5.1.3(a)-(c) - Brønsted–Lowry acids, bases and Ka

The Proton-Transfer Model

In earlier acid-base work, it is useful to say that acids release $\mathrm{H^+}$ ions in aqueous solution and alkalis release $\mathrm{OH^-}$ ions. The Brønsted–Lowry model is more general. It asks what happens to the proton.

This is a model of acid-base behaviour. A model is useful when it explains more situations than the previous one. For example, ammonia does not contain $\mathrm{OH^-}$, but it can still behave as a base because it accepts a proton:

In this equation, $\mathrm{NH_3}$ accepts a proton from water, so $\mathrm{NH_3}$ is the Brønsted–Lowry base. Water donates that proton, so water is the Brønsted–Lowry acid in this reaction.

The same species can behave differently in different reactions. Water can accept a proton from a stronger acid, or donate a proton to a stronger base. The role is decided by the proton transfer in the equation, not just by the formula on its own.

Conjugate Pairs and Basicity

When an acid donates a proton, the species left behind is its conjugate base. When a base accepts a proton, the species formed is its conjugate acid.

For ethanoic acid in water:

$\mathrm{CH_3COOH}$ donates $\mathrm{H^+}$, so it is an acid. After losing $\mathrm{H^+}$, it becomes $\mathrm{CH_3COO^-}$, its conjugate base. Water accepts $\mathrm{H^+}$, so water is a base. After accepting $\mathrm{H^+}$, it becomes $\mathrm{H_3O^+}$, its conjugate acid.

So the two conjugate pairs are:

A common OCR trap is to pair species that look similar but do not differ by exactly one $\mathrm{H^+}$. Charges matter. If a neutral molecule loses $\mathrm{H^+}$, the species left is one unit more negative.

Acids can also be described by how many protons each molecule or formula unit can donate.

Examples:

Do not count every hydrogen atom automatically. Ethanoic acid, $\mathrm{CH_3COOH}$, contains four H atoms but is monobasic because only one of them is donated as an acidic proton in this chemistry.

A polybasic acid loses protons in successive steps. For phosphoric acid, the sequence is $\ce{H3PO4 -> H2PO4- -> HPO4^2- -> PO4^3-}$, with one proton removed at each step. Adjacent species are conjugate pairs; $\ce{H3PO4}$ and $\ce{PO4^3-}$ are not a single conjugate pair because they differ by three protons. Tribasic does not mean all three steps dissociate completely.

$\ce{H+}$ in Acid Reactions

The reacting particle in many acid reactions is $\mathrm{H^+(aq)}$. Full equations show the acid and salt formulae, but ionic equations show the role of $\mathrm{H^+}$ directly.

With reactive metals, acid provides $\mathrm{H^+}$, which is reduced to hydrogen gas:

This is also a redox reaction because magnesium is oxidised to $\mathrm{Mg^{2+}}$, while $\mathrm{H^+}$ is reduced to $\mathrm{H_2}$.

With alkalis, $\mathrm{H^+}$ reacts with hydroxide ions to form water:

Metal oxides are bases. The oxide ion accepts protons overall, forming water. For copper(II) oxide reacting with acid, a useful ionic equation is:

Carbonates also behave as bases in acid reactions. The carbonate ion accepts protons overall, and the products include carbon dioxide and water:

If the carbonate is an insoluble solid, the full ionic equation may show the solid carbonate formula, but the chemical idea is the same: $\mathrm{H^+}$ is consumed, and $\mathrm{CO_2}$ is produced.

Worked example: linking a full equation to an ionic equation

Hydrochloric acid reacts with sodium hydroxide:

The reacting ions are $\mathrm{H^+}$ and $\mathrm{OH^-}$. Sodium and chloride ions are spectators, so the net ionic equation is: