Acid-Base Titrations and Indicators

🧪 Titration Fundamentals

Part 1 of 7 — Setup, Terminology, and Calculations

Titration Essentials

$n_{acid} \times \text{(acid ratio)} = n_{base} \times \text{(base ratio)}$

🔑 Why this matters: Titrations appear on nearly every AP Chemistry exam — both in multiple choice and as multi-part free-response questions.

What You'll Master in Part 1

• Understanding titration setup, terminology, and the equivalence point concept
• Using the moles relationship to find unknown concentrations
• Calculating pH before, at, and after the equivalence point for strong-strong titrations

🧪 Titration Setup

Key Components

The Key Equation

At the equivalence point:

$\boxed{n_{acid} = n_{base}}$

$\boxed{M_{acid} \times V_{acid} = M_{base} \times V_{base}}$

(for 1:1 stoichiometry)

🔑 Key Equation: This relationship lets you find the unknown concentration from the known titrant and measured volumes.

🧪 Strong Acid – Strong Base Titration

The Reaction

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

The net ionic equation:

$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

Before Equivalence Point

Excess $H^+$ remains → acidic

$[H^+] = \frac{\text{mol } H^+ - \text{mol } OH^-}{\text{total volume}}$

At Equivalence Point

All acid and base have reacted. Only $NaCl$ and $H_2O$ remain.

$pH = 7.00$

🔑 Key Fact: Strong acid + strong base always gives pH = 7 at equivalence — neither ion hydrolyzes.

After Equivalence Point

Excess $OH^-$ remains → basic

$[OH^-] = \frac{\text{mol } OH^- - \text{mol } H^+}{\text{total volume}}$

Titration Fundamentals Check 🎯

🧪 Worked Example: Finding Unknown Concentration

Problem: A 25.0 mL sample of $HCl$ of unknown concentration requires 18.5 mL of 0.150 M $NaOH$ to reach the equivalence point. What is $[HCl]$?

Solution:

$n_{NaOH} = 0.150 \text{ M} \times 0.0185 \text{ L} = 2.775 \times 10^{-3} \text{ mol}$

At equivalence: $n_{HCl} = n_{NaOH} = 2.775 \times 10^{-3}$ mol

$[HCl] = \frac{2.775 \times 10^{-3}}{0.0250} = 0.111 \text{ M}$

Titration Calculations 🧮

1) 30.0 mL of 0.200 M $NaOH$ is titrated with 0.100 M $HCl$. What volume (mL) of $HCl$ is needed to reach the equivalence point?

2) After adding 20.0 mL of 0.100 M $NaOH$ to 40.0 mL of 0.100 M $HCl$, what is the pH? (2 decimal places)

3) After the equivalence point, 5.0 mL of excess 0.100 M $NaOH$ has been added to a total volume of 80.0 mL. What is the pH? (2 decimal places)

Titration Setup Reasoning 🔍

Exit Quiz — Titration Fundamentals ✅

📈 Strong Acid–Strong Base Titration Curves

Part 2 of 7 — Analyzing the S-Shaped Curve

The Four Regions of a Strong-Strong Titration

🔑 Why this matters: Understanding each region of the curve is essential — the AP exam asks you to calculate pH at specific volumes and interpret the curve shape.

What You'll Master in Part 2

• Sketching the S-shaped titration curve for strong acid–strong base
• Calculating pH at key points before, at, and after equivalence
• Explaining why the equivalence point pH = 7.00 for strong-strong titrations

📌 Regions of the Curve

Consider titrating 50.0 mL of 0.100 M $HCl$ with 0.100 M $NaOH$:

Region 1: Before Equivalence (0 to ~45 mL)

• Excess $HCl$ present
• $[H^+] = \frac{\text{remaining mol } H^+}{\text{total volume}}$
• pH increases slowly
• Example: After 10.0 mL $NaOH$:
  • Mol $H^+ = 0.0050 - 0.0010 = 0.0040$
  • $[H^+] = 0.0040/0.060 = 0.0667$ M
  • $pH = 1.18$

Region 2: Near Equivalence (~45 to ~55 mL)

• Very little excess acid or base
• pH changes dramatically with each drop
• The steep vertical portion of the curve

Region 3: At Equivalence (50.0 mL)

• $pH = 7.00$ exactly
• All $H^+$ and $OH^-$ have reacted
• Only $NaCl$ + $H_2O$ remain

🔑 Key Fact: Strong acid + strong base → equivalence at pH 7.00 every time.

Region 4: After Equivalence (>50 mL)

• Excess $NaOH$ present
• $[OH^-] = \frac{\text{excess mol } OH^-}{\text{total volume}}$
• pH levels off at high values

🔢 pH Calculations at Key Points

Titrating 50.0 mL of 0.100 M $HCl$ with 0.100 M $NaOH$

Notice: pH jumps from ~4 to ~10 in just 0.2 mL! That's the dramatic equivalence point region.

💡 Tip: On the AP exam, look for the steepest part of the curve — that marks the equivalence point.

Titration Curve Analysis 🎯

Strong Acid–Strong Base Calculations 🧮

Titrating 25.0 mL of 0.200 M $HCl$ with 0.200 M $NaOH$:

1) What is the pH at the start (before adding any $NaOH$)? (2 decimal places)

2) What volume of $NaOH$ is needed to reach the equivalence point? (1 decimal place, in mL)

3) What is the pH after adding 30.0 mL of $NaOH$? (2 decimal places)

📌 Key Features of the Strong–Strong Curve

Shape Analysis

1. Initial pH is low (strong acid) — typically pH 1-2
2. Gradual rise as acid is slowly consumed
3. Steep jump near equivalence (pH ~3 to ~11)
4. Equivalence at pH 7 (always for strong-strong)
5. Gradual leveling after equivalence

Why pH 7 at Equivalence?

The products are water and a salt of a strong acid/strong base (e.g., $NaCl$, $KNO_3$). These salts are neutral — their ions do not react with water (no hydrolysis).

⚠️ Common Mistake: pH 7 at equivalence ONLY applies to strong acid + strong base. Weak acid or weak base titrations have equivalence pH ≠ 7.

Effect of Concentration

Higher concentrations → steeper jump at equivalence, but equivalence point is still at pH 7.

Titration Curve Reasoning 🔍

Exit Quiz — Strong-Strong Curves ✅

📈 Weak Acid–Strong Base Titration Curves

Part 3 of 7 — The Most Important Titration for AP Chemistry

Weak Acid–Strong Base: What Changes

🔑 Why this matters: Weak acid–strong base titrations are the single most tested titration type on the AP exam — understanding the buffer region and half-equivalence point is critical.

What You'll Master in Part 3

• Identifying the four regions of a weak acid–strong base titration curve
• Explaining why the buffer region exists and using Henderson-Hasselbalch there
• Calculating pH at the half-equivalence point using pH = p$K_a$

🧪 Four Regions of the Weak Acid–Strong Base Curve

Consider titrating 50.0 mL of 0.100 M $CH_3COOH$ ($K_a = 1.8 \times 10^{-5}$) with 0.100 M $NaOH$:

Region 1: Initial Point (0 mL added)

Only weak acid present. Use ICE table:

$K_a = \frac{x^2}{0.100 - x} \approx \frac{x^2}{0.100}$

$x = \sqrt{1.8 \times 10^{-5} \times 0.100} = 1.34 \times 10^{-3}$

$pH = -\log(1.34 \times 10^{-3}) = 2.87$

Higher starting pH than strong acid (1.00 vs 2.87)!

Region 2: Buffer Region (0 to 50 mL)

Both $CH_3COOH$ and $CH_3COO^-$ present — this IS a buffer!

Use Henderson-Hasselbalch: $pH = pK_a + \log([A^-]/[HA])$

Region 3: Equivalence Point (50.0 mL)

All $HA$ converted to $A^-$. The conjugate base hydrolyzes:

$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$

$pH > 7$ (basic, NOT neutral!)

⚠️ Critical: The equivalence point of a weak acid + strong base titration is ALWAYS above pH 7 because the conjugate base hydrolyzes.

Region 4: After Equivalence (>50 mL)

Excess $NaOH$ dominates. Calculate $[OH^-]$ from excess.

📌 The Half-Equivalence Point

At exactly half the volume needed for equivalence (25.0 mL in our example):

$\text{Half the acid is neutralized: } [HA] = [A^-]$

$pH = pK_a + \log\frac{[A^-]}{[HA]} = pK_a + \log(1) = pK_a$

$\boxed{\text{At the half-equivalence point: } pH = pK_a}$

This is how you can determine $K_a$ experimentally — read the pH at the half-equivalence point!

🔑 AP Must-Know: Read pH at the half-equivalence point from the titration curve. That pH equals $pK_a$, so $K_a = 10^{-pK_a}$.

For acetic acid: $pH = pK_a = 4.74$ at the half-equivalence point.

Why This Matters on the AP Exam

• Given a titration curve, find the half-equivalence volume (half of equivalence volume)
• Read the pH at that point — that's $pK_a$
• $K_a = 10^{-pK_a}$

Weak Acid Titration Concepts 🎯

🔢 Calculating pH at the Equivalence Point

At the equivalence point, only the conjugate base $A^-$ is present. It hydrolyzes:

$A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq)$

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

ICE Table

Concentration of $A^- = \frac{0.005}{0.100} = 0.050$ M (total volume = 100 mL)

$K_b = \frac{x^2}{0.050}$

$x = \sqrt{5.6 \times 10^{-10} \times 0.050} = 5.3 \times 10^{-6}$

$pOH = -\log(5.3 \times 10^{-6}) = 5.28$

$pH = 14 - 5.28 = 8.72$

The equivalence point is at pH 8.72 — clearly basic, not neutral!

💡 Tip: Whenever the equivalence pH is above 7, you know the original acid was weak (its conjugate base makes the solution basic).

Weak Acid Titration Calculations 🧮

Titrating 40.0 mL of 0.150 M $HCOOH$ ($pK_a = 3.75$) with 0.150 M $NaOH$:

1) What volume of $NaOH$ is needed to reach the equivalence point? (1 decimal place, mL)

2) What volume of $NaOH$ gives the half-equivalence point? (1 decimal place, mL)

3) What is the pH at the half-equivalence point? (2 decimal places)

Curve Feature Identification 🔍

Exit Quiz — Weak Acid Curves ✅

🎯 Special Points on the Titration Curve

Part 4 of 7 — Half-Equivalence, Equivalence, and Beyond

Critical Points Summary

🔑 Why this matters: The AP exam frequently asks you to identify these points on a graph and calculate pH at each — this is high-yield content.

What You'll Master in Part 4

• Finding pH at the half-equivalence, equivalence, and post-equivalence points
• Understanding weak base–strong acid titrations (inverted curves)
• Handling polyprotic acid titrations with multiple equivalence points

📋 Critical Points Summary

For titrating a weak acid $HA$ with strong base $NaOH$:

The Rule for Equivalence Point pH

🔑 Key Pattern: Equivalence pH tells you the titration type: pH 7 = strong+strong, pH > 7 = weak acid+strong base, pH < 7 = strong acid+weak base.

🧪 Weak Base–Strong Acid Titration

When $NH_3$ is titrated with $HCl$:

$NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$

The Curve Is Inverted!

• Initial pH: High (basic — weak base)
• Buffer region: Both $NH_3$ and $NH_4^+$ present (pH decreases)
• Half-equivalence: $pH = pK_a(NH_4^+) = 14 - pK_b = 14 - 4.74 = 9.25$
• Equivalence point: Only $NH_4^+$ (a weak acid) → $pH < 7$
• After equivalence: Excess $HCl$ → strongly acidic

Key Difference

The curve goes from high pH to low pH — a mirror image of the weak acid curve!

💡 Tip: For weak base titrations, $pH = pK_a$ of the conjugate acid at the half-equivalence point. Use $pK_a = 14 - pK_b$ to convert.

Critical Point Identification 🎯

🧪 Polyprotic Acid Titrations

Polyprotic acids (like $H_2SO_3$, $H_3PO_4$) show multiple equivalence points:

Diprotic Acid ($H_2A$) with $NaOH$

First equivalence point: $H_2A + NaOH \rightarrow NaHA + H_2O$

Second equivalence point: $NaHA + NaOH \rightarrow Na_2A + H_2O$

The curve shows two S-shaped jumps!

Key Features

• Volume to second equivalence = 2× volume to first equivalence
• First half-equivalence: $pH = pK_{a1}$
• Midpoint between equivalences: $pH = pK_{a2}$
• Each steep region corresponds to one deprotonation

⚠️ Watch Out: For polyprotic acids, the volume to the second equivalence is always 2× the first. Each proton requires an equal amount of base.

Example: $H_3PO_4$

Three equivalence points (three protons):

• $pK_{a1} = 2.15$, $pK_{a2} = 7.20$, $pK_{a3} = 12.35$

Critical Point Calculations 🧮

50.0 mL of 0.100 M $NH_3$ ($K_b = 1.8 \times 10^{-5}$, $pK_b = 4.74$) is titrated with 0.100 M $HCl$:

1) What volume of $HCl$ is needed for the equivalence point? (1 decimal place, mL)

2) What is the pH at the half-equivalence point? (2 decimal places)

3) At the equivalence point, the $pH$ is less than 7. What is the $pK_a$ of $NH_4^+$? (2 decimal places)

Titration Point Analysis 🔍

Exit Quiz — Special Points ✅

🎨 Acid-Base Indicators

Part 5 of 7 — Choosing the Right Indicator

Indicator Selection at a Glance

🔑 Why this matters: Choosing the wrong indicator gives a false endpoint — the AP exam tests whether you can match an indicator's range to the equivalence point pH.

What You'll Master in Part 5

• Understanding how indicators work as weak acids that change color
• Matching indicator pH range to the equivalence point pH
• Comparing indicators to pH meters for accuracy

🔧 How Indicators Work

An indicator ($HIn$) is itself a weak acid with different colored forms:

$HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$

$\text{Color A} \quad\quad\quad\quad\quad \text{Color B}$

Color Change Rules

• Acidic solution ($[H^+]$ high): Equilibrium shifts left → $HIn$ form dominates → Color A
• Basic solution ($[H^+]$ low): Equilibrium shifts right → $In^-$ form dominates → Color B
• Transition range: Both forms present → intermediate color

The Transition Range

The indicator changes color when:

$\frac{[In^-]}{[HIn]} = \frac{1}{10} \text{ to } \frac{10}{1}$

Using Henderson-Hasselbalch for the indicator:

$\boxed{pH = pK_{In} \pm 1}$

The indicator changes color over approximately 2 pH units centered on its $pK_{In}$.

🔑 Key Rule: An indicator is useful when its $pK_{In}$ is close to the equivalence point pH.

📌 Common Indicators

Choosing the Right Indicator

Match the indicator range to the equivalence point pH!

⚠️ Common AP Error: Students pick phenolphthalein for every titration. It only works when equivalence pH is 8–10 (weak acid + strong base).

Indicator Concepts 🎯

Indicator Selection Practice 🔍

📌 pH Meters vs. Indicators

Advantages of pH Meters

• Continuous pH readings throughout the titration
• More precise than indicators
• Can identify the exact equivalence point
• Can generate a complete titration curve
• No color interpretation needed

When Indicators Are Still Used

• Quick, inexpensive field tests
• Visual demonstration in teaching
• When a pH meter is not available
• For routine quality control with known endpoints

Finding Equivalence with a pH Meter

Plot pH vs. volume. The equivalence point is at the inflection point — where the curve is steepest (maximum $\Delta pH/\Delta V$).

Alternatively, plot the first derivative ($\Delta pH/\Delta V$ vs. $V$). The equivalence point is at the peak of this graph.

💡 Tip: On the AP exam, a pH meter graph with a clear inflection point is a strong clue to identify the equivalence volume.

Indicator Calculations 🧮

1) An indicator has $K_{In} = 1.0 \times 10^{-7}$. What is its $pK_{In}$?

2) What is the lower limit of its transition range? (3 significant figures)

3) What is the upper limit of its transition range? (3 significant figures)

Exit Quiz — Indicators ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Acid-Base Titrations

Workshop Problem Types

🔑 Why this matters: These multi-part problems mirror the exact format of AP Chemistry FRQ #3 (the lab/quantitative question).

What You'll Master in Part 6

• Working through complete titration curve calculations step-by-step
• Identifying unknown acids from titration data
• Integrating indicator selection with curve analysis

🔢 Problem 1: Complete Titration Curve Calculations

Problem: 50.0 mL of 0.200 M $CH_3COOH$ ($K_a = 1.8 \times 10^{-5}$, $pK_a = 4.74$) is titrated with 0.200 M $NaOH$. Find the pH at the initial, half-equivalence, and equivalence points.

Solution:

(a) Initial pH:

$K_a = \frac{x^2}{0.200} = 1.8 \times 10^{-5}$ $x = \sqrt{3.6 \times 10^{-6}} = 1.90 \times 10^{-3}$ $pH = -\log(1.90 \times 10^{-3}) = 2.72$

(b) After 25.0 mL $NaOH$ (half-equivalence):

$pH = pK_a = 4.74$

(c) Equivalence Point (50.0 mL $NaOH$):

All $HA \rightarrow A^-$. $[A^-] = 0.0100/0.100 = 0.100$ M

$K_b = K_w/K_a = 5.56 \times 10^{-10}$ $x = \sqrt{5.56 \times 10^{-10} \times 0.100} = 7.45 \times 10^{-6}$ $pOH = 5.13, \quad pH = 8.87$

Your Turn: Continuing the Titration 🧮

Same titration: 50.0 mL of 0.200 M $CH_3COOH$ with 0.200 M $NaOH$ ($pK_a = 4.74$)

1) After adding 10.0 mL $NaOH$, what is the pH? (Use H-H. 2 decimal places)

2) After adding 40.0 mL $NaOH$, what is the pH? (2 decimal places)

3) After adding 60.0 mL $NaOH$ (past equivalence), what is the pH? (2 decimal places)

🧪 Problem 2: Unknown Acid Identification

A monoprotic weak acid $HA$ (25.0 mL, 0.100 M) is titrated with 0.100 M $NaOH$. The following data is collected:

Analysis:

• Equivalence point is at 25.0 mL (equal $M$ and $V$)
• Half-equivalence is at 12.5 mL → $pH = pK_a = 3.75$
• $K_a = 10^{-3.75} = 1.8 \times 10^{-4}$
• Equivalence pH = 8.26 (> 7, confirms weak acid)

The acid is likely formic acid ($HCOOH$, $K_a = 1.8 \times 10^{-4}$).

🔑 AP Strategy: To identify an unknown acid from a titration curve: (1) find the half-equivalence volume, (2) read the pH there to get $pK_a$, (3) match $K_a = 10^{-pK_a}$ to a known acid.

Unknown Acid Analysis 🎯

Problem 3: Indicator Selection 🧮

A weak acid $HA$ ($pK_a = 6.50$) is titrated with $NaOH$.

1) At the half-equivalence point, pH = ? (2 decimal places)

2) The equivalence point will be at pH approximately (choose: <7, =7, or >7). Enter: less, equal, or greater.

3) Which indicator should be used? Enter the color-change pH range lower bound for an indicator with $pK_{In}$ matching the equivalence pH of ~10. (1 decimal place)

Workshop Synthesis 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Acid-Base Titrations

Everything at a Glance

🔑 Why this matters: AP exam questions can test any titration type — this review prepares you for the full range of possible questions.

What You'll Master in Part 7

• Distinguishing all titration types from curve shape alone
• Solving AP-style multi-step titration problems under timed conditions
• Connecting titrations to buffer chemistry and equilibrium concepts

📋 Complete Titration Summary

Method at Each Point

Equivalence Point pH Summary

Indicator Selection Rule

$\boxed{\text{Choose an indicator whose } pK_{In} \text{ is close to the equivalence point pH.}}$

🔑 Summary: Initial → ICE with $K_a$ | Buffer → Henderson-Hasselbalch | Equivalence → ICE with $K_b$ | Post-equivalence → excess $OH^-$.

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

30.0 mL of 0.150 M weak acid $HA$ ($pK_a = 5.00$) is titrated with 0.150 M $NaOH$.

1) Volume of $NaOH$ at equivalence (mL, 1 decimal place):

2) Volume of $NaOH$ at half-equivalence (mL, 1 decimal place):

3) pH at the half-equivalence point (2 decimal places):

AP-Style Questions — Set 2 🎯

Comprehensive Review 🔍

AP FRQ-Style Question 🎯

Final Exit Quiz — Titrations ✅