Serial Dilution

Introduction

Very dilute solutions are often made in stages when direct measurement is impractical. Each stage transfers a measured aliquot and brings it to a new final volume. The concentration factors multiply; the volumes from different steps must not be paired arbitrarily.

Core explanation

For one stage with conserved solute, c(after) = c(before) × V(aliquot)/V(final). A 1.00 M stock diluted by taking 10.0 mL to 100.0 mL becomes 0.100 M. Taking 10.0 mL of that intermediate to another 100.0 mL gives 0.0100 M. Each stage is a tenfold dilution, so the overall factor is 10 × 10 = 100. The final concentration is 1.00/100 = 0.0100 M.

The second aliquot is taken from the first diluted solution , not from the original stock. If it came from original stock, the result would be another 0.100 M solution, not 0.0100 M. A table with columns for stock concentration, aliquot volume and final volume at each step prevents this confusion.

The overall concentration can be written cₙ = c₀ × (V₁,aliquot/V₁,final) × (V₂,aliquot/V₂,final) and so on. Units of each volume ratio cancel, so consistent mL units can be used within a step. This multiplication assumes the solute survives every step and transfer losses are negligible. A reaction, precipitation or degradation makes a pure dilution model inadequate.

Serial dilution can improve practical measurement. A direct 1000-fold dilution might demand a very tiny aliquot if final volume is modest. Three tenfold stages can use easier aliquots. However, every stage introduces its own measurement uncertainty and demands thorough mixing before the next aliquot is taken. If the intermediate is not uniform, the later calculation's assumed concentration is wrong even if the arithmetic is correct.

There is a difference between an aliquot's moles and the total moles in all vessels. The first transfer leaves most stock behind. The second leaves most intermediate behind. Conservation applies to the particular transferred portion during each stage, not to the total moles across all separate containers being equal.

Step-by-step reasoning

1. Draw a separate arrow for each stage and label its source solution. 2. Calculate each V(aliquot)/V(final) ratio. 3. Multiply the ratios by the original concentration. 4. Confirm that each new concentration falls for a true dilution. 5. Mix each intermediate before sampling the next aliquot.

Visual explanation

Draw three flasks linked by pipette arrows: 1.00 M stock → 0.100 M intermediate → 0.0100 M final. Put “10.0 mL to 100.0 mL” over each arrow and “factor 10” below it.

Real-world analogy

If one part colored syrup is made into ten parts drink, then one part of that drink is again made into ten parts, the color concentration is one hundredth of the original. The analogy assumes thorough mixing at each step.

Real-world example

Calibration experiments may require a range of low concentrations from a single stable stock. Serial dilutions allow measurable pipette volumes; their records must show each stage so a later analyst can reconstruct the final concentration.

Why?

Why multiply dilution factors rather than add them? Each stage scales the concentration already produced by the previous stage. A tenfold reduction applied to a tenfold reduction gives one hundredfold overall.

Common misconception

“Two tenfold dilutions make a twentyfold dilution.” The correct overall factor is 10 × 10 = 100 when each stage uses the previous solution.

Worked example

A 2.00 M stock is diluted by taking 20.0 mL to 200.0 mL, then 5.00 mL of that intermediate to 100.0 mL. First c = 2.00 × 20.0/200.0 = 0.200 M. Second c = 0.200 × 5.00/100.0 = 0.0100 M. Overall factor is (200/20) × (100/5) = 10 × 20 = 200; 2.00/200 = 0.0100 M.

Quick check

1. What concentration follows two successive tenfold dilutions of 0.500 M stock? Answer: The overall factor is one hundred, so final concentration is 0.500/100 = 0.00500 M.

Exam focus

Make a stage table and state the source of each aliquot. Do not mix an aliquot from one stage with a final volume from another.

Advanced insight

Relative uncertainty can accumulate through a dilution chain. Several accurate stages may beat one unmeasurably tiny aliquot, but needless extra stages can add transfer error and contamination risk.

Summary

Serial dilution applies the conservation relation repeatedly. Each concentration equals the preceding one times its aliquot-to-final-volume ratio, and overall factors multiply. Complete mixing and correct stage labels are essential.

Practice questions

1. A 1.00 M stock is diluted 1:5 and then 1:10. Find final c. Answer: Overall factor is 5 × 10 = 50, so c = 1.00/50 = 0.0200 M. 2. Why must the first intermediate be mixed before the second pipette step? Answer: Without mixing, the aliquot may not have the calculated intermediate concentration. 3. If the second aliquot came from original stock instead, would the serial result apply? Answer: No. It would be a separate direct dilution of the original stock, so recalculate from its own aliquot and final volume.