Unit IV: Titrations
The Art of Precision Neutralization.
This is the final chapter of Unit IV. It combines everything we've learned about solutions, stoichiometry, and acids & bases.
← Go back to Unit IV, Chapter 3: Acids & Bases.
The Problem: The Case of the Unknown Concentration
Imagine you work in a lab and find a bottle of hydrochloric acid (HCl) with a smudged label. You know it's HCl, but you have no idea what its molarity is. Using this acid in an experiment would be dangerous and would ruin your results. How can you determine its exact concentration?
This is a common problem in chemistry. The solution is a precise and elegant technique called titration. By carefully reacting our unknown acid with a base of a known concentration, we can figure out the exact molarity of the acid.
The Technique: Controlled Neutralization
An acid-base titration is a controlled neutralization reaction. Here are the key players:
- Analyte: The solution of unknown concentration (in our case, the HCl). It goes in an Erlenmeyer flask.
- Titrant: The solution of known concentration (e.g., 0.1 M NaOH). It goes in a long, graduated tube called a burette.
- Indicator: A chemical (like phenolphthalein) that changes color at a specific pH. We add a few drops to the analyte in the flask.
The goal is to slowly add the titrant to the analyte until the reaction is exactly complete. This point is called the equivalence point, where the moles of acid exactly equal the moles of base. The indicator's color change, which happens very close to this point, tells us when to stop. This is called the endpoint.
The "Aha!" Moment: The Titration Curve
Use the simulator below to perform a virtual titration. We'll add 0.1 M NaOH to 25 mL of HCl. The "Aha!" moment is seeing the indicator's color change happen at the exact same time as the steep, vertical part of the pH curve. This is the equivalence point!
The Calculation: Finding the Unknown
At the equivalence point, we know that moles of acid = moles of base. Since Molarity = moles/Liters, we can say moles = Molarity × Liters. This gives us the master equation for titrations:
From our simulation, we found the equivalence point was at 25.00 mL of NaOH added. We know:
- Macid = ? (This is what we want to find)
- Vacid = 25.0 mL = 0.025 L (We measured this into the flask)
- Mbase = 0.100 M (The known concentration of our titrant)
- Vbase = 25.00 mL = 0.025 L (The volume we added from the burette)
Plugging these in: Macid × (0.025 L) = (0.100 M) × (0.025 L). Solving for Macid gives us 0.100 M. We've found the unknown concentration!
Note: This simple M₁V₁=M₂V₂ formula only works for reactions with a 1:1 mole ratio, like HCl + NaOH. For other reactions (e.g., H₂SO₄ + 2NaOH), you must use the full mole bridge calculation from the solution stoichiometry tutorial.
Congratulations! You've Completed Unit IV.
You have now mastered the fundamentals of chemical arithmetic and reactions in solution. You have the skills to tackle a huge range of chemistry problems.