Why Extrapolating a Calibration Line Can Give an Unreliable Answer

A colourimeter has been calibrated using blue-dye solutions with concentrations between 1.0 and 8.0 mg/L. An unknown sample produces a blank-corrected absorbance of 0.440. The calibration equation gives a concentration of 11.0 mg/L.

Should you report 11.0 mg/L?

Not yet. The calculated concentration is outside the range tested by the standards. Although the equation produces a number, the experiment has not established that the calibration line remains reliable at that concentration.

This is the central problem with extrapolation: mathematics can extend a calibration line beyond the measured data, but it cannot prove that the relationship continues beyond that range.

The response might remain linear. It might curve, level off, or become affected by background signals. Without measurements beyond the highest standard, you cannot know which behaviour applies.

An extrapolated result is not necessarily wrong. It is simply not supported by the calibration evidence.

01What the calibration standards have actually tested

A calibration curve is constructed by measuring solutions with known concentrations and recording their instrument responses.

Suppose a chemist prepares five blue-dye standards. Each solution contains a known concentration of dye, and each is measured using the same colourimeter settings.

If increasing the dye concentration produces a proportional increase in absorbance, the measurements may follow a straight-line relationship.

For the moment, imagine that the line fits the standards perfectly.

What does that tell you?

It tells you that a linear relationship describes the measured standards over their tested concentration range. It does not establish that the same relationship applies to every possible concentration.

Interpolation versus extrapolation

These two operations look similar mathematically, but the experimental evidence supporting them is different.

• Interpolation estimates a concentration between measured calibration standards.
• Extrapolation estimates a concentration beyond the lowest or highest measured standard.

Consider a calibration range of 1.0-8.0 mg/L.

If an unknown produces an absorbance corresponding to 5.0 mg/L, its concentration can be estimated using the standards on either side.

The estimate is supported by measurements below and above the unknown.

However, suppose the unknown corresponds to 11.0 mg/L. There are no standards between 8.0 and 11.0 mg/L, or above 11.0 mg/L, to demonstrate that the relationship remains linear.

The result now depends on an assumption.

A useful way to picture this is to imagine following a straight road. You have checked its direction using several markers over the first 8 km. Extending a ruler beyond the last marker predicts where the road should go at 11 km.

But what if the road bends after the final marker?

The ruler still gives a prediction. It just doesn’t establish where the road actually goes.

Calibration works similarly. The analogy is not exact because chemical and instrumental responses can change for several different physical and chemical reasons, but the important point remains: a mathematical extension is not an experimental observation.

Why a good straight-line fit is not enough

A calibration equation is often written as:

\[ A=mc+b \]

where:

• \(A\) is absorbance, which is dimensionless.
• \(c\) is concentration, commonly measured in mg/L.
• \(m\) is the gradient of the calibration line.
• \(b\) is the vertical intercept.

Rearranging gives:

\[ c=\frac{A-b}{m} \]

This equation can calculate a concentration for almost any absorbance value.

However, the equation assumes that the linear relationship remains appropriate.

Even a very high coefficient of determination, \(R^2\), does not establish this.

An \(R^2\) value describes how closely the calibration measurements follow the fitted line. It does not demonstrate that the relationship continues beyond the concentrations actually tested.

A line can fit every standard extremely well and still give an unreliable extrapolated result.

02What happens during the experiment?

To understand why extrapolation can fail, follow the measurement from the prepared solution to the instrument reading.

Consider a colourimeter measuring the concentration of a blue dye.

First, the chemist prepares calibration standards containing known dye concentrations. A blank is also prepared using the appropriate solvent and reagents, but without the dye.

The colourimeter contains four main components:

1. A light source produces light.
2. A filter selects an appropriate wavelength.
3. Light passes through a cuvette containing the prepared solution.
4. A detector measures the transmitted light.

The instrument calculates absorbance from the incoming and transmitted light intensities:

\[ A=\log_{10}\left(\frac{I_0}{I}\right) \]

Here, \(I_0\) is the incident light intensity and \(I\) is the transmitted light intensity.

As the concentration of absorbing dye increases, less light may reach the detector, producing a higher absorbance.

This relationship is approximately linear under suitable conditions, but it is not guaranteed to remain linear indefinitely.

What should the original calibration experiment show?

If the response is linear, increasing the concentration by a fixed amount should produce a consistent increase in absorbance.

Consider the following illustrative measurements.

All absorbance readings are raw measurements taken without first zeroing the instrument using the blank.

The blank contributes an absorbance of 0.030.

Subtracting this background gives the following corrected results:

For these illustrative data, the calibration equation is:

\[ A_{\mathrm{net}}=0.0400c \]

The gradient is \(0.0400\ \mathrm{L\,mg^{-1}}\).

The standards support this linear relationship between 1.0 and 8.0 mg/L.

Notice something important: the equation may produce a concentration of 11.0 mg/L, but there is no standard at 11.0 mg/L to confirm the predicted absorbance.

The experiment has established the relationship over a particular interval, not everywhere.

Why might the relationship stop being linear?

There are several possible reasons.

1. Instrument limitations at high concentrations

At high dye concentrations, very little light may reach the detector.

Instrumental effects, such as stray light and limitations in detector response, can cause measured absorbance to depart from the expected linear relationship.

The instrument may still display a numerical reading, but that does not mean the reading is suitable for quantitative analysis.

2. Background signals become important at low concentrations

At very low concentrations, the analyte contributes relatively little absorbance.

The solvent, reagents, cuvette, and other background effects may become significant compared with the analyte signal.

A small error in measuring or correcting the background can produce a substantial relative error in the calculated concentration.

3. The sample matrix may affect the measurement

The sample matrix is everything in the sample other than the analyte being measured.

Other dissolved substances may absorb light, scatter light, or interfere with the measurement.

A blank containing only solvent and reagents might not account for these sample-specific effects.

Consequently, an unknown sample may produce a different response from a calibration standard containing the same analyte concentration.

4. Uncertainty in the calibration equation increases the problem

The gradient and intercept are estimated from experimental measurements. They are not perfectly exact.

When you extrapolate, small uncertainties in the fitted relationship can produce increasingly uncertain predictions as you move further from the measured data.

This is another reason why a calculated concentration outside the calibration range should not automatically be treated as a reliable result.

03Blank correction: make sure the signal is consistent

Before using a calibration equation, you need to understand how the blank has been handled.

A blank measures background absorbance from substances other than the target analyte.

If the instrument has not been zeroed using the blank, the corrected absorbance is:

\[ A_{\mathrm{net}} = A_{\mathrm{sample}}-A_{\mathrm{blank}} \]

For example, suppose:

\[ A_{\mathrm{sample}}=0.250 \]

\[ A_{\mathrm{blank}}=0.030 \]

Then:

\[ A_{\mathrm{net}}=0.250-0.030 \]

\[ A_{\mathrm{net}}=0.220 \]

The corrected absorbance is 0.220.

However, if the instrument was already zeroed using the blank, the displayed sample absorbance is already corrected. Do not subtract the blank again.

There is another important situation.

If the calibration equation was fitted using raw readings and includes a non-zero intercept, you must use the raw sample reading with that equation.

For example:

\[ A_{\mathrm{raw}}=mc+b \]

If the intercept already accounts for the background signal, subtracting the blank from the sample while retaining the original intercept would remove the background twice.

The key is consistency: standards and unknowns must be processed in the same way.

For a more detailed explanation of this distinction, see How to Correct Absorbance Readings for a Blank.

04Measurement limits: when should you trust the result?

A calibration range is not necessarily the same thing as a detection limit or quantification limit.

These terms describe different aspects of an analytical method.

For this illustrative colourimeter method, suppose the stated limits are:

• Detection limit: 0.20 mg/L.
• Quantification limit: 0.80 mg/L.
• Validated quantitative calibration range: 1.0-8.0 mg/L.

The detection limit is the smallest concentration that can be distinguished from the background signal under the stated conditions.

The quantification limit is the smallest concentration that can be measured with acceptable quantitative performance.

The calibration range is the interval over which the supplied calibration has been validated for quantitative measurements.

These limits are illustrative, not universal instrument specifications or official HSC numerical requirements.

Consider what each situation means.

Notice the difference between the third and fourth situations.

A sample could be above the quantification limit but still fall outside the calibration range supplied for the experiment.

Being detectable does not automatically mean that a particular calibration equation can quantify the sample reliably.

What should you do when the result is outside the range?

The next step depends on which end of the range the sample falls outside.

If the concentration is too high:

Dilute a measured aliquot of the original sample using an appropriate solvent. Measure the diluted solution again.

The aim is to bring its concentration inside the calibrated range.

If the concentration is too low:

Prepare lower-concentration calibration standards and validate a suitable lower measurement range, or use an appropriately more sensitive analytical method.

Do not simply extend the original calibration line towards zero.

The important distinction is that a dilution changes the concentration being measured, while a new calibration establishes a relationship over a different concentration range.

05From absorbance to the original sample concentration

When a sample has been diluted, there are two separate calculations.

First, determine the concentration of the solution that entered the instrument.

Then, reverse the dilution to calculate the concentration in the original sample.

The dilution factor is:

\[ DF=\frac{V_f}{V_a} \]

where:

• \(DF\) is the dilution factor.
• \(V_f\) is the final volume of the diluted solution.
• \(V_a\) is the original sample volume transferred into the flask.

Both volumes must use the same units.

The original concentration is:

\[ c_{\mathrm{original}} = c_{\mathrm{diluted}}\times DF \]

Remember that the final volume is used, not the volume of solvent added.

If 5.00 mL of sample is diluted to 20.00 mL, the dilution factor is 4.00, not 3.00.

06Questions and solutions

The following questions test more than your ability to rearrange a calibration equation. Pay attention to which measurements are actually supported by the standards, how the blank has been processed, and whether the original sample or diluted solution is being measured.

Question 1

A colourimeter is used to determine the concentration of a dissolved dye.

The instrument has not been zeroed using the blank.

The blank gives an absorbance of 0.025, and the unknown sample gives a raw absorbance of 0.205.

The blank-corrected calibration equation is:

\[ A_{\mathrm{net}}=0.0300c \]

The validated calibration range is 0.50-7.00 mg/L.

Before measurement, 4.00 mL of the original sample was diluted to a final volume of 10.00 mL.

Calculate the concentration of dye in the original sample and explain whether the calibration has been used appropriately.

Solution 1

The original dye concentration is \(15.0\ \mathrm{mg\,L^{-1}}\), and the calibration has been used appropriately because the diluted solution falls within the validated range.

First, correct the absorbance:

\[ \begin{aligned} A_{\mathrm{net}} &=0.205-0.025\\ &=0.180 \end{aligned} \]

Use the calibration equation:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.180}{0.0300}\\ &=6.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The diluted concentration is within the range of 0.50-7.00 mg/L.

Now calculate the dilution factor:

\[ DF=\frac{10.00}{4.00}=2.50 \]

Finally:

\[ \begin{aligned} c_{\mathrm{original}} &=6.00\times2.50\\ &=15.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The original concentration exceeds the calibration range, but this does not invalidate the result. The solution actually measured by the colourimeter was within the calibrated interval.

Question 2

An analytical instrument is calibrated using raw, uncorrected absorbance measurements.

The calibration equation is:

\[ A_{\mathrm{raw}}=0.0300c+0.0250 \]

where \(c\) is concentration in mg/L.

The calibration range is 0.50-6.00 mg/L.

The reagent blank has an absorbance of 0.0250. The instrument has not been zeroed using this blank.

An unknown sample gives a raw absorbance of 0.235.

A 2.00 mL aliquot is then diluted to a final volume of 3.00 mL. The diluted sample produces a raw absorbance of 0.145.

Calculate the original concentration.

Explain why subtracting the blank from the diluted reading and then using the original calibration equation would give an incorrect result.

Solution 2

The original sample concentration is \(6.00\ \mathrm{mg\,L^{-1}}\), determined using the diluted measurement.

The calibration equation includes a non-zero intercept, which already represents the background response in the raw measurements.

First, consider the original sample:

\[ \begin{aligned} c&=\frac{A_{\mathrm{raw}}-b}{m}\\ &=\frac{0.235-0.0250}{0.0300}\\ &=7.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This result exceeds the upper calibration limit of 6.00 mg/L.

It is therefore an extrapolated estimate and should not be used as the reported quantitative result.

Now consider the diluted sample.

Use its raw absorbance directly:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.145-0.0250}{0.0300}\\ &=\frac{0.120}{0.0300}\\ &=4.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This result is within the validated range.

Calculate the dilution factor:

\[ DF=\frac{3.00}{2.00}=1.50 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{original}} &=4.00\times1.50\\ &=6.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The original concentration is \(6.00\ \mathrm{mg\,L^{-1}}\).

Why shouldn’t you subtract the blank first?

Because the original calibration equation already includes the background contribution through its intercept.

If you subtract the blank and then retain the original intercept, you effectively remove the background twice.

If you wanted to use blank-corrected readings instead, you would also need to use the corresponding blank-corrected calibration equation:

\[ A_{\mathrm{net}}=0.0300c \]

The calibration and unknown must always use matching signal-processing methods.

Question 3

A laboratory method has the following stated measurement limits:

• Detection limit: 0.20 mg/L.
• Quantification limit: 0.80 mg/L.
• Validated calibration range: 1.0-8.0 mg/L.

The calibration equation is:

\[ A_{\mathrm{net}}=0.0500c \]

An unknown sample has a raw absorbance of 0.061, while its blank has an absorbance of 0.030.

A student subtracts the blank, uses the calibration equation, and obtains a concentration of 0.62 mg/L.

The student argues that the result should be reported because it is above the detection limit.

Is this conclusion justified? Explain what can actually be concluded and suggest an appropriate next step if a quantitative result is required.

Solution 3

The student is not justified in reporting 0.62 mg/L as a reliable quantitative result.

The analyte may be detected, but the estimated concentration is below the quantification limit and outside the validated calibration range.

First, correct the absorbance:

\[ \begin{aligned} A_{\mathrm{net}} &=0.061-0.030\\ &=0.031 \end{aligned} \]

Substituting into the calibration equation:

\[ \begin{aligned} c&=\frac{0.031}{0.0500}\\ &=0.62\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The mathematical result is 0.62 mg/L.

However, compare it with the stated limits:

\[ 0.20<0.62<0.80 \]

The signal is above the detection limit but below the quantification limit.

This means the analyte may be distinguished from the background, but its concentration cannot be determined with the required quantitative reliability.

There is a second problem: 0.62 mg/L is also below the lowest validated calibration concentration of 1.0 mg/L.

The student has therefore extrapolated below the calibration range.

The appropriate conclusion is that the analyte is detected but is below the quantification limit of 0.80 mg/L.

If a more precise concentration is required, the chemist should prepare lower-concentration standards and validate a suitable calibration range, or use a more sensitive analytical method.

The important distinction is that detection and reliable quantification are not the same thing.

Question 4

A chemist is analysing a concentrated solution using a calibration curve with a validated range of 1.0-8.0 mg/L.

A preliminary estimate suggests that the original sample concentration is somewhere between 18 and 28 mg/L.

The chemist wants to select a single dilution factor that will keep every possible diluted concentration within the calibration range.

(a) Determine the acceptable range of dilution factors.

(b) Select a suitable dilution factor that provides some margin from the upper calibration limit.

(c) The chemist transfers 2.00 mL of the original sample into a volumetric flask and dilutes it to 10.00 mL. The raw absorbance is 0.230, and the blank absorbance is 0.030.

Using the calibration equation

\[ A_{\mathrm{net}}=0.0400c \]

calculate the original sample concentration.

Solution 4

An acceptable dilution factor lies between 3.5 and 18.0. A fivefold dilution is a suitable choice, and the measured original concentration is \(25.0\ \mathrm{mg\,L^{-1}}\).

(a) Determine the acceptable dilution factors.

The diluted concentration is:

\[ c_{\mathrm{diluted}} = \frac{c_{\mathrm{original}}}{DF} \]

To ensure the highest possible original concentration does not exceed 8.0 mg/L:

\[ \frac{28}{DF}\leq8 \]

Therefore:

\[ DF\geq\frac{28}{8} \]

\[ DF\geq3.5 \]

Now consider the lowest possible original concentration.

To ensure it does not fall below 1.0 mg/L:

\[ \frac{18}{DF}\geq1 \]

Therefore:

\[ DF\leq18 \]

Combining both conditions:

\[ 3.5\leq DF\leq18 \]

Any dilution factor within this interval keeps all predicted concentrations inside the calibration range.

(b) Select a suitable dilution factor.

A fivefold dilution is a sensible choice.

It produces the following possible diluted concentrations:

\[ \frac{18}{5}=3.6\ \mathrm{mg\,L^{-1}} \]

\[ \frac{28}{5}=5.6\ \mathrm{mg\,L^{-1}} \]

Both values are comfortably within the calibration range.

Choosing a dilution factor of 3.5 would place the highest possible concentration exactly at 8.0 mg/L, leaving no margin for measurement variation.

A fivefold dilution gives more room.

(c) Calculate the original concentration.

First, correct the absorbance:

\[ A_{\mathrm{net}}=0.230-0.030 \]

\[ A_{\mathrm{net}}=0.200 \]

Calculate the diluted concentration:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.200}{0.0400}\\ &=5.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is within the calibration range.

Calculate the actual dilution factor:

\[ DF=\frac{10.00}{2.00}=5.00 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{original}} &=5.00\times5.00\\ &=25.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The original sample concentration is \(25.0\ \mathrm{mg\,L^{-1}}\).

This question illustrates an important experimental decision: dilution should be selected before measurement so that the resulting signal is likely to fall within the validated range.

Question 5

A laboratory is determining the concentration of a blue dye in a coloured beverage.

The beverage contains the target dye and a second coloured ingredient that also absorbs light.

The colourimeter has been zeroed using a reagent blank. Therefore, all absorbance readings below are already corrected for the reagent blank.

The calibration standards contain the target dye in a clear solvent. Their calibration equation is:

\[ A_{\mathrm{net}}=0.0400c \]

The validated calibration range is 1.0-8.0 mg/L.

The chemist makes two separate dilutions of the beverage and measures a matrix-only control. This control contains the same beverage ingredients but no target dye.

The following illustrative results are obtained:

Assume:

• The matrix contributes additional absorbance that is additive to the dye’s absorbance.
• The matrix does not change the dye’s response factor.
• The matrix background decreases proportionally during dilution.
• The sample is homogeneous, the dilution volumes are accurate, and no dye is lost during preparation.

A student ignores the matrix-only control and calculates the original concentration using each diluted beverage reading.

Both calculations give 14.0 mg/L.

The student argues that the agreement between the two dilution results proves that the concentration is reliable.

Determine the actual dye concentration using the available information. Explain why the two agreeing results are misleading, and decide whether the undiluted measurement can be used.

Solution 5

The correct original dye concentration is \(12.0\ \mathrm{mg\,L^{-1}}\), not 14.0 mg/L.

The student’s two results agree because both calculations include the absorbance contributed by the second coloured ingredient.

The key is that the reagent blank and matrix-only control remove different background contributions.

Step 1: Calculate the dye contribution in the twofold-diluted sample.

The beverage sample has a corrected absorbance of 0.280.

However, the matrix-only control contributes 0.040.

The absorbance attributable to the target dye is therefore:

\[ \begin{aligned} A_{\mathrm{dye}} &=0.280-0.040\\ &=0.240 \end{aligned} \]

Now calculate the dye concentration in the diluted solution:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.240}{0.0400}\\ &=6.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This concentration is within the calibrated range.

Reverse the twofold dilution:

\[ \begin{aligned} c_{\mathrm{original}} &=6.00\times2.00\\ &=12.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

Step 2: Check the fourfold-diluted sample.

The sample absorbance is 0.140, while the matrix-only control contributes 0.020.

Therefore:

\[ \begin{aligned} A_{\mathrm{dye}} &=0.140-0.020\\ &=0.120 \end{aligned} \]

Calculate the diluted concentration:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.120}{0.0400}\\ &=3.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is also within the calibration range.

Reverse the fourfold dilution:

\[ \begin{aligned} c_{\mathrm{original}} &=3.00\times4.00\\ &=12.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

Both corrected measurements give the same original concentration.

This supports the result of \(12.0\ \mathrm{mg\,L^{-1}}\).

Step 3: Explain why the student’s two results agree at 14.0 mg/L.

The student uses the full beverage absorbance rather than subtracting the matrix-only contribution.

For the twofold dilution:

\[ c_{\mathrm{diluted}} = \frac{0.280}{0.0400} =7.00\ \mathrm{mg\,L^{-1}} \]

Multiplying by the dilution factor:

\[ 7.00\times2.00=14.0\ \mathrm{mg\,L^{-1}} \]

For the fourfold dilution:

\[ c_{\mathrm{diluted}} = \frac{0.140}{0.0400} =3.50\ \mathrm{mg\,L^{-1}} \]

Multiplying by the dilution factor:

\[ 3.50\times4.00=14.0\ \mathrm{mg\,L^{-1}} \]

The results agree, but both include the contribution from the second coloured ingredient.

Why does this happen?

Because dilution reduces both the target dye and the interfering coloured ingredient.

When the chemist multiplies the calculated concentration by the dilution factor, the contribution from both substances is effectively restored.

The agreement therefore demonstrates consistency between the two dilution calculations, not necessarily accuracy.

This is an important limitation of using repeated dilutions as a reliability check: a systematic error can remain present in every measurement.

Step 4: Decide whether the undiluted measurement is suitable.

The undiluted sample has a corrected absorbance of 0.560.

The matrix-only background at the original concentration is:

\[ 0.040\times2=0.080 \]

This gives a matrix-corrected absorbance of:

\[ A_{\mathrm{dye}}=0.560-0.080 \]

\[ A_{\mathrm{dye}}=0.480 \]

Using the calibration equation would produce:

\[ c=\frac{0.480}{0.0400} \]

\[ c=12.0\ \mathrm{mg\,L^{-1}} \]

However, the maximum calibrated concentration is 8.0 mg/L.

The corrected undiluted absorbance is therefore still outside the validated range.

The undiluted measurement cannot be used as the quantitative result, even after correcting for the matrix contribution.

The two diluted measurements provide the defensible result because their matrix-corrected concentrations fall within the calibration range and agree after reversing their respective dilution factors.

This final example reveals an important distinction: agreement between repeated measurements does not guarantee that the correct quantity has been measured. You must also establish that the relevant background has been removed and that the calibration is appropriate for the sample.

If the matrix affected the dye’s response factor rather than simply adding absorbance, subtracting a matrix-only control would not necessarily be sufficient. A matrix-matched calibration or another suitable validation method would then be needed.

07The decision to remember

When an unknown sample produces a signal outside the calibrated range, resist the temptation to accept whatever concentration the equation returns.

Instead, work through three decisions:

1. Has the blank been handled correctly, and are the standards and unknown processed consistently?
2. Does the measured signal fall within the validated calibration range?
3. If dilution or additional background correction is required, does the corrected measurement support a reliable concentration for the original sample?

The calibration equation is only one part of the experiment. The quality of the final concentration also depends on the measurement range, sample preparation, and treatment of background signals.

Once the instrument measurement is defensible, the next question is whether the sample itself represents the larger material being investigated. A precise calibration cannot compensate for an unrepresentative sample.