4.4 Molarity and Dilutions
PRE-HEALTH PREP
Concentration Basics
Molarity and Dilutions
General Chemistry 4.4
Molarity, defined as moles of solute per liter of solution (M = n/V), is the most widely used concentration unit in general chemistry. The molarity equation can be rearranged to solve for moles or volume, and proper solution preparation requires dissolving a calculated mass of solute in a volumetric flask filled to the target volume. Dilution calculations rely on the principle that moles of solute remain constant, expressed by the equation M₁V₁ = M₂V₂, which generalizes to C₁V₁ = C₂V₂ for any concentration unit.
What Is Molarity?
The most common way to express concentration in a general chemistry course is molarity — though it is far from the only one. Other units of concentration (such as molality and percent composition) will appear later, but molarity is the workhorse you will use most often. Its definition is refreshingly simple.
Molarity equals the number of moles of solute divided by the total volume of solution in liters. Notice the word solution here: it refers to the combined volume of solvent and solute together, not just the solvent alone.
Important Distinction: The denominator is liters of solution (solvent + solute combined), not liters of solvent. This is a detail that trips up students on exams.
The Molarity Equation (M = n/V)
Think of molarity as a straightforward algebraic expression with three variables:
The Core Molarity Equation:
M = n / V
where M = molarity (mol/L), n = moles of solute, and V = volume of solution in liters.
| Molarity, Moles, and Volume | |||
|---|---|---|---|
| Molarity = moles solute / Liters solution | M = mol / VL | (M)(VL) = mol | VL = mol / M |
3 Ways to Rearrange the Molarity Equation
Because this equation contains three variables, supplying any two allows you to solve for the third:
| Solving For | Rearranged Equation | You Need |
|---|---|---|
| Molarity (M) | M = n / V | Moles and volume |
| Moles (n) | n = M × V | Molarity and volume |
| Volume (V) | V = n / M | Moles and molarity |
A New Way to Find Moles
The middle rearrangement — n = M × V — deserves special attention. You already know how to convert between grams and moles using molar mass, and between moles and particles using Avogadro's number. This equation gives you yet another route to moles: if you know both the molarity and the volume of a solution, you automatically know the moles of solute it contains.
Why This Matters: Once you have moles, every other conversion is open to you — grams (via molar mass), molecules or atoms (via Avogadro's number), or moles of a different substance (via a balanced chemical equation). Molarity and volume together unlock all of those pathways.
How to Prepare a NaOH Solution: 4-Step Example
Example: Preparing 500 mL of 0.25 M NaOH
Given: Molar mass of NaOH = 40 g/mol
Step 1: Recognize What You Have
The problem provides both a volume (500 mL) and a molarity (0.25 M). Whenever you are given both of these, you can immediately determine the moles of solute. Train yourself to spot this pattern — it appears constantly in general chemistry.
Step 2: Calculate the Moles of Solute
Apply the rearranged molarity equation:
n = M × V = 0.25 mol/L × 0.500 L = 0.125 mol NaOH
Unit Check: The volume must be in liters when using the molarity equation. Convert 500 mL to 0.500 L by dividing by 1000.
Step 3: Convert Moles to Grams
You cannot measure out moles directly on a balance — you measure grams. The bridge between moles and grams is always the molar mass:
0.125 mol NaOH × (40 g / 1 mol) = 5.0 g NaOH
Step 4: Dissolve in a Volumetric Flask
To prepare this solution, weigh out 5.0 grams of solid NaOH and place it in a 500 mL volumetric flask. A volumetric flask is calibrated to hold a precise volume at a thin line etched on its narrow neck, giving you an accurate measurement.
Proper Lab Technique: Do not simply dump in the solid and add 500 mL of water. The solid itself occupies a small amount of volume, and dissolving can cause slight volume changes. Instead, add water to about one-quarter of the flask, swirl until the solute dissolves completely, and then fill to the 500 mL mark. This ensures your final volume is exactly 500 mL.
How to Find the Volume of a Solution
Example: What volume of 5.0 M NH₃ contains 0.10 moles?
Here you are given the molarity (5.0 M) and the moles (0.10 mol), so you are solving for volume. Rearrange the molarity equation:
V = n / M = 0.10 mol / 5.0 mol/L = 0.020 L = 20 mL
You can also confirm this using dimensional analysis. Writing 5.0 M as "5.0 mol per liter," the mol units cancel, leaving liters in the numerator — exactly the unit you need.
Tip: If you find yourself uncertain about which variable goes where, treat M = n / V as a simple algebra problem. Identify the two known values, plug them in, and solve. This approach is more reliable than trying to memorize which numbers to multiply or divide.
What Are Dilutions in Chemistry?
A dilution takes a solution from a higher concentration to a lower concentration, typically by adding more solvent. In most general chemistry problems, that solvent is water.
The Kool-Aid Analogy
Imagine you dump 30 packets of Kool-Aid into a pitcher. You taste it — far too concentrated. So you pour that entire pitcher into a bathtub and fill the tub with water. Now it tastes just right.
Here is the key question: how many packets of Kool-Aid ended up in the bathtub? Still 30. You did not add more Kool-Aid; you only added more water. The amount of solute never changed — only the total volume of solution increased.
Core Principle of Dilution: In a dilution, the moles of solute remain constant. You are only adding more solvent, which increases the total volume and decreases the concentration.
The Dilution Equation: M₁V₁ = M₂V₂
Since the moles of solute do not change during a dilution, and since moles equal molarity times volume, you can write:
M₁V₁ = M₂V₂
The initial molarity times the initial volume equals the final molarity times the final volume.
This dilution equation works because both sides represent the same quantity: the moles of solute, which remain unchanged when more solvent is added.
Flexible Volume Units
You do not actually need to convert volumes to liters for dilution problems. Because volume appears on both sides of the equation, the units cancel — as long as you use the same unit on both sides. Milliliters on the left and milliliters on the right works perfectly.
The General Form: C₁V₁ = C₂V₂
The dilution equation is not limited to molarity. It works with any unit of concentration — percent concentration, molality, or others you will encounter later. That is why you may also see it written as C₁V₁ = C₂V₂, where C stands for concentration in any consistent unit. The M₁V₁ = M₂V₂ form is simply the molarity-specific version.
Dilution Calculation Example: HCl
Example: If 50.0 mL of 3.0 M HCl is diluted to 300 mL, what is the final concentration?
Setting Up the M₁V₁ = M₂V₂ Equation
Identify each variable:
- M₁ = 3.0 M (initial concentration)
- V₁ = 50.0 mL (initial volume)
- M₂ = ? (final concentration — solving for this)
- V₂ = 300 mL (total final volume)
Common Mistake: V₂ is the total final volume of the solution, not the amount of water added. In this problem, you started with 50 mL and ended with 300 mL total, meaning 250 mL of water was added. If a problem asks "how much water was added," the answer is V₂ − V₁, not V₂ itself.
Solving for Final Molarity (M₂)
M₂ = (M₁ × V₁) / V₂ = (3.0 × 50.0) / 300 = 150 / 300 = 0.50 M
The milliliter units cancel on both sides, leaving molarity as the unit for the answer.
The Shortcut: Fold Dilutions
You can also reason through this without the equation. The volume increased from 50 mL to 300 mL — a factor of six. A six-fold increase in volume produces a six-fold decrease in concentration: 3.0 M ÷ 6 = 0.50 M. This mental check is a great way to verify your algebra.
Dilution Calculation Example: NaCl
Example: To what final volume must 20 mL of 0.40 M NaCl be diluted to produce a 0.080 M solution?
Identifying the Unknown
The question asks "to what final volume," which means you are solving for V₂. Note that this is the total final volume, not the amount of water to add.
Solving for Final Volume (V₂)
V₂ = (M₁ × V₁) / M₂ = (0.40 × 20) / 0.080 = 8.0 / 0.080 = 100 mL
Checking with Fold Dilutions
The concentration dropped from 0.40 M to 0.080 M — a factor of five. If the concentration decreased five-fold, the volume must have increased five-fold: 20 mL × 5 = 100 mL. The math checks out.
Tip: If the problem had asked "how much water must be added?" instead, the answer would be V₂ − V₁ = 100 mL − 20 mL = 80 mL. Always read the question carefully to determine whether it wants the total final volume or the volume of solvent added.
8 Key Takeaways: Molarity and Dilutions
- Molarity (M) is defined as moles of solute divided by liters of solution — the most common concentration unit in general chemistry.
- The molarity equation M = n / V contains three variables; knowing any two lets you solve for the third.
- The rearrangement n = M × V provides a new pathway to moles, connecting solution chemistry to grams, particles, and stoichiometry.
- When preparing a solution, calculate the required grams of solute and dissolve in a volumetric flask, filling to the target volume — not adding that volume of water.
- In a dilution, the moles of solute remain constant; only the volume of solution changes.
- The dilution equation M₁V₁ = M₂V₂ works with any consistent volume units and generalizes to C₁V₁ = C₂V₂ for any concentration unit.
- V₂ is the total final volume, not the volume of water added. The volume of water added is V₂ − V₁.
- Fold dilutions offer a quick mental check: if the volume increases by a factor of n, the concentration decreases by the same factor.