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# 20.5: Brønsted-Lowry Acids and Bases

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define Brønsted-Lowry acids and bases.
• identify Brønsted-Lowry acids and bases in chemical equations.
• define conjugate acids and bases.
• write the formula for the conjugate acid of any base and for the conjugate base of any acid.
• identify conjugate acids and bases in equations.
• given the strengths of acids and bases, identify the strength the conjugate acids and bases.

## Vocabulary

• amphoteric
• Brønsted-Lowry acid
• Brønsted-Lowry base
• conjugate acid
• conjugate base

## Introduction

Arrhenius provided chemistry with the first definition of acids and bases, but like a lot of scientific ideas, these definitions have been refined over time. Two chemists, named Brønsted and Lowry, derived a more generalized definition for acids and bases that we use in conjunction with the Arrhenius theory. The Brønsted-Lowry theory is the focus of this lesson. As the Brønsted-Lowry definition unfolded, the number of acids and bases that were able to fit into each category increased.

## Brønsted-Lowry Acids and Bases

Arrhenius made great in-roads into the understanding of acids and bases and how they behaved in chemical reactions. Brønsted and Lowry slightly altered the Arrhenius definition and greatly increased the number of compounds that qualify as bases. A Brønsted-Lowry acid is a substance that is a proton donor, and a Brønsted-Lowry base is a proton acceptor. Look at the equation below in which hydrochloric acid is reacting with water:

\begin{align*}\mathrm{HCl}_{(g)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}\end{align*}

The Brønsted-Lowry concept of acids and bases states that the acid donates a proton and the base accepts a proton. Therefore, \begin{align*}\mathrm{HCl}\end{align*} acts as the acid and donates an \begin{align*}\mathrm{H}^+\end{align*} ion to \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} to form \begin{align*}\mathrm{Cl}^-\end{align*}. The \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} acts as the base and accepts an \begin{align*}\mathrm{H}^+\end{align*} ion from \begin{align*}\mathrm{HCl}\end{align*} to form \begin{align*}\mathrm{H}_3\mathrm{O}^+\end{align*}. Look at the equation below for another example of Brønsted-Lowry acid and base:

\begin{align*}\mathrm{H}_2\mathrm{PO}^-_{4(aq)} + \mathrm{OH}^-_{(aq)} \rightleftharpoons \mathrm{HPO}^{2-}_{4(aq)} + \mathrm{H}_2\mathrm{O}_{(l)}\end{align*}

The equation shows \begin{align*}\mathrm{H}_2\mathrm{PO}_4^-\end{align*} donating a proton to \begin{align*}\mathrm{OH}^-\end{align*} and forming \begin{align*}\mathrm{HPO}_4^{2-}\end{align*}, while \begin{align*}\mathrm{OH}^-\end{align*} is accepting the proton to form \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}. Thus, \begin{align*}\mathrm{H}_2\mathrm{PO}_4^-\end{align*} is acting as the acid and \begin{align*}\mathrm{OH}^-\end{align*} is acting as the base.

Example:

Identify the Brønsted-Lowry acids and bases from each of the following equations:

1. \begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)} \rightleftharpoons \mathrm{C}_2\mathrm{H}_3\mathrm{O}^-_{2(aq)} + \mathrm{H}_3\mathrm{O}^+_{(aq)}\end{align*}
2. \begin{align*}\mathrm{HCN}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{CN}^-_{(aq)} + \mathrm{H}_2\mathrm{O}^+_{(aq)}\end{align*}

Solution:

1. \begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\end{align*} is acting as the acid and \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} is acting as the base.
2. \begin{align*}\mathrm{HCN}\end{align*} is acting as the acid and \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} is acting as the base.

If you think about the definition of an Arrhenius acid, it includes substances such as \begin{align*}\mathrm{HCl}\end{align*}, \begin{align*}\mathrm{HNO}_3\end{align*}, \begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\end{align*} - in essence all substances that contain \begin{align*}\mathrm{H}^+\end{align*} ions. This is because according to Arrhenius, the acid dissociates in water to produce \begin{align*}\mathrm{H}^+\end{align*} ions. This definition limits what can fit under the umbrella of the definition of acid. The Brønsted-Lowry definition of the acid is broader in that it defines the acid as a proton donor. With this broader definition there is the ability to include more compounds in the category of acid.

It needs to be pointed out that if a substance is an acid in the Arrhenius definition, it will be an acid in the Brønsted–Lowry definition. The same relationship holds for bases. The reverse, however, is not true. Consider the equation below. The hydroxide ion, \begin{align*}\mathrm{OH}^-\end{align*}, is both an Arrhenius base and a Brønsted–Lowry base. In other words, the Brønsted-Lowry definition can be viewed as an extension to the Arrhenius definition rather than a replacement of it.

\begin{align*}\mathrm{HSO}^-_{4(aq)} + \mathrm{OH}^-_{(aq)} \rightleftharpoons \mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{SO}^{2-}_{4(aq)}\end{align*}

With the Arrhenius theory, water was a necessary part of the equation. Arrhenius said that an acid must produce \begin{align*}\mathrm{H}^+\end{align*} ions in a water solution. Therefore, the following equation would be representative of an Arrhenius acid:

\begin{align*}\mathrm{HCl}_{(g)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}\end{align*}

Now consider the following reaction:

\begin{align*}\mathrm{NH}_{3(aq)} + \mathrm{NH}_{3(aq)} \rightleftharpoons \mathrm{NH}^+_{4(aq)} + \mathrm{NH}^-_{2(aq)}\end{align*}

The first \begin{align*}\mathrm{NH}_3\end{align*} molecule is accepting a proton to form \begin{align*}\mathrm{NH}_4^+\end{align*} and is therefore a Brønsted-Lowry base, the second \begin{align*}\mathrm{NH}_3\end{align*} molecule is donating a proton to form \begin{align*}\mathrm{NH}_2^-\end{align*} and is therefore a Brønsted-Lowry acid. Ammonia molecules, however, do not donate hydrogen ions in water, so they do not qualify as Arrhenius acids. The Brønsted-Lowry theory has provided a broader theory for acid-base chemistry.

It should be noted that \begin{align*}\mathrm{NH}_3\end{align*} is an example of an amphoteric species. Amphoteric species are those that can act as either an acid or a base, depending on the situation. That is, in some circumstances, they donate a proton, and in other circumstances, they accept a proton. Water is another example of an amphoteric species.

## Acid-Base Conjugate Pairs

There is one more aspect of the Brønsted-Lowry theory that was a significant breakthrough to acid-base chemistry. Brønsted and Lowry said that in acid-base reactions, there are actually pairs of acids and bases in the reaction itself. According to Brønsted-Lowry, for every acid there is a conjugate base associated with that acid. The conjugate base is the result of the acid losing (or donating) a proton. Therefore, if you look at the figure below, you can see the acid on the left and the conjugate base on the right.

Notice that the difference between the acid and its conjugate base is simply a proton. Similarly, for every base in the acid-base reaction, there must be a corresponding conjugate acid. The conjugate acid is the result of the base gaining (or accepting) the proton. Look at the figure below to see the difference between the base and the corresponding conjugate acid.

Now that we know what conjugate acids and bases are, let’s try to identify them in acid-base reactions. Consider the reaction between acetic acid and water shown below:

\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{C}_2\mathrm{H}_3\mathrm{O}^-_{(aq)} + \mathrm{H}_3\mathrm{O}^+_{(aq)}\end{align*}

Step 1: Identify the acid and base on the reactants side.

\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\end{align*} is the acid and \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} is the base.

Step 2: Identify the conjugate acid and base on the products side.

Look at the product side to see what product has gained a proton (this is the conjugate acid) and which product has lost a proton (this is the conjugate base).

\begin{align*}\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-\end{align*} is the conjugate base of \begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\end{align*}, and \begin{align*}\mathrm{H}_3\mathrm{O}^+\end{align*} is the conjugate acid of \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}.

As a result, the conjugate acid/base pairs are \begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2\end{align*} / \begin{align*}\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-\end{align*} and \begin{align*}\mathrm{H}_3\mathrm{O}^+\end{align*} / \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}.

Let's consider another example. Identify the conjugate acid-base pairs in the following equation.

\begin{align*}\mathrm{CH}_3\mathrm{NH}_{2(aq)} + \mathrm{HClO}_{(aq)} \rightleftharpoons \mathrm{ClO}^-_{(aq)} + \mathrm{CH}_3\mathrm{NH}^+_{3(aq)}\end{align*}

Step 1: Identify the acid and base on the reactants side.

\begin{align*}\mathrm{CH}_3\mathrm{NH}_2\end{align*} is the base and \begin{align*}\mathrm{HClO}\end{align*} is the acid.

Step 2: Identify the conjugate acid and base on the products side.

\begin{align*}\mathrm{ClO}^-\end{align*} is the conjugate base and \begin{align*}\mathrm{CH}_3\mathrm{NH}_3^+\end{align*} is the conjugate acid.

Hence, the conjugate acid/base pairs are \begin{align*}\mathrm{CH}_3\mathrm{NH}_2\end{align*} / \begin{align*}\mathrm{CH}_3\mathrm{NH}_3^+\end{align*} and \begin{align*}\mathrm{HClO}\end{align*} / \begin{align*}\mathrm{ClO}^-\end{align*}.

Example:

Identify the conjugate acid-base conjugate pairs in each of the following equations:

1. \begin{align*}\mathrm{NH}_{3(aq)} + \mathrm{HCN}_{(aq)} \rightleftharpoons \mathrm{NH}^+_{4(aq)} + \mathrm{CN}^-_{(aq)}\end{align*}
2. \begin{align*}\mathrm{CO}^{2-}_{3(aq)} + \mathrm{H}_2\mathrm{O}_{(l)}~ \rightleftharpoons \mathrm{HCO}^-_{3(aq)} + \mathrm{OH}^-_{(aq)}\end{align*}

Solution:

1. \begin{align*}\mathrm{NH}_3\end{align*} / \begin{align*}\mathrm{NH}_4^+\end{align*} and \begin{align*}\mathrm{HCN}\end{align*} / \begin{align*}\mathrm{CN}^-\end{align*}
2. \begin{align*}\mathrm{CO}_3^{2-}\end{align*} / \begin{align*}\mathrm{HCO}_3^-\end{align*} and \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*} / \begin{align*}\mathrm{OH}^-\end{align*}

## The Strength of Conjugate Acids and Bases

As a general rule, the stronger an acid is, the weaker its conjugate base will be. The same is true for the relationship between bases and their conjugate acids. Overall:

Strong acid \begin{align*}\rightarrow\end{align*} Very weak conjugate base
Weak acid \begin{align*}\rightarrow\end{align*} Weak conjugate base
Weak base \begin{align*}\rightarrow\end{align*} Weak conjugate acid
Strong base \begin{align*}\rightarrow\end{align*} Very weak conjugate acid

Since acid strength and conjugate base strength are inversely related, you might think that strong acids would have weak conjugate bases, while weak acids would have strong conjugate bases. Although this is true in a relative sense, it does not match up to the way we have been defining “strong” acids and bases so far. This can often be confusing for students just beginning their study of chemistry.

Let’s re-examine how we have defined strong and weak acids and bases. This can be illustrated most clearly by looking at some representative examples. Consider the following equilibria:

\begin{align*}\mathrm{HCl}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{Cl}^-_{(aq)} + \mathrm{H}_3\mathrm{O}^+_{(aq)} \ \ \ \ \ K_a = \mathrm{Very \ large}\end{align*}
\begin{align*}\mathrm{CH}^-_{3(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_{4(g)} + \mathrm{OH}^-_{(aq)} \ \ \ \ \ K_b = \mathrm{Very \ large}\end{align*}
\begin{align*}\mathrm{CH}_3\mathrm{CO}_2\mathrm{H}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3\mathrm{CO}^-_{2(aq)} + \mathrm{H}_3\mathrm{O}^+_{(aq)} \ \ \ \ \ K_a = 1.74 \times 10^{-5}\end{align*}
\begin{align*}\mathrm{CH}_3\mathrm{CO}^-_{2(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3\mathrm{CO}_2\mathrm{H}_{(aq)} + \mathrm{OH}^-_{(aq)} \ \ \ \ \ K_b = 5.6 \times 10^{-10} \end{align*}

\begin{align*}\mathrm{HCl}\end{align*} is an example of a strong acid because it completely dissociates in water. Another way to say this is that \begin{align*}\mathrm{HCl}\end{align*} is a very strong acid compared to \begin{align*}H_3O^+\end{align*}, because when \begin{align*}\mathrm{Cl}^-\end{align*} (its conjugate base) and water are competing for a proton, water wins nearly every time. \begin{align*}\mathrm{Cl}^-\end{align*} is such a weak base that, when considered outside the context of an acid/base reaction, you would not even consider it a base. For example, a solution of \begin{align*}\mathrm{NaCl}\end{align*} contains \begin{align*}\mathrm{Cl}^-\end{align*} ions, but you would never say that the solution is basic.

On the opposite side of things, we can consider the reaction of \begin{align*}\mathrm{CH}_3^-\end{align*} with water. (The strong bases you have learned about so far all contain OH-, but using one of those bases as an example makes the following explanation a little confusing.) \begin{align*}\mathrm{CH}_3^-\end{align*} is one of the strongest bases known, and it is at least \begin{align*}10^{25}\end{align*} times stronger than \begin{align*}\mathrm{OH}^-\end{align*}. In the competition for control of a proton, \begin{align*}\mathrm{CH}_3^-\end{align*} beats \begin{align*}\mathrm{OH}^-\end{align*} by a landslide. On the other side of the equation, we see that its conjugate “acid” is methane, \begin{align*}\mathrm{CH}_4\end{align*}. Again, outside this context, you would never consider methane to be an acid, because almost nothing is strong enough to remove one of its protons. Thus, \begin{align*}\mathrm{CH}^-_3\end{align*} is an extremely strong base compared to \begin{align*}OH^-\end{align*}. Notice that the species we are using for comparison changes when we are talking about acids versus bases.

Now let’s look at the dissociation of acetic acid. You can see by the equilibrium constant that the reactants should be favored. This means that when the base and the conjugate base (in this case, acetate ions and water molecules) are competing for a proton, acetate wins most of the time. Compared to \begin{align*}H_3O^+\end{align*}, acetic acid is a much weaker acid. This is equivalent to the statement that acetic acid will not dissociate much in water, which is how we have previously defined weak acids.

However, acetate ion is not considered a strong base. Look at the final equation above. Once again, the equilibrium constant is small, so reactants will be favored. How can this be? Why would acetic acid be favored in the third reaction, while acetate is favored in the fourth one? The reason is that we are changing the base that the acetate ion is competing with. In the third equation, acetate is competing with water. Compared to water, acetate is a good base. However, we generally define “strong” and “weak” bases by comparing them to \begin{align*}\mathrm{OH}-\end{align*}, not \begin{align*}\mathrm{H}_2\mathrm{O}\end{align*}. Compared to \begin{align*}OH^-\end{align*}, acetate is a weak base.

Thus, we can see why a weak acid can still produce a weak conjugate base, and vice versa. It is important to remember that “acidic” and “basic” are relative terms. It is very useful to have standard benchmarks (\begin{align*}\mathrm{H}_3\mathrm{O}^+\end{align*} and \begin{align*}\mathrm{OH}^-\end{align*}) to use for comparison, but figuring out what is “strong” and what is “weak” can be confusing if you don’t realize that the reference point we use when talking about acids is different than the one we use when talking about bases.

## Lesson Summary

• The Brønsted-Lowry concept of acids and bases states that the acid donates a proton and the base accepts a proton.
• A conjugate acid is a substance that results when a base gains (or accepts) a proton.
• A conjugate base is a substance that results when an acid loses (or donates) a proton.
• Strong acids result in very weak conjugate bases when they lose a proton, and weak acids result in very strong conjugate bases when they lose a proton.
• Strong bases result in very weak conjugate acids when they gain a proton, and weak bases result in very strong conjugate acids when they gain a proton.

The learner.org website allows users to view the Annenberg series of chemistry videos. You are required to register before you can watch the videos, but there is no charge to register. The video called “The Proton in Chemistry” explains pH and how it is measured, as well as the important role of acids and bases.

## Review Questions

1. What improvements did Brønsted-Lowry make over the Arrhenius definition for acids-bases?
2. What is the Brønsted-Lowry definition of an acid?
1. A substance that donates protons.
2. A substance that accepts protons.
3. A substance that dissolves in water to form \begin{align*}\mathrm{OH}^-\end{align*} ions.
4. A substance that dissolves in water to form \begin{align*}\mathrm{H}^+\end{align*} ions.
3. If \begin{align*}\mathrm{H}_3\mathrm{O}^+\end{align*} is an acid according to the Brønsted-Lowry theory, what is the conjugate base of this acid?
1. \begin{align*}\mathrm{H}_4\mathrm{O}{^{2+}}_{(aq)}\end{align*}
2. \begin{align*}\mathrm{H}{^{+}}_{(aq)}\end{align*}
3. \begin{align*}\mathrm{H}_2\mathrm{O}_{(l)}\end{align*}
4. \begin{align*}\mathrm{OH}^{-}_{(aq)}\end{align*}
4. What is the conjugate base of \begin{align*}\mathrm{H}_2\mathrm{PO}_4{^{-}}\end{align*}?
1. \begin{align*}\mathrm{H}_3\mathrm{O}{^{+}}_{(aq)}\end{align*}
2. \begin{align*}\mathrm{H}_3\mathrm{PO}4_{(aq)}\end{align*}
3. \begin{align*}\mathrm{HPO}_4{^{2-}}_{(aq)}\end{align*}
4. \begin{align*}\mathrm{PO}_4{^{3-}}_{(aq)}\end{align*}
5. In the following reactions, which are the Brønsted-Lowry acids? \begin{align*}\mathrm{H}_3\mathrm{PO}_{4(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}{^{+}}_{(aq)} + \mathrm{H}_2\mathrm{PO}_4{^{-}}_{(aq)}\end{align*}, \begin{align*}\mathrm{H}_2\mathrm{PO}_4{^{-}}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}{^{+}}_{(aq)} + \mathrm{HPO}_4{^{2-}}_{(aq)}\end{align*}
1. \begin{align*}\mathrm{H}_2\mathrm{PO}{_{4}}^-, \mathrm{H}_2\mathrm{O}, \mathrm{HPO}{_{4}}^{2-}\end{align*}
2. \begin{align*}\mathrm{H}_3\mathrm{PO}^4, \mathrm{H}_2\mathrm{O}, \mathrm{H}_2\mathrm{PO}{_{4}}^-\end{align*}
3. \begin{align*}\mathrm{H}_3\mathrm{O}^+, \mathrm{H}_2\mathrm{O}, \mathrm{HPO}{_{4}}^{2-}\end{align*}
4. \begin{align*}\mathrm{H}_3\mathrm{PO}4, \mathrm{H}_3\mathrm{O}^+, \mathrm{H}_2\mathrm{PO}{_{4}}^-\end{align*}
6. Label the conjugate acid-base pairs in each reaction.
1. \begin{align*}\mathrm{HCO}{_{3}}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_2\mathrm{CO}_3 + \mathrm{OH}^-\end{align*}
2. \begin{align*}\mathrm{H}_2\mathrm{PO}{_{4}}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ + \mathrm{HPO}{_{4}}^{2-}\end{align*}
3. \begin{align*}\mathrm{CN}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-\end{align*}
4. \begin{align*}\mathrm{HF}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \ \rightleftharpoons \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{F}^-_{(aq)}\end{align*}
7. Complete the following reactions. When done, label the conjugate acid/base pairs.
1. \begin{align*}\mathrm{BrO}{_{3}}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons\end{align*}
2. \begin{align*}\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{O} \rightarrow\end{align*}
3. \begin{align*}\mathrm{HSO}{_{4}}^- + \mathrm{C}_2\mathrm{O}_4{^{2-}} \rightleftharpoons \end{align*}
8. For the reactions in question 7, which are the weak conjugate bases and which are the strong conjugate bases?

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