Solution Dilution Calculator (C1V1 = C2V2)

The core principle of dilution is simple: the total amount of solute is conserved before and after dilution. Written as a formula:

C₁ × V₁ = C₂ × V₂

C₁/V₁ are the concentration and volume taken from the stock solution; C₂/V₂ are the target concentration and final volume after dilution. Given any three of the four quantities, the fourth is determined. Volume of solvent to add = V₂ − V₁.

Units do not need to be in SI, but the concentration units on both sides of the equation must match, and the volume units must match — this is the most common source of error.

Serial dilution is a process in which each step uses the product of the previous step as the new stock. For a 10-fold serial dilution, take 1 part from each tube and add 9 parts solvent. Mix thoroughly at every step before taking the next aliquot — otherwise errors compound at each stage. By the sixth tube, a 5% error per step accumulates to over 30%.

A worked example

Preparing 500 mL of 0.5× working solution from a 5× TBE stock:

V₁ = C₂V₂ / C₁ = 0.5 × 500 / 5 = 50 mL stock
Solvent to add = 500 − 50 = 450 mL

What does “1:10 dilution” actually mean?

This phrase is genuinely ambiguous and has caused real experimental errors. Two interpretations exist:

• 1 part stock + 9 parts solvent, final volume = 10 parts → 10-fold dilution (the more common meaning)
• 1 part stock + 10 parts solvent, final volume = 11 parts → 11-fold dilution

When writing up your methods, state the final concentration or “10-fold dilution” explicitly — never write “1:10”. This saves the next person from having to guess.

High-fold dilutions must be done in steps

Diluting from 1 M to 1 μM is a 10⁶-fold dilution. Done in a single step, you would take 1 μL and add it to 999,999 μL — at typical pipetting volumes, a 1 μL transfer can carry errors of several percent, directly determining the final concentration.

The correct approach is to perform six 10-fold steps, each taking 100 μL into 900 μL. The trade-off is that errors accumulate across steps: a 1% error per step gives roughly 6% after six steps (root-sum-square accumulation gives ~2.4%; worst-case linear accumulation gives 6%). More steps are therefore not always better — use as few as are sufficient.