pH During Titration — Compute pH during titration using equilibrium reasoning - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, chemists worked with acids and bases without a precise way to measure their strength in solution. The concept of titration — slowly adding one solution to another to determine concentration — dates back to the 18th century, when French chemist François-Antoine-Henri Descroizilles invented the first burette around 1791. However, early titrations relied on crude color-change indicators, and chemists had no mathematical framework to predict how acidic or basic a solution would be at any given point during the process.

1791 • First Burette Invented

Descroizilles devised the first volumetric glassware for measuring reagent volumes, enabling quantitative acid-base analysis.

1884 • Arrhenius Theory of Acids and Bases

Svante Arrhenius proposed that acids release H⁺ ions and bases release OH⁻ ions in water, providing the first ionic model of acid-base behavior.

1909 • Sørensen Introduces the pH Scale

Søren Sørensen defined pH as the negative logarithm of H⁺ concentration, giving chemists a numerical scale to express acidity.

1923 • Brønsted-Lowry Theory

Johannes Brønsted and Thomas Lowry independently expanded acid-base theory to include proton donors and acceptors, enabling equilibrium reasoning for weak acids and bases.

1930s • Potentiometric Titration & pH Meters

Electronic pH meters allowed continuous measurement of pH during titration, confirming the S-shaped titration curves predicted by equilibrium calculations.

With these breakthroughs, a powerful question could finally be answered: How can we calculate the exact pH of a solution at any point during a titration? Answering this question requires combining stoichiometry with equilibrium reasoning — exactly the skill this lesson builds.

SECTION 2

Core Principles & Definitions

Before diving into pH calculations, you need to understand the key ideas that govern what happens when an acid and a base are mixed during a titration. Each stage of the titration requires a slightly different approach, so recognizing which stage you are in is just as important as knowing the math.

Stoichiometric Neutralization

When acid and base are mixed, they react in a mole-to-mole ratio (for monoprotic species). The reaction goes essentially to completion for strong acid–strong base pairs. For weak acid–strong base pairs, the neutralization still proceeds fully, but the product is the conjugate base of the weak acid.

The Buffer Region

When a weak acid is partially neutralized, the solution contains both the weak acid and its conjugate base — a buffer. In this region, pH changes slowly and is best calculated using the Henderson-Hasselbalch equation.

The Equivalence Point

At the equivalence point, moles of acid equal moles of base. For a strong acid–strong base titration, pH = 7. For a weak acid–strong base titration, the pH is above 7 because the conjugate base hydrolyzes water.

Excess Region

After the equivalence point, the excess strong acid or strong base determines the pH. The equilibrium contribution from weak species becomes negligible, so a simple excess-concentration calculation is used.

Equilibrium vs. Stoichiometry

The general strategy is: first do the stoichiometry (what reacts?), then do the equilibrium (what remains and how does it ionize?). Keeping these two steps separate avoids the most common mistakes in titration problems.

✦

KEY TAKEAWAY

Think of a titration like slowly filling a bucket with water. At first the bucket is nearly empty (excess acid). As you pour, the water level rises steadily (buffer region). When the bucket is full (equivalence point), the behavior changes completely — any additional water overflows (excess base). Calculating pH during titration is about figuring out which "stage of the bucket" you're in and then applying the right equilibrium tool.

SECTION 3

Visualizing the Titration Curve

A titration curve is a graph of pH (y-axis) versus volume of titrant added (x-axis). The shape of the curve reveals the four key regions of a titration and tells you which calculation method to use at each point. Below is a titration curve for the titration of a weak acid (such as acetic acid) with a strong base (NaOH).

The curve shows three distinct regions. Region ① (red–amber) is the buffer region where a weak acid and its conjugate base coexist. At the half-equivalence point, exactly half the acid has been neutralized, so pH = pK_a. Region ② (violet) marks the equivalence point where the steep jump occurs. Region ③ (cyan) is dominated by excess strong base.

Notice how the curve is relatively flat in the buffer region and then rises steeply near the equivalence point. This steep rise is why a single drop of titrant can change the pH by several units at the equivalence point, making indicator choice critical. After the equivalence point, the curve flattens again as excess OH⁻ ions dominate the solution.

SECTION 4

Mathematical Framework

Calculating pH during a titration requires different equations depending on which region of the titration curve you are in. Below are the key equations and when to apply each one. In every case, the first step is always the same: use stoichiometry to find out how many moles of acid and base remain after the neutralization reaction.

GENERAL pH DEFINITION

pH = −log[H⁺]

[H⁺] is the molar concentration of hydrogen ions in solution. This equation applies everywhere, but the method for finding [H⁺] differs by region.

HENDERSON-HASSELBALCH (BUFFER REGION)

pH = pKₐ + log([A⁻] / [HA])

pK_a = −log(K_a); [A⁻] = moles of conjugate base / total volume; [HA] = moles of remaining weak acid / total volume. Because both species share the same total volume, you can also use the mole ratio directly: pH = pK_a + log(mol A⁻ / mol HA).

WEAK BASE HYDROLYSIS (EQUIVALENCE POINT)

Kᵦ = Kw / Kₐ ; [OH⁻] = √(Kᵦ × C)

At the equivalence point of a weak acid–strong base titration, only the conjugate base A⁻ remains. C is the concentration of A⁻ (moles of conjugate base / total volume). Use the K_b expression to find [OH⁻], then convert: pOH = −log[OH⁻], pH = 14 − pOH.

EXCESS STRONG BASE (PAST EQUIVALENCE)

[OH⁻] = (mol excess NaOH) / (total volume) ; pH = 14 − pOH

After the equivalence point, the excess strong base completely dissociates. The contribution of the conjugate base hydrolysis is negligible compared to the excess OH⁻, so the pH depends only on the moles of excess NaOH divided by total volume.

⚠

STOICHIOMETRY FIRST!

Always start by writing the neutralization reaction and calculating how many moles of acid and base react. Only then decide which equilibrium equation applies. A common mistake is to jump straight to Henderson-Hasselbalch without first accounting for the neutralization reaction.

SECTION 5

Detailed Breakdown of Titration Regions

A titration can be broken into four distinct calculation regions. The diagram below summarizes which mathematical approach to use in each region and what species dominate the solution. Being able to classify the region is the single most important skill in titration pH problems.

The four cards show each region's dominant species and the calculation method. The green flowchart at the bottom summarizes the three-step decision process you should follow for every titration pH problem.

Region	Volume Added	Dominant Species	Calculation Method	pH for Weak Acid–Strong Base
Initial	0 mL	HA	ICE table with K_a	< 7 (acidic)
Buffer	0 < V < V_eq	HA + A⁻	Henderson-Hasselbalch	Near pK_a
Half-equiv.	V = ½V_eq	HA = A⁻ (equal)	H-H with log(1) = 0	pH = pK_a exactly
Equivalence	V = V_eq	A⁻ only	K_b hydrolysis	> 7 (basic)
Excess base	V > V_eq	OH⁻ (excess)	Direct [OH⁻] → pOH → pH	>> 7 (strongly basic)

Summary of titration regions for a weak acid–strong base titration

SECTION 6

Worked Example: Weak Acid–Strong Base Titration

Let's work through a complete problem: 25.0 mL of 0.100 M acetic acid (CH₃COOH, K_a = 1.8 × 10⁻⁵) is titrated with 0.100 M NaOH. Find the pH after adding (a) 0 mL, (b) 10.0 mL, (c) 12.5 mL, (d) 25.0 mL, and (e) 30.0 mL of NaOH.

Part (a): Initial pH — 0 mL NaOH added (Region 0)

1
Step 1 — Identify Region

No base has been added, so we have a pure weak acid solution. This is Region 0. We use an ICE table with K_a.
2
Step 2 — Set Up Equilibrium

CH₃COOH ⇌ CH₃COO⁻ + H⁺. Initial: [HA] = 0.100 M, [A⁻] = 0, [H⁺] = 0. Change: −x, +x, +x. Equilibrium: 0.100 − x, x, x.
3
Step 3 — Solve for x

K_a = x² / (0.100 − x) ≈ x² / 0.100 = 1.8 × 10⁻⁵. So x² = 1.8 × 10⁻⁶, and x = 1.34 × 10⁻³ M.
4
Step 4 — Calculate pH

pH = −log(1.34 × 10⁻³) = 2.87.

pH = 2.87

Part (b): Buffer Region — 10.0 mL NaOH added (Region 1)

1
Step 1 — Stoichiometry

mol HA (initial) = 0.0250 L × 0.100 M = 0.00250 mol. mol OH⁻ added = 0.0100 L × 0.100 M = 0.00100 mol. The neutralization reaction: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. After reaction: mol HA = 0.00250 − 0.00100 = 0.00150 mol. mol A⁻ = 0.00100 mol.
2
Step 2 — Identify Region

Both HA and A⁻ are present → buffer region → Henderson-Hasselbalch.
3
Step 3 — Apply Henderson-Hasselbalch

pH = pK_a + log(mol A⁻ / mol HA) = −log(1.8 × 10⁻⁵) + log(0.00100 / 0.00150) = 4.74 + log(0.667) = 4.74 + (−0.18) = 4.56.

pH = 4.56

Part (c): Half-Equivalence Point — 12.5 mL NaOH added

1
Step 1 — Stoichiometry

mol OH⁻ = 0.0125 L × 0.100 M = 0.00125 mol. After reaction: mol HA = 0.00250 − 0.00125 = 0.00125 mol. mol A⁻ = 0.00125 mol. Notice mol HA = mol A⁻.
2
Step 2 — Apply Henderson-Hasselbalch

pH = pK_a + log(0.00125 / 0.00125) = 4.74 + log(1) = 4.74 + 0 = 4.74.

pH = pKₐ = 4.74

Part (d): Equivalence Point — 25.0 mL NaOH added (Region 2)

1
Step 1 — Stoichiometry

mol OH⁻ = 0.0250 L × 0.100 M = 0.00250 mol. mol HA = 0.00250 mol. They are exactly equal, so all HA has been converted to A⁻. mol A⁻ = 0.00250 mol.
2
Step 2 — Find Concentration of A⁻

Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.0500 L. [A⁻] = 0.00250 mol / 0.0500 L = 0.0500 M.
3
Step 3 — Calculate Kᵦ

K_b = K_w / K_a = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰.
4
Step 4 — Find [OH⁻]

[OH⁻] = √(K_b × [A⁻]) = √(5.56 × 10⁻¹⁰ × 0.0500) = √(2.78 × 10⁻¹¹) = 5.27 × 10⁻⁶ M.
5
Step 5 — Convert to pH

pOH = −log(5.27 × 10⁻⁶) = 5.28. pH = 14.00 − 5.28 = 8.72.

pH = 8.72 (above 7, as expected for weak acid–strong base)

Part (e): Excess Base — 30.0 mL NaOH added (Region 3)

1
Step 1 — Stoichiometry

mol OH⁻ = 0.0300 L × 0.100 M = 0.00300 mol. mol HA = 0.00250 mol. After reaction: excess OH⁻ = 0.00300 − 0.00250 = 0.000500 mol.
2
Step 2 — Find [OH⁻]

Total volume = 25.0 + 30.0 = 55.0 mL = 0.0550 L. [OH⁻] = 0.000500 / 0.0550 = 9.09 × 10⁻³ M.
3
Step 3 — Calculate pH

pOH = −log(9.09 × 10⁻³) = 2.04. pH = 14.00 − 2.04 = 11.96.

pH = 11.96

SECTION 7

Comparing Strong vs. Weak Acid Titrations

Not all titrations look the same. The shape of the curve and the pH at the equivalence point depend heavily on whether the acid being titrated is strong or weak. Understanding these differences helps you predict results and choose appropriate indicators.

Feature	Strong Acid–Strong Base	Weak Acid–Strong Base
Initial pH	Very low (depends only on [H⁺] from strong acid)	Moderate (K_a ICE table needed)
Buffer region	Does NOT exist — no weak acid/conjugate base pair	Exists — pH changes slowly; Henderson-Hasselbalch applies
Equivalence point pH	Exactly 7.00 (neutral salt formed)	Above 7 (conjugate base hydrolyzes)
Steep jump	Very large and symmetric (pH ≈ 3 to pH ≈ 11)	Smaller and shifted upward
Best indicator	Any indicator changing near pH 7 (e.g., bromothymol blue)	Indicator changing near pH 8–10 (e.g., phenolphthalein)
Post-equivalence	Identical — excess OH⁻ dominates	Identical — excess OH⁻ dominates

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

✦

KEY TAKEAWAY

Think of a strong acid titration like a light switch — pH jumps sharply from acidic to basic in one quick flip. A weak acid titration is more like a dimmer switch — there is a gradual transition through the buffer region, and the "sweet spot" (equivalence point) is shifted toward the basic side. Choosing the wrong indicator is like trying to read the dimmer position in the dark — you need one that changes color at the right pH.

SECTION 8

Connection to Polyprotic & Advanced Titrations

The principles you have learned for monoprotic acid titrations extend directly to more complex systems you may encounter in AP Chemistry or college courses. Polyprotic acids like H₂SO₃ or H₃PO₄ have multiple protons to donate, which means their titration curves show multiple equivalence points — one for each proton. Between each equivalence point lies a buffer region with its own pK_a value. The same logic applies at each stage: do stoichiometry first, then apply the appropriate equilibrium calculation.

Concept in This Lesson	Advanced Extension
Henderson-Hasselbalch for monoprotic buffer	Applied to each K_a of polyprotic acid (K_a1, K_a2, K_a3)
Single equivalence point	Multiple equivalence points (one per proton)
K_b hydrolysis at equivalence	Amphoteric species at intermediate equivalence points (both K_a and K_b matter)
Indicator selection for one jump	Different indicator for each equivalence point
K_w = 1.0 × 10⁻¹⁴ at 25 °C	K_w varies with temperature, shifting equivalence point pH

How this lesson's concepts extend to advanced titration topics

Even though these advanced scenarios have more moving parts, the fundamental three-step strategy remains the same: calculate moles, determine what reacts and what remains, and then apply the appropriate equilibrium expression. Mastering the monoprotic case in this lesson gives you all the tools you need to tackle any titration problem.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

During a weak acid–strong base titration, the pH at the equivalence point is above 7. Explain why this happens in terms of the species present in solution at that moment.

Show answer

At the equivalence point, all the weak acid (HA) has been converted to its conjugate base (A⁻). This conjugate base is itself a weak base — it can accept a proton from water in a hydrolysis reaction: A⁻ + H₂O ⇌ HA + OH⁻. This reaction produces hydroxide ions, making the solution basic and pushing the pH above 7. In contrast, a strong acid–strong base titration produces a neutral salt that does not hydrolyze, so its equivalence point pH is exactly 7.

Question 2: BASIC CALCULATION

A buffer is prepared during a titration and contains 0.020 mol of formic acid (HCOOH, K_a = 1.8 × 10⁻⁴) and 0.015 mol of sodium formate (HCOONa). Calculate the pH of this buffer solution.

Show answer

Using Henderson-Hasselbalch: pH = pK_a + log(mol A⁻ / mol HA). First, pK_a = −log(1.8 × 10⁻⁴) = 3.74. Then pH = 3.74 + log(0.015 / 0.020) = 3.74 + log(0.75) = 3.74 + (−0.12) = 3.62. The pH is slightly below pK_a because there is more acid than conjugate base, which is consistent with Le Chatelier's principle.

Question 3: INTERMEDIATE

50.0 mL of 0.200 M HF (K_a = 6.8 × 10⁻⁴) is titrated with 0.200 M NaOH. Calculate the pH after 50.0 mL of NaOH has been added (the equivalence point).

Show answer

At the equivalence point, mol HF = mol NaOH = 0.0500 L × 0.200 M = 0.0100 mol. All HF is converted to F⁻. Total volume = 100.0 mL = 0.1000 L. [F⁻] = 0.0100 / 0.1000 = 0.100 M. K_b for F⁻ = K_w / K_a = (1.0 × 10⁻¹⁴) / (6.8 × 10⁻⁴) = 1.47 × 10⁻¹¹. [OH⁻] = √(1.47 × 10⁻¹¹ × 0.100) = √(1.47 × 10⁻¹²) = 1.21 × 10⁻⁶ M. pOH = −log(1.21 × 10⁻⁶) = 5.92. pH = 14.00 − 5.92 = 8.08. The pH is above 7 because F⁻ is a weak base that hydrolyzes water to produce OH⁻.

Question 4: APPLIED

A chemist titrates 30.0 mL of 0.150 M acetic acid (K_a = 1.8 × 10⁻⁵) with 0.150 M NaOH. After adding 35.0 mL of NaOH, the chemist notices the phenolphthalein indicator has turned pink. Calculate the pH and explain whether the titration has passed the equivalence point.

Show answer

mol CH₃COOH = 0.0300 L × 0.150 M = 0.00450 mol. mol NaOH = 0.0350 L × 0.150 M = 0.00525 mol. Since mol NaOH > mol CH₃COOH, this is past the equivalence point (Region 3). Excess OH⁻ = 0.00525 − 0.00450 = 0.000750 mol. Total volume = 65.0 mL = 0.0650 L. [OH⁻] = 0.000750 / 0.0650 = 0.01154 M. pOH = −log(0.01154) = 1.94. pH = 14.00 − 1.94 = 12.06. The titration has indeed passed the equivalence point. Phenolphthalein turns pink around pH 8.2–10, so it turned pink when the pH first exceeded about 8.2 — which occurred near the equivalence point (pH ≈ 8.7) or just past it.

Question 5: CRITICAL THINKING

Two titrations are performed: Titration A uses 25.0 mL of 0.100 M HCl with 0.100 M NaOH. Titration B uses 25.0 mL of 0.100 M CH₃COOH (K_a = 1.8 × 10⁻⁵) with 0.100 M NaOH. Both reach their equivalence point after exactly 25.0 mL of NaOH is added. Compare and explain: (i) the initial pH of each, (ii) the equivalence point pH of each, and (iii) which titration has a larger steep-jump region near the equivalence point.

Show answer

(i) Initial pH: Titration A starts at pH = −log(0.100) = 1.00 because HCl is a strong acid that fully dissociates. Titration B starts at approximately pH 2.87 because acetic acid is weak and only partially ionizes (calculated via ICE table: [H⁺] = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³). (ii) Equivalence point pH: Titration A has pH = 7.00 because NaCl is a neutral salt. Titration B has pH ≈ 8.72 because the acetate ion (CH₃COO⁻) is a weak base that hydrolyzes water to produce OH⁻. (iii) Titration A has a larger steep-jump region because there is no buffer region to moderate pH changes. In Titration B, the presence of the conjugate base/weak acid buffer resists pH changes throughout the middle of the titration, so the steep jump is narrower and shifted toward higher pH. This analysis demonstrates why the choice of indicator matters: phenolphthalein works for Titration B but not ideally for A, while bromothymol blue works for A but would change color too early for B.

SUMMARY

Lesson Summary

Summary

Computing pH during a titration requires a two-step strategy: first perform stoichiometry to determine which species remain after the neutralization reaction, then apply the correct equilibrium equation based on which region of the titration curve you are in. Before any base is added, treat the solution as a pure weak acid using an ICE table. In the buffer region (where both HA and A⁻ coexist), apply the Henderson-Hasselbalch equation: pH = pK_a + log([A⁻]/[HA]). At the half-equivalence point, pH equals pK_a exactly.

At the equivalence point of a weak acid–strong base titration, only the conjugate base remains, and its hydrolysis makes the solution basic (pH > 7). Past the equivalence point, excess strong base dominates and pH is found from the excess [OH⁻] concentration. The key to success is always identifying the correct region first — then the right equation follows naturally.