5.1.3 (a,b) Brønsted–Lowry acids and bases

Syllabus

Brønsted–Lowry acids and bases

(a) (i) a Brønsted–Lowry acid as a species that donates a proton and a Brønsted–Lowry base as a species that accepts a proton

{Learners should be able to identify acid–base pairs in equations for acid–base equilibria.}

(ii) use of the term conjugate acid–base pairs

(iii) monobasic, dibasic and tribasic acids

(b) the role of H⁺ in the reactions of acids with metals and bases (including carbonates, metal oxides and alkalis), using ionic equations

What does this mean?

Brønsted–Lowry definitions

You may remember from GCSE that Acids are often defined as being substances that dissociate in water (break up) to form H⁺ ions.

This is a perfectly adequate definition in most (but not all) instances, and the Brønsted–Lowry definition differs only in that it uses the phrase "proton-donor".

But an H⁺ ion is a proton - so anything that breaks up to produce H⁺ ions is clearly a proton donor.

A GCSE definition of a base is generally "something which neutralises an acid", which is clearly true.

But how?

Well, if acids donate protons (H⁺) the base must be able to accept them.

Mono-, di- and tri-basic acids

Most GCSEs require you to learn the formula of three strong acids: Hydrochloric, Nitric and Sulphuric.

1. Hydrochloric acid: HCl à H⁺_(aq) + Cl^- _(aq)
2. Nitric acid: HNO₃ à H⁺_(aq) + NO₃^- _(aq)
3. Sulphuric acid: H₂SO₄ à 2H⁺_(aq) + SO₄^2- _(aq)

All the above are considered to be Strong Acids because they "Completely Dissociate" - 100% of the acid molecules break up on dissolving in water (not strictly true but close enough).

Hydrochloric and Nitric Acid are considered monobasic because when their molecules dissociate they make only 1 H⁺ ion/ proton.

Sulphuric Acid is considered dibasic because when its molecules dissociate they make only 2 H⁺ ions/ protons.

You should now learn the formula of Phosphoric Acid (H₃PO₄) - this is tri-basic because its molecules can dissociate to make theree H⁺ ions (although this is certainly not 100% the case).

Phosphoric acid: H₃PO₄ à 3H⁺_(aq) + PO₄^3- _(aq)

Having studied some Organic Chemistry you will know about Carboxyllic Acids.

Standard Carboxyllic Acids (like Ethanoic) are mono-basic.

CH₃COOH ⇌ CH₃COO-_(aq) + H⁺_(aq)

But they are weak acids because they only partly dissociate - most of the molecules dissolve without breaking up.

There is no reason why an organic molecule cannot have more than one Carboxyllic Acid group. So Ethanedioc acid would be considered di-basic (but still weak).

HOOCCOOH ⇌2H⁺_(aq) + ^-OOCCOO^-_(aq)

Conjugate Acids and Bases.

In reality H⁺ ions rarely exist in water in isolation. They form co-ordinate(dative) bonds with water molecules.

HCl + H₂O ⇌ H₃O+ Cl^-

In this instance the HCl is clearly donating a proton (H⁺) so is the acid.

The H₂O has clearly accepted it and so is the base.

However, this is an equilibrium.

The Hydronium ion (H₃O⁺) may donate its "spare" H⁺ to the Cl^-.

If this happens the Hydronium ion would be acting as an acid and the Chloride ion as a base (H⁺ acceptor)

So we would say that the Hydronium ion is the conjugate acid of the water, and the Chloride ion is the conjugate base of the HCl.

You may well be asked to label the acid-base pairs in a neutralisation equation.

eg

In this case the water accepts a proton (base) the Hydronium ion formed is the conjugate acid again because it may doante the proton it gained.

This time ethanoic acid is the acid, the ethanoate ion is the conjugate base because it may re-accept the proton it lost.

Here Ammonia is the base (accepts a proton.

Ammonium ions are the conjugate acid (may re-donate the proton).

Water is the acid (donate the proton).

OH^- is the base (may re-accept the proton)

The role of H⁺

With Metals

Acid + Metal à Salt + Hydrogen

Using the example of the reaction of Calcium and HCl

2HCl(aq) + Ca(s) à CaCl_2(aq) + H_2(g)

Since dissolved ionic substances dissociate we could have written the above as:

2H⁺(aq) + 2Cl^-(aq) + Ca(s) à Ca²⁺(aq)+ 2Cl^-_(aq) + H_2(g)

To make this into an Ionic Equation we remove any ions that have not changed Oxidation Number or state.

2H⁺(aq) + Ca(s) à Ca²⁺(aq)+ H_2(g)

H⁺ ions are becoming elemental Hydrogen atoms (Hydrogen as an element).

This is reduction because the Oxidation Number changes from +1 for H⁺ to 0 for H₂.

And since the Hydrogen ions are reduced they must be an oxidising agent.

The Calcium is Oxidised from 0 to +2

With Alkalis

Alkalis are soluble bases that dissociate in water making OH^-

OH^- ions readily accept H+ to make water.

H⁺(aq) + OH^- (aq) à H₂O_(l)

or more likely:

H₃O⁺(aq) + OH^- (aq) à H₂O_(l)

This is the ionic equation for neutralisation of an acid with an alkali.

There are no changes in Oxidation Number - it is not a Redox reaction.

With Metal Oxides

You'll probably remember that Metal Oxides are bases, and the general equation:

Acid + Base à Salt + Water

Usually GCSEs focus on:

Sulphuric acid + Copper Oxide à Copper Sulphate + Water

or

H₂SO_4(aq) + CuO_(s) à CuSO_4(aq) + H₂O_(l)

As an ionic equation we would remove all the ions that are unchanged:

2H⁺_(aq) + CuO_(s) à Cu²⁺ + H₂O_(l)

Clearly this time it is the Oxide ions in the Copper Oxide that are accepting the protons.

We might think of this happening in two stages.

O^2- + H⁺ àOH^- (although this isn't the ionic equation for this step)

Then

OH^- + H⁺ àH₂O (but this is the ionic equation for this step)

Again, there are no changes in Oxidation Number - it is not a Redox reaction.

It is straightforward neutralisation.

With Carbonates

Carbonate + Acid à Salt + Water + Carbon Dioxide

eg

Na₂CO_3(aq) + H₂SO_4(aq) à Na₂SO_4(aq) + CO_2(g) + H₂O_(l)

2Na⁺_(aq) + CO₃^2-_(aq) + 2H⁺_(aq) + SO₄^2- _(aq) à 2Na⁺_(aq) + SO₄^2- _(aq) + CO_2(g) + H₂O_(l)

Or as an Ionic Equation:

CO₃^2-_(aq) + 2H⁺_(aq) à CO_2(g) + H₂O_(l)

This is another two step process.

CO₃^2-_(aq) + H⁺_(aq) à HCO₃^-_(aq)

_{Followed by}

HCO₃^-_(aq) + H⁺_(aq) à CO_2(g) + H₂O_(l)

_{At no time do Oxidation Numbers change so both steps are neutralisation.}