Introduction to acids and bases — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Arrhenius, Brønsted-Lowry, and Lewis acid-base definitions, conjugate acid-base pairs, autoionization of water, the acid dissociation constant $K_a$, base dissociation constant $K_b$, pH and pOH calculations, and introductory amphoteric species behavior for AP exam assessment.

You should already know: Equilibrium constant expression rules for aqueous solutions, properties of pure liquids in equilibrium, basic logarithm calculations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Introduction to acids and bases?

This foundational topic introduces core definitions and quantitative relationships that underpin all subsequent acid-base chemistry in AP Chemistry, falling within Unit 8 (Acids and Bases), which makes up 10-15% of the overall AP Chemistry exam score. As an introductory topic, it is tested on both multiple-choice (MCQ) and free-response (FRQ) sections, often as the opening part of a longer FRQ or as a standalone MCQ testing definitions or basic calculations. Unlike more advanced topics that focus on titrations or buffers, this topic establishes the shared language and core quantitative rules that all other acid-base concepts rely on. Synonyms for the core relationships here include acid-base equilibrium constants, water autoionization, and pH scale fundamentals. All acid-base problems on the AP exam, from simple pH calculation to buffer pH and titration endpoint prediction, start with the definitions and relationships covered in this introductory topic.

2. Acid-Base Definitions and Conjugate Pairs

The first core set of ideas for acid-base chemistry is a hierarchy of definitions, all of which are tested on the AP exam. The simplest, earliest definition is the Arrhenius definition, which applies only to aqueous solutions: an Arrhenius acid is any substance that increases the concentration of hydronium ion (${\text{H}}^{+}$ or ${\text{H}}_3{\text{O}}^{+}$) when dissolved in water, while an Arrhenius base increases the concentration of hydroxide ion (${\text{OH}}^{-}$) when dissolved. This definition is limited to aqueous systems, so it is rarely the primary definition tested but is a common starting point for understanding.

The most widely used definition on the AP exam is the Brønsted-Lowry definition, which generalizes beyond aqueous solutions: a Brønsted-Lowry acid is a proton (${\text{H}}^{+}$ nucleus) donor, and a Brønsted-Lowry base is a proton acceptor. This definition naturally leads 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 relationship for AP is: the stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid.

The most general definition is the Lewis definition, which focuses on electron pairs rather than protons: a Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor. Lewis acids are often electron-deficient species like ${\text{BF}}_3$, while Lewis bases are species with lone pairs to donate, like ${\text{NH}}_3$. This definition covers acid-base reactions that do not involve proton transfer, though it is less commonly tested in introductory problems than Brønsted-Lowry.

Worked Example

For the reaction ${\text{NH}}_3(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{NH}}_4^{+}(aq)+{\text{OH}}^{-}(aq)$, 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: ${\text{NH}}_3$ gains a proton to become ${\text{NH}}_4^{+}$, so ${\text{NH}}_3$ is the base. ${\text{H}}_2\text{O}$ loses a proton to become ${\text{OH}}^{-}$, so ${\text{H}}_2\text{O}$ is the acid.
3. On the product side: ${\text{NH}}_4^{+}$ can lose a proton to reform ${\text{NH}}_3$, so it is the conjugate acid. ${\text{OH}}^{-}$ can gain a proton to reform ${\text{H}}_2\text{O}$, so it is the conjugate base.
4. Pair species that differ by exactly one proton: Pair 1 = Base ${\text{NH}}_3$ / Conjugate acid ${\text{NH}}_4^{+}$; Pair 2 = Acid ${\text{H}}_2\text{O}$ / Conjugate base ${\text{OH}}^{-}$.

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 or matching charge but different proton counts, so counting H atoms will eliminate wrong answers quickly.

3. Autoionization of Water and the pH Scale

Water is amphoteric, meaning it can act as either a Brønsted-Lowry acid or base depending on the reaction partner. In pure water, two water molecules react with each other in a reversible reaction called 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 called the ion product of water, $K_w$. Pure liquid water has an activity of 1, so it is omitted from the equilibrium expression, leaving: $$K_w=[{\text{H}}_3{\text{O}}^{+}][{\text{OH}}^{-}]$$ At 25°C, $K_w=1.0\times 10^{-14}$, so $-\log K_w=14$. The pH scale was invented to simplify working with very small hydronium concentrations, and is defined as $\text{pH}=-\log [{\text{H}}_3{\text{O}}^{+}]$, with pOH defined analogously as $\text{pOH}=-\log [{\text{OH}}^{-}]$. Combining these definitions gives the key relationship at 25°C: $\text{pH}+\text{pOH}=14$.

A neutral solution has equal concentrations of hydronium and hydroxide: $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]$, which gives $\text{pH}=7$ only at 25°C. Acidic solutions have $[{\text{H}}_3{\text{O}}^{+}]>[{\text{OH}}^{-}]$ so $\text{pH}<7$, while basic solutions have $[{\text{H}}_3{\text{O}}^{+}]<[{\text{OH}}^{-}]$ so $\text{pH}>7$. Importantly, $K_w$ changes with temperature like all equilibrium constants: autoionization is endothermic, so higher temperatures increase $K_w$, lower $\text{p}{K}_w$, and lower the pH of a neutral solution.

Worked Example

At 50°C, $K_w$ for water is $5.48\times 10^{-14}$. 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 $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]$, so substitute into the $K_w$ expression: $K_w=[{\text{H}}_3{\text{O}}^{+}{]}^2$.
2. Solve for $[{\text{H}}_3{\text{O}}^{+}]$: $[{\text{H}}_3{\text{O}}^{+}]=\sqrt{5.48\times 10^{-14}}\approx 2.34\times 10^{-7}M$.
3. Calculate pH: $\text{pH}=-\log (2.34\times 10^{-7})\approx 6.63$.
4. A solution is neutral if and only if $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]$, 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.

4. Acid and Base Dissociation Constants ($K_a$ and $K_b$)

Strong acids and bases dissociate completely in dilute aqueous solution, so their equilibrium lies fully toward products, and 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 $\text{HA}$, the dissociation reaction 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 $\text{B}$, the reaction with water is: $$\text{B}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{BH}}^{+}(aq)+{\text{OH}}^{-}(aq)$$ The base dissociation constant $K_b$ is: $$K_b=\frac{[{\text{BH}}^{+}][{\text{OH}}^{-}]}{[\text{B}]}$$ Larger values of $K_a$ correspond to stronger weak acids, and larger values of $K_b$ correspond to stronger weak bases. A key relationship can be derived for conjugate acid-base pairs: multiplying $K_a$ of an acid by $K_b$ of its conjugate base gives $K_w$: $$K_a\times K_b=\left(\frac{[{\text{H}}_3{\text{O}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}\right)\times \left(\frac{[\text{HA}][{\text{OH}}^{-}]}{[{\text{A}}^{-}]}\right)=[{\text{H}}_3{\text{O}}^{+}][{\text{OH}}^{-}]=K_w$$ This relationship is used constantly in acid-base calculations to find $K_b$ of a conjugate base from $K_a$ of the parent acid, and vice versa.

Worked Example

Formic acid ($\text{HCOOH}$), the active component in ant stings, has $K_a=1.8\times 10^{-4}$ at 25°C. What is $K_b$ for its conjugate base, formate ion (${\text{HCOO}}^{-}$) at 25°C? Is formate a stronger or weaker base than fluoride ion ($K_b=1.6\times 10^{-11}$)?

1. Use the conjugate pair relationship: $K_a(\text{acid})\times K_b(\text{conjugate base})=K_w$. At 25°C, $K_w=1.0\times 10^{-14}$.
2. Rearrange to solve for $K_b$: $K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-4}}\approx 5.6\times 10^{-11}$.
3. Recall that a larger $K_b$ means a stronger base. Compare the two values: $5.6\times 10^{-11}>1.6\times 10^{-11}$, 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.

5. Common Pitfalls (and how to avoid them)

• 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 $\text{p}{K}_w$. Correct move: Always confirm the temperature and given $K_w$ value; a solution is neutral only when $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]$, 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 ${\text{H}}^{+}$.
• 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 $\log [{\text{H}}_3{\text{O}}^{+}]$ instead of $-\log [{\text{H}}_3{\text{O}}^{+}]$, 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 $[{\text{H}}_3{\text{O}}^{+}]$ 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 ${\text{H}}_2\text{O}$ 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 ${\text{H}}_2\text{O}$ gains or loses a proton to assign its role correctly.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following pairs is correctly labeled as a Brønsted-Lowry conjugate acid-base pair? A) ${\text{H}}_2{\text{CO}}_3$ and ${\text{CO}}_3^{2-}$ B) ${\text{H}}_2{\text{PO}}_4^{-}$ and ${\text{HPO}}_4^{2-}$ C) ${\text{H}}_2\text{S}$ and ${\text{S}}^{2-}$ D) ${\text{H}}_2\text{O}$ and ${\text{O}}^{2-}$

Worked Solution: By definition, Brønsted-Lowry conjugate acid-base pairs differ by exactly one proton (${\text{H}}^{+}$). Check each option: Option A has a difference of two protons, so it is incorrect. Option B: ${\text{H}}_2{\text{PO}}_4^{-}$ has 2 H and a -1 charge, ${\text{HPO}}_4^{2-}$ has 1 H and a -2 charge, so they differ by exactly one proton, making them a conjugate pair. Options C and D both differ by two protons, so they are incorrect. The correct answer is B.

Question 2 (Free Response)

The autoionization of water is an endothermic process: $2{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}(aq)+{\text{OH}}^{-}(aq)\Delta H>0$ (a) Explain how increasing temperature affects the value of $K_w$. Justify your answer using Le Chatelier's principle. (b) At 0°C, $K_w=1.14\times 10^{-15}$. Calculate $[{\text{H}}_3{\text{O}}^{+}]$ and $[{\text{OH}}^{-}]$ for a neutral solution at 0°C. (c) Calculate the pH and pOH of this neutral solution, and classify the solution as acidic, basic, or neutral. Justify your answer.

Worked Solution: (a) For an endothermic reaction, heat is a reactant. Increasing temperature adds heat to the system, so equilibrium shifts right to consume the added heat per Le Chatelier's principle. Shifting right increases product concentrations, so $K_w=[{\text{H}}_3{\text{O}}^{+}][{\text{OH}}^{-}]$ increases. (b) For neutral solution, $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]$, so $K_w=[{\text{H}}_3{\text{O}}^{+}{]}^2$. Solving gives: $$[{\text{H}}_3{\text{O}}^{+}]=\sqrt{1.14\times 10^{-15}}=\sqrt{11.4\times 10^{-16}}\approx 3.38\times 10^{-8}M$$ Thus, $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]=3.38\times 10^{-8}M$. (c) Calculate pH and pOH: $$\text{pH}=-\log (3.38\times 10^{-8})\approx 7.47,\text{pOH}=-\log (3.38\times 10^{-8})\approx 7.47$$ This solution is neutral, because neutrality is defined as $[{\text{H}}_3{\text{O}}^{+}]=[{\text{OH}}^{-}]$, not a pH of 7.

Question 3 (Application / Real-World Style)

Baking soda (sodium bicarbonate) dissociates completely in water to give ${\text{Na}}^{+}$ and bicarbonate ion (${\text{HCO}}_3^{-}$), which is the conjugate base of carbonic acid (${\text{H}}_2{\text{CO}}_3$). Carbonic acid has $K_a=4.3\times 10^{-7}$ at 25°C. For a 0.10 M solution of sodium bicarbonate, calculate $K_b$ of ${\text{HCO}}_3^{-}$, and calculate the pH of the solution (assume only the reaction of ${\text{HCO}}_3^{-}$ as a base contributes to $[{\text{OH}}^{-}]$, and that $K_b$ is small enough that $0.10-x\approx 0.10$). Explain why baking soda solution is mildly basic, which makes it useful for treating acid spills on skin.

Worked Solution: First, find $K_b$ for ${\text{HCO}}_3^{-}$ (conjugate base of ${\text{H}}_2{\text{CO}}_3$): $$K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{4.3\times 10^{-7}}\approx 2.3\times 10^{-8}$$ The base reaction is ${\text{HCO}}_3^{-}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_2{\text{CO}}_3(aq)+{\text{OH}}^{-}(aq)$, so $K_b=\frac{x^2}{0.10}=2.3\times 10^{-8}$, where $x=[{\text{OH}}^{-}]$. Solving gives $x^2=2.3\times 10^{-9}$, so $x\approx 4.8\times 10^{-5}M=[{\text{OH}}^{-}]$. $\text{pOH}=-\log (4.8\times 10^{-5})\approx 4.32$, so $\text{pH}=14-4.32=9.68$. This pH is mildly basic, which neutralizes excess acid from spills on skin, making baking soda a safe emergency treatment for acid exposure.

7. Quick Reference Cheatsheet

8. What's Next

This introductory chapter is the foundation for all subsequent acid-base topics in AP Chemistry Unit 8. Next, you will learn to calculate pH for strong and weak acid and base solutions, apply the equilibrium relationships you learned here to solve for pH of mixtures, and extend core concepts to polyprotic acids. Without mastering the definitions of conjugate pairs, the $K_a$-$K_w$ relationship, and pH scale rules covered here, more advanced topics like buffer pH calculations and titration curve interpretation will be impossible to solve correctly. This topic also reinforces core equilibrium concepts from Unit 7, showing how general equilibrium rules apply to specific aqueous acid-base reactions. Follow-up topics to study next:

• pH of strong and weak acids and bases
• Acid-base titrations
• Buffer solutions and buffer pH
• Polyprotic acid-base equilibria