🧪 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

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

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

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

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

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

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

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

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.

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).

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

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 $NaOH$:

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!)

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 ✅

🎯 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 $NaOH$:

The Rule for Equivalence Point pH

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!

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

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

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:

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

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

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:

$pH = pK_{In} \pm 1$

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

Common Indicators

Choosing the Right Indicator

Match the indicator range to the equivalence point pH!

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 ✅

🛠️ 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 $CH_3COOH$ ($K_a = 1.8 \times 10^{-5}$, $pK_a = 4.74$) is titrated with 0.200 M $NaOH$.

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

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

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

Equivalence Point pH Summary

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 ✅