Acid-base reactions and buffers — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Bronsted-Lowry acid-base conjugate pairs, neutralization reaction stoichiometry, buffer action and composition, the Henderson-Hasselbalch equation, buffer pH calculation, buffer capacity, and preparation of buffered solutions aligned to AP CED objectives.

You should already know: Bronsted-Lowry acid-base definitions. $K_a$ and $p{K}_a$ notation for weak acids. Basic equilibrium calculation fundamentals.

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 Acid-base reactions and buffers?

Acid-base reactions are proton-transfer reactions, defined under the Bronsted-Lowry model that dominates AP Chemistry, between a proton-donor acid and proton-acceptor base. A common subclass of acid-base reactions is neutralization, where a strong acid and strong base react to form a neutral salt and water, or a weak acid reacts with a strong base (or vice versa) to form a conjugate salt. Buffers are specialized solutions that resist dramatic changes in pH when small quantities of strong acid or strong base are added, a core application of acid-base equilibrium. This subtopic makes up approximately 8-10% of the total AP Chemistry exam, and accounts for roughly a third of the 25-30% exam weight allocated to Unit 8: Acids and Bases. It is tested in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically assess conceptual understanding of buffer action, buffer capacity, and pH prediction, while FRQ require calculation of buffer pH, analysis of buffer preparation, and connection to titration curves. Mastery of this topic is required for almost all long-form acid-base FRQ on the AP exam.

2. Acid-Base Reactions and Neutralization Stoichiometry

All acid-base reactions follow the general proton transfer form: $$HA+B\rightleftharpoons A^{-}+H{B}^{+}$$ where $HA$ is the acid, $B$ is the base, $A^{-}$ is the conjugate base of $HA$, and $H{B}^{+}$ is the conjugate acid of $B$. For all reactions involving a strong acid or strong base (which dissociate completely in water), the reaction proceeds to completion. This means we always solve for limiting reactant stoichiometry first before any equilibrium pH calculation, regardless of whether we are making a buffer or doing a titration.

For example, when mixing a weak acid with a strong base, the strong base will deprotonate as much weak acid as it has moles to react with. Moles of weak acid decrease by the moles of strong base added, and moles of conjugate base increase by the same amount. This step is non-negotiable, and skipping it is the most common source of error on acid-base problems.

Worked Example

25.0 mL of 0.150 M acetic acid ($p{K}_a=4.76$) is mixed with 15.0 mL of 0.200 M NaOH. Calculate the moles of acetic acid and acetate after the neutralization reaction goes to completion.

1. Calculate initial moles of each reactant: Moles of acetic acid = $0.0250\text{L}\times 0.150\text{mol/L}=0.00375\text{mol}$ Moles of ${\text{OH}}^{-}$ from NaOH = $0.0150\text{L}\times 0.200\text{mol/L}=0.00300\text{mol}$
2. NaOH is a strong base that dissociates completely, so the neutralization reaction goes to completion in a 1:1 stoichiometry: ${\text{CH}}_3\text{COOH}+{\text{OH}}^{-}\to {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}_2\text{O}$.
3. ${\text{OH}}^{-}$ is the limiting reactant, so moles of acetic acid remaining = $0.00375-0.00300=0.00075\text{mol}$.
4. Moles of acetate (conjugate base) formed equal moles of ${\text{OH}}^{-}$ consumed, so moles of acetate = $0.00300\text{mol}$.

Exam tip: Always complete the stoichiometry step before any equilibrium pH calculation when mixing acids and bases. Even in buffer problems, the composition of the buffer changes when you add strong acid or base, so you must adjust moles first.

3. Buffer Composition and the Henderson-Hasselbalch Equation

A buffer is a solution that contains appreciable amounts of a weak conjugate acid-base pair: this is either (1) a weak acid plus its conjugate base (added as a soluble salt) or (2) a weak base plus its conjugate acid (added as a soluble salt). Buffers resist pH change because the acid component neutralizes added strong base, and the base component neutralizes added strong acid, as long as the amount added is much smaller than the total moles of buffer components.

The Henderson-Hasselbalch equation is derived directly from the $K_a$ expression for a weak acid: Start with $K_a=\frac{[H^{+}][A^{-}]}{[HA]}$ Take the negative log of both sides and rearrange to get: $$pH=p{K}_a+\log \left(\frac{[A^{-}]}{[HA]}\right)$$ A key simplification: because both $A^{-}$ and $HA$ are in the same total volume of solution, the volume terms cancel out in the ratio. This means you can use moles of $A^{-}$ and $HA$ directly instead of concentrations, which saves time and avoids calculation errors. This equation only applies to buffer solutions where the x-is-small approximation holds, which is almost always the case for AP Chemistry problems.

Worked Example

Using the resulting moles from the previous example (0.00075 mol acetic acid, 0.00300 mol acetate, $p{K}_a=4.76$), calculate the pH of the final solution.

1. Confirm this is a valid buffer: we have appreciable amounts of both a weak acid (acetic acid) and its conjugate base (acetate), so buffer conditions hold and the Henderson-Hasselbalch equation applies.
2. We can use moles directly in the ratio because volume cancels: $\frac{[A^{-}]}{[HA]}=\frac{\text{moles }{A}^{-}/V}{\text{moles }HA/V}=\frac{\text{moles }{A}^{-}}{\text{moles }HA}$.
3. Plug values into the equation: $pH=4.76+\log \left(\frac{0.00300}{0.00075}\right)=4.76+\log (4)=4.76+0.60=5.36$.
4. Check for consistency: the concentration of conjugate base is higher than the weak acid, so pH should be higher than $p{K}_a$, which matches our result.

Exam tip: You can always use moles instead of concentrations in the Henderson-Hasselbalch equation for buffers. This eliminates errors from forgetting to update total volume after adding strong acid or base.

4. Buffer Capacity

Buffer capacity is a measure of how much strong acid or strong base a buffer can absorb before the pH changes by a large, unacceptable amount. It depends on two key factors:

1. Total concentration of buffer components: Higher total concentration = higher buffer capacity, because there are more moles of the acid and base components to neutralize added strong acid/base.
2. Ratio of $[A^{-}]$ to $[HA]$: Maximum buffer capacity occurs when $[A^{-}]=[HA]$, so $pH=p{K}_a$. Buffers are considered effective for pH values within $\pm 1$ unit of the weak acid's $p{K}_a$. Outside this range, the ratio of components is more than 10:1, so adding a small amount of strong acid/base changes the ratio drastically, leading to a large pH change.

When comparing two buffers to identify which has higher capacity, always check total moles first, then how close the pH is to $p{K}_a$.

Worked Example

Which of the following 1.0 L buffers has the highest capacity to resist pH change after addition of 0.10 moles of strong acid? Buffer X: 0.10 M acetic acid / 0.10 M acetate ($p{K}_a=4.76$) Buffer Y: 0.50 M acetic acid / 0.50 M acetate Buffer Z: 0.05 M acetic acid / 0.50 M acetate

1. The capacity to absorb added strong acid depends on the moles of conjugate base (acetate) available to neutralize the added $H^{+}$, and how close the component ratio is to 1:1.
2. Calculate moles of acetate for each buffer: X = 0.10 mol, Y = 0.50 mol, Z = 0.50 mol. Both Y and Z have enough acetate to absorb 0.10 mol of $H^{+}$, while X does not.
3. Compare Y and Z: Y has a 1:1 ratio of acetate to acetic acid, which is optimal for maximum capacity, while Z has a 10:1 ratio, far from optimal.
4. Y has the highest buffer capacity for the addition of strong acid.

Exam tip: When asked to select the best buffer for a target pH, the closest $p{K}_a$ to the target pH (within 1 unit) is always the correct choice, all else equal.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the Henderson-Hasselbalch equation when only one member of the conjugate pair is present (e.g., only weak acid, no conjugate base left after neutralization). Why: Students memorize the equation and reach for it automatically regardless of solution composition. Correct move: Always confirm that both the weak acid and its conjugate base are present in appreciable concentrations before using the equation.
• Wrong move: Using the original buffer component moles to calculate pH after adding strong acid or base. Why: Students forget that added acid/base reacts with buffer components to change their amounts. Correct move: Always adjust moles of HA and A- after adding strong acid/base: subtract added H+ from A- and add to HA, or subtract added OH- from HA and add to A- before calculating pH.
• Wrong move: Claiming a solution of strong acid and its conjugate base (e.g., HCl + NaCl) is a buffer. Why: Students think any conjugate pair forms a buffer, but strong acid conjugate bases are too weakly basic to neutralize added H+. Correct move: Only weak acid/conjugate base or weak base/conjugate acid pairs form valid buffers; strong acid/base pairs never do.
• Wrong move: Flipping the ratio in the Henderson-Hasselbalch equation, writing $\log \left(\frac{[HA]}{[A^{-}]}\right)$. Why: Students misremember the derivation and reverse the terms. Correct move: If pH > pKa, the log term must be positive, so the numerator must be larger than the denominator, meaning conjugate base is always on top. Use this check every time you use the equation.
• Wrong move: Assuming higher buffer capacity corresponds to lower pH. Why: Students confuse buffer capacity (how much acid/base can be absorbed) with pH (the current pH of the buffer). Correct move: Separate the two concepts: capacity depends on total moles of buffer components and ratio, not pKa or pH.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following combinations of solutions will produce a buffer solution when mixed in equal volumes at 25°C? A) 0.1 M HCl and 0.1 M NH₄Cl B) 0.1 M HCl and 0.2 M NH₃ C) 0.1 M NaOH and 0.1 M CH₃COOH D) 0.1 M NaOH and 0.1 M HCl

Worked Solution: A valid buffer requires appreciable amounts of a weak conjugate pair after mixing is complete. Option A has no weak base, so cannot form a buffer. Option C mixes equal moles of strong base and weak acid, which completely neutralizes the weak acid to form only conjugate base, with no weak acid left, so no buffer. Option D is a neutralization of strong acid and strong base, forming only a neutral salt, so no buffer. Option B: mixing equal volumes of 0.1 M HCl and 0.2 M NH₃ means half the NH₃ reacts with HCl to form NH₄+, leaving half the NH₃ unreacted, resulting in equal moles of weak base (NH₃) and conjugate acid (NH₄+), which is a valid buffer. Correct answer: B

Question 2 (Free Response)

A student prepares a buffer by mixing 100.0 mL of 0.300 M hydrocyanic acid (HCN, $p{K}_a=9.21$) and 50.0 mL of 0.300 M KOH. (a) Calculate the pH of the resulting buffer solution. (b) Explain why the pH of this buffer changes by less than 0.5 pH units when 1.0 mL of 0.10 M HCl is added, while the pH of 150 mL of pure water changes by more than 4 pH units when the same HCl is added. (c) A student wants to prepare a buffer with pH = 9.00. Would this HCN/KCN buffer system be appropriate for this pH? Justify your answer.

Worked Solution: (a) Calculate initial moles: Moles HCN = $0.1000\text{L}\times 0.300\text{mol/L}=0.0300\text{mol}$ Moles ${\text{OH}}^{-}$ = $0.0500\text{L}\times 0.300\text{mol/L}=0.0150\text{mol}$ After 1:1 neutralization: moles HCN = $0.0300-0.0150=0.0150\text{mol}$, moles ${\text{CN}}^{-}$ = $0.0150\text{mol}$ Use Henderson-Hasselbalch: $pH=9.21+\log \left(\frac{0.0150}{0.0150}\right)=9.21+0=9.21$

(b) The buffer contains comparable amounts of weak acid (HCN) and conjugate base (CN-). The added $H^{+}$ from HCl reacts completely with CN- to form HCN: $H^{+}+C{N}^{-}\to HCN$. Since total moles of buffer components are much larger than moles of added H+, the ratio $\frac{[C{N}^{-}]}{[HCN]}$ changes only slightly, so pH changes very little. In pure water, there are no buffer components to neutralize added H+, so H+ concentration increases from $1\times 10^{-7}\text{M}$ to ~$6.6\times 10^{-4}\text{M}$, leading to a pH drop from 7 to ~3.2, a change of ~3.8 pH units.

(c) A buffer is effective when the target pH is within 1 pH unit of the $p{K}_a$ of the weak acid component. The target pH of 9.00 is only 0.21 pH units away from the $p{K}_a$ of 9.21, which falls within the effective range. So this buffer system is appropriate.

Question 3 (Application / Real-World Style)

Blood plasma is buffered by the carbonic acid/bicarbonate buffer system, with $p{K}_{a1}$ of $H_{2}C{O}_3=6.37$. Normal blood pH is 7.40, and the total concentration of carbonic acid plus bicarbonate in blood is approximately 0.025 M. Calculate the concentration of bicarbonate ($HC{O}_3^{-}$) in normal blood, and explain what this means for buffering against excess metabolic acid.

Worked Solution: Use Henderson-Hasselbalch: $7.40=6.37+\log \left(\frac{[HC{O}_3^{-}]}{[H_{2}C{O}_3]}\right)$ Rearrange to get: $\log \left(\frac{[HC{O}_3^{-}]}{[H_{2}C{O}_3]}\right)=1.03$, so $\frac{[HC{O}_3^{-}]}{[H_{2}C{O}_3]}=10^{1.03}\approx 10.7$ We know $[HC{O}_3^{-}]+[H_{2}C{O}_3]=0.025\text{M}$, so substitute $[H_{2}C{O}_3]=\frac{[HC{O}_3^{-}]}{10.7}$: $[HC{O}_3^{-}]+\frac{[HC{O}_3^{-}]}{10.7}=0.025⟹[HC{O}_3^{-}]\approx 0.023\text{M}$ This means ~92% of the buffer is in the bicarbonate form, so the buffer has a high capacity to absorb excess H+ from metabolic acid production, keeping blood pH within the narrow range required for enzyme function.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for acid-base titrations and solubility equilibria, which make up the remaining parts of Unit 8 (Acids and Bases) and Unit 9 (Applications of Thermodynamics), respectively. When solving titration problems, you will use the exact same stoichiometry-first approach and Henderson-Hasselbalch calculation you learned here to find the pH at any point along a titration curve, including the buffer region before the equivalence point. Buffers are also critical for understanding biological acid-base homeostasis, a common real-world context for AP FRQ questions. Without mastering the stoichiometry step and buffer pH calculation, you will not be able to correctly interpret titration data or solve solubility problems that involve pH-dependent dissolution of ionic compounds.

• Acid-base titrations
• Solubility equilibria
• pH-dependent solubility