Conjugate Acids and Bases

In Bronsted-Lowry acid-base theory, an acid is defined as a proton (\(H^+\)) donor and a base is a proton acceptor. Whenever an acid donates a proton, whatever is left behind is still capable of accepting a proton back, so it acts as a base. That leftover species is called the conjugate base of the original acid. In the same way, when a base accepts a proton, the new species formed (with the extra \(H^+\) attached) can donate that proton back, so it behaves as an acid. That new species is the conjugate acid of the original base.

A conjugate acid and its conjugate base always come as a matched pair. They are identical in formula except for one single proton. This is the easiest way to check your work: if two species do not differ by exactly one \(H^+\), they are not a conjugate pair.

Every Bronsted-Lowry reaction is really just a proton handoff between two conjugate pairs at once. Consider the general reaction:

Here \(HA\) is acid 1 and its conjugate base is \(A^-\). \(B\) is base 2 and its conjugate acid is \(HB^+\). Notice both pairs differ by exactly one proton, and the reaction is really two conjugate pairs trading that proton back and forth.

Follow these steps whenever you are asked to name a conjugate acid or conjugate base:

1. Decide whether the species is acting as an acid (donating \(H^+\)) or a base (accepting \(H^+\)) in the reaction.
2. If it donates a proton, remove one \(H^+\) and adjust the charge by \(-1\) to write its conjugate base.
3. If it accepts a proton, add one \(H^+\) and adjust the charge by \(+1\) to write its conjugate acid.
4. Check that the two formulas differ by exactly one \(H\) and that the charges balance correctly.

Each row is a conjugate acid-base pair: the left column loses one proton to become the right column, and the right column can gain that proton back to return to the left column.

Ammonia, \(NH_3\), can act as a base. When it accepts a proton it forms its conjugate acid, \(NH_4^+\). If instead \(NH_3\) donates a proton (a much rarer role, but possible), it forms its conjugate base, \(NH_2^-\). Most of the time, when people ask for "the conjugate acid of \(NH_3\)," they mean \(NH_4^+\), since ammonia typically behaves as a base in water:

\(NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\)

Here \(NH_3\)/\(NH_4^+\) is one conjugate pair, and \(H_2O\)/\(OH^-\) is the other.

Sulfuric acid, \(H_2SO_4\), donates a proton to become its conjugate base, \(HSO_4^-\):

\(H_2SO_4 \)→\( H^+ + HSO_4^-\)

Since \(HSO_4^-\) still has an ionizable proton, it can act as an acid again, donating a second proton to form \(SO_4^{2-}\), the conjugate base of \(HSO_4^-\). This is why sulfuric acid is described as diprotic, and the strength of each ionization step is different, which connects directly to how you use the acid dissociation constant for each step.

Acetic acid, \(CH_3COOH\), donates a proton to form its conjugate base, the acetate ion \(CH_3COO^-\):

\(CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+\)

Because acetic acid is a weak acid, this reaction does not go to completion, and a mix of all four species exists at equilibrium.

A useful pattern to remember: the stronger the acid, the weaker its conjugate base, and vice versa. A strong acid like \(HCl\) ionizes almost completely, which means its conjugate base, \(Cl^-\), is extremely weak and has almost no tendency to accept a proton back. On the other hand, a weak acid like acetic acid leaves behind a conjugate base (\(CH_3COO^-\)) that is noticeably more reactive, since it did not fully give up its hold on the proton in the first place. This inverse relationship is the basis for many equilibrium calculations you will see when working with \(K_a\) and \(K_b\) values.

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