Weak Acid-Strong Base Titrations — Interpret weak acid–strong base titration curves - General Chemistry 2

SECTION 1

Historical Context & Motivation

The analytical technique of titration — the controlled addition of a reagent of known concentration to an analyte of unknown concentration — has roots stretching back to the eighteenth century, when French chemist François-Antoine-Henri Descroizilles devised the first graduated measuring tubes for volumetric analysis. Over the following two centuries, chemists refined both the glassware and the underlying theory until titration became the cornerstone of quantitative acid–base chemistry. Understanding how pH evolves during a weak acid–strong base titration, in particular, requires concepts drawn from equilibrium theory, buffer chemistry, and hydrolysis of conjugate bases — all of which matured in the early twentieth century.

1795 • Birth of Volumetric Analysis

François Descroizilles introduced the berthollimeter, a graduated tube for measuring reagent volumes, laying the groundwork for modern titration technique.

1884 • Arrhenius Acid–Base Theory

Svante Arrhenius proposed that acids produce H⁺ and bases produce OH⁻ in aqueous solution, providing the first theoretical lens through which to view neutralization reactions quantitatively.

1909 • The pH Scale

Søren Sørensen defined the pH scale as −log[H⁺], giving chemists a compact logarithmic metric with which to track hydrogen-ion concentration during a titration.

1916 • Henderson–Hasselbalch Equation

Lawrence Henderson and Karl Hasselbalch independently formalized the relationship pH = pK_a + log([A⁻]/[HA]), enabling rapid buffer-region pH calculations that are central to interpreting weak acid titration curves.

1930s • Potentiometric Titration

The development of glass pH electrodes allowed continuous, instrument-based monitoring of pH, replacing color-change indicators and revealing the full sigmoidal shape of titration curves with high precision.

With these theoretical and instrumental advances in place, a central question crystallized: How does the pH of a weak acid solution change as we progressively add a strong base, and why does the resulting curve look so different from a strong acid–strong base titration? Answering this question requires a synthesis of acid dissociation equilibria, buffer behavior, stoichiometric equivalence, and the hydrolysis of conjugate bases — concepts we will develop in the sections that follow.

SECTION 2

Core Principles & Definitions

Before dissecting the titration curve itself, it is essential to internalize several foundational ideas that govern weak acid–strong base systems. A weak acid (HA) only partially dissociates in water; its degree of ionization is characterized by the acid dissociation constant K_a. When a strong base such as NaOH is added, the hydroxide ion reacts essentially to completion with HA, converting it to the conjugate base A⁻. The interplay between the remaining HA and the formed A⁻ defines a buffer system that dominates the middle portion of the curve.

Ka and pKa

The acid dissociation constant K_a quantifies the equilibrium position of HA ⇌ H⁺ + A⁻. Its negative logarithm, pK_a, directly appears on the titration curve at the half-equivalence point.

Buffer Region

When appreciable amounts of both HA and A⁻ coexist in solution, the system resists pH changes upon addition of acid or base. The Henderson–Hasselbalch equation describes pH throughout this region.

Equivalence Point

At the equivalence point, moles of OH⁻ added equal the original moles of HA. All HA has been converted to A⁻, whose hydrolysis makes the solution basic — the pH is above 7.

Half-Equivalence Point

When exactly half the acid has been neutralized, [HA] = [A⁻] and the Henderson–Hasselbalch equation simplifies to pH = pK_a. This provides a direct experimental route to K_a.

Post-Equivalence Region

Beyond the equivalence point, excess OH⁻ overwhelms the conjugate base hydrolysis, and the pH is determined almost entirely by the concentration of surplus hydroxide.

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KEY TAKEAWAY

Think of a weak acid–strong base titration like filling a bathtub that has a drain partially open (the equilibrium). At first, water rises slowly because some drains away (the buffer region resists pH change). Once the tub overflows (equivalence point), the dynamics shift entirely — the excess water pours over the edge and the level shoots up (pH rises sharply). The buffer region is the slow-rise phase, the equivalence point is the overflow threshold, and the post-equivalence region is the rapid rise of excess base.

SECTION 3

Visual Explanation — The Titration Curve

The titration curve for a 0.10 M acetic acid solution titrated with 0.10 M NaOH. Note the initial region starting above pH 1, the buffer region with its gradual slope, the equivalence point above pH 7, and the post-equivalence rise governed by excess OH⁻.

The diagram above illustrates the four distinct regions of a typical weak acid–strong base titration curve. The initial region (before any NaOH is added) reflects the pH of the weak acid alone, determined by K_a and the initial concentration [HA]₀. Because the acid is weak, the starting pH is considerably higher than it would be for a strong acid of the same concentration — typically between 2 and 5 depending on the acid's strength and concentration.

As NaOH is added, the solution enters the buffer region, where HA and A⁻ coexist. The pH rises slowly and nearly linearly on a logarithmic scale; this is the hallmark of buffer action. At the half-equivalence point — exactly halfway to the equivalence point — [HA] = [A⁻], so pH = pK_a. This is the flattest portion of the curve and the point of maximum buffer capacity. Continuing to add base pushes the pH upward more sharply until the steep inflection at the equivalence point, where all HA has been consumed and the solution contains only the conjugate base A⁻ in water. The hydrolysis of A⁻ generates OH⁻, placing the equivalence-point pH above 7 — a critical distinction from strong acid–strong base titrations, where the equivalence pH is exactly 7.

SECTION 4

Mathematical Framework

Quantitative interpretation of each region of the titration curve requires its own mathematical treatment. We present the key equations in the order they appear along the curve, from zero added base to post-equivalence excess.

INITIAL pH (NO BASE ADDED)

Kₐ = x² / (C₀ − x) → pH = −log(x)

Here x = [H⁺] at equilibrium, C₀ = initial weak acid concentration, and K_a is the acid dissociation constant. If C₀/K_a > 100, simplify to x ≈ √(K_a × C₀).

BUFFER REGION (HENDERSON–HASSELBALCH)

pH = pKₐ + log([A⁻] / [HA])

[A⁻] = moles of conjugate base formed (equal to moles of OH⁻ added) divided by total volume; [HA] = moles of acid remaining divided by total volume. Because the volumes cancel in the ratio, one may equivalently use mole quantities directly: pH = pK_a + log(n_A⁻ / n_HA).

EQUIVALENCE POINT — CONJUGATE BASE HYDROLYSIS

Kb = Kw / Kₐ ; [OH⁻] = √(Kb × Ceq) ; pH = 14 − pOH

At the equivalence point, all HA has been converted to A⁻ with concentration C_eq = n_HA,initial / V_total. The base hydrolysis constant K_b is derived from K_w = 1.0 × 10⁻¹⁴ at 25 °C.

POST-EQUIVALENCE REGION

[OH⁻]excess = (nOH⁻ − nHA,initial) / Vtotal ; pOH = −log[OH⁻] ; pH = 14 − pOH

Beyond the equivalence point, the excess strong base dominates the pH. The contribution of A⁻ hydrolysis becomes negligible compared to the unreacted OH⁻.

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Important Distinction

Unlike a strong acid–strong base titration, the equivalence-point pH of a weak acid–strong base titration is always above 7. This is because the conjugate base A⁻ undergoes hydrolysis (A⁻ + H₂O ⇌ HA + OH⁻), producing excess hydroxide. The weaker the acid (smaller K_a), the more basic the equivalence-point pH.

SECTION 5

Detailed Breakdown of the Four Regions

A systematic region-by-region analysis is the most effective way to interpret any weak acid–strong base titration curve. The second diagram below focuses on the distinct chemical species present in each region, along with the dominant equilibrium or calculation strategy.

Schematic overview of the four titration regions showing the dominant species, calculation method, and approximate pH range for each. The lower panel highlights the special points and the indicator-selection rule.

Region	Volume Range	Dominant Species	Calculation Strategy
1 — Initial	V = 0	HA	ICE table with K_a
2 — Buffer	0 < V < V_eq	HA + A⁻	Henderson–Hasselbalch
3 — Equivalence	V = V_eq	A⁻ only	K_b hydrolysis
4 — Post-equivalence	V > V_eq	A⁻ + excess OH⁻	Excess [OH⁻] / V_total

Summary of regions, species, and calculation strategies

SECTION 6

Worked Example — Acetic Acid Titrated with NaOH

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 points: (a) before any base is added, (b) after 12.50 mL of NaOH (half-equivalence), (c) after 25.00 mL of NaOH (equivalence), and (d) after 30.00 mL of NaOH (post-equivalence).

Comprehensive Titration Curve Calculation

1
Step 1 — Initial pH (V = 0 mL NaOH)

With no base added, the solution contains only 0.100 M CH₃COOH. Set up the equilibrium: K_a = x²/(0.100 − x). Since C₀/K_a = 0.100/1.8 × 10⁻⁵ = 5556 ≫ 100, approximate: 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 (V = 12.50 mL NaOH)

Moles of HA initially = 0.02500 L × 0.100 M = 2.50 × 10⁻³ mol. Moles of OH⁻ added = 0.01250 L × 0.100 M = 1.25 × 10⁻³ mol. After reaction: moles HA remaining = 2.50 × 10⁻³ − 1.25 × 10⁻³ = 1.25 × 10⁻³ mol; moles A⁻ 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 (V = 25.00 mL NaOH)

All 2.50 × 10⁻³ mol HA have been converted to A⁻. Total volume = 25.00 + 25.00 = 50.00 mL = 0.05000 L. [A⁻] = 2.50 × 10⁻³ / 0.05000 = 0.0500 M. Calculate K_b = K_w/K_a = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. Then [OH⁻] = √(5.56 × 10⁻¹⁰ × 0.0500) = √(2.78 × 10⁻¹¹) = 5.27 × 10⁻⁶ M. pOH = −log(5.27 × 10⁻⁶) = 5.28.

pH = 14.00 − 5.28 = 8.72
4
Step 4 — Post-Equivalence (V = 30.00 mL NaOH)

Moles OH⁻ added = 0.03000 L × 0.100 M = 3.00 × 10⁻³ mol. Moles OH⁻ excess = 3.00 × 10⁻³ − 2.50 × 10⁻³ = 5.00 × 10⁻⁴ mol. Total volume = 25.00 + 30.00 = 55.00 mL = 0.05500 L. [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

💡

Check Your Results

Verify that your calculated values are internally consistent: the initial pH should be acidic (2–5), the half-equivalence pH should equal pK_a, the equivalence-point pH should be above 7 (typically 7.5–10 for common weak acids), and the post-equivalence pH should rise sharply toward 12–13. If any value violates these expectations, recheck your stoichiometry and algebra.

SECTION 7

Weak Acid vs. Strong Acid Titration — Key Differences

A common source of confusion in acid–base chemistry is conflating the titration curve of a weak acid with that of a strong acid. While both produce sigmoidal pH-vs-volume plots, the two curves differ in several critical respects that directly impact indicator selection, buffer utility, and the interpretation of experimental data.

Feature	Strong Acid–Strong Base	Weak Acid–Strong Base
Initial pH	Very low (pH ≈ 1 for 0.1 M HCl)	Higher (pH ≈ 2–5, depends on K_a)
Buffer region	Absent — pH rises steadily	Present — pH rises slowly around pK_a
Equivalence-point pH	Exactly 7.00 (at 25 °C)	Above 7 (typically 8–10)
Steep-jump range	Wide (pH ≈ 3–11)	Narrower (pH ≈ 7–10)
Suitable indicator	Any with endpoint pH 4–10 (e.g., bromothymol blue)	Must change color above pH 7 (e.g., phenolphthalein, pH 8.2–10)
Post-equivalence	Same — excess OH⁻ governs pH	Same — excess OH⁻ governs pH

Comparison of strong acid and weak acid titrations with a strong base

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KEY TAKEAWAY

The two most distinctive signatures of a weak acid–strong base curve compared to a strong acid–strong base curve are the buffer region (which has no analog in strong acid titrations) and the elevated equivalence-point pH (above 7 instead of at 7). Think of the buffer region like the shock absorbers on a car — they dampen the pH "bumps" caused by added base, keeping the ride smooth until the system is overwhelmed at the equivalence point.

SECTION 8

Connection to Polyprotic Acids & Advanced Theory

The monoprotic weak acid–strong base titration is the simplest case of a broader category that extends to polyprotic acid titrations (e.g., H₂CO₃, H₃PO₄), where multiple equivalence points appear on the curve, each preceded by its own buffer region and half-equivalence point. For a diprotic acid H₂A, the curve exhibits two inflection points corresponding to the sequential loss of the first and second protons, with plateau regions governed by K_a1 and K_a2, respectively. Each half-equivalence point gives pH = pK_a,n for the corresponding deprotonation step.

Feature	Monoprotic Weak Acid + Strong Base	Diprotic Weak Acid + Strong Base
Number of equivalence points	1	2 (at V_eq,1 and V_eq,2 = 2 × V_eq,1)
Buffer regions	1 (around pK_a)	2 (around pK_a1 and pK_a2)
pH at 1st equivalence	> 7 (K_b hydrolysis)	pH = (pK_a1 + pK_a2) / 2 (amphoteric HA⁻)
Mathematical complexity	Moderate — one K_a	High — two coupled equilibria, often requires alpha-fraction approach

Monoprotic vs. diprotic weak acid titrations

Beyond polyprotic acids, advanced courses treat the exact titration equation — a single expression derived from charge balance, mass balance, and equilibrium constants — that yields the pH at any point without the need to divide the curve into regions. While algebraically demanding (it leads to a cubic or quartic in [H⁺]), this approach eliminates the approximations inherent in the Henderson–Hasselbalch treatment and is essential when K_a is very small or the solution is very dilute. Mastering the four-region approach presented here, however, provides the conceptual framework upon which these advanced treatments are built.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the equivalence-point pH of a weak acid–strong base titration is above 7, while that of a strong acid–strong base titration is exactly 7. Reference the specific chemical species responsible for the elevated pH.

Show answer

At the equivalence point of a weak acid–strong base titration, all the weak acid HA has been stoichiometrically converted to its conjugate base A⁻. This conjugate base undergoes hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻. The production of OH⁻ raises the pH above 7. In contrast, a strong acid–strong base titration produces only water and a spectator salt (e.g., NaCl) at equivalence; neither ion hydrolyzes appreciably, so the pH remains at 7. The weaker the original acid (smaller K_a), the larger K_b of the conjugate base, and the higher the equivalence-point pH.

Question 2: BASIC CALCULATION

A 50.00 mL sample of 0.200 M formic acid (HCOOH, K_a = 1.8 × 10⁻⁴) is titrated with 0.200 M NaOH. Calculate the pH after 25.00 mL of NaOH has been added (the half-equivalence point).

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At the half-equivalence point, exactly half of the formic acid has been neutralized, so [HCOOH] = [HCOO⁻]. The Henderson–Hasselbalch equation gives pH = pK_a + log([HCOO⁻]/[HCOOH]) = pK_a + log(1) = pK_a. Therefore pH = −log(1.8 × 10⁻⁴) = 3.74. This is also the point of maximum buffer capacity. Note that the identity holds regardless of the concentrations or volumes, as long as exactly half the acid has been neutralized.

Question 3: INTERMEDIATE

For the same titration in Problem 2, calculate the pH at the equivalence point. (V_eq = 50.00 mL NaOH)

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At equivalence, moles HCOOH initially = 0.05000 L × 0.200 M = 0.01000 mol, all converted to HCOO⁻. Total volume = 50.00 + 50.00 = 100.00 mL = 0.10000 L. [HCOO⁻] = 0.01000 / 0.10000 = 0.1000 M. K_b = K_w/K_a = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁴ = 5.56 × 10⁻¹¹. [OH⁻] = √(5.56 × 10⁻¹¹ × 0.1000) = √(5.56 × 10⁻¹²) = 2.36 × 10⁻⁶ M. pOH = 5.63, so pH = 14.00 − 5.63 = 8.37. The equivalence pH is above 7 as expected, confirming the basic hydrolysis of the formate ion.

Question 4: APPLIED

A biochemistry student is preparing a phosphate buffer at pH 7.20 using NaH₂PO₄ (K_a2 = 6.2 × 10⁻⁸ for H₂PO₄⁻) and NaOH. If she starts with 100.0 mL of 0.150 M NaH₂PO₄ and adds 0.150 M NaOH, what volume of NaOH should she add to achieve pH 7.20? (Hint: treat this as a point in the buffer region of an H₂PO₄⁻/HPO₄²⁻ system.)

Show answer

First, pK_a2 = −log(6.2 × 10⁻⁸) = 7.21. Using Henderson–Hasselbalch: 7.20 = 7.21 + log([HPO₄²⁻]/[H₂PO₄⁻]). So log([HPO₄²⁻]/[H₂PO₄⁻]) = −0.01, giving [HPO₄²⁻]/[H₂PO₄⁻] = 10⁻⁰·⁰¹ = 0.977. Let x = moles OH⁻ added. Initial moles H₂PO₄⁻ = 0.1000 L × 0.150 M = 0.01500 mol. After reaction: moles HPO₄²⁻ = x and moles H₂PO₄⁻ = 0.01500 − x. So x/(0.01500 − x) = 0.977. Solving: x = 0.977(0.01500 − x) = 0.01466 − 0.977x; 1.977x = 0.01466; x = 0.00741 mol. Volume NaOH = 0.00741 / 0.150 = 0.0494 L = 49.4 mL. She should add approximately 49.4 mL of the 0.150 M NaOH.

Question 5: CRITICAL THINKING

Two monoprotic weak acids, HA₁ (K_a = 1 × 10⁻³) and HA₂ (K_a = 1 × 10⁻⁹), are each titrated with NaOH under identical conditions (same concentration, same volume). Compare their titration curves qualitatively: (a) Which has the higher initial pH? (b) Which has the higher equivalence-point pH? (c) Which would have a larger steep-jump region around the equivalence point? (d) For which acid would phenolphthalein be a more reliable indicator, and why?

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(a) HA₂ has the higher initial pH because its K_a is smaller, meaning less dissociation and lower [H⁺]. (b) HA₂ also has the higher equivalence-point pH because its conjugate base A₂⁻ has a larger K_b (= K_w/10⁻⁹ = 10⁻⁵), producing more OH⁻ upon hydrolysis than A₁⁻ (K_b = 10⁻¹¹). (c) HA₁ has the larger steep-jump region. With K_a = 10⁻³, the buffer region is centered at pH 3, so the jump from pH ≈ 5 to pH ≈ 11 is dramatic. For HA₂ (pK_a = 9), the buffer region extends up to pH ≈ 10 and the equivalence pH may be near 11, leaving a much smaller vertical jump. (d) Phenolphthalein (color change pH 8.2–10) is more reliable for HA₁, because its equivalence-point pH (≈ 8) sits within the steep-jump range where a single drop causes a large pH change. For HA₂, the equivalence pH is higher (≈ 11) and the jump is compressed, so phenolphthalein may change color before the true equivalence point is reached, within the buffer zone. A higher-range indicator would be more appropriate for HA₂.

SUMMARY

Lesson Summary

Summary

A weak acid–strong base titration curve is divided into four regions, each governed by different chemistry. The initial region (before any base is added) is treated as a standard weak acid equilibrium problem using an ICE table and K_a. The buffer region (0 < V < V_eq) is quantified by the Henderson–Hasselbalch equation, with pH = pK_a at the half-equivalence point where [HA] = [A⁻]. The equivalence point occurs when moles OH⁻ = moles HA; the pH is above 7 because the conjugate base A⁻ undergoes hydrolysis to produce OH⁻. In the post-equivalence region, excess NaOH controls the pH directly.

Compared to a strong acid–strong base titration, the weak acid curve starts at a higher initial pH, exhibits a prominent buffer plateau, has a basic equivalence-point pH, and shows a narrower steep-jump region. Indicator selection must match the basic equivalence pH — phenolphthalein (pH 8.2–10) is typically the best choice. These principles extend naturally to polyprotic acid titrations, where multiple equivalence points and buffer regions appear, each corresponding to a successive deprotonation step.