The pH of Solutions of Weak Acids

When any weak acidAn acid that ionizes only partially in a given solvent., which we will denote by the general formula HA, is dissolved in water, the reaction

      (1)

proceeds to only a limited extent, and we must allow for this in calculating the hydronium-ion concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated. and hence the pHA logarithmic measure of the concentration of hydrogen (hydronium) ion; pH = -log10([H+]) or pH = -log10([H3O+]). of such a solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture.. In general the pH of a solution of a weak acid depends on only two factors, the concentration of the acid, c_a, and the magnitude of an equilibrium constantThe value of the equilibrium constant expression when equilibrium concentrations are substituted; a value greater than one indicates the position of equilibrium lies toward products (product-favored), and a value less than one indicates the position of equilibrium lies toward reactants (reactant-favored). K_a, called the acid constant, which measures the strength of the acid. The acid constant is defined by the relationship:

      (2)

(The acid constant is K_c multiplied by the constant concentration of water, as already defined in the section on the law of chemical equilibrium. It is also called the ionizationA process in which an atom, molecule, or negative ion loses an electron; a process in which a covalent molecule reacts with a solvent to form positive and negative ions; for example, a weak acid reacting with water to form its conjugate base (an anion) and a hydrogen (hydronium) ion. constant or the dissociationThe breaking apart of one species into two or more smaller species; often applied to ions in a crystal lattice, which dissociate when the ionic solid dissolves in water. Dissociation refers to separation of particles that already exist; ionization refers to the formation of ions from neutral species, as in the ionization of a weak acid in aqueous solutoin. constant of the acid.)

In Example 5 on The Law of Chemical Equilibrium we showed how measurements of the conductivity of acetic acid solutions could be used to find the acid constant for acetic acid, K_a(CH₃COOH). You will recall from that example that K_a was only approximately a constant, varying from a value of 1.81 × 10^–5 mol dm^–3 in very dilute solutions to a value of 1.41 × 10^–7 mol dm^–3 in a 1M solution. A similar variation is found for other weak acids, so that most of the calculations we do using K_a are only approximate. Only two or possibly three significant figures should be retained. The table below gives the K_a values for a few selected acids arranged in order of their strength. This table is part of our larger collection of acid-base resources. It is at once apparent from this table that the larger the K_a value, the stronger the acid. The strongest acids, like HCl and H₂SO₄ have K_a values which are too large to measure, while another strong acidAn acid that ionizes completely in a particular solvent., HNO₃, has K_a value close to 20 mol dm^–3. Typical weak acids such as HF and CH₃COOH have acid constants with a value of 10^–4 or 10^–5 mol dm^–3. Acids like the ammonium ion, NH₄⁺, and hydrogen cyanide, HCN, for which K_a is less than 10^–9 mol dm^–3 are very weakly acidic.

The Acid Constants for Some Acids at 25°C.

Acid	Formula and Ionization Equation	Ka
Acetic	CH3COOH + H2O H3O+ + CH3COO–	1.8 × 10–5
Aluminum ion	Al(H2O)63+ + H2O H3O+ + Al(H2O)5OH2+	7.2 × 10–6
Ammonium ion	NH4+ + H2O H3O+ + NH3	5.6 × 10–10
Arsenic	H3AsO4 + H2O H3O+ + H2AsO–4 H2AsO–4 + H2O H3O+ + HAsO2–4 HAsO2–4 + H2O H3O+ + AsO3–4	K1 = 6.17 × 10–3 K2 = 1.17 × 10–7 K3 = 3.09 × 10–12
Benzoic	C6H5COOH + H2O H3O+ + C6H5COO–	1.2 × 10–4
Boric	B(OH)3(H2O) + H2O H3O+ + B(OH)–4	5.8 × 10–10
Carbonic	H2CO3 + H2O H3O+ + HCO–3 HCO–3 + H2O H3O+ + CO2–3	K1 = 4.3 × 10–7 K2 = 4.7 × 10–11
Citric	H3C6H5O7 + H2O H3O+ + H2C6H5O–7 H2C6H5O–7 + H2O H3O+ + HC6H5O2–7 HC6H5O2–7 + H2O H3O+ + C6H5O3–7	K1 = 1.4 × 10–3 K2 = 4.5 × 10–5 K3 = 1.5 × 10–6
Formic	HCOOH + H2O H3O+ + HCOO–	3.0 × 10–4
Hexaaquaaluminum ion		1.0 × 10–5
Hexaaquacopper (II) ion		4.57 × 10–8
Hexaaquairon (II) ion		2.63 × 10–10
Hexaaquairon (III) ion		6.76 × 10–3
Hexaaqualead (II) ion		1.38 × 10–8
Hexaaquanickel (II) ion		1.38 × 10–10
Hydrazoic	HN3 + H2O H3O+ + N–3	1.0 × 10–5
Hydrochloric	HCl + H2O H3O+ + Cl–	very large
Hydrocyanic	HCN + H2O H3O+ + CN–	3.3 × 10–10
Hydrofluoric	HF + H2O H3O+ + F–	6.8 × 10–4
Hydrogen peroxide	H2O2 + H2O H3O+ + HO–2	2.1 × 10–12
Hydrosulfuric	H2S + H2O H3O+ + HS– HS– + H2O H3O+ + S2–	K1 = 1 × 10–7 K2 = 1 × 10–19
Hypochlorous	HOCl + H2O H3O+ + OCl–	6.8 × 10–8
Nitric	HNO3 + H2O H3O+ + NO3–	27.5
Lactic		1.48 × 10–4
Nitrous		7.11 × 10–4
Oxalic	H2C2O4 + H2O H3O+ + HC2O–4 HC2O–4 + H2O H3O+ + C2O2–4	K1 = 5.5 × 10–2 K2 = 1.4 × 10–4
Perchloric		131.8
Phenol	HC6H5O + H2O H3O+ + C6H5O–	1.7 × 10–10
Phosphoric	H3PO4 + H2O H3O+ + H2PO–4 H2PO–4 + H2O H3O+ + HPO2–4 HPO2–4 + H2O H3O+ + PO3–4	K1 = 7.2 × 10–3 K2 = 6.3 × 10–8 K3 = 4.6 × 10–13
Phosphorous	H3PO3 + H2O H3O+ + H2PO–3 H2PO–3 + H2O H3O+ + HPO2–3	K1 = 2.4 × 10–2 K2 = 2.9 × 10–7
Propanoic		1.32 × 10–5
Selenic	H2SeO4 + H2O H3O+ + HSeO–4 HSeO–4 + H2O H3O+ + SeO2–4	K1 = very large K2 = 2.2 × 10–2
Selenous	H2SeO3 + H2O H3O+ + HSeO–3 HSeO–3 + H2O H3O+ + SeO2–3	K1 = 2.5 × 10–3 K2 = 1.6 × 10–9
Sulfuric	H2SO4 + H2O H3O+ + HSO–4 HSO–4 + H2O H3O+ + SO2–4	K1 = very large K2 = 1.1 × 10–2
Sulfurous	H2SO3 + H2O H3O+ + HSO–3 HSO–3 + H2O H3O+ + SO2–3	K1 = 1.7 × 10–2 K2 = 6.3 × 10–8
Tellurous	H2TeO3 + H2O H3O+ + HTeO–3 HTeO–3 + H2O H3O+ + TeO2–3	K1 = 1.9 × 10–3 K2 = 4.6 × 10–10

• Taken from Hogfelt, E. Perrin, D. D. Stability Constants of MetalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. Ion Complexes, 1^st ed. Oxford; New Pergamon, 1979-1982. International Union of Pure and Applied Chemistry, Commission on Equilibrium. ISBN: 0080209580

Before we can go on to discuss how the hydronium-ion concentration and the pH of a solution of a weak acid depend on the concentration of the acid, we need to clarify a point of terminology. In order to do this let us take as an example a 0.0010 M solution of acetic acid. Conductivity measurements discussed in the section on weak acids in reactions in aqueous solutions show that only about 10 percent of the acid molecules have donated protons to water at any given time. We thus have a situation which can be summarized schematically in the following way:

      (3)

In such a solution there is some ambiguity as to what we mean by the phrase concentration of acetic acid. Do we mean 0.0010 mol dm^–3, or do we mean 90 percent of this value, namely, 0.0009 mol dm^–3? In order to resolve this difficulty, we will use the term stoichiometric concentration of acid and the symbol c_a to indicate the quantity 0.0010 mol dm^–3, that is, to indicate the total amount of acetic acid originally added per unitA particular measure of a physical quantity that is used to express the magnitude of the physical quantity; for example, the meter is the unit of the physical quantity, length. volume of solution. On the other hand we will use the term equilibrium concentration

and the symbol [CH₃COOH] indicate the quantity 0.0009 mol dm^–3 that is, the final concentration of this species in the equilibrium mixtureA combination of two or more substances in which the substances retain their chemical identity..

Let us now consider the general problem of finding [H₃O⁺]in a solution of a weak acid HA whose acid constant is K_a and whose stoichiometric concentration is c_a. According to the equation for the equilibrium,

for every moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. of H₃O⁺ produced, there must also be a mole of A^– produced. At the same time a mole of HA and a mole of H₂O must be consumed. Since the volume which all these species occupy is the same, any increase in [H₃O⁺] must be accompanied by an equal increase in [A^–] and an equal decrease in [HA]. Consequently we can draw up the following table (in which equilibrium concentrations of all species have been expressed in terms of [H₃O⁺]:

^aThe hydronium-ion concentration actually increases from 10⁷ mol dm^-3 to the equilibrium concentration, and so the change in each of the concentrations is ± ([H₃O⁺ ] –10^-7 mol dm^-3). However, the concentration of hydronium ions produced by the weak acid is usually so much larger than –10^-7 mol dm^-3 that the latter quantity can be ignored. In the case of 0.0010 M acetic acid, for example, [H₃O⁺ ] ≈ 1 × 10^-4 mol dm^-3. Subtracting gives:

   0.000 100 0 mol dm^-3

– 0.000 000 1 mol dm^-3

   0.000 099 9 mol dm^-3

which is very close to 1 × 10^-4 mol dm^-3.

We can now substitute the equilibrium concentrations into the expression

      (4)

This could be solved for [H₃O⁺] by means of the quadratic formula, but in most cases a quicker approximate method is available. Since the acid is weak, only a small fraction of the HA molecules will have donated protons to form H₃O⁺ ions. Therefore [H₃O⁺] is only a small fraction of c_a and can be ignored when we calculate c_a – [H₃O⁺]. That is,

c_a – [H₃O⁺] ≈ c_a       (5)

Equation (4) then becomes

which rearranges to

[H₃O⁺]² ≈ K_ac_a

Taking the square root of both sides gives an important approximate formula:

[H₃O⁺] ≈       (6)

EXAMPLE 1 Use Eq. (6) to calculate the pH of a 0.0200-M solution of acetic acid. Compare this with the pH obtained using the [H₃O⁺] of 5.92 × 10^–4 mol dm^–3 derived from accurate conductivity measurements.

Solution From first table above, K_a = 1.8 × 10^–5 mol dm^–3. Since c_a = 2.00 × 10^–2 mol dm^–3, we have

Using the accurate [H₃O⁺] from conductivity measurements,

pH = –log(5.92 × 10^–4) = –(0.772 – 4) = 3.228

Note that the approximate equation gives an [H₃O⁺] which differs by 1 in the second significant digit from the accurate value. The calculated pH differs by 1 in the second place to the right of the decimal―roughly the same as the accuracyThe extent to which an experimental value agrees with true value for a quantity. of simple pH measurements.

In a few cases, if the acid is quite strong or the solution very dilute, it turns out that Eq. (6) is too gross an approximation. A convenient rule of thumb for determining when this is the case is to take the ratio [H₃O⁺]/ c_a. If this is larger than 5 percent or so, we need to make a second approximation, and then the rules for successive approximations can be applied. Equation (4) can be converted to a convenient form for successive approximations by multiplying both sides by c_a – [H₃O⁺]:

         [H₃O⁺]² = K_a(c_a - [H₃O⁺])

or      [H₃O⁺] = ...(7)

To get a second approximation for [H₃O⁺] we can feed the first approximation into the right side of this equation. The exact procedure is detailed in the following example.

EXAMPLE 2 Find the pH of 0.0200 M HF using the table of Acid Constants on this page.

Solution Using Eq. (6) we find

Checking we find

A second approximation is thus necessary. We feed our first approximation of [H₃O⁺]= 3.69 × 10^–3 mol dm^–3 into Eq. (7):

Taking a third approximation we find

Since this differs from the second approximation by less than 5 percent, we take it as the final result. The pH is

pH = –log(3.37 × 10^–3) = 2.47

Cross check: Since HF is a stronger acid than acetic acid, we expect this solution to have a lower pH than 0.02 M CH₃COOH and indeed it does.