Dilution Calculations

10 min read Updated Mar 26, 2026

Dilution is simple: you add more solvent to a solution to decrease its concentration. The amount of solute does not change - you are just spreading it through more liquid. This one idea gives you the most useful equation in solution chemistry.

The Dilution Equation

Why it works: Molarity times volume equals moles (mol/L x L = mol). When you dilute, you do not add or remove solute - only solvent. So the moles before dilution equal the moles after dilution.

Dilution diagram showing two graduated cylinders: before dilution with a small volume of concentrated solution containing many solute particles, and after adding solvent with a larger volume and the same number of solute particles spread out, with the equation c1V1 = c2V2 — Dilution in action. The same number of solute particles (red triangles) are spread through a larger volume of solvent. Concentration decreases but total moles remain constant: c₁V₁ = c₂V₂. *Credit: Wikimedia Commons, CC BY-SA 4.0*

Worked Example

Problem: You have 50.0 mL of 6.0 M HCl. How much water must you add to make a 1.0 M solution?

Step 1: Identify variables.

M1 = 6.0 M, V1 = 50.0 mL
M2 = 1.0 M, V2 = ?

Step 2: Solve for V2.

M1V1 = M2V2
(6.0)(50.0) = (1.0)(V2)
V2 = 300 mL

Step 3: Find water added.

Water added = V2 - V1 = 300 - 50.0 = 250 mL

Note the question asked how much water to add, not the final volume. This is a common MCAT trap - read carefully.

When M1V1 = M2V2 Does NOT Apply

This equation only works when the moles of solute stay constant. It does not apply when:

A chemical reaction occurs - if mixing two solutions produces a reaction, you cannot use M1V1 = M2V2 for the reacting species
You mix two solutions of the same solute - this is a mixing problem, not a dilution. Use: $M_{\text{final}}$ = (M1V1 + M2V2) / (V1 + V2)
Solute is added or removed - the equation assumes only solvent changes

Serial Dilutions

A serial dilution is a sequence of dilutions, each using the product of the previous step as the starting solution. They are used to achieve very low concentrations efficiently.

Example: Starting with 1.0 M NaCl, perform three consecutive 1:10 dilutions.

After dilution 1: 1.0 / 10 = 0.10 M
After dilution 2: 0.10 / 10 = 0.010 M
After dilution 3: 0.010 / 10 = 0.0010 M = $1.0 \times 10^{-3}$ M

Each 1:10 dilution divides the concentration by 10. Three such dilutions give a 1:1000 overall dilution factor.

Preparing Standard Solutions

Labs often prepare standard solutions by diluting a concentrated stock solution. The process:

Calculate the volume of stock solution needed using M1V1 = M2V2
Measure that volume with a pipette
Transfer to a volumetric flask
Add solvent until the solution reaches the calibration mark on the flask

Example: Prepare 250 mL of 0.10 M NaOH from a 2.0 M stock solution.

V1 = M2V2 / M1 = (0.10)(250) / (2.0) = 12.5 mL of stock solution
Add 12.5 mL of 2.0 M NaOH to a 250 mL volumetric flask, then fill to the mark with water

You dilute 25 mL of 12 M HCl to a final volume of 300 mL. What is the final concentration?

Click to reveal answer

1.0 M. M2 = M1V1 / V2 = (12)(25) / (300) = 1.0 M. The concentration dropped by a factor of 12 because the volume increased by a factor of 12 (from 25 to 300 mL).

Why can you not use M1V1 = M2V2 when mixing 50 mL of 1.0 M HCl with 50 mL of 1.0 M NaOH?

Click to reveal answer

Because a chemical reaction occurs. HCl and NaOH undergo a neutralization reaction (HCl + NaOH -> NaCl + H2O). The moles of HCl and NaOH are consumed - they do not simply spread through more volume. M1V1 = M2V2 requires that the solute moles remain unchanged, which is not the case when a reaction destroys the solute.