Topic 4.8 Notes – Introduction to Acid-Base Reactions

1. Brønsted-Lowry Acid-Base Reactions

At this level, acid-base reactions are proton-transfer reactions.

• A proton (H⁺) is a hydrogen atom with no electrons.
• A Brønsted-Lowry acid donates a proton.
• A Brønsted-Lowry base accepts a proton.

General pattern:

$$\text{acid}_{1} + \text{base}_{2} \rightleftharpoons \text{base}_{1} + \text{acid}_{2}$$

Notice there’s an acid and base on both sides. That’s because when one species donates H⁺, it turns into something that could accept it back.

Conjugate Acid-Base Pairs

A conjugate acid-base pair differs by exactly one H⁺.

If you look at:

$$\ce{NH3(aq) + H2O(l) <=> NH4^{+}(aq) + OH^{-}(aq)}$$

Match them up:

• $\ce{NH3}$ and $\ce{NH4^{+}}$ → differ by one H⁺
• $\ce{H2O}$ and $\ce{OH^{-}}$ → differ by one H⁺

Now decide roles:

• $\ce{NH3}$ gained H⁺ → it’s the base
• $\ce{H2O}$ lost H⁺ → it’s the acid

Quick check that saves points on MCQs:

• More H = conjugate acid
• Fewer H = conjugate base

If it doesn’t differ by one H⁺, it’s not a conjugate pair.

2. The Role of Water in Acid-Base Reactions

AP Chemistry focuses on aqueous acid-base reactions. Water is not just a solvent. It actively participates.

Water Is Amphiprotic

Water can:

• Accept H⁺ → becomes $\ce{H3O^{+}}$
• Donate H⁺ → becomes $\ce{OH^{-}}$

Water molecule showing bent geometry and lone pairs

Water’s two lone pairs on oxygen allow it to accept H⁺ (forming H₃O⁺), and its O-H bonds allow it to donate H⁺ (leaving OH⁻).

Because free H⁺ does not float around in solution, acids in water form hydronium:

$$\ce{HCl(aq) + H2O(l) -> H3O^{+}(aq) + Cl^{-}(aq)}$$

And bases can pull H⁺ from water:

$$\ce{CH3NH2(aq) + H2O(l) <=> CH3NH3^{+}(aq) + OH^{-}(aq)}$$

How to Spot Amphiprotic Species

They must:

• Have a removable H
• Have a lone pair to accept H⁺

Examples:

• $\ce{H2O}$
• $\ce{HCO3^{-}}$

If a question asks which species can act as both acid and base, look for both features.

Important boundary: Lewis acid-base theory is not tested here. Stay focused on proton transfer in water.

3. Strength of Acids and Bases and Conjugate Pairs

Here’s the relationship students mix up:

• Strong acid → weak conjugate base
• Weak acid → stronger conjugate base
• Strong base → weak conjugate acid

Why?
If an acid gives up H⁺ easily, its conjugate base has little desire to grab it back.

Example:

$$\ce{HNO3} \text{ (strong)} \rightarrow \ce{NO3^{-}} \text{ (extremely weak base)}$$

What “Strong” Means

• Strong acids/bases completely ionize in water.
• Weak acids/bases establish equilibrium.

Reaction direction favors the side with the weaker acid and weaker base.

If you’re asked which way equilibrium lies, compare strengths. The stronger acid reacts with the stronger base to make weaker partners.

Students often think strong reacts to make strong. It’s the opposite. The system moves toward stability.

4. Acid-Base Neutralization Reactions

A neutralization reaction forms water and a salt.

$$\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}$$

At the particle level (for strong acid + strong base):

$$\ce{H^{+}(aq) + OH^{-}(aq) -> H2O(l)}$$

That net ionic equation is worth memorizing.

Writing a Net Ionic Equation

Suppose:

$$\ce{HBr(aq) + NaOH(aq) -> NaBr(aq) + H2O(l)}$$

Steps:

1. Dissociate strong electrolytes:
   $$\ce{H^{+} + Br^{-} + Na^{+} + OH^{-} -> Na^{+} + Br^{-} + H2O}$$
2. Cancel spectator ions ($\ce{Na^{+}}$, $\ce{Br^{-}}$).
3. Final:
   $$\ce{H^{+} + OH^{-} -> H2O}$$

Do not dissociate:

• Weak acids (like $\ce{HF}$)
• Weak bases
• Solids

That mistake shows up constantly on tests.

5. Stoichiometry of Acid-Base Reactions

These are limiting reactant problems with dilution at the end.

Suppose you mix:

• 0.150 M $\ce{HCl}$, 40.0 mL
• 0.200 M $\ce{KOH}$, 25.0 mL

Step 1. Convert to moles

$$\text{mol HCl} = 0.150 \times 0.0400 = 0.00600$$
$$\text{mol KOH} = 0.200 \times 0.0250 = 0.00500$$

1:1 ratio → KOH is limiting.

Step 2. Subtract reacted moles

0.00600 − 0.00500 = 0.00100 mol excess $\ce{HCl}$

Step 3. Divide by total volume

Total volume = 0.0650 L

$$[H^{+}] = \frac{0.00100}{0.0650} = 0.0154 \text{ M}$$

Since acid is in excess, solution is acidic.

Always:

• Add volumes
• Track moles first, not concentration
• Only one of $[H^{+}]$ or $[OH^{-}]$ is in excess after strong acid-strong base neutralization (both are always nonzero due to water autoionization)