3: Conjugate Acid-Base Pairs and pH

One of the more useful aspects of the Brönsted-Lowry definition of acids and bases in helping us deal with the pH of solutions is the concept of the conjugate acid-base pair. We argued qualitatively in the section on conjugate acid-base pairs in aqueous reactions that the strength of an acid and its conjugate base are inversely related. The stronger one is, the weaker the other will be. This relationship can be expressed quantitatively in terms of a very simple mathematical equation involving the appropriate acid and base constants.

Suppose in the general case we have a weak acid HA whose conjugate base is A^–. If either or both of these species are dissolved in H₂O we will have both the following equilibria set up simultaneously.

$\text{HA} + \text{ H}_{\text{2}}\text{O} \rightleftharpoons$ $\text{ H}_{\text{3}}\text{O}^{\text{+}}+ \text{ A}^{-}$ in which HA acts as acid

and $\text{ A}^{-} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons$ $\text{HA} + \text{ OH}^{-}$ in which A^– acts as base

To the first of these equilibria we can apply the equilibrium constant K_a(HA):

$$K_{a}\text{(HA)}=\frac{[\text{ H}_{\text{3}}\text{O}^{\text{+}}][\text{ A}^{-}\text{ }]}{[\text{ HA }]}$$

while to the second we can apply the equilibrium constant K_b(A^–):

$$K_{b}\text{(A}^{-}\text{)}=\frac{[\text{ HA }][\text{ OH}^{-}\text{ }]}{[\text{ A}^{-}\text{ }]\text{ }}$$

Multiplying these two constants together, we obtain a simple relationship between them.

$\begin{align}K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}=\frac{[\text{H}_{\text{3}}\text{O}^{\text{+}}][\text{A}^{-}]}{[\text{HA }]}\times \frac{[\text{HA}][\text{OH}^{-}]}{[\text{A}^{-}]}\\\\\text{ }=[\text{H}_{\text{3}}\text{O}^{\text{+}}][\text{ OH}^{-}]\end{align}$

$$K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}=K_{w}\label{6}$$

If we divide both sides of this equation by the units and take negative logarithms of both sides, we obtain

$$\begin{align}\text{p}K_{a}=-\text{log}\frac{K_{a}\text{(HA)}}{\text{mol L}^{-\text{1}}}-\text{log}\frac{K_{b}\text{(A}^{-}\text{)}}{\text{mol L}^{-\text{1}}}\\\text{ }\\\text{ }=-\text{log}\frac{\text{10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{mol}^{\text{2}}\text{ L}^{-2}}\end{align}$$

$$\text{p}K_{a}\left(\text{HA}\right) + \text{ }\text{p}K_{b}\text{(A}^{-}\text{)}=\text{p}K_{w}\label{10}$$

Thus the product of the acid constant for a weak acid and the base constant for the conjugate base must be K_w, and the sum of pK_a and pK_b for a conjugate acid-base pair is 14.

Equation $\ref{6}$ or $\ref{10}$ enables us to calculate the base constant of a conjugate base from the acid constant of the acid, and vice versa. Given the acid constant for a weak acid like HOCl, for instance, we are able to calculate not only the pH of HOCl solutions but also the pH of solutions of salts like NaOCl or KOCl which are, in effect, solutions of the conjugate base of HOCl, namely, the hypochlorite ion, OCl^–.

Example $\PageIndex{1}$: pH Calculations with K_a

Find the pH of (a) 0.1 M HOCl (hypochlorous acid) and (b) 0.1 M NaOCl (sodium hypochlorite) from the value for K_a given in the table of K_a values.

Solution

a) For 0.1 M HOCl, we find in the usual way that

$\begin{align}\left[\text{H}_{3}\text{O}^{+}\right]=\sqrt{K_{a}c_{a}}\\\text{ }=\sqrt{\text{3}\text{.1 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}}\times \text{ 0}\text{.1 mol}^{\text{2}}\text{ L}^{-\text{2}}}\\\text{ }=\text{5}\text{.57 }\times \text{ 10}^{-5}\text{ mol L}^{-\text{1}}\end{align}$

so that pH = 4.25

b) For 0.1 M NaOCl we must first calculate K_b:

$\begin{align}K_{b}\text{(OCl}^{-}\text{)}=\frac{K_{w}}{K_{a}\text{(HOCl)}}\\\text{ }=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{3}\text{.1 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}}}\\\text{ }=\text{3}\text{.22 }\times \text{ 10}^{-\text{7}}\text{ mol L}^{-\text{1}}\end{align}$

Thus

$\begin{align}\left[\text{OH}^{-}\right]=\sqrt{K_{b}c_{b}}\\\text{ }=\sqrt{\text{3}\text{.22 }\times \text{ 10}^{-\text{7}}\times \text{ 0}\text{.1 mol}^{\text{2}}\text{ L}^{-\text{1}}}\\\text{ }=\text{1}\text{.79 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\end{align}$

From which

pOH = 3.75

and pH = 14.00 – pOH = 10.25

In this, as in all pH problems, it is worth checking that the answers obtained are not wildly unreasonable. A pH of 4 for a weak acid is reasonable, though a little high, but then HOCl is among the weaker acids in table. A pH of 10 corresponds to a mildly basic solution―reasonable enough, for a weak base like OCl^–.

Not only can we use Eq. $\ref{6}$ to find the value of K_b for the base conjugate to a given acid, we can also employ it in the reverse sense to find the value of K_a for the acid conjugate to a given base, as the following example shows.

Example $\PageIndex{2}$: pH Calculation with K_b

Find the pH of 0.05 M NH₄Cl (ammonium chloride), using the value K_b(NH₃) = 1.8 × 10^–5 mol L^–1_.

Solution

We regard this solution as a solution of the weak acid NH₄⁺ and start by finding K_a for this species:

$\begin{align}K_{a}\text{(NH}_{\text{4}}^{\text{+}}\text{)}=\frac{K_{w}}{K_{b}\text{(NH}_{\text{3}}\text{)}}=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}}}\\;\text{ }=\text{5}\text{.56 }\times \text{ 10}^{-\text{10}}\text{ mol L}^{-\text{1}}\end{align}$

We can now evaluate the hydronium-ion concentration with the usual approximation:

$\begin{align}\left[\text{ H}_{3}\text{O}^{+}\right]=\sqrt{K_{a}c_{a}}\\\text{ }=\sqrt{\text{5}\text{.56 }\times \text{ 10}^{-\text{10}}\text{ mol L}^{-\text{1}}\times \text{ 0}\text{.05 mol L}^{-\text{1}}}\\\text{ }=\text{5}\text{.27 }\times \text{ 10}^{-6}\text{ mol L}^{-\text{1}}\end{align}$

whence pH = –log(5.27 × 10^–6) = 5.28

Note

The ammonium ion is a very weak acid (as seen in the Tables of Ka and and Kb values). A solution of NH₄⁺ ions will thus not produce a very acidic solution. A pH of 5 is about the same pH as that of black coffee, not very acidic.

Before the Brönsted-Lowry definition of acids and bases and the idea of conjugate acid-base pairs became generally accepted, the interpretation of acid-base behavior revolved very much around the equation

Acid + base → salt + water

In consequence the idea prevailed that when an acid reacted with a base, the resultant salt should be neither acidic or basic, but neutral. In order to explain why a solution of sodium acetate was basic or a solution of ammonium chloride was acidic, a special term called hydrolysis had to be invoked. Thus, for instance, sodium acetate was said to be hydrolyzed because the acetate ion reacted with water according to the reaction

CH₃COO^- + H₂O $\rightleftharpoons$ CH₃COOH + OH^-

From the Brönsted-Lowry point of view there is, of course, nothing special about such a hydrolysis. It is a regular proton transfer. Nevertheless you should be aware of the existence of the term hydrolysis since it is still often used in this context.

Because the Brönsted-Lowry definition is so successful at explaining why some salt solutions are acidic and some basic, one must beware of making the mistake of assuming that no salt solutions are neutral. Many are. A good example is 0.10 M NaNO₃. This solution is neutral because neither the Na⁺ ion nor the NO₃^– ion shows any appreciable acidic or basic properties. Since NO₃^– is the conjugate base of HNO₃ we might expect it to produce a basic solution, but NO₃^– is such a weak base that it is almost impossible to detect such an effect. Just how weak a base NO₃^– is can be demonstrated using the value of K_a (HNO₃) = 20 mol L^–1 obtained from the Tables of Ka and and Kb values.

$\begin{align}K_{b}\text{(NO}_{\text{3}}^{-}\text{)};=\frac{K_{w}}{K_{a}\text{(HNO}_{\text{3}}\text{)}}\\\text{ }=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{20 mol L}^{-\text{1}}}\\\text{ }=\text{5}\text{.0 }\times \text{ 10}^{-\text{16}}\text{ mol L}^{-\text{1}}\end{align}$

If we now apply the conventional formula from equation 4 from the section on the pH of weak base solutions to calculate [OH^–] in 0.10 M NaNO₃, we obtain

$\begin{align}\left[\text{OH}^{-}\right]=\sqrt{K_{b}c_{b}}\\\text{ }=\sqrt{\text{5}\text{.0 }\times \text{ 10}^{-\text{16}}\times \text{ 1}\text{.0 }\times \text{ 10}^{-\text{1}}\text{ mol L}^{-\text{1}}}\\\text{ }=\text{7}\text{.1 }\times \text{ 10}^{-\text{9}}\text{ mol L}^{-\text{1}}\end{align}$

But this is less than one-tenth the concentration of OH^– ion which would have been present in pure H₂O, with no added NaNO₃. Essentially all the OH^– ions are produced by H₂O, and the pH turns out to be only slightly above 7.00. (Note also that the derivation of equation 4 from the pH of weak base solutions section assumed that the [OH^–] produced by H₂O was negligible. To get an accurate result in this case requires a completely different equation.)

In general all salts in which group I and group II cations are combined with anions which are the conjugate bases of strong acids yield neutral solutions when dissolved in water. Examples are CaI₂, LiNO₃, KCl, Mg(ClO₄)₂.

There is only one exception to this rule. The hydrated beryllium ion, Be(H₂O)₄²⁺, is a weak acid (K_a = 3.2 × 10^–7 mol L^–1) so that solutions of beryllium salts are acidic.

Table $\PageIndex{1}$ The Acid-Base Properties of Some Common Ions

	Cations	Anions
Acidic	Cr^3+, Fe^3+, Al^3+ Hg^2+, Be^2+ NH4^+, H3O^+	HSO4^–
Neutral	Mg^2+, Ca^2+, Sr^2+, Ba^2+ Li^+, Na^+, K^+ Ag^+	NO3^–, ClO4^– Cl^–, Br^–, I^– SO4^2– (very weakly basic)
Basic	None	PO4^3–, CO3^2–, SO3^2– F^–, CN^–, OH^–, S^2– CH3COO^–, HCO3^–

The table lists the acid-base properties of some of the more frequently encountered ions and provides a quick reference for deciding whether a given salt will be acidic, basic, or neutral in solution. Note that the table tells us nothing about the strength of any acid or base. If we need to know more about the pH, other than whether it is above, below, or equal to 7, we need information about the actual value of the acid or base constant. The table also lists the SO₄^2–ion as neutral, though classifying it as very feebly basic would be more accurate.

Example $\PageIndex{3}$: Acid, Base, or Neutral

Classify the following solutions as acidic, basic, or neutral: (a) 1 M KBr; (b) 1 M calcium acetate; (c)1 M MgF₂; (d) 1 M Al(NO₃)₃; (f) 1 M KHSO₄; (f) 1 M NH₄I.

Solution

1. Both cation and anion are neutral: neutral.
2. Cation is neutral but anion basic: basic.
3. Cation is neutral but anion basic: basic.
4. Cation is acidic and anion neutral: acidic.
5. Cation is neutral but anion acidic: acidic.
6. Cation is acidic but anion neutral: acidic.

Example $\PageIndex{4}$: pH Matching

Without actually doing any calculations, match the following solutions and pH values, using the Tables of Ka and and Kb values, and the table on this page.

Aqueous Solution	pH
1 M NH4NO3	8.0
1 M KCN	11.7
1 M Ca(NO3)2	9.4
1 M MgSO4	7.0
1 M CH3COONa(sodium acetate)	1.0
1 M KHSO4	4.6

Solution

The pH of 7.0 is easiest to pick. Only one of the salt solutions given has both a neutral anion and a neutral cation. This is Ca(NO₃)₂. In the case of MgSO₄ the Mg²⁺ ion is neutral but theSO₄^2– ion is very feebly basic; this would agree with a pH of 8.0, only slightly basic. The SO₄^2– ion is such a feeble base because its conjugate acid, HSO₄^–, is quite a strong acid, certainly the most acidic of all the ions featured. Accordingly we expect 1 M KHSO₄ to correspond to the lowest pH, namely, 1.0. The only other acidic solution is 4.6, and this must correspond to 1 M NH₄NO₃ since NO₄⁺ is the only other acidic ion present. Among basic ions the cyanide ion, CN^–, is the strongest. The most basic pH, 11.7, thus corresponds to 1 M KCN. Only one solution is left: 1 M CH₃COONa. This should be feebly basic and so matches the remaining pH of 9.4 rather well.

Contributors

• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.