# Acid-base reactions and buffers

> AP Chemistry · AP Chemistry CED Unit 8
> Source: https://www.owlsprep.com/study/ap-chemistry-u8-acid-base-reactions-and-buffers/

This aligned study guide covers Bronsted-Lowry conjugate pairs, neutralization stoichiometry, buffer action, Henderson-Hasselbalch calculations, buffer capacity, and buffer preparation for AP Chemistry exam preparation.

**Prerequisites:** Bronsted-Lowry acid-base definitions; $K_a$ and $pK_a$ notation for weak acids; Basic equilibrium calculation fundamentals

## Learning objectives

- Explain Bronsted-Lowry proton transfer in acid-base reactions
- Calculate neutralization stoichiometry for mixed acid-base solutions
- Identify valid buffer compositions and explain buffer action
- Use the Henderson-Hasselbalch equation to calculate buffer pH
- Analyze buffer capacity and select appropriate buffers for a target pH

## Acid-Base Reactions and Neutralization Stoichiometry

Acid-base reactions are proton-transfer reactions under the Bronsted-Lowry model: an acid donates a proton, and a base accepts a proton. Neutralization reactions between acids and bases go to completion whenever a strong acid or strong base is involved, meaning you must always solve limiting reactant stoichiometry first before any equilibrium pH calculation.

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

**Worked example:** 25.0 mL of 0.150 M acetic acid ($pK_a = 4.76$) 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:
2. $$Moles of acetic acid = 0.0250\ \text{L} \times 0.150\ \text{mol/L} = 0.00375\ \text{mol}\Moles of \text{OH}^- from NaOH = 0.0150\ \text{L} \times 0.200\ \text{mol/L} = 0.00300\ \text{mol}$$
3. NaOH is a strong base that dissociates completely, so the 1:1 neutralization reaction proceeds to completion:
4. $$\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$$
5. $\text{OH}^-$ is the limiting reactant, so subtract consumed moles from acetic acid:
6. $$Moles of acetic acid remaining = 0.00375 - 0.00300 = 0.00075\ \text{mol}$$
7. Moles of acetate conjugate base formed equal moles of $\text{OH}^-$ consumed:
8. $$Moles of acetate = 0.00300\ \text{mol}$$

> **Exam tip:** Always complete the stoichiometry step before any equilibrium pH calculation when mixing acids and bases, even for buffer problems.

## Buffer Composition and the Henderson-Hasselbalch Equation

A buffer is a solution that resists large pH changes when small amounts of strong acid or base are added. Valid buffers contain appreciable amounts of a weak conjugate acid-base pair: either a weak acid plus its conjugate base (as a soluble salt), or a weak base plus its conjugate acid.

**Buffer** — A solution containing appreciable amounts of a weak conjugate acid-base pair that resists large pH changes when small amounts of strong acid or base are added.

*Example:* 0.1 M acetic acid + 0.1 M sodium acetate

**Derivation:** Derive the Henderson-Hasselbalch equation from the $K_a$ expression for a weak acid

*Starting from:* K_a = \frac{[H^+][A^-]}{[HA]}

1. Take the negative base-10 logarithm of both sides:
2. $$-\log(K_a) = -\log([H^+]) - \log\left(\frac{[A^-]}{[HA]}\right)$$
3. Rearrange using the definitions $pK_a = -\log(K_a)$ and $pH = -\log([H^+])$:
4. $$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$$

*Conclusion:* The volume terms cancel in the ratio, so moles of $A^-$ and $HA$ can be used directly instead of concentrations, simplifying calculations and avoiding common errors.

**Worked example:** Using the moles from the previous neutralization example (0.00075 mol acetic acid, 0.00300 mol acetate, $pK_a = 4.76$), calculate the pH of the final solution.

1. Confirm this is a valid buffer: we have appreciable amounts of both weak acid (acetic acid) and its conjugate base (acetate), so the Henderson-Hasselbalch equation applies.
2. Since both components share the same total volume, the ratio of moles equals the ratio of concentrations:
3. $$\frac{[A^-]}{[HA]} = \frac{\text{moles } A^- / V}{\text{moles } HA / V} = \frac{\text{moles } A^-}{\text{moles } HA}$$
4. Substitute values into the Henderson-Hasselbalch equation:
5. $$pH = 4.76 + \log\left(\frac{0.00300}{0.00075}\right) = 4.76 + 0.60 = 5.36$$
6. Check for consistency: since conjugate base concentration is higher than weak acid, pH should be higher than $pK_a$, which matches our result.

> **Exam tip:** You can always use moles instead of concentrations in the Henderson-Hasselbalch equation, eliminating errors from forgetting to update total volume after mixing.

## Buffer Capacity

Buffer capacity is a measure of how much strong acid or strong base a buffer can absorb before pH changes by a large, unacceptable amount. It depends on two key factors: (1) total concentration of buffer components: higher total concentration gives higher buffer capacity, and (2) the ratio of conjugate base to weak acid: maximum buffer capacity occurs when $[A^-] = [HA]$, so $pH = pK_a$.

Buffers are considered effective for pH values within $\pm 1$ unit of the weak acid's $pK_a$. 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.

**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 ($pK_a = 4.76$); Buffer Y: 0.50 M acetic acid / 0.50 M acetate; Buffer Z: 0.05 M acetic acid / 0.50 M acetate.

1. Capacity to absorb added strong acid depends on the moles of conjugate base (acetate) available to neutralize added $H^+$, and how close the component ratio is to 1:1 (optimal for maximum capacity).
2. Calculate moles of acetate for each buffer: X = 0.10 mol, Y = 0.50 mol, Z = 0.50 mol. Only Y and Z have enough acetate to absorb 0.10 mol of $H^+$.
3. Compare Y and Z: Y has a 1:1 ratio of acetate to acetic acid, which is optimal for maximum buffer capacity, while Z has a 10:1 ratio far from optimal.
4. Conclusion: Buffer 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 weak acid with $pK_a$ closest to the target pH (within 1 unit) is always the correct choice, all else equal.

## AP-Style Practice

**Check your understanding**

Test your understanding of buffer composition with this AP-style multiple choice question:

1. 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

   *Why:* A valid buffer requires appreciable amounts of a weak conjugate pair after mixing. Half of the NH₃ reacts with HCl to form NH₄+, leaving equal moles of NH₃ (weak base) and NH₄+ (conjugate acid), forming a valid buffer.

**Worked example:** A student prepares a buffer by mixing 100.0 mL of 0.300 M hydrocyanic acid (HCN, $pK_a = 9.21$) and 50.0 mL of 0.300 M KOH. (a) Calculate the pH of the resulting buffer solution. (b) Explain why pH changes very little when small amounts of strong acid are added. (c) Is this system appropriate for a buffer of pH = 9.00? Justify your answer.

1. Part (a): Calculate initial moles of each reactant:
2. $$Moles HCN = 0.1000\ \text{L} \times 0.300\ \text{mol/L} = 0.0300\ \text{mol}\Moles \text{OH}^- = 0.0500\ \text{L} \times 0.300\ \text{mol/L} = 0.0150\ \text{mol}$$
3. After 1:1 neutralization, adjust moles of buffer components:
4. $$Moles HCN remaining = 0.0300 - 0.0150 = 0.0150\ \text{mol}\Moles CN^- formed = 0.0150\ \text{mol}$$
5. Substitute into the Henderson-Hasselbalch equation:
6. $$pH = 9.21 + \log\left(\frac{0.0150}{0.0150}\right) = 9.21 + 0 = 9.21$$
7. Part (b): The buffer contains comparable amounts of weak acid (HCN) and conjugate base (CN⁻). Added H⁺ reacts completely with CN⁻ to form HCN. Since total moles of buffer components are much larger than moles of added H⁺, the ratio $\frac{[CN^-]}{[HCN]}$ changes only slightly, so pH changes very little. Pure water has no buffer components to neutralize added H⁺, so pH changes drastically.
8. Part (c): A buffer is effective when the target pH is within 1 pH unit of the weak acid's $pK_a$. 9.00 is only 0.21 pH units away from 9.21, so this buffer system is appropriate.

## Common pitfalls

- **Wrong:** 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 it fails: Students memorize the equation and reach for it automatically regardless of the actual composition of the solution.
  - Correct: Always confirm that both the weak acid and its conjugate base are present in appreciable concentrations before using the Henderson-Hasselbalch equation.
- **Wrong:** Using original buffer component moles to calculate pH after adding strong acid or base
  - Why it fails: Students forget that added strong acid/base reacts with buffer components to change the amount of each component.
  - Correct: 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:** Claiming a solution of strong acid and its conjugate base (e.g., HCl + NaCl) is a buffer
  - Why it fails: Students incorrectly assume any conjugate acid-base pair forms a buffer.
  - Correct: Only weak acid/conjugate base or weak base/conjugate acid pairs form valid buffers; strong acid/base pairs never form buffers.
- **Wrong:** Flipping the ratio in the Henderson-Hasselbalch equation, writing $\log\left(\frac{[HA]}{[A^-]}\right)$
  - Why it fails: Students misremember the derivation and reverse the terms.
  - Correct: If pH > pKa, the log term must be positive, so conjugate base must be in the numerator. Use this check every time you use the equation.
- **Wrong:** Assuming higher buffer capacity corresponds to lower pH
  - Why it fails: Students confuse buffer capacity (how much acid/base can be absorbed) with the current pH of the buffer.
  - Correct: Separate the two concepts: buffer capacity depends on total moles of buffer components and their ratio, not pKa or pH.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| General Bronsted-Lowry reaction | $HA + B \rightleftharpoons A^- + HB^+$ | HA = acid, B = base, $A^-$ = conjugate base |
| Neutralization stoichiometry | Always complete before equilibrium calculations | Reactions with strong acids/bases go to completion |
| Henderson-Hasselbalch equation | $pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$ | Only for buffers; moles substitute for concentrations |
| Weak base buffer equation | $pOH = pK_b + \log\left(\frac{[HB^+]}{[B]}\right)$ | Convert pOH to pH at the end |
| Maximum buffer capacity | $pH = pK_a$ | Occurs when $[A^-] = [HA]$ |
| Effective buffer pH range | $pK_a \pm 1$ | Buffers are ineffective outside this range |
| Valid buffer composition | Weak acid + conjugate base *or* weak base + conjugate acid | Strong acid/base conjugate pairs do not form buffers |
| Buffer capacity trend | Higher total concentration = higher capacity | Same ratio: more concentrated buffers absorb more acid/base |

## 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 pH-dependent solubility problems.

- [pH of weak acids](https://www.owlsprep.com/study/ap-chemistry-u8-ph-of-weak-acids/)
- [pH of weak bases](https://www.owlsprep.com/study/ap-chemistry-u8-ph-of-weak-bases/)
- [Molecular structure of acids and bases](https://www.owlsprep.com/study/ap-chemistry-u8-molecular-structure-of-acids-and/)

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