Unit IV: Solution Stoichiometry
Applying Stoichiometry to Reactions in Water.
This tutorial is the second chapter of Unit IV. It builds on our knowledge of Molarity.
← Go back to Unit IV, Chapter 1: Solutions & Molarity.
The Problem: The Liquid Recipe
We've mastered stoichiometry for solids (grams) and we understand concentration (molarity). But how do we put it all together? Most real-world chemistry happens in solutions, not with neat piles of powder.
Imagine you need to react a specific amount of silver nitrate (AgNO₃) with sodium chloride (NaCl). You don't have solid AgNO₃, you have a 0.5 M solution of it. How many milliliters of that solution do you need to get the right amount of reactant?
This is the core of solution stoichiometry: using volume and molarity to work with the same "mole bridge" we used for grams.
The Solution: A New On-Ramp to the Mole Bridge
The "mole bridge" is the heart of all stoichiometry, allowing us to relate one substance to another. We've already learned one on-ramp: converting grams to moles using molar mass.
Now, we have a second on-ramp. The molarity formula gives us a direct path from a measurable liquid quantity (volume) to moles.
That's it! This is the key. Once you have moles, you can use the mole ratio from the balanced equation just like you did before. You can then convert the moles of your target substance back to grams (using molar mass) or even back to a volume (if you know its molarity).
Interactive Solution Stoichiometry Calculator
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Let's find out how many grams of solid silver chloride (AgCl) we can make if we start with 50 mL of 0.5 M AgNO₃ and excess NaCl.
Application: Titrations
One of the most common uses of solution stoichiometry is in a lab technique called titration. This is used to find the unknown concentration of a solution (the "analyte") by reacting it with a solution of known concentration (the "titrant").
At the "equivalence point" of a titration, the moles of the titrant added are just enough to completely react with the moles of the analyte in the flask. By knowing the volumes of both solutions and the concentration of the titrant, we can use solution stoichiometry to calculate the unknown concentration of the analyte.
The Titration Calculation Path
- Volume of Titrant (Known) → Moles of Titrant: Use M = mol/L.
- Moles of Titrant → Moles of Analyte (Unknown): Use the mole ratio from the balanced acid-base reaction.
- Moles of Analyte → Molarity of Analyte: Use M = mol/L with the known volume of the analyte.
You've Mastered Reactions in Solution. What's Next?
You now have the complete toolkit for stoichiometry. The next step is to dive deeper into a specific and very important type of solution chemistry: Acids and Bases.