Serial Dilution and Calibration Curves: From Stock to Unknown

By Aniket Bhardwaj · 24 October 2026 · Updated 24 October 2026 · Analytical Chemistry

Almost every quantitative measurement in an analytical or biochemical laboratory starts with a dilution. A stock solution is too concentrated to measure, so it is diluted in steps, and the readings of the diluted samples are converted back to the original. The arithmetic is only M₁V₁ = M₂V₂ repeated, but exam questions, and real laboratory records, go wrong in the bookkeeping. This article covers serial dilution and its main use, the calibration curve. It stays at the level of method and principle; it does not rely on any particular published study.

A five tube ten fold serial dilution lowers 1000 mg L⁻¹ stock to 0.01 mg L⁻¹, with total DF 100 000. — Each tube comes from the previous one, so the dilution factors multiply.

The ideas and formulas

Single step: C₁V₁ = C₂V₂
Dilution factor, DF = V_final ÷ V_aliquot = C_start ÷ C_end
Serial dilution: DF_total = DF₁ × DF₂ × … × DF_n
Calibration line: A = m c + b → c_diluted = (A − b) ÷ m, c_original = c_diluted × DF

C is concentration in any one unit (mol L⁻¹, mg L⁻¹ or ppm for dilute water solutions) and V is volume in any one unit. V_aliquot is the volume of solution you transfer, and V_final is the total volume after adding diluent, not the volume of diluent alone. A “1:10 dilution” in most laboratory writing means 1 part sample in 10 parts total (DF = 10). Some texts write “1:9” for 1 part sample plus 9 parts diluent. Both mean the same solution; always write the dilution factor, not just the ratio, in your working.

In a serial dilution each tube is made from the previous one, not from the stock. That is the reason the factors multiply. It also means any pipetting error in an early tube carries through to every later tube.

Worked example 1: a 10-fold series

Question. Stock = 1000 mg L⁻¹. Transfer 1.00 mL into 9.00 mL of diluent, mix, and repeat from each new tube to make five tubes. Find each concentration and the total dilution factor.

Each step: V_final = 1.00 + 9.00 = 10.00 mL, so DF = 10.00 ÷ 1.00 = 10.

Tube 1: 1000 ÷ 10 = 100 mg L⁻¹
Tube 2: 10 mg L⁻¹
Tube 3: 1 mg L⁻¹
Tube 4: 0.1 mg L⁻¹
Tube 5: 0.01 mg L⁻¹

Total DF for tube 5 = 10⁵ = 100 000. Check: 1000 ÷ 100 000 = 0.01. Correct.

Worked example 2: a two-fold series

Question. Start with 800 units mL⁻¹. At each step mix equal volumes of the previous tube and diluent (for example 1.00 mL + 1.00 mL). Give the first six concentrations.

Each step has DF = 2.00 ÷ 1.00 = 2. Concentrations: 400, 200, 100, 50, 25, 12.5.

The total DF after 6 tubes is 2⁶ = 64, and 800 ÷ 64 = 12.5. Two-fold series are useful when you need many points between two limits, because the concentrations change slowly from tube to tube.

Worked example 3: preparing a standard from stock

Question. Prepare 250.0 mL of a 5.0 mg L⁻¹ standard from a 1000 mg L⁻¹ stock.

V₁ = C₂V₂ ÷ C₁ = (5.0 × 250.0) ÷ 1000 = 1.25 mL.

That volume is small, and a small pipette error is a large share of 1.25 mL. A better plan is two steps. First make an intermediate of 100 mg L⁻¹ (10.0 mL of stock diluted to 100.0 mL). Then V₁ = (5.0 × 250.0) ÷ 100 = 12.5 mL of the intermediate, diluted to 250.0 mL. The total DF is the same (1000/5 = 200 = 10 × 20), but each transfer is a more measurable volume.

Worked example 4: reading an unknown from a calibration curve

Practice data. Six standards were measured at one wavelength after zeroing with the blank:

c (mg L⁻¹)	0	2	4	6	8	10
A	0.002	0.121	0.238	0.362	0.479	0.601

Least-squares line. Mean c = 5.00, mean A = 0.3005. S_xx = Σ(c − 5)² = 70.0. S_xy = Σ(c − 5)(A − 0.3005) = 4.193. Slope m = 4.193 ÷ 70.0 = 0.0599 per mg L⁻¹. Intercept b = 0.3005 − 0.0599 × 5.00 = 0.0010. The fit has R² = 0.9999.

Unknown. A sample was diluted by taking 1.00 mL and making it up to 25.00 mL (DF = 25.00). Its absorbance was 0.315, which lies inside the range of the standards (0.002 to 0.601).

c_diluted = (0.315 − 0.0010) ÷ 0.0599 = 5.242 mg L⁻¹.
c_original = 5.242 × 25.00 = 131.05, so about 131 mg L⁻¹.

Check: predicted A at 5.242 mg L⁻¹ = 0.0599 × 5.242 + 0.0010 = 0.3150. Correct. (These numbers are made-up practice data, not a real measurement.)

Common mistakes

Using the diluent volume instead of the final volume in the dilution factor. 1 mL in 9 mL of diluent is DF = 10, not 9.
Forgetting to multiply back by the dilution factor after reading the concentration from the curve. The curve gives the concentration in the measured solution, not in the original sample.
Adding dilution factors in a series. They multiply.
Extrapolating the line. A reading above the highest standard may lie outside the linear range. Dilute the sample further and measure again.
Ignoring the blank. Standards must be corrected for the blank, or the intercept will be wrong.
Treating each tube as made from stock when it is made from the previous tube, so the error accumulates.
Mixing units such as mg L⁻¹ and µg mL⁻¹ (these two are numerically equal, but mol L⁻¹ is not).

Why a straight line, and when it fails

The calibration line is usually based on the Beer–Lambert law, A = εlc, which predicts absorbance proportional to concentration at a fixed path length. This holds over a limited range. At high concentration, interactions between solute molecules, stray light in the instrument and chemical equilibria such as dimerisation can bend the line. The correct response is to stay inside the range that your standards cover, never to stretch the line beyond it.

Exam relevance

Question type	What to do
Concentration after n serial steps	Divide the stock by the product of the dilution factors
Volume of stock for a standard	V₁ = C₂V₂ ÷ C₁; use an intermediate if V₁ is tiny
Unknown from calibration data	Fit A = mc + b, solve for c, multiply by DF
Why standards should bracket the sample	The line is valid only inside the measured range
Why is the blank subtracted?	Removes reagent and cuvette absorbance from the signal

Details such as the number of standards or acceptable R² depend on the method and the question; check the instructions given in your paper or the official syllabus.

Use the dilution tool to check each single-step volume in Examples 1 and 3 before you plan the series.

Open the Dilution Calculator (M₁V₁ = M₂V₂) →

Physical Chemistry · CSIR-NET / GATE

Practise Physical Chemistry questions

Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.

Practise CSIR-NET questions →Take a timed quiz Open Dilution Calculator (M₁V₁ = M₂V₂)