Acid-base reactions and buffers

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

1. Acid-Base Reactions and Neutralization Stoichiometry★★☆☆☆⏱ 4 min

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 ( $p K_{a} = 4.76$ ) 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:
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
$OH^{-}$ 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 $OH^{-}$ 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.

2. Buffer Composition and the Henderson-Hasselbalch Equation★★★☆☆⏱ 5 min

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.

📘 Definition

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

Goal:

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 $p K_{a} = - lo g (K_{a})$ and $p H = - lo g ([H^{+}])$ :
4
$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$

Result:

The volume terms cancel in the ratio, so moles of $A^{-}$ and $H A$ 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, $p K_{a} = 4.76$ ), calculate the pH of the final solution.

1
Confirm this is a valid buffer: we have appreciable amounts of both 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 $p K_{a}$ , 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.

3. Buffer Capacity★★★☆☆⏱ 3 min

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 $[A^{-}] = [H A]$ , so $p H = p K_{a}$ .

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.

📐 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
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 $p K_{a}$ closest to the target pH (within 1 unit) is always the correct choice, all else equal.

4. AP-Style Practice★★★★☆⏱ 2 min

✓ Quick check

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

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
Reveal answer
B —
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, $p K_{a} = 9.21$ ) and 50.0 mL of 0.300 M KOH. (a) Calculate the pH of the resulting buffer solution. (b) Explain why 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 $\frac{[ C N ^{-} ]}{[ H C N ]}$ 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 $p K_{a}$ . 9.00 is only 0.21 pH units away from 9.21, so this buffer system is appropriate.

5. Common Pitfalls

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 the actual composition of the solution.

Correct move:

Always confirm that both the weak acid and its conjugate base are present in appreciable concentrations before using the Henderson-Hasselbalch equation.

Wrong move:

Using original buffer component moles to calculate pH after adding strong acid or base

Why:

Students forget that added strong acid/base reacts with buffer components to change the amount of each component.

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 incorrectly assume any conjugate acid-base pair forms a buffer.

Correct move:

Only weak acid/conjugate base or weak base/conjugate acid pairs form valid buffers; strong acid/base pairs never form buffers.

Wrong move:

Flipping the ratio in the Henderson-Hasselbalch equation, writing $lo g (\frac{[ H A ]}{[ A ^{-} ]})$

Why:

Students misremember the derivation and reverse the terms.

Correct move:

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 move:

Assuming higher buffer capacity corresponds to lower pH

Why:

Students confuse buffer capacity (how much acid/base can be absorbed) with the current pH of the buffer.

Correct move:

Separate the two concepts: buffer capacity depends on total moles of buffer components and their ratio, not pKa or pH.

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
General Bronsted-Lowry reaction	$H A + B ⇌ A^{-} + H B^{+}$	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	$p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$	Only for buffers; moles substitute for concentrations
Weak base buffer equation	$pO H = p K_{b} + lo g (\frac{[ H B ^{+} ]}{[ B ]})$	Convert pOH to pH at the end
Maximum buffer capacity	$p H = p K_{a}$	Occurs when $[A^{-}] = [H A]$
Effective buffer pH range	$p K_{a} \pm 1$	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

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
Buffer capacity comparison
2022 · FRQ
Buffer pH calculation

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.