in parenthesis in 23, x – Kw/x, becomes exactly zero at pH = 7.0, i.e. when x = 10-7. Thus, we always have CT = 0 at pH = 7.0.

Proton Balance Equation (Proton Condition)

The proton balance is of great significance in acid-base systems, and it was already used in 22e above. It is a balance between the species that have excess protons versus those that are deficient in protons relative to a defined proton reference level (PRL).

Example 1. The simplest case is pure water with its three species H+, OH-, and H2O. Choosing H2O as the reference level, the species H+ (or H3O+) is richer in 1 proton (excess proton), while OH- is 1 proton poorer (deficient proton). The proton balance equation becomes:4

PRL excess protons = deficient protons
(26) H2O [H+] = [OH-]

Because water is ever-present in any acid-base system, H+ and OH- always enter the proton balance, one on the left- and the other on the right-hand site of the equation.

Example 2. The carbonic acid system has three distinct reference levels:5

PRL excess protons = deficient protons
(27a) H2CO3 [H+] = [HCO3-] + 2 [CO3-2] + [OH-]
(27b) HCO3- [H+] + [H2CO3] = [CO3-2] + [OH-]
(27c) CO3-2 [H+] + 2 [H2CO3] + [HCO3-] = [OH-]

How to write down these equations? First, the two species H+ and OH- that appear in each equation trace back from the H2O-reference level in 26.6 They have their permanent place on opposite sides in any proton balance. Thus, all we have to do is to add the carbonic-acid species (H2CO3, HCO3-, CO3-2) to the correct side of the equation.

In 27a, H2CO3 is the reference level. There are no carbonate species that have more protons than H2CO3, hence, there is nothing to add to the left-hand side. Conversely, HCO3- is deficient by 1 proton and CO3-2 by 2 protons; therefore, both species enter the right-hand side.7

In 27b, HCO3- is the reference level. From this perspective, H2CO3 has 1 excess proton (species enters the left-hand side), while CO3-2 is deficient by 1 proton (species enters the right-hand side).

In 27c, CO3-2 is the reference level. Now, H2CO3 has 2 excess protons and HCO3- has 1 excess proton (both species enter the left-hand side); but there are no species that have less protons than CO3-2 (i.e. no carbonate species enters the right-hand side).

General Case. Given the proton-reference level by H2-n A-n, the proton balance equation becomes (for n = 0, 1, 2):

PRL 0 = excess protons – deficient protons
(28) H2-n A-n   0 = [H+] + n [H2A] + (n-1) [HA-] + (n-2) [A-2] – [OH-]

This one-liner comprises all three equations of Example 2. Equation (28) was adopted in 22e above.

The proton reference level (PRL) is closely related to the concept of alkalinity and equivalence points (often both terms used as synonyms).

Remarks & Footnotes

1. A detailed mathematical description is provided as pdf

2. In organic acids, the second and third acidity constant may be similar.

3. The best introduction to the tableaux method is given in the classical textbook of F.M.M. Morel and J.G. Hering (Principles and Applications of Aquatic Chemistry, John Wiley, 1993).

4. Square brackets denote molar concentrations.

5. In hydrochemistry, instead of H2CO3 the composite carbonic acid H2CO3* is used. [In the program, H2CO3* is abbreviated by CO2, because almost all of H2CO3* is just dissolved CO2.]

6. The reference level “H2O” is not extra indicated in the table’s PRL column. But keep in mind that it is always present (in addition to H2CO3, HCO3- or CO3-2).

7. If a species has lost 2 protons relative to PRL, its concentration is multiplied by 2.