🎯⭐ INTERACTIVE LESSON

Acid-Base Titrations and Indicators

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Acid-Base Titrations and Indicators - Complete Interactive Lesson

Part 1: Titration Basics

🧪 Titration Fundamentals

Part 1 of 7 — Setup, Terminology, and Calculations

Acid-base titrations are quantitative experiments where a solution of known concentration (the titrant) is gradually added to a solution of unknown concentration (the analyte) until the reaction is complete. This technique is fundamental to analytical chemistry and AP Chemistry.

🧪 Titration Setup

Key Components

Component	Role
Titrant	Solution of known concentration in the buret
Analyte	Solution of unknown concentration in the flask
Buret	Delivers titrant precisely
Indicator	Changes color near equivalence point
Equivalence point	Stoichiometrically exact amount of titrant added
End point	Where indicator changes color (ideally ≈ equivalence point)

The Key Equation

At the equivalence point:

$n_{acid} = n_{base}$

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

(for 1:1 stoichiometry)

This allows you to calculate the unknown concentration!

🧪 Strong Acid – Strong Base Titration

The Reaction

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

Titration Fundamentals Check 🎯

🧪 Worked Example: Finding Unknown Concentration

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

$n_{N a O H} = 0.150 M \times 0.0185 L = 2.775 \times 10^{- 3}$

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 , what is the pH? (2 decimal places)

Titration Setup Reasoning 🔍

Exit Quiz — Titration Fundamentals ✅

Part 2: Strong Acid–Strong Base

📈 Strong Acid–Strong Base Titration Curves

Part 2 of 7 — Analyzing the S-Shaped Curve

The titration curve plots pH vs. volume of titrant added. For a strong acid–strong base titration, the curve has a characteristic S-shape with a sharp jump at the equivalence point. Understanding each region is essential for AP Chemistry.

📌 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

Part 3: Weak Acid–Strong Base

📈 Weak Acid–Strong Base Titration Curves

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

When a weak acid is titrated with a strong base, the curve is dramatically different from the strong-strong case. Understanding every region of this curve is essential — it combines equilibrium, buffers, and stoichiometry.

🧪 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 :

Part 4: Titration Curves

🎯 Special Points on the Titration Curve

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

AP Chemistry frequently asks you to identify and calculate pH at specific points on a titration curve. This lesson focuses on the critical points that earn you maximum credit on free-response questions.

📋 Critical Points Summary

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

Point	Volume of $NaOH$

Part 5: Indicators & Equivalence Point

🎨 Acid-Base Indicators

Part 5 of 7 — Choosing the Right Indicator

An indicator is a weak acid (or base) that changes color in a specific pH range. Choosing the right indicator is critical — its color change should occur as close to the equivalence point as possible.

🔧 How Indicators Work

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

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

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Acid-Base Titrations

This workshop takes you through complete titration calculations — the kind that appear as multi-part free-response questions on the AP Chemistry exam. Practice the full workflow: stoichiometry, equilibrium, buffer calculations, and curve analysis.

🔢 Problem 1: Complete Titration Curve Calculations

50.0 mL of 0.200 M $CH_3COOH$ ( $K_a = 1.8 \times 10^{-5}$ , ) is titrated with 0.200 M .

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Acid-Base Titrations

This comprehensive review integrates all titration concepts: setup, calculations at every point, curve analysis, indicator selection, and polyprotic systems. Master these for AP Chemistry success!

📋 Complete Titration Summary

Method at Each Point

Region	What's Present	Calculation Method
Initial (weak acid)	Only $HA$	ICE table with $K_a$

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.

$pH = 7.00$ (neutral — neither ion hydrolyzes)

After Equivalence Point

Excess $OH^-$ remains → basic

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

n_{N a O H} = 0.150 M \times 0.0185 L = 2.775 \times 1 0^{- 3} mol

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

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

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)

[H^{+}] = \frac{remaining mol H ^{+}}{total volume}

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

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 $NaOH$

Volume $NaOH$ (mL)	Calculation	pH
0.0	$[H^+] = 0.100$ M	1.00
25.0	$[H^+] = 0.00250/0.075 = 0.0333$	1.48
49.0	$[H^+] = 0.0001/0.099 = 0.00101$	3.00
49.9	$[H^+] = 0.00001/0.0999 = 1.0 \times 10^{-4}$	4.00
50.0	Equivalence point	7.00
50.1	$[OH^-] = 0.00001/0.1001 = 1.0 \times 10^{-4}$	10.00
51.0	$[OH^-] = 0.0001/0.101 = 9.9 \times 10^{-4}$	11.00
75.0	$[OH^-] = 0.00250/0.125 = 0.0200$	12.30

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

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

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

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 ✅

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!)

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!

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!

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 ✅

The Rule for Equivalence Point pH

Titration Type	Equivalence pH
Strong acid + Strong base	$= 7$
Weak acid + Strong base	$> 7$
Strong acid + Weak base	$< 7$
Weak acid + Weak base	Depends on relative $K_a$ and $K_b$

🧪 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:

Key Difference

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

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: $pH = pK_{a2}$

Example: $H_3PO_4$

Three equivalence points (three protons):

$pK_{a1} = 2.15$ , $pK_{a2} = 7.20$ ,

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 ✅

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

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:

$pH = pK_{In} \pm 1$

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

📌 Common Indicators

Indicator	$pK_{In}$	pH Range	Acid Color	Base Color
Thymol blue	1.7	1.2 – 2.8	Red	Yellow
Methyl orange	3.4	3.1 – 4.4	Red	Yellow
Methyl red	5.0	4.4 – 6.2	Red	Yellow
Bromothymol blue	7.1	6.0 – 7.6	Yellow	Blue
Phenolphthalein	9.1	8.2 – 10.0	Colorless	Pink
Alizarin yellow	11.0	10.1 – 12.0	Yellow	Red

Choosing the Right Indicator

Match the indicator range to the equivalence point pH!

Titration Type	Equivalence pH	Best Indicator
Strong acid + Strong base	7	Bromothymol blue
Weak acid + Strong base	8 – 10	Phenolphthalein
Strong acid + Weak base	3 – 5	Methyl orange or methyl red

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.

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 ✅

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

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

(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:

Volume $NaOH$ (mL)	pH
0.0	2.37
12.5	3.75
25.0	8.26
37.5	12.52

Analysis:

Equivalence point is at 25.0 mL (equal $M$ and $V$ )
Half-equivalence is at 12.5 mL → $pH = pK_a = 3.75$

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

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 ✅

Equivalence Point pH Summary

Titration	Equivalence pH	Why
Strong + Strong	= 7	Neutral salt, no hydrolysis
Weak acid + Strong base	> 7	$A^-$ hydrolyzes (basic)
Strong acid + Weak base	< 7	$BH^+$ hydrolyzes (acidic)

Indicator Selection Rule

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

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 ✅

1.8×10−51.0×10−14=

p H = p K_{a} (N H_{4}^{+}) = 14 - p K_{b} = 14 - 4.74 = 9.25

Equivalence point: Only

NH_4^+

(a weak acid) →

pH < 7

After equivalence: Excess

HCl

→ strongly acidic

Each steep region corresponds to one deprotonation

K_a = 10^{-3.75} = 1.8 \times 10^{-4}

Equivalence pH = 8.26 (> 7, confirms weak acid)

Initial	0 mL	Only $HA$	ICE table with $K_a$
Buffer region	$0 < V < V_{eq}$	$HA + A^-$	Henderson-Hasselbalch
Half-equivalence	$V_{eq}/2$	$[HA] = [A^-]$
Equivalence	$V_{eq}$	Only $A^-$	ICE with $K_b = K_w/K_a$
After equivalence	$V > V_{eq}$	$A^-$ + excess $OH^-$

Acid-Base Titrations and Indicators

Acid-Base Titrations and Indicators - Complete Interactive Lesson

Part 1: Titration Basics

🧪 Titration Fundamentals

🧪 Titration Setup

Key Components

The Key Equation

🧪 Strong Acid – Strong Base Titration

The Reaction

🧪 Worked Example: Finding Unknown Concentration

Solution

Part 2: Strong Acid–Strong Base

📈 Strong Acid–Strong Base Titration Curves

📌 Regions of the Curve

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

Part 3: Weak Acid–Strong Base

📈 Weak Acid–Strong Base Titration Curves

🧪 Four Regions of the Weak Acid–Strong Base Curve

Part 4: Titration Curves

🎯 Special Points on the Titration Curve

📋 Critical Points Summary

Part 5: Indicators & Equivalence Point

🎨 Acid-Base Indicators

🔧 How Indicators Work

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

🔢 Problem 1: Complete Titration Curve Calculations

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

📋 Complete Titration Summary

Method at Each Point

Before Equivalence Point

At Equivalence Point

After Equivalence Point

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

Region 3: At Equivalence (50.0 mL)

Region 4: After Equivalence (>50 mL)

🔢 pH Calculations at Key Points

Titrating 50.0 mL of 0.100 M HClHClHCl with 0.100 M NaOHNaOHNaOH

📌 Key Features of the Strong–Strong Curve

Shape Analysis

Why pH 7 at Equivalence?

Effect of Concentration

Region 1: Initial Point (0 mL added)

Region 2: Buffer Region (0 to 50 mL)

Region 3: Equivalence Point (50.0 mL)

Region 4: After Equivalence (>50 mL)

📌 The Half-Equivalence Point

Why This Matters on the AP Exam

🔢 Calculating pH at the Equivalence Point

ICE Table

The Rule for Equivalence Point pH

🧪 Weak Base–Strong Acid Titration

The Curve Is Inverted!

Key Difference

🧪 Polyprotic Acid Titrations

Diprotic Acid (H2AH_2AH2​A) with NaOHNaOHNaOH

Key Features

Example: H3PO4H_3PO_4H3​PO4​

Color Change Rules

The Transition Range

📌 Common Indicators

Choosing the Right Indicator

📌 pH Meters vs. Indicators

Advantages of pH Meters

When Indicators Are Still Used

Finding Equivalence with a pH Meter

(a) Initial pH

(b) After 25.0 mL NaOHNaOHNaOH (half-equivalence)

(c) Equivalence Point (50.0 mL NaOHNaOHNaOH)

🧪 Problem 2: Unknown Acid Identification

Analysis:

Equivalence Point pH Summary

Indicator Selection Rule

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

Diprotic Acid ( $H_2A$ ) with $NaOH$

Example: $H_3PO_4$

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

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