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The Brønsted–Lowry model: an acid donates a proton, a base accepts one, and every acid–base reaction pairs each with its conjugate.
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By the end of this lesson you will be able to identify the acid, the base and the two conjugates in a proton-transfer equation, write the conjugate base or conjugate acid of a species, balance a proton transfer with charges, and work out how many moles of hydroxide an acid reacts with from the number of protons it can give.
From Chemistry 1 you can write ionic equations and balance charge, and you know that an atom of hydrogen is one proton and one electron. From the equilibrium unit you know that a reversible reaction runs in both directions at once. This lesson uses both: it defines acids and bases by what their particles do in a reaction, and it reads every acid–base equation in both directions.
A proton, in acid–base chemistry, means a hydrogen ion, $\mathrm{H^+}$: a hydrogen atom that has lost its only electron, so it is nothing but a proton. A Brønsted–Lowry acid is a particle that donates a proton; a Brønsted–Lowry base is a particle that accepts one. The hydronium ion, $\mathrm{H_3O^+}$, is a water molecule that has accepted a proton. A conjugate pair is two species that differ by exactly one proton: the conjugate base is the acid after it has given its proton away, and the conjugate acid is the base after it has taken one. A substance that can act as either is amphiprotic.
The oldest definitions of acids — sour, corrosive, turning litmus red — describe what acids do to people and to dyes. They cannot explain why ammonia, $\mathrm{NH_3}$, which contains no hydroxide at all, makes a solution alkaline. The Brønsted–Lowry model explains it by looking at particles: an acid is a proton donor and a base is a proton acceptor, and in every acid–base reaction a proton moves from one to the other.
Hydrogen chloride dissolving in water is the simplest case:
$$\mathrm{HCl + H_2O \rightarrow H_3O^+ + Cl^-}$$
Follow the proton. Hydrogen chloride has one hydrogen before the arrow and the chloride ion has none after it, so $\mathrm{HCl}$ gave a proton away: it is the acid. Water has two hydrogens before and the hydronium ion has three after, so water took the proton: here it is the base. There is no free $\mathrm{H^+}$ floating in the solution; a bare proton always attaches to something, and in water that something is a water molecule.
Now ammonia:
$$\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$$
Ammonia gains a hydrogen and becomes ammonium, so it is the base. Water loses one and becomes hydroxide, so this time water is the acid. The hydroxide ions that make the solution alkaline were never in the ammonia; they are what is left of the water molecules that gave their protons away.
Those two equations settle the most important point in the unit. Acid and base are roles, not substances. Water is a base in the first reaction and an acid in the second. A label belongs to a particle in one direction of one reaction.
Another way: picture
Think of a ball being passed. The player who throws is the acid and the player who catches is the base. After the pass, the thrower is empty-handed and could catch it back — that is the conjugate base — and the catcher is holding the ball and could throw it back — that is the conjugate acid. The same player can throw in one game and catch in the next.
Another way: steps
To label an acid–base equation:
Most acid–base reactions are reversible, and the reverse reaction is also a proton transfer. In
$$\mathrm{CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-}$$
ethanoic acid gives a proton to water going forwards. Going backwards, hydronium gives a proton to the ethanoate ion. So there are two acids in the equation — one on each side — and two bases.
| Pair | Acid | Conjugate base |
|---|---|---|
| 1 | $\mathrm{CH_3COOH}$ | $\mathrm{CH_3COO^-}$ |
| 2 | $\mathrm{H_3O^+}$ | $\mathrm{H_2O}$ |
The word conjugate means joined: each species on the left is joined to one on the right by the loss or gain of a single proton. To write a conjugate base, take one H off the formula and lower the charge by one: $\mathrm{H_2SO_4}$ becomes $\mathrm{HSO_4^-}$, not $\mathrm{SO_4^{2-}}$. To write a conjugate acid, add one H and raise the charge by one: $\mathrm{NH_3}$ becomes $\mathrm{NH_4^+}$.
A species that sits in the middle of such a chain is amphiprotic. The hydrogencarbonate ion, $\mathrm{HCO_3^-}$, can accept a proton to become carbonic acid or give one to become carbonate — which is how it helps hold the pH of blood steady.
Counting hydrogen atoms in a formula does not tell you how many protons an acid can give. In ethanoic acid, $\mathrm{CH_3COOH}$, three of the four hydrogens are bonded to carbon. Carbon–hydrogen bonds are strong and hardly polar, and those hydrogens stay where they are. Only the hydrogen on the oxygen of the $\mathrm{-COOH}$ group comes off, so ethanoic acid gives one proton per molecule.
| Acid | Hydrogen atoms | Protons given to hydroxide |
|---|---|---|
| hydrochloric, $\mathrm{HCl}$ | 1 | 1 |
| sulfuric, $\mathrm{H_2SO_4}$ | 2 | 2 |
| ethanoic, $\mathrm{CH_3COOH}$ | 4 | 1 |
| citric, $\mathrm{C_6H_8O_7}$ | 8 | 3 |
This is what decides the ratio in a neutralisation. One mole of sulfuric acid needs two moles of sodium hydroxide; one mole of ethanoic acid needs one, whatever its formula looks like.
An acid is a dangerous substance. Some acids are corrosive, and so are some bases. Citric acid is in lemonade and hydrogencarbonate is in baking powder. The model is about what the particles do, not about harm.
A base must contain hydroxide. Ammonia contains none and is a base, because it accepts a proton. Hydroxide ions in its solution come from the water.
The conjugate base has every proton removed. It has exactly one removed. $\mathrm{SO_4^{2-}}$ is the conjugate base of $\mathrm{HSO_4^-}$, not of $\mathrm{H_2SO_4}$.
A substance is always an acid or always a base. Water is either, depending on what it meets, and so is hydrogencarbonate. The role belongs to the reaction.
Every hydrogen in the formula is acidic. Only the hydrogens that can leave as a proton count, and in organic acids that is usually just the one in each $\mathrm{-COOH}$ group.
$\mathrm{HCO_3^- + OH^- \rightarrow CO_3^{2-} + H_2O}$
Four species, and one proton moves between the two on the left.
Hydrogencarbonate goes from one H to none: it is the acid, and carbonate is its conjugate base.
One H fewer and a charge one lower, from $-1$ to $-2$.
Hydroxide goes from one H to two: it is the base, and water is its conjugate acid.
One H more and a charge one higher, from $-1$ to $0$.
$0.050$ mol of phosphoric acid, $\mathrm{H_3PO_4}$, reacts completely with hydroxide ions.
All three hydrogens are on oxygen atoms, and all three can be given up.
Each proton is accepted by one hydroxide ion, making one water molecule.
One proton, one acceptor.
$0.050 \times 3 = 0.15$ mol of hydroxide ions, and the equation is $\mathrm{H_3PO_4 + 3OH^- \rightarrow PO_4^{3-} + 3H_2O}$.
The charges check: $-3$ on each side.
Nitric acid goes from one H to none, so it is the acid and nitrate is its conjugate base.
It lost the proton.
Ammonia goes from three H to four, so it is the $\ldots$
…base, and ammonium is its conjugate acid. No water appears, and it is still an acid–base reaction.
For the reaction $\mathrm{HNO_3 + NH_3 \rightarrow NH_4^+ + NO_3^-}$, give the role of each species in the forward reaction.
| role in the forward reaction | |
|---|---|
| $\mathrm{HNO_3}$ | |
| $\mathrm{NH_3}$ | |
| $\mathrm{NH_4^+}$ | |
| $\mathrm{NO_3^-}$ |
Which is the conjugate base of $\mathrm{H_3PO_4}$?
Match each acid to its conjugate base.
| $\mathrm{Cl^-}$ | $\mathrm{NH_3}$ | $\mathrm{OH^-}$ | $\mathrm{SO_4^{2-}}$ | |
|---|---|---|---|---|
| $\mathrm{HCl}$ | ||||
| $\mathrm{NH_4^+}$ | ||||
| $\mathrm{H_2O}$ | ||||
| $\mathrm{HSO_4^-}$ |
Balance the ionic equation for hydronium ions giving protons to a carbonate ion, which falls apart into carbon dioxide and water. Both the atoms and the total charge must match.
This task has no paper form; do it on a device.
$\mathrm{HCOOH}$ is methanoic acid, the acid in an ant sting. Each molecule gives up $1$ proton(s) to hydroxide ions. How many moles of hydroxide ions react completely with $0.05$ mol of it?
Answer: mol
A learner explains why a solution of ammonia in water turns universal indicator blue. Select every sentence that is wrong.
This task has no paper form; do it on a device.
An indigestion tablet contains $0.4$ g of calcium carbonate, $\mathrm{CaCO_3}$, with a molar mass of $100$ g/mol. In the stomach each carbonate ion accepts protons from hydronium ions until it becomes carbon dioxide and water. What amount of hydronium ions can one tablet take up?
Answer: unit: mol / mmol
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Read $\mathrm{HCl + H_2O \rightarrow H_3O^+ + Cl^-}$ from right to left, as the reverse reaction. Which species is the base in the reverse reaction?
You can follow one proton across an equation and name every species by its role. Say out loud why water can be the base in one reaction and the acid in the next. Next: why some acids give up their protons completely and others hardly at all — strong and weak acids.
10. Your turn: label $\mathrm{HNO_3 + NH_3 \rightarrow NH_4^+ + NO_3^-}$., step 3