Conjugate Acid-Base Pairs
Through examples found in the sections on acids and bases protonThe positively charged particle in an atomic nucleus; its mass is similar to the mass of a hydrogen atom.-transfer processes are broken into two hypothetical steps: (1) donation of a proton by an acid, and (2) acceptance of a proton by a base. (Water served as the base in the acid example and as the acid in the base example) The hypothetical steps are useful because they make it easy to see what species is left after an acid donated a proton and what species is formed when a base accepted a proton. We shall use hypothetical steps or half-equations in this section, but you should bear in mind that free protons never actually exist in aqueous solution.
Suppose we first consider a weak acid, the ammonium ion. When it donates a proton to any other species, we can write the half-equation
NH4+ → H+ + NH3
But NH3 is one of the compounds we know as a weak base. In other words, when it donates a proton, the weak acid NH4+ is transformed into a weak base NH3. Another example, this time starting with a weak base, is provided by fluoride ion:
F– + H+ → HF
The proton converts a weak base into a weak acid.
The situation just described for NH4+ and NH3or for F– and HF applies to all acids and bases. Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a conjugate acidThe acid formed when a base accepts a hydrogen ion (proton).-base pair. Thus NH3 is called the conjugate baseThe base formed when an acid releases a hydrogen ion (proton). of NH4+, and NH4+ is the conjugate acid of NH3. Similarly, HF is the conjugate acid of F–, and F– the conjugate base of HF.
EXAMPLE 1 What is the conjugate acid or the conjugate base of (a) HCl; (b) CH3NH2; (c) OH–; (d) HCO3–.
a) HCl is a strong acidAn acid that ionizes completely in a particular solvent.. When it donates a proton, a Cl– ion is produced, and so Cl– is the conjugate base.
b) CH3NH2 is an amineAn organic compound formally derived from ammonia by the replacement of one or more hydrogen atoms by alkyl groups. Examples are primary amine: RNH2; secondary amine: R2NH; tertiary amine: R3N. and therefore a weak base. Adding a proton gives CH3NH3+, its conjugate acid.
c) Adding a proton to the strong baseA base that dissociates completely or ionizes completely in a particular solvent. OH– gives H2O its conjugate acid.
d) Hydrogen carbonate ion, HCO3–, is derived from a diproticDescribes an acid that can donate two hydrogen ions (protons) to a base. acid and is amphiproticA molecule or ion that can both gain a proton and lose a proton; that is, a species that can serve as either an Bronsted-Lowry acid or a Bronsted-Lowry base.. Its conjugate acid is H2CO3, and its conjugate base is CO32–.
The use of conjugate acid-base pairs allows us to make a very simple statement about relative strengths of acids and bases. The stronger an acid, the weaker its conjugate base, and, conversely, the stronger a base, the weaker its conjugate acid.
TABLE 1 Some Important Conjugate Acid-Base Pairs Arranged in Order of Increasing Strength of the Base.
A strong acid like HCl donates its proton so readily that there is essentially no tendency for the conjugate base Cl– to reaccept a proton. Consequently, Cl– is a very weak base. A strong base like the H– ion accepts a proton and holds it so firmly that there is no tendency for the conjugate acid H2 to donate a proton. Hence, H2 is a very weak acid.
Table 1 gives a list of some of the more important conjugate acid-base pairs in order of increasing strength of the base. This table enables us to see how readily a given acid will react with a given base. The reactions with most tendency to occur are between the strong acids in the top left-hand comer of the table and the strong bases in the bottom right-hand comer. If a line is drawn from acid to base for such a reaction, it will have a downhill slope. By contrast, reactions with little or no tendency to occur (between the weak acids at the bottom left and the weak bases at the top right) correspond to a line from acid to base with an uphill slope. When the slope of the line is not far from horizontal, the conjugate pairs are not very different in strength, and the reaction goes only part way to completion. Thus, for example, if the acid HF is compared with the base CH3COO–, we expect the reaction to go part way to completion since the line is barely downhill.
EXAMPLE 2 Write a balanced equationA representation of a chemical reaction that has values of the stoichiometric coefficients of reactants and products such that the number of atoms of each element is the same before and after the reaction. to describe the reaction which occurs when a solution of potassium hydrogen sulfate, KHSO4, is mixed with a solution of sodium bicarbonate, NaHCO3.
Solution The Na+ ions and K+ ions have no acid-base properties and function purely as spectator ions. Therefore any reaction which occurs must be between the hydrogen sulfate ion, HSO4– and the hydrogen carbonate ion, HCO3–. Both HSO4– and HCO3– are amphiprotic, and either could act as an acid or as a base. The reaction between them is thus either
HCO3– + HSO4– → CO32– + H2SO4
or HSO4– + HCO3– → SO42– + H2CO3
Table 1 tells us immediately that the second reaction is the correct one. A line drawn from HSO4– as an acid to HCO3– as a base is downhill. The first reaction cannot possibly occur to any extent since HCO3– is a very weak acid and HSO4– is a base whose strength is negligible.
EXAMPLE 3 What reactions will occur when an excess of acetic acid is added to a solution of potassium phosphate, K3PO4?
Solution The line joining CH3COOH to PO43– in Table 1 is downhill, and so the reaction
CH3COOH + PO43– → CH3COO– + HPO42–
should occur. There is a further possibility because HPO42– is itself a base and might accept a second proton. The line from CH3COOH to HPO42– is also downhill, but just barely, and so the reaction
CH3COOH + HPO42– → CH3COO– + H2PO4–
can occur, but it does not go to completion. Hence double arrows are used. Although H2PO4– is a base and might be protonated to yield phosphoric acid, H3PO4, a line drawn from CH3COOH to H2PO4– is uphill, and so this does not happen.