Acid-Base Titrations - AP Chemistry

SECTION 1

Historical Context & Motivation

The need to determine the exact concentration of an acid or base in solution has driven chemical analysis for centuries. Before modern instrumentation, chemists relied on volumetric methods—carefully measuring the volume of a solution of known concentration required to react completely with an unknown. Acid-base titration emerged as the most widely practiced of these volumetric techniques, enabling quantitative work in pharmacy, water treatment, food science, and industrial chemistry. The method's elegance lies in its simplicity: a measured volume of titrant (a solution of precisely known concentration) is added to an analyte (the unknown) until the reaction reaches its stoichiometric equivalence point.

1729 • First Volumetric Analysis

Claude-Joseph Geoffroy used a measured volume of potassium carbonate to test vinegar strength, establishing the principle of volumetric neutralization.

1835 • Gay-Lussac Standardizes Titrations

Joseph Louis Gay-Lussac refined the buret and introduced the term 'titre,' formalizing the technique for silver assays and acid-base work.

1894 • Ostwald's Indicator Theory

Wilhelm Ostwald explained why indicators change color by applying the ionization theory, allowing chemists to choose indicators rationally based on pH range.

1909 • Sørensen Defines pH

Søren Sørensen introduced the pH scale, giving titrations a quantitative framework for tracking hydrogen-ion concentration throughout the procedure.

Today, acid-base titrations remain a cornerstone of analytical chemistry—not merely as a historical curiosity, but as a practical tool that directly illustrates stoichiometry, equilibrium, and pH in a single experiment. The central question the technique addresses is deceptively simple: How much acid or base is actually present in a sample? Answering it requires mastery of neutralization stoichiometry, buffer chemistry, and the interpretation of titration curves.

SECTION 2

Core Principles & Definitions

An acid-base titration is built on the reaction between a Brønsted–Lowry acid (proton donor) and a Brønsted–Lowry base (proton acceptor). The titrant is delivered from a buret into a flask containing the analyte. A standard solution is one whose concentration has been determined to high precision, often by primary-standard titration. The point at which stoichiometrically equivalent amounts of acid and base have been combined is the equivalence point, while the end point is the experimentally observed signal—usually an indicator color change or a sharp jump on a pH meter—that approximates it.

Equivalence Point

The exact point where moles of titrant equal the stoichiometric requirement of the analyte. The pH at equivalence depends on the strength of the acid and base involved.

Half-Equivalence Point

The point where exactly half the analyte has been neutralized. For a weak acid titrated with a strong base, pH = pK_a at this point, because [HA] = [A⁻].

Indicator Selection

An acid-base indicator is a weak acid whose conjugate base differs in color. It must have a color-change range (pK_In ± 1) that overlaps the steep region of the titration curve near equivalence.

Buffer Region

Before equivalence in a weak acid/strong base titration, the solution contains both HA and A⁻, forming a buffer. The Henderson–Hasselbalch equation governs pH throughout this region.

SECTION 3

Titration Curve — Visual Explanation

The most informative representation of a titration is the titration curve, a plot of solution pH versus volume of titrant added. Two canonical cases—strong acid/strong base and weak acid/strong base—illustrate fundamentally different curve shapes and equivalence-point pH values.

The curve shows pH rising slowly at first (excess strong acid), then surging through the steep region near the equivalence point at 20.0 mL (pH = 7.00 for strong acid–strong base), and finally leveling off as excess NaOH dominates. The yellow dashed box highlights the region where a single drop causes a pH change of several units.

Several features of this curve are critical for the AP exam. Before any titrant is added, the pH reflects the initial concentration of the strong acid. As NaOH is added, the excess H⁺ concentration decreases gradually, producing a relatively flat region. Near the equivalence point, the curve steepens dramatically—a hallmark that allows indicators to signal the end point. After equivalence, excess OH⁻ controls the pH, and the curve flattens again. For strong acid–strong base titrations, the equivalence-point pH is always 7.00 at 25 °C because the salt produced (NaCl) does not hydrolyze.

SECTION 4

Mathematical Framework

The quantitative backbone of every titration calculation is the mole relationship at the equivalence point. For a monoprotic acid–monoprotic base reaction, the stoichiometry is 1 : 1, and the central equation is straightforward.

EQUIVALENCE-POINT STOICHIOMETRY

n(acid) = n(base) → M_a × V_a = M_b × V_b

M_a = molarity of acid, V_a = volume of acid, M_b = molarity of base, V_b = volume of base. For polyprotic acids or polyhydroxy bases, multiply each side by the appropriate stoichiometric coefficient.

HENDERSON–HASSELBALCH EQUATION

pH = pKₐ + log([A⁻] / [HA])

Applies in the buffer region of a weak acid / strong base titration. At the half-equivalence point, [A⁻] = [HA], so log(1) = 0 and pH = pKₐ.

pH AT EQUIVALENCE (WEAK ACID / STRONG BASE)

Kᵦ = Kw / Kₐ ; [OH⁻] = √(Kᵦ × C_salt) ; pOH = −log[OH⁻] ; pH = 14 − pOH

At equivalence, only the conjugate base A⁻ remains in solution at concentration C_salt. It hydrolyzes, so the equivalence-point pH is above 7 for weak acid / strong base titrations.

ℹ

AP Exam Tip

SECTION 5

Titration Curve Classification

The shape and key features of a titration curve depend on whether the analyte and titrant are strong or weak. Recognizing these curve types at a glance is essential for the AP exam. The diagram below contrasts the weak acid / strong base curve with the strong acid / strong base curve to highlight the buffer region and the shifted equivalence-point pH.

The red curve (strong acid / strong base) begins at a much lower initial pH and has its equivalence point at pH 7.00. The violet curve (weak acid / strong base) starts higher, displays a flat buffer region (green dashed box), passes through the half-equivalence point where pH = pKₐ (yellow dot), and reaches an equivalence-point pH above 7 due to conjugate base hydrolysis.

Titration Type	Initial pH	Equiv. Pt. pH	Suitable Indicator
Strong acid / Strong base	Very low (≈ 1)	7.00	Bromothymol blue (6.0–7.6)
Weak acid / Strong base	Moderate (≈ 2–5)	> 7 (basic)	Phenolphthalein (8.2–10.0)
Strong acid / Weak base	High (≈ 10–13)	< 7 (acidic)	Methyl red (4.4–6.2)
Weak acid / Weak base	Variable	≈ 7 (if Kₐ ≈ K_b)	No sharp break; not titrated with indicators

Summary of titration curve characteristics by acid-base strength

SECTION 6

Worked Example — Weak Acid / Strong Base

A 25.00 mL sample of 0.200 M acetic acid (CH₃COOH, Kₐ = 1.8 × 10⁻⁵) is titrated with 0.100 M NaOH. Determine the pH (a) initially, (b) after adding 25.00 mL of NaOH (half-equivalence), (c) at the equivalence point, and (d) after adding 55.00 mL of NaOH.

Worked example

1
Step 1 — Determine the equivalence-point volume

Moles of CH₃COOH = 0.02500 L × 0.200 M = 0.00500 mol. At equivalence, moles NaOH = moles acid = 0.00500 mol. Volume of NaOH = 0.00500 mol ÷ 0.100 M = 0.05000 L = 50.00 mL.

V(eq) = 50.00 mL NaOH
2
Step 2 — Initial pH (0 mL NaOH added)

Only weak acid is present. Set up the ICE table: Kₐ = x² / (0.200 − x) ≈ x² / 0.200. Solving: x = √(1.8 × 10⁻⁵ × 0.200) = √(3.6 × 10⁻⁶) = 1.90 × 10⁻³ M. pH = −log(1.90 × 10⁻³) = 2.72. The 5% approximation check: (1.90 × 10⁻³ / 0.200) × 100% = 0.95%, valid.

pH = 2.72
3
Step 3 — Half-equivalence point (25.00 mL NaOH)

Moles NaOH added = 0.02500 L × 0.100 M = 0.00250 mol, which neutralizes exactly half the acid. Remaining CH₃COOH = 0.00250 mol; CH₃COO⁻ produced = 0.00250 mol. Since [HA] = [A⁻], the Henderson–Hasselbalch equation gives pH = pKₐ + log(1) = pKₐ = −log(1.8 × 10⁻⁵) = 4.74.

pH = pKₐ = 4.74
4
Step 4 — Equivalence point (50.00 mL NaOH)

All acid has been converted to CH₃COO⁻. Total volume = 25.00 + 50.00 = 75.00 mL = 0.07500 L. [CH₃COO⁻] = 0.00500 mol / 0.07500 L = 0.0667 M. The acetate ion hydrolyzes: K_b = Kw / Kₐ = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. [OH⁻] = √(5.56 × 10⁻¹⁰ × 0.0667) = √(3.71 × 10⁻¹¹) = 6.09 × 10⁻⁶ M. pOH = 5.22; pH = 14.00 − 5.22 = 8.78.

pH = 8.78 (basic, as expected)
5
Step 5 — Post-equivalence (55.00 mL NaOH)

Excess NaOH = (0.05500 − 0.05000) L × 0.100 M = 0.000500 mol. Total volume = 0.08000 L. [OH⁻] = 0.000500 / 0.08000 = 6.25 × 10⁻³ M. pOH = −log(6.25 × 10⁻³) = 2.20. pH = 14.00 − 2.20 = 11.80. In the post-equivalence region, the excess strong base dominates and the hydrolysis contribution is negligible.

pH = 11.80

SECTION 7

Indicators vs. pH Meter Methods

Two primary methods exist for detecting the end point of a titration: visual indicators and potentiometric (pH meter) measurements. Each has distinct advantages and limitations depending on the system being titrated and the precision required.

Feature	Indicator Method	pH Meter Method
Equipment	Buret, flask, indicator solution	Buret, flask, calibrated pH electrode
Precision	± 0.5–1 pH unit (color transition range)	± 0.01 pH unit
Weak/weak titrations	Unreliable—no sharp color break	Can locate inflection point on derivative plot
Polyprotic acids	Can detect first endpoint only if ΔpH is large	Can resolve multiple equivalence points
Data output	Single endpoint volume	Complete titration curve; pKₐ directly readable

Comparison of end-point detection methods

One complete titration setup A buret delivers measured titrant into a flask containing the solution being analyzed.

SECTION 8

Connection to Advanced Theory

Acid-base titrations in the AP course rely on simplifying assumptions—dilute solutions, activity coefficients of unity, and negligible autoionization contributions. In more advanced analytical chemistry, these assumptions are relaxed, and the full alpha (α) fraction approach or exact proton-balance equations become necessary.

Feature	AP-Level Treatment	Advanced / Analytical Treatment
Activity	Concentrations used directly as activities	Activity coefficients (γ) from Debye–Hückel; a = γ × C
Polyprotic acids	Treated as sequential monoprotic systems	Alpha fraction (α₀, α₁, α₂) derived from coupled equilibria
Exact pH calculation	ICE table or Henderson–Hasselbalch	Charge balance + mass balance solved simultaneously (proton balance)
Curve fitting	Qualitative sketch or point-by-point	Gran plot linearization to determine Kₐ and V(eq) precisely

AP vs. advanced approaches to titration analysis

Understanding the AP-level framework thoroughly prepares you to step into these more rigorous models. The Henderson–Hasselbalch equation, for example, is a special case of the full alpha-fraction treatment, and the ICE-table approach is an approximation to the charge-balance method. If you continue into quantitative analysis or biochemistry, titration theory extends to amino acid zwitterion titrations, EDTA complexometric titrations, and redox titrations—all of which share the same underlying principle of stoichiometric equivalence.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student titrates a weak base with a strong acid. Which of the following correctly describes the pH at the equivalence point?

A. pH = 7.00, because equal moles of acid and base have reacted
B. pH > 7, because the conjugate base hydrolyzes in solution
C. pH < 7, because the conjugate acid of the weak base hydrolyzes in solution
D. pH < 7, because excess strong acid remains unreacted

Show answer (C)

C is correct. At the equivalence point, all the weak base has been converted to its conjugate acid, which donates protons to water (hydrolyzes), making the solution acidic (pH < 7). A is wrong because pH = 7 only applies to strong acid / strong base titrations where the salt does not hydrolyze. B is wrong because it describes the equivalence point of a weak acid / strong base titration, not weak base / strong acid. D is wrong because at the equivalence point, stoichiometrically equivalent amounts have reacted—there is no excess strong acid yet.

Question 2: BASIC CALCULATION

What volume of 0.250 M NaOH is required to reach the equivalence point when titrating 30.0 mL of 0.150 M HCl?

A. 12.0 mL
B. 18.0 mL
C. 20.0 mL
D. 50.0 mL

Show answer (B)

B is correct. At equivalence, moles acid = moles base: (0.0300 L)(0.150 M) = (V)(0.250 M). V = 0.00450 / 0.250 = 0.0180 L = 18.0 mL. A (12.0 mL) likely results from an arithmetic error such as dividing by the wrong molarity. C (20.0 mL) would be correct only if both concentrations were equal. D (50.0 mL) confuses the molarity ratio and gives far too much NaOH.

Question 3: INTERMEDIATE

A 20.0 mL sample of 0.100 M formic acid (HCOOH, Kₐ = 1.8 × 10⁻⁴) is titrated with 0.100 M NaOH. What is the pH after 10.0 mL of NaOH has been added?

A. 2.37
B. 3.74
C. 4.74
D. 7.00

Show answer (B)

B is correct. After 10.0 mL NaOH, moles NaOH = 0.00100 mol. Initial moles HCOOH = 0.00200 mol. Remaining HCOOH = 0.00100 mol; HCOO⁻ formed = 0.00100 mol. This is the half-equivalence point, so pH = pKₐ = −log(1.8 × 10⁻⁴) = 3.74. A (2.37) is the initial pH of the formic acid before any base is added. C (4.74) is the pKₐ of acetic acid, not formic acid—a common mix-up. D (7.00) applies only at the equivalence point of a strong acid–strong base titration.

Question 4: APPLIED

A 50.0 mL sample of an unknown monoprotic weak acid HA is titrated with 0.120 M KOH. The equivalence point is reached after 40.0 mL of KOH is added, and the pH at equivalence is measured to be 8.95. (a) Calculate the concentration of HA in the original sample. (b) Calculate the Kₐ of HA. Show all work. (c) A student selects methyl orange (color change range pH 3.1–4.4) as the indicator for this titration. Would this indicator give an accurate result? Justify your answer. (d) Sketch and label a titration curve for this experiment. Indicate the initial pH region, the buffer region, the half-equivalence point, and the equivalence point.

Show answer

(a) Moles KOH = 0.0400 L × 0.120 M = 0.00480 mol = moles HA. [HA] = 0.00480 / 0.0500 = 0.0960 M. (b) At equivalence, total volume = 50.0 mL + 40.0 mL = 90.0 mL. [A⁻] = 0.00480 / 0.0900 = 0.05333 M. The conjugate base hydrolyzes: A⁻ + H₂O ⇌ HA + OH⁻. From pH 8.95, pOH = 14.00 − 8.95 = 5.05, so [OH⁻] = 10⁻⁵·⁰⁵ = 8.91 × 10⁻⁶ M. Kb = x² / (0.05333 − x) ≈ x² / 0.05333 = (8.91 × 10⁻⁶)² / 0.05333 = 7.94 × 10⁻¹¹ / 0.05333 = 1.49 × 10⁻⁹. The approximation is valid since x ≪ 0.05333. Kₐ = Kw / Kb = 1.0 × 10⁻¹⁴ / 1.49 × 10⁻⁹ = 6.7 × 10⁻⁶. (c) Methyl orange changes color in the pH 3.1–4.4 range. The equivalence-point pH is 8.95, which is far above this range. The indicator would change color long before equivalence (during initial titrant additions), causing the student to underestimate the volume of KOH needed. A better choice would be phenolphthalein (8.2–10.0), whose range brackets the equivalence-point pH. (d) The sketch should show: initial pH around 2–3 (consistent with a weak acid of moderate Kₐ), a gradually rising buffer region from 0 to 40 mL, a half-equivalence point at 20.0 mL where pH = pKₐ = −log(6.7 × 10⁻⁶) ≈ 5.2, a steep rise near 40 mL, the equivalence point at 40.0 mL and pH 8.95, and a leveling-off in the post-equivalence region around pH 12–13.

Question 5: CRITICAL THINKING

A student performs a titration of 25.0 mL of a diprotic acid H₂A with 0.100 M NaOH. The following data are collected: Volume NaOH (mL): 0, 5, 10, 15, 20, 25, 30, 35, 40, 45 pH: 1.52, 1.80, 2.15, 2.69, 4.27, 5.90, 6.35, 6.73, 7.21, 9.85 (a) Identify the volume at the first equivalence point. Justify using the data. (b) Calculate the initial concentration of H₂A. (c) Estimate pKₐ₁ from the data. Explain your reasoning. (d) At what volume would you expect the second equivalence point? Explain. (e) A classmate claims the pH at the second equivalence point will be 7.00. Evaluate this claim.

Show answer

(a) The first equivalence point occurs near 20 mL. The data show the largest single-step pH increase between 15 mL (pH 2.69) and 20 mL (pH 4.27), a jump of 1.58 units, which is characteristic of the steep inflection at an equivalence point. The pH then begins rising more gradually, indicating the start of the second buffer region. (b) At the first equivalence point (≈ 20.0 mL NaOH), moles NaOH = 0.0200 L × 0.100 M = 0.00200 mol. Since H₂A loses one proton: moles H₂A = 0.00200. [H₂A] = 0.00200 / 0.0250 = 0.0800 M. (c) The half-equivalence point for the first dissociation occurs at half of 20.0 mL = 10.0 mL. At this point, pH = pKₐ₁. From the data, pH at 10.0 mL = 2.15, so pKₐ₁ ≈ 2.15 (Kₐ₁ ≈ 7.1 × 10⁻³). (d) The second equivalence point requires another 0.00200 mol of NaOH (for the second proton), so an additional 20.0 mL beyond the first equivalence point, giving 40.0 mL total. This is consistent with the data showing a second steep rise between 35 and 45 mL. (e) The claim is incorrect. At the second equivalence point, the solution contains only A²⁻, the fully deprotonated conjugate base. A²⁻ undergoes hydrolysis (A²⁻ + H₂O ⇌ HA⁻ + OH⁻), producing OH⁻ and raising the pH above 7. The data support this: the pH at 40 mL is 7.21 (close to or at the second equivalence point) and by 45 mL (post-equivalence) it reaches 9.85 due to excess NaOH. A pH of 7 would only be expected if the conjugate base did not hydrolyze, which would require A²⁻ to be the conjugate base of a strong acid—contradicting the given diprotic weak acid.