# The pH of Solutions of Weak Acids

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:52

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

$\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{ A}^{-}$      (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, ca, 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). Ka, called the acid constant, which measures the strength of the acid. The acid constant is defined by the relationship:

$K_{c}\text{ }\times \text{ 55.5 mol dm}^{-3}=K_{a}=\frac{[\text{ H}_{3}\text{O}^{+}][\text{ A}^{-}]}{[\text{ HA }] }$      (2)

(The acid constant is Kc 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, Ka(CH3COOH). You will recall from that example that Ka 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 Ka are only approximate. Only two or possibly three significant figures should be retained. The table below gives the Ka 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 Ka value, the stronger the acid. The strongest acids, like HCl and H2SO4 have Ka values which are too large to measure, while another strong acidAn acid that ionizes completely in a particular solvent., HNO3, has Ka 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 Ka 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 $\rightleftharpoons$ H3O+ + CH3COO– 1.8 × 10–5 Aluminum ion Al(H2O)63+ + H2O $\rightleftharpoons$ H3O+ + Al(H2O)5OH2+ 7.2 × 10–6 Ammonium ion NH4+ + H2O $\rightleftharpoons$ H3O+ + NH3 5.6 × 10–10 Arsenic H3AsO4 + H2O $\rightleftharpoons$ H3O+ + H2AsO–4 H2AsO–4 + H2O $\rightleftharpoons$ H3O+ + HAsO2–4 HAsO2–4 + H2O $\rightleftharpoons$ H3O+ + AsO3–4 K1 = 6.17 × 10–3 K2 = 1.17 × 10–7 K3 = 3.09 × 10–12 Benzoic C6H5COOH + H2O $\rightleftharpoons$ H3O+ + C6H5COO– 1.2 × 10–4 Boric B(OH)3(H2O) + H2O $\rightleftharpoons$ H3O+ + B(OH)–4 5.8 × 10–10 Carbonic H2CO3 + H2O $\rightleftharpoons$ H3O+ + HCO–3 HCO–3 + H2O $\rightleftharpoons$ H3O+ + CO2–3 K1 = 4.3 × 10–7 K2 = 4.7 × 10–11 Citric H3C6H5O7 + H2O $\rightleftharpoons$ H3O+ + H2C6H5O–7 H2C6H5O–7 + H2O $\rightleftharpoons$ H3O+ + HC6H5O2–7 HC6H5O2–7 + H2O $\rightleftharpoons$ H3O+ + C6H5O3–7 K1 = 1.4 × 10–3 K2 = 4.5 × 10–5 K3 = 1.5 × 10–6 Formic HCOOH + H2O $\rightleftharpoons$ 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 $\rightleftharpoons$ H3O+ + N–3 1.0 × 10–5 Hydrochloric HCl + H2O $\rightarrow$ H3O+ + Cl– very large Hydrocyanic HCN + H2O $\rightleftharpoons$ H3O+ + CN– 3.3 × 10–10 Hydrofluoric HF + H2O $\rightleftharpoons$ H3O+ + F– 6.8 × 10–4 Hydrogen peroxide H2O2 + H2O $\rightleftharpoons$ H3O+ + HO–2 2.1 × 10–12 Hydrosulfuric H2S + H2O $\rightleftharpoons$ H3O+ + HS– HS– + H2O $\rightleftharpoons$ H3O+ + S2– K1 = 1 × 10–7 K2 = 1 × 10–19 Hypochlorous HOCl + H2O $\rightleftharpoons$ H3O+ + OCl– 6.8 × 10–8 Nitric HNO3 + H2O $\rightleftharpoons$ H3O+ + NO3– 27.5 Lactic 1.48 × 10–4 Nitrous 7.11 × 10–4 Oxalic H2C2O4 + H2O $\rightleftharpoons$ H3O+ + HC2O–4 HC2O–4 + H2O $\rightleftharpoons$ H3O+ + C2O2–4 K1 = 5.5 × 10–2 K2 = 1.4 × 10–4 Perchloric 131.8 Phenol HC6H5O + H2O $\rightleftharpoons$ H3O+ + C6H5O– 1.7 × 10–10 Phosphoric H3PO4 + H2O $\rightleftharpoons$ H3O+ + H2PO–4 H2PO–4 + H2O $\rightleftharpoons$ H3O+ + HPO2–4 HPO2–4 + H2O $\rightleftharpoons$ H3O+ + PO3–4 K1 = 7.2 × 10–3 K2 = 6.3 × 10–8 K3 = 4.6 × 10–13 Phosphorous H3PO3 + H2O $\rightleftharpoons$ H3O+ + H2PO–3 H2PO–3 + H2O $\rightleftharpoons$ H3O+ + HPO2–3 K1 = 2.4 × 10–2 K2 = 2.9 × 10–7 Propanoic 1.32 × 10–5 Selenic H2SeO4 + H2O $\rightleftharpoons$ H3O+ + HSeO–4 HSeO–4 + H2O $\rightleftharpoons$ H3O+ + SeO2–4 K1 = very large K2 = 2.2 × 10–2 Selenous H2SeO3 + H2O $\rightleftharpoons$ H3O+ + HSeO–3 HSeO–3 + H2O $\rightleftharpoons$ H3O+ + SeO2–3 K1 = 2.5 × 10–3 K2 = 1.6 × 10–9 Sulfuric H2SO4 + H2O $\rightleftharpoons$ H3O+ + HSO–4 HSO–4 + H2O $\rightleftharpoons$ H3O+ + SO2–4 K1 = very large K2 = 1.1 × 10–2 Sulfurous H2SO3 + H2O $\rightleftharpoons$ H3O+ + HSO–3 HSO–3 + H2O $\rightleftharpoons$ H3O+ + SO2–3 K1 = 1.7 × 10–2 K2 = 6.3 × 10–8 Tellurous H2TeO3 + H2O $\rightleftharpoons$ H3O+ + HTeO–3 HTeO–3 + H2O $\rightleftharpoons$ 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, 1st 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:

$\underset{\text{0.0009 mol dm}^{-3}}{\overset{\text{90 }\!\!%\!\!\text{ }}{\mathop{\text{CH}_{3}\text{COOH}}}}\,\text{ + H}_{2}\text{O }\rightleftharpoons \text{ }\underset{\text{0.0001 mol dm}^{-3}}{\overset{\text{10 }\!\!%\!\!\text{ }}{\mathop{\text{CH}_{3}\text{COO}^{-}}}}\,\text{ + }\underset{\text{0.0001 mol dm}^{-3}}{\mathop{\text{H}_{3}\text{O}^{\text{+}}}}\,$      (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 ca 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 Ka and whose stoichiometric concentration is ca. According to the equation for the equilibrium,

$\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{ A}^{-}$

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+]:

 Species HA H3O+ A– Initial concentration (mol dm-3) ca 10–7 0 Change in concentrationa (mol dm-3) –[H3O+] [H3O+] [H3O+] Equilibrium concentration (mol dm-3) ca –[H3O+] [H3O+] [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

$K_{a}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ A}^{-}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HA }\!\!]\!\!\text{ }}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }^{\text{2}}}{c_{a}-\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}\text{ }$      (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 ca and can be ignored when we calculate ca – [H3O+]. That is,

ca – [H3O+] ≈ ca       (5)

Equation (4) then becomes

$K_{a}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }^{\text{2}}}{c_{a}}$

which rearranges to

[H3O+]2Kaca

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

[H3O+] ≈ $\sqrt{K_{a}c_{a}}$      (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, Ka = 1.8 × 10–5 mol dm–3. Since ca = 2.00 × 10–2 mol dm–3, we have

\begin{align} & \text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }=\sqrt{K_{a}c_{a}} \ & \text{ }=\sqrt{\text{(1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol dm}^{-\text{3}}\text{)(2}\text{.00 }\times \text{ 10}^{-\text{2}}\text{ mol dm}^{-\text{3}}\text{)}} \ & \text{ }=\sqrt{\text{3}\text{.6 }\times \text{ 10}^{-\text{7}}\text{ mol}^{\text{2}}\text{ dm}^{-6}}=\sqrt{\text{36 }\times \text{ 10}^{-8}\text{ mol dm}^{-\text{3}}} \ & \text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }=\text{6}\text{.0 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}} \ & \text{ pH}=-\text{log(6}\text{.0 }\times \text{ 10}^{-\text{4}}\text{)}=-\text{(0}\text{.78}-\text{4)}=\text{3}\text{.22} \ \end{align}

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+]/ ca. 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 ca – [H3O+]:

[H3O+]2 = Ka(ca - [H3O+])

or      [H3O+] = $\sqrt{K_{\text{a}}\text{(}c_{\text{a}}-\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ )}}$...(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

\begin{align} & \text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }=\sqrt{K_{a}c_{a}} \ & \text{ }=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\times \text{ 0}\text{.0200 mol dm}^{-\text{3}}} \ & \text{ }=\text{3}\text{.69 }\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \ \end{align}

Checking we find

$\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{c_{a}}=\frac{\text{0}\text{.003 69}}{\text{0}\text{.0200}}=\text{0}\text{.185, that is, 18.5 percent}$

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

\begin{align} & \text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }=\sqrt{K_{a}c_{a}} \ & \text{ }=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\text{(0}\text{.02}-\text{0}\text{.003 69) mol dm}^{-\text{3}}} \ & \text{ }=\text{3}\text{.33 }\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \ \end{align}

Taking a third approximation we find

\begin{align} & \text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}\text{(0}\text{.02}-\text{0}\text{.003 33) mol dm}^{-\text{3}}} \ & \text{ }=\text{3}\text{.37 }\times \text{ 10}^{-3}\text{ mol dm}^{-\text{3}} \ \end{align}

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.