Library / Solution chemistry, equilibria and kinetics / SCH-03
Solution chemistry, equilibria and kinetics
Concentration after serial dilutions
Concentration after up to three dilution stages.
- Ref
- SCH-03
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- cn [mol/L]
Each dilution stage multiplies the concentration by the ratio of aliquot to final volume, v/V, so after several stages c = c₀ × (v₁/V₁)(v₂/V₂)(v₃/V₃). Use v = V for an unused stage. The two volumes of one stage must be in the same unit, and each aliquot must be at most the final volume.
Concentration after up to three dilution stages. At each stage an aliquot v is taken from the previous well-mixed solution and made up to the final volume V. Set v = V for an unused stage. The volumes in one pair must share a unit.
Calculator
- Overall dilution factor500
- Concentration after stage 10.05 mol/L
- Concentration after stage 20.001 mol/L
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The series dilutes the solution by a factor of 500, from 0.5 to 0.001 mol/L.
- Errors multiply through the stages, so a single larger dilution with a good pipette is often more accurate than several small ones.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol/L) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V2 Final volume, stage 2 | 2 to 1,000 mL | 1e-4 | 0.05 | 0.0499 | +10% | −9.09 % |
| c0 Starting concentration | 1e-6 to 10 mol/L | 2e-9 | 0.02 | 0.02 | +10% | +10 % |
| V1 Final volume, stage 1 | 5 to 1,000 mL | 5e-5 | 0.01 | 0.00995 | +10% | −9.09 % |
| V3 Final volume, stage 3 | 1 to 1,000 mL | 1e-6 | 0.001 | 9.99e-4 | +10% | −9.09 % |
| v1 Aliquot transferred, stage 1 | 0.01 to 5 mL | 2e-6 | 0.001 | 9.98e-4 | +10% | +10 % |
| v2 Aliquot transferred, stage 2 | 0.01 to 2 mL | 5e-6 | 0.001 | 9.95e-4 | +10% | +10 % |
| v3 Aliquot transferred, stage 3 (equal to V₃ if unused) | 0.01 to 1 mL | 1e-5 | 0.001 | 9.9e-4 | +10% | — |
Sweep of Aliquot transferred, stage 2
| Aliquot transferred, stage 2 (mL) | Final concentration (mol/L) |
|---|---|
| 0.01 | 5e-6 |
| 0.2587 | 1.294e-4 |
| 0.5075 | 2.538e-4 |
| 0.7562 | 3.781e-4 |
| 1.005 | 5.025e-4 |
| 1.254 | 6.269e-4 |
| 1.502 | 7.513e-4 |
| 1.751 | 8.756e-4 |
| 2 | 0.001 |
Sweep one variable
Variables and how each one contributes
Final concentration
What the equation gives you. Everything else feeds it.
Starting concentration
The starting concentration, in mol/L. Raises the final concentration in direct proportion. Raising it by 10 % (to 0.55) moves the result by +10 %, from 0.001 to 0.0011 mol/L.
Aliquot transferred, stage 1
The aliquot transferred in stage 1. A larger aliquot raises the final concentration in direct proportion. Raising it by 10 % (to 5.5) moves the result by +10 %, from 0.001 to 0.0011 mol/L.
Final volume, stage 1
The final volume of stage 1, in the same unit as the aliquot. A larger final volume lowers the concentration in inverse proportion. Raising it by 10 % (to 55) moves the result by −9.09 %, from 0.001 to 9.091×10−4 mol/L.
Aliquot transferred, stage 2
The aliquot transferred in stage 2. A larger aliquot raises the final concentration in direct proportion. Raising it by 10 % (to 2.2) moves the result by +10 %, from 0.001 to 0.0011 mol/L.
Final volume, stage 2
The final volume of stage 2. A larger final volume lowers the concentration in inverse proportion. Raising it by 10 % (to 110) moves the result by −9.09 %, from 0.001 to 9.091×10−4 mol/L.
Aliquot transferred, stage 3 (equal to V₃ if unused)
The aliquot transferred in stage 3 (equal to V₃ for an unused stage). A larger aliquot raises the final concentration in direct proportion. Raising it by 10 % (to 1.1) gives no valid result at the other default values.
Final volume, stage 3
The final volume of stage 3. A larger final volume lowers the concentration in inverse proportion. Raising it by 10 % (to 1.1) moves the result by −9.09 %, from 0.001 to 9.091×10−4 mol/L.
Worked example with the values above
- Stage ratios: 5/50 = 0.1, 2/100 = 0.02, 1/1 = 1.
- Product of ratios: 0.002.
- c = 0.5 × 0.002 = 0.001 mol/L.
Assumptions and limits
- Ideal dilute solutions, with concentrations in mol/L and masses in grams unless stated.
- Complete mixing and no evaporation or contraction on mixing unless the inputs say otherwise.
Common mistakes
- Mixing molality (per kg of solvent) with molarity (per litre of solution).
- Entering a mass percentage where a fraction between 0 and 1 is needed.
- Leaving out counterions when adding up charges.
Statistics and uncertainty
Masses and volumes are usually known to 0.1 % to 1 %, and concentrations from them carry the same relative error. Volumetric glassware tolerances add up through serial dilutions, so each extra stage widens the uncertainty.
Where it is used
- Preparing standards, stocks and dilution series in the laboratory.
- Checking charge balance and ionic strength in water analysis.
Maths you need
- Amount of substance: n = m/M, in mol
- Concentration: c = n/V, in mol/L
- Electroneutrality: total positive charge equals total negative charge