3.1 Brønsted-Lowry theory and acid-base conjugate pairs

Brønsted-Lowry Theory and Acid-Base Conjugate Pairs

Brønsted-Lowry theory redefines acids and bases around one central event: proton transfer. Instead of focusing on what a substance is, this theory focuses on what it does in a reaction. That shift in thinking is what makes it so useful for predicting how acid-base reactions behave and which direction equilibrium favors.

Brønsted-Lowry Acids and Bases

A Brønsted-Lowry acid is any species that donates a proton ($H^+$), and a Brønsted-Lowry base is any species that accepts one. The acid doesn't have to contain $OH^-$ or look like a "classic" acid. If it hands off a proton during a reaction, it's acting as an acid.

• $HCl$ donates a proton to water, so it's an acid
• $NH_3$ accepts a proton from water, so it's a base
• $H_2O$ can do either, depending on the reaction partner (this is called being amphoteric)

Notice that this definition is broader than the Arrhenius model you learned in Gen Chem I. Ammonia ($NH_3$) doesn't contain $OH^-$, yet it clearly acts as a base under Brønsted-Lowry because it accepts a proton.

Conjugate Acid-Base Pairs

When a proton transfers, it creates a conjugate pair: two species that differ by exactly one $H^+$.

• An acid loses a proton and becomes its conjugate base
  • $HF \rightarrow F^-$ ($HF$ lost a proton, so $F^-$ is its conjugate base)
• A base gains a proton and becomes its conjugate acid
  • $CN^- \rightarrow HCN$ ($CN^-$ gained a proton, so $HCN$ is its conjugate acid)

Every Brønsted-Lowry reaction has two conjugate pairs. In the reaction $CH_3COOH + H_2O \rightarrow CH_3COO^- + H_3O^+$, the two pairs are:

• Pair 1: $CH_3COOH$ / $CH_3COO^-$ (acid and its conjugate base)
• Pair 2: $H_2O$ / $H_3O^+$ (base and its conjugate acid)

Being able to identify both pairs in any reaction is a skill you'll use constantly in this unit.

Strength Relationship Between Conjugates

There's an inverse relationship between the strength of an acid and the strength of its conjugate base. A stronger acid holds onto its proton less tightly, which means its conjugate base has very little tendency to grab that proton back.

• Stronger acid → weaker conjugate base
  • $HCl$ is a strong acid; $Cl^-$ is such a weak base that it essentially doesn't act as a base in water
• Weaker acid → stronger conjugate base
  • $HF$ is a weak acid; $F^-$ is a reasonably strong conjugate base that can accept protons in solution
• Stronger base → weaker conjugate acid
  • $NH_2^-$ (amide ion) is a very strong base; $NH_3$ is a very weak conjugate acid
• Weaker base → stronger conjugate acid
  • $H_2O$ is a weak base; $H_3O^+$ is a strong conjugate acid

This pattern is how you predict which side of an equilibrium is favored: the reaction proceeds in the direction that produces the weaker acid and the weaker base.

Writing Equations for Acid-Base Reactions

To write a Brønsted-Lowry equation, follow these steps:

1. Identify the acid and base. Which species can donate a proton? Which can accept one?
2. Transfer one proton from the acid to the base.
3. Write the products. The acid becomes its conjugate base (lost $H^+$), and the base becomes its conjugate acid (gained $H^+$).
4. Label both conjugate pairs to confirm the equation makes sense.

The general form is:

$HA + B \rightarrow A^- + HB^+$

where $HA$ is the acid, $B$ is the base, $A^-$ is the conjugate base, and $HB^+$ is the conjugate acid. Keep in mind that the charges shown here are generic. The actual charges depend on the species involved.

Example: Acetic acid reacting with water

$CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+$

• $CH_3COOH$ donates a proton → conjugate base is $CH_3COO^-$
• $H_2O$ accepts a proton → conjugate acid is $H_3O^+$

Note the equilibrium arrow ($\rightleftharpoons$). Because acetic acid is a weak acid, this reaction doesn't go to completion. The equilibrium lies to the left, meaning most of the acetic acid stays undissociated. You'll connect this directly to $K_a$ values later in the unit.