Introduction to Acids and Bases

AP Chemistry· AP Chemistry CED — Acids and Bases· 14 min read

1. Acid-Base Definitions and Conjugate Pairs★★☆☆☆⏱ 4 min

The AP Chemistry exam expects you to know three hierarchical definitions of acids and bases, each with a different scope. The Arrhenius definition, limited to aqueous solutions, defines acids as substances that increase $[H_{3} O^{+}]$ and bases as substances that increase $[OH^{-}]$ when dissolved in water.

The most commonly tested definition on the AP exam is Brønsted-Lowry: acids are proton ( $H^{+}$ ) donors, and bases are proton acceptors. This leads directly to the concept of conjugate acid-base pairs: every acid donates a proton to form its conjugate base, and every base accepts a proton to form its conjugate acid. By definition, conjugate pairs differ by exactly one proton. A key rule: the stronger an acid, the weaker its conjugate base, and vice versa.

The most general definition is Lewis: Lewis acids accept an electron pair, and Lewis bases donate an electron pair. This covers reactions without proton transfer, but is less commonly tested in introductory problems.

📘 Definition

Conjugate Acid-Base Pair

Two species that differ by exactly one proton ( $H^{+}$ ), where one acts as an acid (donates proton) to form the other (the conjugate base), or a base accepts a proton to form its conjugate acid.

Example:

$NH_{3}$ (base) and $NH_{4}^{+}$ (conjugate acid)

📐 Worked Example

For the reaction $NH_{3} (a q) + H_{2} O (l) ⇌ NH_{4}^{+} (a q) + OH^{-} (a q)$ , identify all Brønsted-Lowry acids and bases, and label all conjugate acid-base pairs.

1
Recall the Brønsted-Lowry rule: acids donate a proton, bases accept a proton.
2
On the reactant side: $NH_{3}$ gains a proton to become $NH_{4}^{+}$ , so $NH_{3}$ is the base. $H_{2} O$ loses a proton to become $OH^{-}$ , so $H_{2} O$ is the acid.
3
On the product side: $NH_{4}^{+}$ can lose a proton to reform $NH_{3}$ , so it is the conjugate acid. $OH^{-}$ can gain a proton to reform $H_{2} O$ , so it is the conjugate base.
4
Pair species that differ by exactly one proton: Pair 1 = Base $NH_{3}$ / Conjugate acid $NH_{4}^{+}$ ; Pair 2 = Acid $H_{2} O$ / Conjugate base $OH^{-}$ .

Exam tip:

When asked to identify conjugate pairs on the AP exam, always confirm the two species differ by exactly one proton. Common distractors use pairs differing by two protons, so counting H atoms will eliminate wrong answers quickly.

2. Autoionization of Water and the pH Scale★★★☆☆⏱ 4 min

Water is amphoteric, meaning it can act as either a Brønsted-Lowry acid or base depending on its reaction partner. In pure water, two water molecules undergo reversible autoionization:

2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)

The equilibrium constant for this reaction is the ion product of water, $K_{w}$ . Pure liquid water is omitted from the equilibrium expression, giving:

K_w = [\text{H}_3\text{O}^+][\text{OH}^-]

At 25°C, $K_{w} = 1.0 \times 1 0^{- 14}$ . The pH scale simplifies working with small hydronium concentrations, defined as $pH = - lo g [H_{3} O^{+}]$ , and pOH is $pOH = - lo g [OH^{-}]$ . At 25°C, this gives the key relationship $pH + pOH = 14$ .

A neutral solution has equal concentrations of hydronium and hydroxide: $[H_{3} O^{+}] = [OH^{-}]$ , which only gives $pH = 7$ at 25°C. Acidic solutions have $pH < 7$ , basic solutions have $pH > 7$ at 25°C. $K_{w}$ increases with temperature (autoionization is endothermic), so neutral pH decreases as temperature increases.

📐 Worked Example

At 50°C, $K_{w}$ for water is $5.48 \times 1 0^{- 14}$ . Calculate the pH of a neutral aqueous solution at 50°C, and classify the solution as acidic, basic, or neutral.

1
By definition, a neutral solution has $[H_{3} O^{+}] = [OH^{-}]$ , so substitute into the $K_{w}$ expression:
2
$K_w = [\text{H}_3\text{O}^+]^2$
3
Solve for $[H_{3} O^{+}]$ :
4
$[\text{H}_3\text{O}^+] = \sqrt{5.48 \times 10^{-14}} \approx 2.34 \times 10^{-7} \; M$
5
Calculate pH:
6
$\text{pH} = -\log(2.34 \times 10^{-7}) \approx 6.63$
7
A solution is neutral if and only if $[H_{3} O^{+}] = [OH^{-}]$ , regardless of pH, so this solution is neutral.

Exam tip:

Never automatically assume neutral solutions have pH = 7. Always check if the problem gives a non-room temperature or a different $K_{w}$ value, and apply the definition of neutrality correctly.

3. Acid and Base Dissociation Constants ($K_a$ and $K_b$)★★★☆☆⏱ 4 min

Strong acids and bases dissociate completely in dilute aqueous solution, so no equilibrium constant is needed for introductory calculations. Weak acids and bases only partially dissociate, so we use equilibrium constants to describe their strength.

For a general weak acid $HA$ , dissociation in water is:

\text{HA}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{A}^-(aq) + \text{H}_3\text{O}^+(aq)

The acid dissociation constant $K_{a}$ is:

K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}

For a general weak base $B$ , reaction with water is:

\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq)

K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}

Larger $K_{a}$ corresponds to a stronger weak acid, and larger $K_{b}$ corresponds to a stronger weak base. For any conjugate acid-base pair, the key relationship is:

K_a \times K_b = K_w

This relationship lets you calculate $K_{b}$ of a conjugate base from $K_{a}$ of the parent acid, and vice versa, and it only applies to conjugate pairs.

📐 Worked Example

Formic acid ( $HCOOH$ ), the active component in ant stings, has $K_{a} = 1.8 \times 1 0^{- 4}$ at 25°C. What is $K_{b}$ for its conjugate base, formate ion ( $HCOO^{-}$ ) at 25°C? Is formate a stronger or weaker base than fluoride ion ( $K_{b} = 1.6 \times 1 0^{- 11}$ )?

1
Use the conjugate pair relationship: $K_{a} (acid) \times K_{b} (conjugate base) = K_{w}$ . At 25°C, $K_{w} = 1.0 \times 1 0^{- 14}$ .
2
Rearrange to solve for $K_{b}$ :
3
$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} \approx 5.6 \times 10^{-11}$
4
A larger $K_{b}$ means a stronger base. Comparing values: $5.6 \times 1 0^{- 11} > 1.6 \times 1 0^{- 11}$ , so formate ion is a stronger base than fluoride ion.

Exam tip:

The $K_{a} \times K_{b} = K_{w}$ relationship only applies to conjugate pairs. Never use it to relate an acid and an unrelated base, as this will always give an incorrect result.

4. AP Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding of core concepts with these AP-style multiple-choice questions:

Which of the following pairs is correctly labeled as a Brønsted-Lowry conjugate acid-base pair?
- A) $H_{2} CO_{3}$ and $CO_{3}^{2 -}$
- B) $H_{2} PO_{4}^{-}$ and $HPO_{4}^{2 -}$
- C) $H_{2} S$ and $S^{2 -}$
- D) $H_{2} O$ and $O^{2 -}$
Reveal answer
B —
Conjugate pairs differ by exactly one proton. Only option B fits this definition; all other options differ by two protons.
The autoionization of water is endothermic. What happens to $K_{w}$ and the pH of a neutral solution when temperature increases?
- A) $K_{w}$ increases, pH of neutral solution decreases
- B) $K_{w}$ increases, pH of neutral solution increases
- C) $K_{w}$ decreases, pH of neutral solution decreases
- D) $K_{w}$ decreases, pH of neutral solution increases
Reveal answer
A —
Endothermic reactions shift right when temperature increases, increasing product concentrations and $K_{w}$ . Higher $K_{w}$ means lower pH for neutral solutions, which still have equal $[H_{3} O^{+}]$ and $[OH^{-}]$ .

5. Common Pitfalls

Wrong move:

Assuming all neutral solutions have pH = 7 at any temperature

Why:

Students memorize the 25°C value and forget that $K_{w}$ changes with temperature, changing $p K_{w}$ .

Correct move:

Always confirm the temperature and given $K_{w}$ value; a solution is neutral only when $[H_{3} O^{+}] = [OH^{-}]$ , not when pH = 7.

Wrong move:

Identifying conjugate acid-base pairs that differ by more or less than one proton

Why:

Students confuse charge change with proton change, or forget that one proton adds both one H atom and +1 charge.

Correct move:

Always count the number of hydrogen atoms and check the charge difference to confirm the two species differ by exactly one $H^{+}$ .

Wrong move:

Writing $K_{a}$ or $K_{b}$ expressions that include pure liquid water (the solvent) in the denominator

Why:

Students forget that pure liquids have an activity of 1 in equilibrium expressions and are omitted.

Correct move:

Omit all pure solids and pure liquids (including water solvent) from $K_{a} / K_{b}$ equilibrium expressions.

Wrong move:

Calculating pH as $lo g [H_{3} O^{+}]$ instead of $- lo g [H_{3} O^{+}]$ , leading to a negative pH for acidic solutions when it should be positive

Why:

Students misremember the definition of the pH scale and forget the negative sign.

Correct move:

Always double-check the relationship: high $[H_{3} O^{+}]$ should give low pH, so the negative sign is required.

Wrong move:

Using $K_{a} \times K_{b} = K_{w}$ for an acid and a base that are not a conjugate pair

Why:

Students memorize the relationship but forget the requirement that it only applies to pairs that differ by one proton.

Correct move:

Only apply the $K_{a}$ - $K_{b}$ relationship to the acid and its own conjugate base, or the base and its own conjugate acid.

Wrong move:

Calling $H_{2} O$ exclusively an acid or exclusively a base, forgetting it is amphoteric

Why:

Students learn water as the solvent first, so they forget it can act as either proton donor or acceptor depending on the reaction partner.

Correct move:

When identifying acids/bases in a reaction, always check if $H_{2} O$ gains or loses a proton to assign its role correctly.

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Brønsted-Lowry conjugate pairs	Difference of exactly 1 $H^{+}$	Stronger acid = weaker conjugate base, and vice versa
Ion product of water	$K_{w} = [H_{3} O^{+}] [OH^{-}]$	$K_{w} = 1.0 \times 1 0^{- 14}$ at 25°C only; increases with temperature
pH definition	$pH = - lo g [H_{3} O^{+}]$	Valid for dilute aqueous solutions
pOH definition	$pOH = - lo g [OH^{-}]$	Valid for dilute aqueous solutions
pH + pOH relationship	$pH + pOH = p K_{w}$	Equals 14 at 25°C only
Weak acid dissociation constant	$K_{a} = \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ]}$	Larger $K_{a}$ = stronger weak acid
Weak base dissociation constant	$K_{b} = \frac{[ BH ^{+} ] [ OH ^{-} ]}{[ B ]}$	Larger $K_{b}$ = stronger weak base
Conjugate pair $K_{a}$ - $K_{b}$ relationship	$K_{a} \times K_{b} = K_{w}$	Only applies to an acid and its conjugate base

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Identify conjugate acid-base pairs
2022 · FRQ
Calculate neutral pH at non-25°C
2021 · MCQ
Relate Ka and Kb for conjugate pairs

What's Next

This introductory sub-topic is the foundation for all subsequent acid-base chemistry topics in AP Chemistry Unit 8, which makes up 10-15% of your total exam score. The definitions, equilibrium relationships, and calculation rules you learned here will be applied to more complex problems including pH calculations for weak acids and bases, buffer solutions, acid-base titrations, and solubility equilibria involving acidic/basic ions. Mastering these core fundamentals now will make more advanced topics much easier to understand, as almost every acid-base FRQ on the AP exam starts with a question that tests these introductory concepts.