Acid–Base Titrations - College Chemistry

SECTION 1

Historical Context & Motivation

The practice of titration — the incremental addition of a reagent of known concentration to determine the concentration of an unknown — has roots stretching back to the early days of quantitative chemistry. Before modern instrumentation, chemists needed a reliable, reproducible method for measuring the amount of a dissolved substance, and volumetric analysis provided exactly that. The technique answered a fundamental analytical question: how much of a particular solute is present in a given solution? Over several centuries, innovations in glassware, indicator chemistry, and electrochemistry transformed titration from an imprecise craft into one of the most rigorous quantitative tools available to the working chemist.

1729 • Claude-Joseph Geoffroy

Geoffroy performed one of the earliest recorded volumetric analyses, measuring the strength of vinegar by adding potash until effervescence ceased. This established the core idea of reacting a known reagent against an unknown.

1767 • François-Antoine-Henri Descroizilles

Descroizilles invented the berthollimeter, a precursor to the modern burette, enabling precise volume delivery of a titrant for the first time.

1835 • Gay-Lussac Coins 'Titration'

Joseph Louis Gay-Lussac introduced the term titre (from French, meaning 'title' or 'standard') to describe the process of determining the concentration of a solution, thereby giving the technique its modern name.

1894 • Ostwald & Indicator Theory

Wilhelm Ostwald applied the theory of ionic equilibria to explain how acid–base indicators function, showing that indicators are themselves weak acids or bases whose conjugate forms differ in color. This allowed rational indicator selection.

1909 • Sørensen Defines pH

Søren Sørensen introduced the pH scale, enabling chemists to plot titration curves quantitatively and predict equivalence-point pH values with precision, bridging titration practice with thermodynamic theory.

These historical developments collectively shaped a central question that acid–base titration answers: given a solution of unknown concentration, how can we exploit the stoichiometry of a neutralization reaction and a visual or electrochemical signal to determine the exact amount of acid or base present? Understanding the theoretical framework behind titration curves, indicator selection, and buffer regions is essential for any chemist performing quantitative analysis.

SECTION 2

Core Principles & Definitions

An acid–base titration is built on the stoichiometric neutralization of an acid with a base (or vice versa). A solution of precisely known concentration, the titrant (also called the standard solution), is delivered from a burette into a flask containing the analyte — the solution of unknown concentration. As the titrant reacts with the analyte, the pH of the solution changes. The point at which stoichiometrically equivalent amounts of acid and base have been mixed is the equivalence point. The experimentally detected signal change (color shift, pH meter reading, conductivity change) marks the end point, which ideally coincides with the equivalence point within acceptable analytical error.

Equivalence Point

The theoretical point at which the moles of titrant exactly equal the stoichiometric requirement to neutralize the analyte. At this point, n_acid × valence = n_base × valence. The pH at equivalence depends on the nature of the salt formed.

Half-Equivalence Point

The point at which exactly half of the analyte has been neutralized. For a weak acid titrated with a strong base, pH = pK_a at the half-equivalence point, because [HA] = [A⁻]. This provides a direct experimental route to K_a.

Buffer Region

The region of the titration curve between about 10% and 90% neutralization, where both the weak acid (or base) and its conjugate are present in appreciable quantities. pH changes slowly here, governed by the Henderson–Hasselbalch equation.

Indicator Selection

An acid–base indicator is a weak acid (HIn) whose conjugate base (In⁻) has a distinctly different color. The indicator's transition range (pK_In ± 1) must overlap the steep portion of the titration curve near the equivalence point.

Titration Curve

A plot of pH versus volume of titrant added. Its shape — initial pH, buffer plateau, steep inflection at equivalence, and post-equivalence rise or fall — encodes all thermodynamic and stoichiometric information about the reaction.

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KEY TAKEAWAY

Think of a titration like filling a swimming pool to a precise water level using a calibrated hose. The pool (analyte) has an unknown volume deficit, and you know the flow rate (concentration) of the hose (titrant). You watch for the exact moment the water reaches the marked line (equivalence point). An indicator is like a float sensor that triggers an alarm (color change) right when the target level is hit. If you choose a sensor calibrated for the wrong depth, your reading will be off — just as choosing an indicator with the wrong pK_In introduces systematic error.

SECTION 3

Titration Curve: Visual Explanation

The titration curve is the most information-rich representation of an acid–base titration. The diagram below compares the curves for a strong acid–strong base titration and a weak acid–strong base titration. Notice how the weak-acid curve begins at a higher initial pH, displays a buffer region where pH changes gradually, and has an equivalence-point pH above 7 because the conjugate base of the weak acid is itself basic.

Comparison of titration curves: the cyan curve represents 25.0 mL of 0.100 M HCl titrated with 0.100 M NaOH (equivalence pH = 7.00). The pink curve represents 25.0 mL of 0.100 M acetic acid (K_a = 1.8 × 10⁻⁵) titrated with 0.100 M NaOH. Note the buffer region and the elevated equivalence pH (≈ 8.72) in the weak-acid case.

Several features of the curves deserve attention. First, the strong acid curve starts near pH 1 and remains relatively flat until very close to the equivalence point, where it undergoes a dramatic, nearly vertical jump through pH 7. The steep vertical region spans roughly from pH 3 to pH 11, making indicator selection relatively forgiving — phenolphthalein, methyl orange, or bromothymol blue all transition within this steep zone. In contrast, the weak acid curve starts at a higher pH (because acetic acid is only partially ionized), displays a broad buffer plateau centered on pK_a = 4.74 at the half-equivalence point, and its steep region is narrower and shifted upward (approximately pH 7 to pH 11). For this titration, phenolphthalein (transition range 8.2–10.0) is appropriate, whereas methyl orange would change color far too early, in the buffer region, leading to a significant titration error.

SECTION 4

Mathematical Framework

Quantitative treatment of acid–base titrations relies on a set of interconnected equations. The stoichiometric core determines the equivalence point, while equilibrium expressions govern the pH at every stage of the titration.

STOICHIOMETRIC EQUIVALENCE

C_acid × V_acid = C_base × V_base

At the equivalence point, the moles of acid equal the moles of base (for monoprotic systems). C = molar concentration (mol/L), V = volume (L). This relationship is the foundation of every titration calculation.

HENDERSON–HASSELBALCH EQUATION

pH = pKₐ + log([A⁻] / [HA])

Valid in the buffer region of a weak acid–strong base titration. [A⁻] is the concentration of conjugate base formed, and [HA] is the remaining weak acid. At the half-equivalence point, [A⁻] = [HA], so pH = pK_a.

EQUIVALENCE-POINT pH (WEAK ACID–STRONG BASE)

pH = 7 + ½(pKₐ + log C_salt)

At equivalence, the solution contains only the conjugate base A⁻ at concentration C_salt. This base hydrolyzes water: A⁻ + H₂O ⇌ HA + OH⁻. The resulting pH is always > 7.

WEAK BASE $K_b$ RELATIONSHIP

Kₐ × K_b = K_w = 1.0 × 10⁻¹⁴ (at 25 °C)

K_a and K_b are the acid and base dissociation constants of a conjugate pair. This relationship is essential when calculating the pH at the equivalence point of a weak acid–strong base or weak base–strong acid titration.

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When Does Henderson–Hasselbalch Apply?

The Henderson–Hasselbalch equation is an approximation valid when both [HA] and [A⁻] are much larger than K_a (i.e., the 5% approximation holds). It breaks down very close to the start of the titration (where [A⁻] ≈ 0) and at the equivalence point (where [HA] ≈ 0). At those extremes, use the full equilibrium expression or the quadratic formula.

SECTION 5

Classification of Acid–Base Titrations

The shape of a titration curve and the pH at the equivalence point depend critically on the strength classification of the acid and base involved. Four fundamental categories exist, each with distinctive curve features and indicator requirements. The diagram below provides a concise visual comparison of the four titration types, emphasizing the location of the equivalence point pH on the vertical axis.

Summary of the four acid–base titration types. The strong–strong case is the simplest, with an equivalence pH of exactly 7. Weak acid–strong base and strong acid–weak base titrations have equivalence points displaced from 7. The weak–weak case lacks a sharp equivalence inflection, making visual indicators unreliable.

Titration Type	Equivalence pH	Recommended Indicator	Buffer Region?
Strong acid + Strong base	7.00	Any (broad steep region)	No
Weak acid + Strong base	> 7 (basic)	Phenolphthalein (8.2–10.0)	Yes
Strong acid + Weak base	< 7 (acidic)	Methyl red (4.4–6.2)	Yes
Weak acid + Weak base	Depends on Kₐ vs $K_b$	None reliable; use pH meter	Yes (both sides)

Summary of titration types, equivalence-point pH, and indicator recommendations

SECTION 6

Worked Example: Weak Acid–Strong Base Titration

Consider the titration of 25.00 mL of 0.100 M acetic acid (CH₃COOH, K_a = 1.8 × 10⁻⁵) with 0.100 M NaOH. We will calculate the pH at four key stages: (a) before any NaOH is added, (b) after 12.50 mL NaOH (half-equivalence), (c) after 25.00 mL NaOH (equivalence), and (d) after 30.00 mL NaOH (post-equivalence).

pH at Four Key Points in the Titration of Acetic Acid with NaOH

1
Step 1 — Initial pH (0 mL NaOH added)

Before any base is added, we have a 0.100 M weak acid solution. Set up the equilibrium: CH₃COOH ⇌ CH₃COO⁻ + H⁺. Using the ICE table: K_a = x² / (0.100 − x) ≈ x² / 0.100. Solving: x = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M.

pH = −log(1.34 × 10⁻³) = 2.87
2
Step 2 — Half-Equivalence Point (12.50 mL NaOH)

Moles of NaOH added = 0.01250 L × 0.100 mol/L = 1.25 × 10⁻³ mol. Initial moles of CH₃COOH = 0.02500 L × 0.100 mol/L = 2.50 × 10⁻³ mol. After reaction: remaining CH₃COOH = 2.50 × 10⁻³ − 1.25 × 10⁻³ = 1.25 × 10⁻³ mol; CH₃COO⁻ formed = 1.25 × 10⁻³ mol. Since [HA] = [A⁻], the Henderson–Hasselbalch equation gives pH = pK_a + log(1) = pK_a.

pH = pK_a = −log(1.8 × 10⁻⁵) = 4.74
3
Step 3 — Equivalence Point (25.00 mL NaOH)

All acetic acid has been converted to acetate: moles CH₃COO⁻ = 2.50 × 10⁻³ mol in a total volume of 50.00 mL, so [CH₃COO⁻] = 0.0500 M. Acetate hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. K_b = K_w / K_a = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰. x = √(5.56 × 10⁻¹⁰ × 0.0500) = √(2.78 × 10⁻¹¹) = 5.27 × 10⁻⁶ M = [OH⁻]. pOH = 5.28, so pH = 14.00 − 5.28.

pH = 8.72
4
Step 4 — Post-Equivalence (30.00 mL NaOH)

Excess moles NaOH = (0.03000 − 0.02500) L × 0.100 mol/L = 5.00 × 10⁻⁴ mol in 55.00 mL total volume. [OH⁻] = 5.00 × 10⁻⁴ / 0.05500 = 9.09 × 10⁻³ M. pOH = −log(9.09 × 10⁻³) = 2.04.

pH = 14.00 − 2.04 = 11.96

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Key Observation

Notice that the equivalence-point pH of 8.72 is well above 7. This confirms that phenolphthalein (transition range 8.2–10.0) is an appropriate indicator for this titration, whereas methyl orange (transition 3.1–4.4) would have changed color at roughly 20 mL of NaOH added — approximately 5 mL before the true equivalence point — introducing a large negative titration error.

SECTION 7

Strengths, Limitations & Sources of Error

Acid–base titration is one of the most widely used techniques in analytical chemistry, but it is not without limitations. Understanding its strengths and weaknesses allows the analyst to choose the right tool for the job and to minimize systematic errors when titrations are appropriate.

Strengths	Limitations
High precision (±0.1% relative error achievable with careful technique)	Requires a sharp equivalence inflection; weak acid–weak base titrations lack this
Inexpensive equipment — burette, flask, indicator	The analyte must react rapidly and completely with the titrant
Directly yields moles and concentration without calibration curves	CO₂ absorption from air can introduce carbonate error when using strong base titrants
Well-understood theory for selecting indicators and predicting curve shape	Colored or turbid solutions obscure indicator color changes
Applicable to polyprotic acids (multiple equivalence points)	Very dilute solutions (< 10⁻³ M) produce flattened curves with ambiguous end points

Comparison of strengths and limitations of acid–base titrations

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KEY TAKEAWAY

In engineering quality control, a titration is analogous to a go/no-go gauge test: it is simple, rapid, and highly reliable for the intended measurement, but it doesn't tell you everything about the sample. Just as a go/no-go gauge confirms a dimension is within tolerance without providing a full surface profile, a titration confirms concentration with high accuracy but reveals nothing about non-acidic or non-basic solutes in the mixture. Recognizing when to supplement a titration with spectroscopic or chromatographic methods is a hallmark of mature analytical reasoning.

SECTION 8

Connection to Advanced Theory: Polyprotic & Non-Aqueous Titrations

The monoprotic strong- and weak-acid titrations discussed so far represent the simplest cases. In more advanced settings, chemists encounter polyprotic acid titrations (e.g., H₃PO₄, H₂SO₃) where multiple equivalence points appear on the curve, each corresponding to the successive deprotonation of one ionizable hydrogen. The key condition for observing distinct endpoints is that the ratio K_a1 / K_a2 must exceed approximately 10³; otherwise, the two deprotonation steps overlap and the separate equivalence points merge into one broad, poorly defined transition.

Another frontier is non-aqueous titration, where the solvent is changed to a medium such as glacial acetic acid, dimethylformamide, or tert-butanol. In these solvents, species that are too weak to titrate in water (e.g., amines with pK_b > 8 in water) can be made to behave as strong bases, enabling sharp equivalence points. The pharmaceutical industry relies heavily on non-aqueous titrations for assaying amine-containing drugs.

Feature	Monoprotic Titration (This Lesson)	Polyprotic / Non-Aqueous (Advanced)
Number of equivalence points	1	2 or more (polyprotic); 1 (non-aqueous)
Solvent	Water	Water (polyprotic) or organic solvents
Equilibrium constant used	Kₐ, $K_b$ , $K_w$	Kₐ₁, Kₐ₂, …; solvent autoprotolysis constant
End-point detection	Indicator or pH meter	Potentiometric or visual (crystal violet in acetic acid)
Typical applications	General acid/base analysis, water quality	Pharmaceutical assays, geochemistry, food science

Monoprotic versus advanced titration approaches

Courses in quantitative analysis and instrumental analysis will extend these ideas, introducing Gran plots for locating equivalence points in poorly defined curves, second-derivative methods for automated end-point detection, and thermometric titration, where the temperature change of the solution signals the equivalence point. Each method builds on the same stoichiometric foundation established in this lesson.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the equivalence-point pH of a weak acid–strong base titration is above 7, while the equivalence-point pH of a strong acid–weak base titration is below 7. Relate your answer to the hydrolysis of the conjugate species formed at equivalence.

Show answer

At the equivalence point of a weak acid–strong base titration, the solution contains only the conjugate base (A⁻) of the weak acid. This conjugate base hydrolyzes water: A⁻ + H₂O ⇌ HA + OH⁻, generating hydroxide ions and driving the pH above 7. Conversely, at the equivalence point of a strong acid–weak base titration, the solution contains only the conjugate acid (BH⁺) of the weak base, which hydrolyzes: BH⁺ + H₂O ⇌ B + H₃O⁺, producing hydronium ions and a pH below 7. The extent of the pH deviation from 7 depends on the magnitude of K_b (or K_a of BH⁺) and the concentration of the salt at equivalence.

Question 2: BASIC CALCULATION

A 20.00 mL sample of an unknown monoprotic acid is titrated with 0.150 M NaOH. The equivalence point is reached after 28.50 mL of NaOH has been added. Calculate the concentration of the acid.

Show answer

At equivalence, moles of NaOH = moles of acid (monoprotic, 1:1 stoichiometry). Moles NaOH = 0.02850 L × 0.150 mol/L = 4.275 × 10⁻³ mol. Therefore, moles of acid = 4.275 × 10⁻³ mol in 20.00 mL. Concentration of acid = 4.275 × 10⁻³ mol / 0.02000 L = 0.214 M. This calculation uses only the stoichiometric equivalence relation C_acid × V_acid = C_base × V_base and does not require knowledge of whether the acid is strong or weak.

Question 3: INTERMEDIATE

During the titration described in Problem 2, the pH at the half-equivalence point is measured to be 4.20. (a) Determine K_a for the unknown acid. (b) Calculate the pH after 10.00 mL of NaOH has been added.

Show answer

(a) At the half-equivalence point, pH = pK_a. Therefore pK_a = 4.20 and K_a = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵. (b) After 10.00 mL NaOH: moles NaOH = 0.01000 × 0.150 = 1.500 × 10⁻³ mol. Remaining HA = 4.275 × 10⁻³ − 1.500 × 10⁻³ = 2.775 × 10⁻³ mol. A⁻ formed = 1.500 × 10⁻³ mol. Using Henderson–Hasselbalch: pH = 4.20 + log(1.500 × 10⁻³ / 2.775 × 10⁻³) = 4.20 + log(0.5405) = 4.20 + (−0.267) = 3.93. Note that volume cancels in the ratio, so we can use moles directly.

Question 4: APPLIED

A food scientist titrates 50.00 mL of a lemon juice sample with 0.200 M NaOH. The titration requires 18.75 mL of NaOH to reach the equivalence point. Assuming the acidity is due entirely to citric acid (H₃C₆H₅O₇, triprotic, M = 192.12 g/mol), calculate (a) the molar concentration of citric acid and (b) the mass of citric acid per liter of juice. (Assume all three protons are titrated.)

Show answer

(a) Moles NaOH = 0.01875 L × 0.200 mol/L = 3.750 × 10⁻³ mol. Because citric acid is triprotic and all three protons react with NaOH, moles citric acid = 3.750 × 10⁻³ / 3 = 1.250 × 10⁻³ mol. Concentration = 1.250 × 10⁻³ / 0.05000 L = 0.0250 M. (b) Mass per liter = 0.0250 mol/L × 192.12 g/mol = 4.80 g/L. This result is within the typical range for lemon juice (approximately 4–8 g citric acid per 100 mL). The key insight is properly accounting for the stoichiometric factor of 3 in the mole ratio.

Question 5: CRITICAL THINKING

A student titrates 25.00 mL of 0.100 M NH₃ (K_b = 1.8 × 10⁻⁵) with 0.100 M HCl. (a) Calculate the pH at the equivalence point. (b) The student uses phenolphthalein as the indicator and observes the color change at 22.5 mL of HCl. Explain why this introduces a systematic error and state whether the reported concentration of NH₃ will be too high or too low. (c) Recommend a more appropriate indicator.

Show answer

(a) At equivalence, 2.50 × 10⁻³ mol NH₄⁺ is in 50.00 mL, so [NH₄⁺] = 0.0500 M. K_a of NH₄⁺ = K_w / K_b = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. Hydrolysis: x = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ M = [H₃O⁺]. pH = −log(5.27 × 10⁻⁶) = 5.28. (b) Phenolphthalein transitions at pH 8.2–10.0. Since this is a weak base being titrated with a strong acid, the pH starts high and decreases. Phenolphthalein will decolorize when the pH drops below ≈8.2, which occurs well before the equivalence point at pH 5.28. The student stops adding HCl too early (22.5 mL vs. the true equivalence volume of 25.0 mL), underestimating the moles of HCl needed. Because moles of NH₃ = moles HCl at equivalence, the reported concentration of NH₃ will be too low. (c) Methyl red (transition range 4.4–6.2) or methyl orange (3.1–4.4) would be appropriate because their transition ranges overlap the steep portion of the curve around pH 5.28.

SUMMARY

Acid–Base Titrations: Key Concepts Review

Summary

An acid–base titration determines the concentration of an unknown acid or base by reacting it with a standard solution (titrant) of known concentration. The equivalence point is reached when moles of acid equal stoichiometric moles of base. Its pH depends on the nature of the salt formed: pH = 7 for strong–strong combinations, pH > 7 for weak acid–strong base (conjugate base hydrolyzes), and pH < 7 for strong acid–weak base (conjugate acid hydrolyzes). The Henderson–Hasselbalch equation governs pH in the buffer region, and at the half-equivalence point pH = pK_a, providing a direct experimental route to the dissociation constant.

Proper indicator selection requires that the indicator's transition range (pK_In ± 1) falls within the steep portion of the titration curve near the equivalence point. For weak acid–weak base titrations, no sharp inflection exists, and potentiometric methods (pH meter) are preferred. Advanced extensions include polyprotic titrations with multiple equivalence points and non-aqueous titrations that enable analysis of very weak acids or bases by changing the solvent medium.