# Defining Acids and Bases

## Defining acids and bases

One of the first things that was noted about acids is that they have a sour taste. Bases were noted to have a soapy feel and a bitter taste. However you cannot go around tasting and feeling unknown substances since they may be harmful. Also when chemists started to write down chemical reactions more practical definitions were needed.

A number of definitions for acids and bases have developed over the years. One of the earliest was the Arrhenius definition. Arrhenius (1887) noticed that water dissociates (splits up) into hydronium $$(\text{H}_{3}\text{O}^{+})$$ and hydroxide $$(\text{OH}^{-})$$ ions according to the following equation:

$$2\text{H}_{2}\text{O (l)} \rightarrow \text{H}_{3}\text{O}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)}$$

#### Tip:

Arrhenius described an acid as a compound that increases the concentration of $$\text{H}_{3}\text{O}^{+}$$ ions in solution and a base as a compound that increases the concentration of $$\text{OH}^{-}$$ ions in solution.

Look at the following examples showing the dissociation of hydrochloric acid and sodium hydroxide (a base) respectively:

1. $$\text{HCl (aq)} + \text{H}_{2}\text{O}\text{(l)} \rightarrow \text{H}_{3}\text{O}^{+}\text{(aq)} + \text{Cl}^{-}\text{(aq)}$$

Hydrochloric acid in water increases the concentration of $$\text{H}_{3}\text{O}^{+}$$ ions and is therefore an acid.

2. $$\text{NaOH (s)} \stackrel{\text{H}_{2}\text{O}}{\longrightarrow} \text{Na}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)}$$

Sodium hydroxide in water increases the concentration of $$\text{OH}^{-}$$ ions and is therefore a base.

Note that we write $$\stackrel{\text{H}_{2}\text{O}}{\longrightarrow}$$ to indicate that water is needed for the dissociation.

However, this definition could only be used for acids and bases in water. Since there are many reactions which do not occur in water it was important to come up with a much broader definition for acids and bases.

In 1923, Lowry and Bronsted took the work of Arrhenius further to develop a broader definition for acids and bases. The Bronsted-Lowry model defines acids and bases in terms of their ability to donate or accept protons.

### Definition: Acids

A Bronsted-Lowry acid is a substance that gives away protons (hydrogen cations $$\text{H}^{+}$$), and is therefore called a proton donor.

### Definition: Bases

A Bronsted-Lowry base is a substance that takes up protons (hydrogen cations $$\text{H}^{+}$$), and is therefore called a proton acceptor.

Below are some examples:

1. $$\text{HCl (aq)} + \text{NH}_{3}\text{(aq)} \rightarrow \text{NH}_{4}^{+}\text{(aq)} + \text{Cl}^{-}\text{(aq)}$$

We highlight the chlorine and the nitrogen so that we can follow what happens to these two elements as they react. We do not highlight the hydrogen atoms as we are interested in how these change. This colour coding is simply to help you identify the parts of the reaction and does not represent any specific property of these elements.

$$\text{H}{\color{red}{\text{Cl}}} \text{ (aq)} + {\color{blue}{\text{N}}}\text{H}_{3}\text{(aq)} \rightarrow {\color{blue}{\text{N}}}\text{H}_{4}^{+}\text{(aq)} + {\color{red}{\text{Cl}}}^{-}\text{(aq)}$$

In order to decide which substance is a proton donor and which is a proton acceptor, we need to look at what happens to each reactant. The reaction can be broken down as follows:

$$\text{H}{\color{red}{\text{Cl}}}\text{ (aq)} \rightarrow {\color{red}{\text{Cl}}}^{-}\text{(aq)}$$ and

$${\color{blue}{\text{N}}}\text{H}_{3}\text{(aq)} \rightarrow {\color{blue}{\text{N}}}\text{H}_{4}^{+}\text{(aq)}$$

From these reactions, it is clear that $$\text{HCl}$$ is a proton donor and is therefore an acid, and that $$\text{NH}_{3}$$ is a proton acceptor and is therefore a base.

2. $$\text{CH}_{3}\text{COOH (aq)} + \text{H}_{2}\text{O (l)} \rightarrow \text{H}_{3}\text{O}^{+}\text{(aq)} + \text{CH}_{3}\text{COO}^{-}\text{(aq)}$$

Again we highlight the parts of the reactants that we want to follow in this reaction:

$${\color{red}{\text{CH}_{3}\text{COO}}}\text{H (aq)} + \text{H}_{2}{\color{blue}{\text{O}}}\text{ (l)} \rightarrow \text{H}_{3}{\color{blue}{\text{O}}}^{+}\text{(aq)} + {\color{red}{\text{CH}_{3}\text{COO}}}^{-}\text{(aq)}$$

The reaction can be broken down as follows:

$${\color{red}{\text{CH}_{3}\text{COO}}}\text{H (aq)} \rightarrow {\color{red}{\text{CH}_{3}\text{COO}}}^{-}\text{(aq)}$$ and

$$\text{H}_{2}{\color{blue}{\text{O}}}\text{ (l)} \rightarrow \text{H}_{3}{\color{blue}{\text{O}}}^{+}\text{(aq)}$$

In this reaction, $$\text{CH}_{3}\text{COOH}$$ (acetic acid or vinegar) is a proton donor and is therefore the acid. In this case, water acts as a base because it accepts a proton to form $$\text{H}_{3}\text{O}^{+}$$.

3. $$\text{NH}_{3}\text{(aq)} + \text{H}_{2}\text{O (l)} \rightarrow \text{NH}_{4}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)}$$

Again we highlight the parts of the reactants that we want to follow in this reaction:

$${\color{blue}{\text{N}}}\text{H}_{3}\text{(aq)} + \text{H}_{2}{\color{red}{\text{O}}}\text{ (l)} \rightarrow {\color{blue}{\text{N}}}\text{H}_{4}^{+}\text{(aq)} + {\color{red}{\text{O}}}\text{H}^{-}\text{(aq)}$$

The reaction can be broken down as follows:

$$\text{H}_{2}{\color{red}{\text{O}}}\text{ (l)} \rightarrow {\color{red}{\text{O}}}\text{H}^{-}\text{(aq)}$$ and

$${\color{blue}{\text{N}}}\text{H}_{3}\text{(aq)} \rightarrow {\color{blue}{\text{N}}}\text{H}_{4}^{+}\text{(aq)}$$

Water donates a proton and is therefore an acid in this reaction. Ammonia accepts the proton and is therefore the base.

Notice in these examples how we looked at the common elements to break the reaction into two parts. So in the first example we followed what happened to chlorine to see if it was part of the acid or the base. And we also followed nitrogen to see if it was part of the acid or the base. You should also notice how in the reaction for the acid there is one less hydrogen on the right hand side and in the reaction for the base there is an extra hydrogen on the right hand side.

### Amphoteric substances

In examples $$\text{2}$$ and $$\text{3}$$ above we notice an interesting thing about water. In example $$\text{2}$$ we find that water acts as a base (it accepts a proton). In example $$\text{3}$$ however we see that water acts as an acid (it donates a proton)!

Depending on what water is reacting with it can either react as a base or as an acid. Water is said to be amphoteric. Water is not unique in this respect, several other substances are also amphoteric.

### Definition: Amphoteric

An amphoteric substance is one that can react as either an acid or base.

When we look just at Bronsted-Lowry acids and bases we can also talk about amphiprotic substances which are a special type of amphoteric substances.

### Definition: Amphiprotic

An amphiprotic substance is one that can react as either a proton donor (Bronsted-Lowry acid) or as a proton acceptor (Bronsted-Lowry base). Examples of amphiprotic substances include water, hydrogen carbonate ion ($$\text{HCO}_{3}^{-}$$) and hydrogen sulfate ion ($$\text{HSO}_{4}^{-}$$).

Note: You may also see the term ampholyte used to mean a substance that can act as both an acid and a base. This term is no longer in general use in chemistry.

### Polyprotic acids

A polyprotic (many protons) acid is an acid that has more than one proton that it can donate. For example sulfuric acid can donate one proton to form the hydrogen sulfate ion:

$\text{H}_{2}\text{SO}_{4}\text{(aq)} + \text{OH}^{-}\text{(aq)} \rightarrow \text{HSO}_{4}^{-}\text{(aq)} + \text{H}_{2}\text{O (l)}$

Or it can donate two protons to form the sulfate ion:

$\text{H}_{2}\text{SO}_{4}\text{(aq)} + 2\text{OH}^{-}\text{(aq)} \rightarrow \text{SO}_{4}^{2-}\text{(aq)} + 2\text{H}_{2}\text{O (l)}$

In this tutorial we will mostly consider monoprotic acids (acids with only one proton to donate). If you do see a polyprotic acid in a reaction then write the resulting reaction equation with the acid donating all its protons.

Some examples of polyprotic acids are: $$\text{H}_{2}\text{SO}_{4}$$, $$\text{H}_{2}\text{SO}_{3}$$, $$\text{H}_{2}\text{CO}_{3}$$ and $$\text{H}_{3}\text{PO}_{4}$$.