# 14.6: 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 Ka(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 Kb(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 Kw, and the sum of pKa and pKb 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 Ka

Find the pH of (a) 0.1 M HOCl (hypochlorous acid) and (b) 0.1 M NaOCl (sodium hypochlorite) from the value for Ka given in the table of Ka 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 Kb:

\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 Kb for the base conjugate to a given acid, we can also employ it in the reverse sense to find the value of Ka for the acid conjugate to a given base, as the following example shows.

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

Find the pH of 0.05 M NH4Cl (ammonium chloride), using the value Kb(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 Ka 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 Ka (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 (Ka = 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 Cr3+, Fe3+, Al3+ Hg2+, Be2+ NH4+, H3O+ HSO4– Neutral Mg2+, Ca2+, Sr2+, Ba2+ Li+, Na+, K+ Ag+ NO3–, ClO4– Cl–, Br–, I– SO42– (very weakly basic) Basic None PO43–, CO32–, SO32– F–, CN–, OH–, S2– 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 SO42–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 theSO42– ion is very feebly basic; this would agree with a pH of 8.0, only slightly basic. The SO42– 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.