🎯⭐ 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

Titration Essentials

Term	Definition
Titrant	Solution of known concentration (in the buret)
Analyte	Solution of unknown concentration (in the flask)
Equivalence point	Moles of acid = moles of base
Endpoint	Indicator changes color (ideally ≈ equivalence point)

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

🔑 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

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:

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

🧪 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

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:

$n_{N a O H} =$

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 Four Regions of a Strong-Strong Titration

Region	What's Happening	pH Determined By
Before equivalence	Excess acid remains	$[H^+]$ from unreacted acid
Near equivalence	Rapid pH change	Very small excess of acid/base
At equivalence	Complete neutralization	pH = 7.00 (strong-strong only)
After equivalence	Excess base added	from excess base

Part 3: Weak Acid–Strong Base

📈 Weak Acid–Strong Base Titration Curves

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

Weak Acid–Strong Base: What Changes

Feature	Strong-Strong	Weak-Strong
Initial pH	Very low	Higher (partial dissociation)
Buffer region	None	Yes! (before equivalence)
Half-equivalence	No special significance	pH = p $K_a$
Equivalence pH	7.00	> 7 (conjugate base is basic)
After equivalence

Part 4: Titration Curves

🎯 Special Points on the Titration Curve

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

Critical Points Summary

Point	Volume of Base	How to Find pH	Key Feature
Initial	0 mL	ICE table with $K_a$	Weak acid equilibrium
Half-equivalence	½ $V_{eq}$

Part 5: Indicators & Equivalence Point

🎨 Acid-Base Indicators

Part 5 of 7 — Choosing the Right Indicator

Indicator Selection at a Glance

Indicator	Color Change	pH Range	Best For
Methyl orange	Red → Yellow	3.1–4.4	Strong base titrating strong acid
Bromothymol blue	Yellow → Blue	6.0–7.6	Strong-strong titrations
Phenolphthalein	Colorless → Pink	8.2–10.0	Weak acid–strong base
Alizarin yellow R	Yellow → Red	10.1–12.0	Very basic equivalence points

🔑 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 () is itself a weak acid with different colored forms:

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Acid-Base Titrations

Workshop Problem Types

Problem Type	What You'll Practice
Complete titration curve	Calculate pH at 4+ points, sketch curve
Unknown acid ID	Use equivalence volume + pH to find $K_a$ and molar mass
Indicator selection	Match indicator to titration type
Multi-step free-response	Combine all skills in AP format

🔑 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

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Acid-Base Titrations

Everything at a Glance

Titration Type	Equivalence pH	Special Features
Strong acid + Strong base	= 7.00	Sharp, symmetric curve
Weak acid + Strong base	> 7	Buffer region, pH = p $K_a$ at half-eq
Weak base + Strong acid	< 7	Inverted curve
Polyprotic acid	Multiple eq pts	Separate steps for each proton

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

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

(for 1:1 stoichiometry)

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

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.

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

After Equivalence Point

Excess $OH^-$ remains → basic

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

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)

🔑 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 $NaOH$ $N a O H$ :
- Mol $H^+ = 0.0050 - 0.0010 = 0.0040$
- $[H^{+}] =$ M

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 + 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 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.

💡 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

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).

⚠️ 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 ✅

🔑 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 ✅

🔑 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 $NaOH$ :

Point	Volume of $NaOH$	What's Present	How to Find pH
Initial	0 mL	Only $HA$	ICE table with $K_a$
Buffer region	$0 < V < V_{eq}$	$HA + A^-$
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 $OH^-$

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 and

🔑 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:

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: $pH = pK_{a2}$

⚠️ 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):

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

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 ✅

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

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

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:

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

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

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!

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

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.

💡 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 ✅

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:

$K_a = \frac{x^2}{0.200} = 1.8 \times 10^{-5}$

(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

$K_b = K_w/K_a = 5.56 \times 10^{-10}$

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 ( $HCOOH$ , $K_a = 1.8 \times 10^{-4}$ ).

🔑 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 ✅

Why this matters:

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

Region	What's Present	Calculation Method
Initial (weak acid)	Only $HA$	ICE table with $K_a$
Buffer region	$HA + A^-$	$pH = pK_a + \log([A^-]/[HA])$
Half-equivalence	$[HA] = [A^-]$	$pH = pK_a$
Equivalence	Only $A^-$	ICE with $K_b = K_w/K_a$
After equivalence	$A^-$ + excess $OH^-$	$[OH^-]$ from excess

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

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

🔑 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 ✅

[H^{+}] = 0.0040/0.060 = 0.0667

pH levels off at high values

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

x = \sqrt{3.6 \times 10^{-6}} = 1.90 \times 10^{-3}

pH = -\log(1.90 \times 10^{-3}) = 2.72

x = \sqrt{5.56 \times 10^{-10} \times 0.100} = 7.45 \times 10^{-6}

pOH = 5.13, \quad pH = 8.87

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

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

Acid-Base Titrations and Indicators

Acid-Base Titrations and Indicators - Complete Interactive Lesson

Part 1: Titration Basics

🧪 Titration Fundamentals

Titration Essentials

What You'll Master in Part 1

🧪 Titration Setup

Key Components

The Key Equation

🧪 Strong Acid – Strong Base Titration

The Reaction

🧪 Worked Example: Finding Unknown Concentration

Part 2: Strong Acid–Strong Base

📈 Strong Acid–Strong Base Titration Curves

The Four Regions of a Strong-Strong Titration

Part 3: Weak Acid–Strong Base

📈 Weak Acid–Strong Base Titration Curves

Weak Acid–Strong Base: What Changes

Part 4: Titration Curves

🎯 Special Points on the Titration Curve

Critical Points Summary

Part 5: Indicators & Equivalence Point

🎨 Acid-Base Indicators

Indicator Selection at a Glance

What You'll Master in Part 5

🔧 How Indicators Work

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Workshop Problem Types

What You'll Master in Part 6

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Everything at a Glance

Before Equivalence Point

At Equivalence Point

After Equivalence Point

What You'll Master in Part 2

📌 Regions of the Curve

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

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

What You'll Master in Part 3

🧪 Four Regions of the Weak Acid–Strong Base Curve

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

What You'll Master in Part 4

📋 Critical Points Summary

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

🔢 Problem 1: Complete Titration Curve Calculations

🧪 Problem 2: Unknown Acid Identification

Analysis:

What You'll Master in Part 7

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

Diprotic Acid ( $H_2A$ ) with $NaOH$

Example: $H_3PO_4$