Two-Step Dilution: Multiply the Factors

A student adds a little coloured liquid to water, then transfers a small portion of the pale mixture into a new cup. The second cup looks even paler. But when the student writes ‘ten times weaker plus ten times weaker equals twenty times weaker’, the arithmetic no longer matches the physical process.

Serial dilution means taking a measured part of one solution and diluting that part again in one or more stages. For a conserved dissolved substance under an ideal mixing model, each step multiplies the previous concentration by the fraction of the final volume that came from the previous solution. Two successive one-in-ten steps therefore give one-in-one-hundred of the original concentration, not one-in-twenty.

Did you know that the single most important phrase in many dilution questions is make up to? ‘Add water until the total is 20 mL’ is different from ‘add 20 mL of water’. Teach that distinction before asking for a number.

Begin with the physical meaning, not the formula

Imagine a transparent cup of coloured water. Adding more water increases the total volume while leaving the amount of dissolved colourant unchanged in the idealised exercise. Concentration decreases because the same amount is distributed through more solution. Moving only part of the mixed liquid into another container moves only the corresponding share of dissolved material.

For this educational demonstration, use clean water and ordinary food colouring, with adult supervision appropriate to the learners. Do not use hazardous laboratory chemicals, taste unidentified samples, or pretend a visual colour match supplies a calibrated concentration measurement. These are invented mathematical mixtures, not measurements of a real product.

Worked example: a genuine one-hundredfold result

Stage 1: take 1 mL of the starting coloured solution and add water until the total solution volume is 10 mL. Under an ideal additive-volume assumption, that involves about 9 mL of water. The fraction of original solution in the new mixture is 1/10, so the concentration is 10% of the starting concentration.

Stage 2: mix Stage 1 thoroughly. Take 2 mL of that first mixture and add water until the new total volume is 20 mL—about 18 mL added water in our simplified model. The second stage retains 2/20 = 1/10 of the intermediate concentration.

Combine the stages: original fraction = (1/10) × (2/20) = (1/10) × (1/10) = 1/100. The second solution has 1% of the original concentration, or is diluted by an overall factor of 100. For a cross-check, 2 mL of the first mixture contains the equivalent of 0.2 mL of the original in our concentration model. That 0.2 mL equivalent in a final 20 mL is 0.2/20 = 0.01 = 1%.

This is a concentration calculation, not a statement that the original liquid occupies a visibly separate 0.2 mL layer. After thorough mixing, the colourant is distributed through the solution. The ‘equivalent volume’ is only a way to check the amount based on the starting concentration.

Why adding dilution factors gives the wrong answer

A factor of ten means a ratio between concentrations before and after one step. The next ratio acts on the already changed concentration. If the original concentration is represented by C, after Stage 1 it is C/10 and after Stage 2 it is (C/10)/10 = C/100. Adding ten and ten would answer a different arithmetic question; it ignores the sequence.

Notice also that a ‘one in ten’ mixture can mean different things in casual speech. In this article it means one volume of starting solution in ten total volumes of final mixture. Other contexts may use a ratio of one part concentrate to ten parts water, yielding eleven total parts. Write the recipe unambiguously.

The volume trap: adding water versus making up to a mark

Suppose you transfer 2 mL of intermediate liquid into a cup and add exactly 20 mL water. If volumes are additive, the new total is 22 mL, so the second-step fraction is 2/22 = 1/11, not 1/10. If the first step was 1/10, the combined fraction is (1/10) × (1/11) = 1/110. A learner who obtains 1/100 here has silently replaced the given water volume with a target final volume.

For real chemical solutions, volume additivity can fail slightly and accurate preparation may require volumetric glassware and an appropriate protocol. The school example is intended to teach ratios, not to certify an analytical solution. The general conservation idea is explained in OpenStax Chemistry’s dilution chapter, which uses the relationship C₁V₁ = C₂V₂ when the amount of solute is conserved.

Check the logic before accepting any number

Name the material: which liquid is the original stock and which is the intermediate mixture?
Identify the transferred volume: the fraction at each stage uses the amount transferred from the immediately preceding solution.
Identify the final volume: do not confuse final mixture volume with the additional water volume.
Multiply the fractions: a 1/5 stage followed by a 1/10 stage becomes 1/50.
Cross-check direction: dilution without added solute should not make the new mixture more concentrated than the starting one.

A second worked example: uneven dilution steps

Suppose the first mixture uses 4 mL original stock, made up to 20 mL final: fraction 4/20 = 1/5. Then take 3 mL of that mixture and make up to 30 mL final: fraction 3/30 = 1/10. The combined fraction is 1/5 × 1/10 = 1/50 = 2% of the original concentration. The overall factor is 50, not 15, not 30 and not the sum of the added water volumes.

Check through equivalent volumes: 3 mL of the 1/5 intermediate contains the same amount of colourant as 0.6 mL of original stock would in our model. Divide 0.6 mL equivalent by 30 mL final = 0.02. The two methods agree.

Guided practice that becomes independent

First step — match words with containers

Give learners two labelled cups: ‘2 mL concentrate plus water to 10 mL total’ and ‘2 mL concentrate plus 10 mL water’. Ask for their final volumes: 10 mL for the first and, under the simple volume-additive assumption, 12 mL for the second. Do not allow a percentage calculation until they can explain the difference.

Second step — a one-step calculation

Mix 5 mL of a safe coloured solution to 25 mL total. Fraction = 5/25 = 1/5. Ask the child to predict whether the resulting solution has more or less colourant per millilitre; it has less.

Third step — combine two different ratios

Take a 1/5 intermediate, then use 4 mL of it in a final 16 mL. The second fraction is 4/16 = 1/4. Overall concentration fraction = 1/5 × 1/4 = 1/20 = 5% of original. A learner should say why the fractions multiply, not just perform the multiplication.

Fourth step — detect missing information

A note says only ‘add 5 mL of water to a diluted sample’. Ask for the final concentration. Without the amount and concentration of sample being diluted, the answer cannot be determined. A sensible question sometimes has no unique numerical answer.

What teachers and parents can observe

Teachers can test transfer by changing the transferred volumes while keeping the ratios equivalent, and by including one misleading ‘add water’ statement. Observe whether pupils identify the correct final volume, preserve units and show an independent check. Parents can use labelled water cups or drawn rectangles at home and ask the learner to narrate what is moved to the next stage before solving.

A student who only gets the familiar one-in-ten example right may have memorised a pattern. A student who can explain the 1/5 followed by 1/4 case, and reject a problem with insufficient information, has made a more meaningful step toward independent understanding.

Limits and responsible scientific language

The model assumes complete mixing, no colourant loss, unchanged dissolved amount during each dilution and appropriate volume handling. Actual mixing, volumetric uncertainty, contamination and changes in absorbance can alter a physical observation. Visual colour darkness is influenced by lighting and path length, so it should not be presented as proof of a numerical concentration without calibration.

Never perform classroom dilution with substances requiring special handling merely because the mathematics looks easy. A teacher should choose materials and precautions appropriate to the pupils and the activity.

Frequently asked questions

Why not add ten and ten to get twenty?

Because the second tenfold step changes the concentration that remained after the first step. Tenfold followed by tenfold is 100-fold overall.

Does a one-in-ten dilution always need exactly 9 mL water?

Not necessarily. In this simple example 1 mL stock is made up to 10 mL total, often approximated by adding 9 mL water. A real preparation may need to adjust to the final volume rather than assume exact additivity.

Is 1% of the original the same as a final 1% mass concentration?

No. It means one-hundredth of whatever the original concentration was. If the starting concentration is unknown, the final absolute mass percentage is also unknown.

Does the same principle work with three stages?

Yes, when the stated conservation and mixing assumptions hold. Multiply all three stage fractions in order and independently check the overall direction and units.

The Core Aim of Education | How to Combine Two Dilution Steps Without Adding Their Factors