Mastering Acid-Base Titrations: A Comprehensive Guide

Acid-base titrations may sound intimidating, but they bring together all the key concepts of equilibrium, pH, pKa values, and stoichiometry in a single, fascinating procedure. In this guide, we’ll explore what acid-base titrations are, how they work, and how to tackle various types of titrations with confidence. By the end, you’ll have a clear understanding of titration curves, the equivalence point, and how to work with weak and strong acids and bases in titration problems.

What is a Titration?

A titration is a laboratory technique used to determine the concentration of an unknown solution, or “analyte,” by reacting it with a solution of known concentration, or “titrant.” The titrant is added from a burette until the equivalence point is reached, where the moles of the analyte equal the moles of the titrant.

Equation for the Equivalence Point:

$n_a \times M_a \times V_a = m_b \times M_b \times V_b$

where $n$ and $m$ are the stoichiometric coefficients, and $M$ and $V$ are the molarity and volume of the solutions.

Image from Pixabay

Understanding Titration Curves

During a titration, we measure how the pH changes as we add the titrant. This gives us a titration curve, which shows how pH varies with the volume of titrant added.

Example: Titrating 1M NaOH into 25 mL of 1M HCl

1. Pre-Titration:

   • Initial pH of 1M HCl = 0 (since pH = $-\log[1]$).

2. Pre-Equivalence Point:

   • As NaOH is added, it reacts with HCl: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$
     
   • As long as there is excess H+, the pH will gradually rise.
     

     

     Image From ASDL

3. Equivalence Point:

   • At this point, the moles of HCl and NaOH are equal. For this example:
     • $M_a \times V_a = M_b \times V_b$
     • 25 mmol of HCl = 25 mmol of NaOH.
   • The pH at the equivalence point is 7 for strong acid-strong base titrations.
     

     

     Image From LibreTexts/14%3A_Acid-Base_Equilibria/14.07%3A_Acid-Base_Titrations)

4. Post-Equivalence Point:

   • Adding more NaOH results in excess OH-, raising the pH.
     

     

     Image From LibreTexts/14%3A_Acid-Base_Equilibria/14.07%3A_Acid-Base_Titrations)/14%3A_Acid-Base_Equilibria/14.07%3A_Acid-Base_Titrations)/14%3A_Acid-Base_Equilibria/14.07%3A_Acid-Base_Titrations)

Titration Curve Visualization

A typical titration curve for a strong acid and a strong base shows a steep increase in pH at the equivalence point.

Image From LibreTexts/21%3A_Acids_and_Bases/21.19%3A_Titration_Curves)

Weak Acid-Strong Base Titrations

When a weak acid is titrated with a strong base, the process is similar but with important differences:

• Before the equivalence point, the solution contains a buffer (a mixture of the weak acid and its conjugate base).
• The pH at the half-equivalence point (when half of the acid has reacted) equals the pKa of the acid.
• The pH at the equivalence point is greater than 7 due to the presence of the conjugate base.

Example Problem: Calculate the pH when 10 mL of 0.1M NaOH is added to 25 mL of 0.1M acetic acid (CH₃COOH). $K_a = 1.8 \times 10^{-5}$

1. Reaction Equation: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$
   

2. Calculate Millimoles:

   • Acetic acid: $25\times 0.1=\mathrm{2.5\; mmol<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"> </span>}$
   • NaOH: $10\times 0.1=\mathrm{1.0\; mmol<span\; class="katex"\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"></span>}$

3. Stoichiometry:

   • Remaining CH₃COOH: $\mathrm{1.5\; mmol<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"> </span>}$
   • Produced CH₃COO-: $1.0$ mmol.

4. Calculate pH using the Henderson-Hasselbalch Equation:

   $$\text{pH} = \text{p}K_a + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right)$$
   $$\text{pH} = 4.74 + \log \left( \frac{1.0}{1.5} \right) = 4.56$$
   

Titrations with Weak Bases and Strong Acids

The process for a weak base-strong acid titration follows the same principles, but the pH at the equivalence point will be acidic due to the presence of the conjugate acid.

Quick Tips for Titrations

• Identify the type of titration (strong vs. weak acids/bases).
• Calculate millimoles and use stoichiometry.
• Recognize buffer regions and half-equivalence points.
• Use the Henderson-Hasselbalch equation for buffer solutions.
• Know that pH at the equivalence point varies based on the acid/base strength.

Key Takeaways

• Titrations help determine unknown concentrations by neutralizing an acid or base.
• Titration curves graph pH changes, revealing key points like equivalence.
• Weak acids/bases create buffers during titrations, resulting in unique curves and pH values.
• Mastering stoichiometry, pH calculations, and buffer concepts is essential for tackling titration problems.