How to Calculate Original Concentration After Dilution

A colourimeter gives a reading for a diluted sample of cordial, but you need the dye concentration in the original bottle. Should you multiply the colourimeter reading by the dilution factor?

Not yet. The colourimeter measures absorbance, not concentration. You must first correct for the blank, use a calibration curve to find the concentration of the solution actually measured, and then reverse every dilution.

The calculation follows one direction:

Measured signal → blank correction → calibration → diluted concentration → original concentration.

If the calibration gives a concentration of \(5.00\ \mathrm{mg\,L^{-1}}\) for a sample diluted tenfold, the original concentration was \(50.0\ \mathrm{mg\,L^{-1}}\), provided no analyte was lost or chemically changed in a way the method cannot account for.

01Why measure a diluted sample at all?

Imagine trying to measure the amount of blue food colouring in a concentrated cordial.

A simple approach would be to place the original liquid in a colourimeter. The darker the liquid, the more light it absorbs, so perhaps the instrument could tell us how much dye is present.

But there is a problem.

If the cordial is too concentrated, very little light may reach the detector. The reading can fall outside the range over which the instrument has been shown to give reliable quantitative results.

A high absorbance does not automatically mean a more accurate measurement.

The solution is to dilute a measured portion of the original sample until the colourimeter can analyse it within a suitable calibration range. Then we calculate backwards to the original concentration.

Follow the sample through the apparatus

The experimental arrangement has two parts: preparing a known dilution and measuring its absorbance.

First, a volumetric pipette transfers an accurately measured volume of the original sample into a volumetric flask. Water is added until the bottom of the meniscus reaches the calibration mark. The flask is stoppered and mixed thoroughly.

A portion of this diluted solution is transferred into a clean cuvette.

Inside the colourimeter, light travels from a source through a wavelength-selecting filter, passes through the cuvette, and reaches a detector.

The detector measures the transmitted light relative to a reference and the instrument calculates absorbance.

A colourimeter works because light absorption depends on the concentration of absorbing substances.

Under suitable conditions, absorbance increases approximately linearly with concentration. This relationship is described by the Beer-Lambert law.

For a single absorbing analyte, with a fixed wavelength, optical path length, and chemical environment, we can express the calibrated relationship as:

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

where:

• \(A_{\mathrm{net}}\) is the absorbance attributable to the analyte, after the appropriate blank correction. Absorbance has no units.
• \(m\) is the gradient of the calibration line.
• \(c\) is the analyte concentration in the solution represented by the calibration, usually in \(\mathrm{mg\,L^{-1}}\) or \(\mathrm{mol\,L^{-1}}\).

The gradient depends on the analyte, wavelength, optical path length, and concentration units. It must come from the relevant calibration, not from a universal constant.

This simple linear model is useful, but it is not guaranteed for every concentration or every sample. At high concentrations, instrumental limitations, chemical interactions, or changes in the absorbing species can make the response depart from the calibrated relationship.

That is why the experiment uses standards and a stated measurement range.

02Step 1: Correct the instrument signal using a blank

A sample may produce an absorbance even when the target analyte is absent.

The solvent, added reagents, or other materials used in the procedure can contribute to the reading.

A reagent blank contains the same solvent and analytical reagents as the test solution, but no target analyte. It is processed and measured under matching conditions.

Suppose a diluted dye sample gives:

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

and its reagent blank gives:

\[ A_{\mathrm{blank}}=0.050 \]

The analyte-related absorbance is:

\[ \begin{aligned} A_{\mathrm{net}} &=A_{\mathrm{sample}}-A_{\mathrm{blank}}\\ &=0.250-0.050\\ &=0.200 \end{aligned} \]

The value \(0.200\), not \(0.250\), is the signal we use with a blank-corrected calibration.

There is an important assumption here: the blank contributes the same background signal to the standards and sample. A reagent blank will not necessarily correct for unrelated coloured substances already present in the sample.

What if the calibration already includes the blank?

This is a common source of incorrect answers.

Some questions provide a calibration equation based on raw absorbance, including a non-zero intercept:

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

Here, \(b\) is the fitted intercept, representing the modelled signal at zero analyte concentration. It is dimensionless.

Rearranging gives:

\[ c=\frac{A_{\mathrm{raw}}-b}{m} \]

If you use this raw calibration equation, do not separately subtract the blank when the intercept already accounts for that background.

Likewise, if the instrument has already been zeroed using the reagent blank, its displayed absorbance may already be blank-corrected.

The rule is to remove each background contribution once, not twice.

03Step 2: Find the concentration from the calibration

A colourimeter cannot identify a concentration from an absorbance value alone. It needs measurements from solutions of known concentration.

These are calibration standards.

Imagine preparing a series of blue dye standards, each measured in the same type of cuvette and at the same wavelength. Their concentrations are known, and they contain the same background reagents as the unknown.

The table below contains illustrative data, not results from an actual experiment.

For this illustrative method, assume the quantitative working range has been validated from \(2.00\) to \(8.00\ \mathrm{mg\,L^{-1}}\). The zero-concentration blank establishes a baseline; it does not mean concentrations approaching zero can be measured accurately.

The relationship is linear across the stated range.

Between \(2.00\) and \(6.00\ \mathrm{mg\,L^{-1}}\), the raw absorbance increases by \(0.160\) while concentration increases by \(4.00\ \mathrm{mg\,L^{-1}}\).

The gradient is therefore:

\[ \begin{aligned} m&=\frac{\Delta A}{\Delta c}\\ &=\frac{0.160}{4.00\ \mathrm{mg\,L^{-1}}}\\ &=0.0400\ \mathrm{L\,mg^{-1}} \end{aligned} \]

We now have two equivalent calibration equations:

Using raw absorbance:

\[ A_{\mathrm{raw}}=0.0400c+0.050 \]

Using blank-corrected absorbance:

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

In both equations, the numerical value of \(c\) is expressed in \(\mathrm{mg\,L^{-1}}\).

Either equation gives the same concentration when applied correctly.

For example, if the raw absorbance is \(0.250\), the corrected absorbance is \(0.200\).

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

The solution in the cuvette contains \(5.00\ \mathrm{mg}\) of dye per litre of solution.

That result is valid within the stated calibration range. But it is not necessarily the original sample concentration.

We still need the dilution history.

04Step 3: Reverse the dilution factor

Picture adding water to a small amount of cordial.

The amount of dye stays the same, but the dye spreads through a larger volume.

If you increase the solution’s volume tenfold, its concentration becomes one-tenth of its original value.

This assumes the dye is conserved, the solution is mixed evenly, and no dye is lost to the container or removed by a reaction.

The underlying relationship is:

\[ c_1V_1=c_2V_2 \]

where \(c_1\) and \(V_1\) refer to the concentration and volume before dilution, and \(c_2\) and \(V_2\) refer to those after dilution.

Rearranging:

\[ c_1=c_2\frac{V_2}{V_1} \]

The dilution factor is:

\[ F=\frac{V_{\mathrm{final}}}{V_{\mathrm{taken}}} \]

Here, \(V_{\mathrm{taken}}\) is the volume of solution transferred, and \(V_{\mathrm{final}}\) is the total volume after dilution. Both volumes must use the same units, such as mL.

So:

\[ c_{\mathrm{original}}=F\,c_{\mathrm{diluted}} \]

Notice the direction. We multiply by \(F\) when recovering the original concentration because the original was more concentrated.

A useful mental picture is a fixed handful of coloured beads spread through ten times as much water. The total bead count stays the same, but fewer beads occupy each millilitre.

Real dissolved molecules are not tiny coloured beads, and absorbance is not a direct particle count. The model also breaks if the analyte is lost or undergoes an uncontrolled chemical change during preparation.

What if there are several dilutions?

Suppose a laboratory makes two successive dilutions.

The first reduces concentration by a factor of \(F_1\), and the second by a factor of \(F_2\).

The total dilution factor is:

\[ F_{\mathrm{total}}=F_1F_2 \]

Then:

\[ c_{\mathrm{original}} =c_{\mathrm{analysed}}F_1F_2 \]

Never add dilution factors. Each dilution acts on the concentration produced by the previous one.

This is the same conservation principle used in pH calculations involving dilution and acid-base mixing, although here we must first determine concentration from an experimental signal.

05How to avoid counting the same correction twice

Consider the second worked example, where the sample’s raw absorbance was \(0.290\).

We could analyse it using either the raw or corrected calibration. Both must give the same result.

The third calculation removes the same background contribution twice. It produces a result that is too small.

The correct method depends on how the calibration was constructed.

Before calculating, inspect the vertical and horizontal axes of the calibration graph. A graph labelled raw absorbance is not the same as one labelled blank-corrected absorbance.

The horizontal axis matters just as much. Some analytical procedures develop a colour by adding reagents to standards and samples. A calibration may report concentration before those reagents were added or concentration in the final measurement mixture.

If both standards and samples undergo the same preparation, the calibration may already account for a particular dilution step.

We will test that distinction in Question 2.

06What if the reading is outside the measurement limits?

A numerical answer is not automatically a reliable concentration.

For the illustrative calibration above, the validated quantitative range was \(2.00\)-\(8.00\ \mathrm{mg\,L^{-1}}\).

Suppose your diluted sample appears to contain \(1.00\ \mathrm{mg\,L^{-1}}\) when you extrapolate the line.

That is below the stated quantitative range. You cannot confidently report \(1.00\ \mathrm{mg\,L^{-1}}\) simply because a calculator produces it.

The analyte might still be detectable, but detection and reliable quantification are different claims. A detection limit would need to be established separately.

Similarly, a result above \(8.00\ \mathrm{mg\,L^{-1}}\) cannot be justified by extending the straight line beyond the validated range.

This changes how you choose a dilution.

If the analysed sample is too concentrated, prepare a greater dilution.

If the analysed sample is below the quantifiable range, prepare a smaller dilution from the original sample, provided the method allows it.

A second important limitation is interference.

A reagent blank can remove the contribution from reagents, but it cannot necessarily remove an unrelated coloured substance already present in the sample.

Two different compounds may absorb light at the same wavelength. If both dilute by the same factor, repeating the measurement at another dilution can produce beautifully consistent results that are still chemically misleading.

A good analytical method therefore checks two different things:

• Is the signal quantitatively reliable? The measured solution must be within the method’s validated range, and preparation errors must be controlled.
• Is the signal specific to the target analyte? Background substances must be controlled, separated, or accounted for by the calibration or measurement procedure.

The following questions explore what happens when these conditions become harder to satisfy.

07Questions and solutions

Question 1

A sample containing a blue dye is analysed using a colourimeter.

A \(2.00\ \mathrm{mL}\) aliquot of the original sample is diluted to \(20.0\ \mathrm{mL}\).

The diluted sample gives a raw absorbance of \(0.310\), and the reagent blank gives \(0.070\).

The blank-corrected calibration equation is:

\[ A_{\mathrm{net}}=0.0600c \]

where \(c\) is in \(\mathrm{mg\,L^{-1}}\), and the validated range is \(1.00\)-\(8.00\ \mathrm{mg\,L^{-1}}\).

Calculate the original dye concentration.

Solution 1

The original dye concentration is \(40.0\ \mathrm{mg\,L^{-1}}\).

First, subtract the blank contribution:

\[ \begin{aligned} A_{\mathrm{net}}&=0.310-0.070\\ &=0.240 \end{aligned} \]

Use the calibration to calculate the diluted concentration:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.240}{0.0600\ \mathrm{L\,mg^{-1}}}\\ &=4.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is within the stated calibration range.

The dilution factor is:

\[ F=\frac{20.0}{2.00}=10.0 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{original}} &=(4.00)(10.0)\\ &=40.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

The tempting mistake is to multiply the absorbance \(0.310\) directly by ten. That would confuse an instrument signal with a concentration and ignore the blank and calibration.

Only the concentration obtained from the calibration should be corrected for dilution.

Question 2

An iron analysis uses a colour-forming reagent under controlled conditions. The iron produces a red complex, allowing its concentration to be determined using absorbance.

Every calibration standard is prepared by mixing \(10.0\ \mathrm{mL}\) of a known iron solution with the reagent, then making the final mixture up to \(25.0\ \mathrm{mL}\).

The calibration graph’s horizontal axis is labelled:

Iron concentration in the standard before adding the colour reagent (\(\mathrm{mg\,L^{-1}}\)).

The blank-corrected calibration is:

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

where \(c\) represents the iron concentration before reagent addition, expressed in \(\mathrm{mg\,L^{-1}}\). The graph is validated for \(2.00\)-\(12.0\ \mathrm{mg\,L^{-1}}\) on that axis.

An unknown iron-containing concentrate is prepared as follows:

• \(4.00\ \mathrm{mL}\) of original concentrate is diluted to \(100.0\ \mathrm{mL}\).
• \(10.0\ \mathrm{mL}\) of this diluted solution is mixed with the same reagent and made up to \(25.0\ \mathrm{mL}\).

The final coloured solution gives a raw absorbance of \(0.340\). The matching reagent blank gives \(0.040\).

Calculate the iron concentration in the original concentrate and the total iron mass concentration in the final coloured solution.

A student argues that the result must be multiplied by both \(100.0/4.00\) and \(25.0/10.0\) after using the calibration equation.

Explain whether that argument is correct.

Assume identical iron response in standards and unknowns, no losses, and the stated linear calibration.

Solution 2

The original concentrate contains \(250\ \mathrm{mg\,L^{-1}}\) of iron, while the final coloured solution contains \(4.00\ \mathrm{mg\,L^{-1}}\) of total iron. Multiplying by the reagent dilution factor a second time would be incorrect when using the supplied calibration.

First, calculate the net absorbance:

\[ A_{\mathrm{net}}=0.340-0.040=0.300 \]

The calibration gives:

\[ \begin{aligned} c_{\mathrm{before}} &=\frac{0.300}{0.0300\ \mathrm{L\,mg^{-1}}}\\ &=10.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This value is inside the stated calibration range.

The crucial detail is what \(c_{\mathrm{before}}\) represents.

Every standard underwent the same \(10.0\)-to-\(25.0\ \mathrm{mL}\) reagent preparation as the unknown. The calibration therefore reports the concentration in the solution before that preparation.

The calculated \(10.0\ \mathrm{mg\,L^{-1}}\) is already the iron concentration in the first \(100.0\ \mathrm{mL}\) dilution of the unknown.

We only need to reverse the earlier dilution:

\[ F=\frac{100.0}{4.00}=25.0 \]

\[ \begin{aligned} c_{\mathrm{original}} &=(10.0)(25.0)\\ &=250\ \mathrm{mg\,L^{-1}} \end{aligned} \]

To find the total iron concentration actually present in the final coloured \(25.0\ \mathrm{mL}\) mixture, we do apply the additional dilution in the forward direction:

\[ \begin{aligned} c_{\mathrm{final}} &=10.0\times\frac{10.0}{25.0}\\ &=4.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is total iron expressed as mass per unit volume, not necessarily the concentration of free iron ions after complex formation.

The student’s proposed second multiplication would give \(625\ \mathrm{mg\,L^{-1}}\), because it would reverse a reagent dilution already built into the calibration.

The two correct approaches are consistent. You can either use the supplied calibration to get the pre-reagent concentration directly, or use a calibration defined for the final coloured mixture and reverse that reagent dilution explicitly.

What you cannot do is use one concentration convention and then silently switch to the other.

Question 3

A colourimetric procedure is used to determine phosphate in an orange-coloured liquid fertiliser concentrate.

The phosphate itself contributes negligible absorbance at the selected wavelength before adding the colour reagent. The reagent produces a coloured species whose absorbance is proportional to phosphate concentration under the stated conditions.

The original concentrate is prepared by diluting \(2.00\ \mathrm{mL}\) to \(40.0\ \mathrm{mL}\).

Four \(10.0\ \mathrm{mL}\) test mixtures are then prepared. Each uses either \(5.00\ \mathrm{mL}\) of the diluted sample or \(5.00\ \mathrm{mL}\) of phosphate-free water, combined with \(5.00\ \mathrm{mL}\) of active colour reagent or matched inert diluent.

Their raw absorbances are:

Assume the sample’s existing colour and the reagent background make additive contributions, and that adding the active reagent does not change the sample’s unrelated background absorption.

The calibration relates the absorbance produced specifically by the phosphate reaction to the phosphate mass concentration in the final \(10.0\ \mathrm{mL}\) test mixture:

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

Here \(c\) is in \(\mathrm{mg\,L^{-1}}\), and the validated range is \(2.00\)-\(10.0\ \mathrm{mg\,L^{-1}}\).

Calculate the original phosphate concentration.

Explain why subtracting only the water-and-reagent blank would produce an incorrect result.

Solution 3

The original fertiliser concentrate contains \(360\ \mathrm{mg\,L^{-1}}\) of phosphate under the stated assay model.

The correct calculation isolates the additional absorbance produced by phosphate reacting with the reagent. Simply subtracting the water-and-reagent blank would leave the sample’s original colour in the measurement.

First, find the effect of adding active reagent to the sample:

\[ \begin{aligned} \Delta A_{\mathrm{sample}} &=A_{\mathrm{SR}}-A_{\mathrm{SN}}\\ &=0.480-0.160\\ &=0.320 \end{aligned} \]

But this change includes the reagent’s own background contribution.

Measure that contribution using phosphate-free water:

\[ \begin{aligned} \Delta A_{\mathrm{reagent}} &=A_{\mathrm{WR}}-A_{\mathrm{WN}}\\ &=0.070-0.020\\ &=0.050 \end{aligned} \]

The difference between these two changes isolates the reaction signal:

\[ \begin{aligned} A_{\mathrm{product}} &=0.320-0.050\\ &=0.270 \end{aligned} \]

Equivalently:

\[ A_{\mathrm{product}} =0.480-0.160-0.070+0.020 \]

The final \(+0.020\) restores the common baseline that would otherwise be subtracted twice.

Use the calibration:

\[ \begin{aligned} c_{\mathrm{final}} &=\frac{0.270}{0.0300\ \mathrm{L\,mg^{-1}}}\\ &=9.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is within the validated range.

Now reverse the preparation stages.

The final mixture contains \(5.00\ \mathrm{mL}\) of the diluted sample in \(10.0\ \mathrm{mL}\):

\[ F_2=\frac{10.0}{5.00}=2.00 \]

The first dilution factor was:

\[ F_1=\frac{40.0}{2.00}=20.0 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{original}} &=(9.00)(2.00)(20.0)\\ &=360\ \mathrm{mg\,L^{-1}} \end{aligned} \]

Why not just subtract the water-and-reagent blank?

That would give:

\[ 0.480-0.070=0.410 \]

But the sample’s unrelated colour contributes:

\[ 0.160-0.020=0.140 \]

So \(0.410\) contains both the phosphate reaction signal \(0.270\) and the sample background \(0.140\).

It would overestimate phosphate and place the apparent result beyond the stated calibration range.

The more general lesson is that a reagent blank and a sample-background control answer different experimental questions. Neither automatically replaces the other.

Question 4

A dye analysis uses the following blank-corrected calibration:

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

Here \(c\) is in \(\mathrm{mg\,L^{-1}}\), and the validated quantitative range is \(2.00\)-\(8.00\ \mathrm{mg\,L^{-1}}\).

A student first prepares the unknown by diluting \(1.00\ \mathrm{mL}\) of the original sample to \(100.0\ \mathrm{mL}\).

The raw sample absorbance is \(0.103\), and the reagent blank is \(0.053\).

Three students make different suggestions:

• Student A reports an original concentration of \(100\ \mathrm{mg\,L^{-1}}\).
• Student B says the sample contains no detectable dye.
• Student C suggests diluting the already diluted sample by another factor of five.

A fourth student instead prepares a new sample using \(5.00\ \mathrm{mL}\) of the original solution made up to \(100.0\ \mathrm{mL}\).

This new preparation gives a raw absorbance of \(0.298\), with the same reagent blank of \(0.053\).

Determine which concentration result is defensible.

Explain what the first measurement can and cannot establish, compare the concentration ranges accessible through both preparations, and assess Student C’s suggestion.

Solution 4

The defensible concentration is \(98.0\ \mathrm{mg\,L^{-1}}\), calculated from the second preparation. The first preparation is below the validated quantitative range, so its extrapolated result cannot be reported as an established concentration.

Start with the first measurement.

The net absorbance is:

\[ A_{\mathrm{net}}=0.103-0.053=0.050 \]

If we extrapolate the calibration line:

\[ \begin{aligned} c_{\mathrm{apparent}} &=\frac{0.050}{0.0500\ \mathrm{L\,mg^{-1}}}\\ &=1.00\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is below the method’s \(2.00\ \mathrm{mg\,L^{-1}}\) lower quantitative limit.

Multiplying by the dilution factor of \(100\) gives a provisional estimate of \(100\ \mathrm{mg\,L^{-1}}\), but the original measurement was outside the validated range.

Student A’s arithmetic is understandable. The problem is not the dilution calculation; it is treating an extrapolated instrument response as a reliable concentration.

Student B’s conclusion is also unjustified. A result below the quantitative range does not prove the absence of analyte. The question does not provide a detection limit, so a formal detection claim cannot be made from these numbers alone.

Now examine what concentrations the first dilution can quantify.

With \(F=100\):

\[ \begin{aligned} c_{\mathrm{original,min}}&=2.00\times100\\ &=200\ \mathrm{mg\,L^{-1}} \end{aligned} \]

and:

\[ \begin{aligned} c_{\mathrm{original,max}}&=8.00\times100\\ &=800\ \mathrm{mg\,L^{-1}} \end{aligned} \]

So the first preparation is suited to original concentrations from \(200\) to \(800\ \mathrm{mg\,L^{-1}}\), assuming the method is otherwise valid.

The provisional estimate of \(100\ \mathrm{mg\,L^{-1}}\) suggests the sample was diluted too much.

Student C’s proposed further dilution would reduce the concentration and signal even more. It moves in the wrong direction.

The fourth student’s new preparation is more suitable.

Its dilution factor is:

\[ F=\frac{100.0}{5.00}=20.0 \]

The corrected absorbance is:

\[ A_{\mathrm{net}}=0.298-0.053=0.245 \]

Therefore:

\[ \begin{aligned} c_{\mathrm{diluted}} &=\frac{0.245}{0.0500\ \mathrm{L\,mg^{-1}}}\\ &=4.90\ \mathrm{mg\,L^{-1}} \end{aligned} \]

This is inside the validated range.

Recover the original concentration:

\[ \begin{aligned} c_{\mathrm{original}} &=(4.90)(20.0)\\ &=98.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

For this second preparation, the quantifiable original concentration interval is:

\[ 2.00\times20.0=40.0\ \mathrm{mg\,L^{-1}} \]

to:

\[ 8.00\times20.0=160\ \mathrm{mg\,L^{-1}} \]

The result of \(98.0\ \mathrm{mg\,L^{-1}}\) falls comfortably within that interval.

There is another useful consistency check.

The second preparation is five times more concentrated than the first. If the linear model applies, the analyte-related absorbance should be approximately five times larger.

The first net absorbance was \(0.050\), suggesting a second net absorbance near \(0.250\).

The observed second net absorbance was \(0.245\), which is close.

We would not expect the raw absorbances themselves to have a fivefold ratio because the reagent background remains present in both measurements. Also, the first response is below the quantitative range, so this comparison is only a useful plausibility check, not a validation of the first result.

The important experimental decision is to change the sample preparation so the measured concentration falls inside the calibrated range.

Question 5

An analytical laboratory is investigating a liquid containing two dissolved dyes, X and Y.

The aim is to determine the original concentration of dye X.

Both dyes absorb at the wavelength initially selected for analysis, but they absorb different relative amounts of light at a second wavelength.

Assume the absorbance contributions of the two dyes are additive, both dyes remain chemically unchanged during dilution, and the calibrations are valid over every concentration measured.

At wavelength \(\lambda_1\), the blank-corrected calibration is:

\[ A_1=0.050c_X+0.020c_Y \]

At wavelength \(\lambda_2\), the calibration is:

\[ A_2=0.010c_X+0.040c_Y \]

In these equations, \(c_X\) and \(c_Y\) are the numerical mass concentrations of dyes X and Y in the measured solution, expressed in \(\mathrm{mg\,L^{-1}}\). The coefficients are in \(\mathrm{L\,mg^{-1}}\).

The reagent blank gives absorbances of \(0.040\) at \(\lambda_1\) and \(0.030\) at \(\lambda_2\).

Two independent dilutions of the same original liquid are prepared.

All measurements have the same optical path length, and no other substances contribute to absorbance beyond the stated blank.

(a) A student uses only wavelength \(\lambda_1\), assumes dye X is the sole absorber, and calculates the original dye X concentration from each dilution. The two results agree exactly.

Calculate that student’s result and explain whether agreement proves that dye Y is absent.

(b) Use both wavelengths to determine the true original concentrations of dyes X and Y. Check whether the second dilution supports your answer.

(c) Suppose the response coefficients at \(\lambda_2\) had been exactly twice those at \(\lambda_1\) for both dyes. Would measuring at two wavelengths still allow the individual dye concentrations to be determined? Explain what additional experimental information would be needed.

Solution 5

The original mixture contains \(40.0\ \mathrm{mg\,L^{-1}}\) of dye X and \(30.0\ \mathrm{mg\,L^{-1}}\) of dye Y. The student’s one-wavelength estimate of \(52.0\ \mathrm{mg\,L^{-1}}\) for X is too high, even though it agrees perfectly between the two dilutions.

The central issue is that dilution reduces the concentrations of both dyes by the same factor. Agreement between dilutions tests whether the combined signal scales consistently. It does not prove that only one substance produced the signal.

Part (a): Why the matching results are misleading

At wavelength \(\lambda_1\), preparation A has a corrected absorbance of:

\[ A_1=0.300-0.040=0.260 \]

If Y is incorrectly assumed absent:

\[ \begin{aligned} c_X&=\frac{0.260}{0.050\ \mathrm{L\,mg^{-1}}}\\ &=5.20\ \mathrm{mg\,L^{-1}} \end{aligned} \]

Correcting for the tenfold dilution gives:

\[ c_{X,\mathrm{original}}=5.20\times10.0 =52.0\ \mathrm{mg\,L^{-1}} \]

Now consider preparation B.

Its corrected absorbance at \(\lambda_1\) is:

\[ A_1=0.170-0.040=0.130 \]

Again assuming Y is absent:

\[ \begin{aligned} c_X&=\frac{0.130}{0.050\ \mathrm{L\,mg^{-1}}}\\ &=2.60\ \mathrm{mg\,L^{-1}} \end{aligned} \]

Correcting for the twentyfold dilution:

\[ c_{X,\mathrm{original}}=2.60\times20.0 =52.0\ \mathrm{mg\,L^{-1}} \]

The two estimates agree exactly.

But this is what we should expect if both dye concentrations halve between the preparations.

Dye Y’s contribution also halves. The colourimeter cannot distinguish its contribution from X’s using that wavelength alone.

Part (b): Use both wavelengths to separate the contributions

Use preparation A, which has a tenfold dilution.

At \(\lambda_1\):

\[ A_1=0.300-0.040=0.260 \]

At \(\lambda_2\):

\[ A_2=0.190-0.030=0.160 \]

Substitute into the two calibration equations:

\[ 0.050c_X+0.020c_Y=0.260 \]

\[ 0.010c_X+0.040c_Y=0.160 \]

Dividing both equations by \(0.010\) gives a simpler simultaneous system:

\[ 5c_X+2c_Y=26 \]

\[ c_X+4c_Y=16 \]

Here the concentration symbols represent their numerical values in \(\mathrm{mg\,L^{-1}}\).

From the second equation:

\[ c_X=16-4c_Y \]

Substitute into the first:

\[ \begin{aligned} 5(16-4c_Y)+2c_Y&=26\\ 80-18c_Y&=26\\ 18c_Y&=54\\ c_Y&=3.00 \end{aligned} \]

Therefore:

\[ \begin{aligned} c_X&=16-4(3.00)\\ &=4.00 \end{aligned} \]

These are the concentrations in preparation A:

\[ c_X=4.00\ \mathrm{mg\,L^{-1}} \]

\[ c_Y=3.00\ \mathrm{mg\,L^{-1}} \]

Because preparation A was diluted tenfold:

\[ \begin{aligned} c_{X,\mathrm{original}} &=4.00\times10.0\\ &=40.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

and:

\[ \begin{aligned} c_{Y,\mathrm{original}} &=3.00\times10.0\\ &=30.0\ \mathrm{mg\,L^{-1}} \end{aligned} \]

Now test the answer against preparation B.

Its dilution factor is twenty, so the expected concentrations are:

\[ c_X=\frac{40.0}{20.0}=2.00\ \mathrm{mg\,L^{-1}} \]

\[ c_Y=\frac{30.0}{20.0}=1.50\ \mathrm{mg\,L^{-1}} \]

Predict the corrected absorbance at \(\lambda_1\):

\[ \begin{aligned} A_1&=0.050(2.00)+0.020(1.50)\\ &=0.100+0.030\\ &=0.130 \end{aligned} \]

This matches the measured corrected absorbance:

\[ 0.170-0.040=0.130 \]

At \(\lambda_2\):

\[ \begin{aligned} A_2&=0.010(2.00)+0.040(1.50)\\ &=0.020+0.060\\ &=0.080 \end{aligned} \]

Again, the prediction matches:

\[ 0.110-0.030=0.080 \]

Both preparations are therefore consistent with the same two-dye composition.

The two-wavelength method succeeds because the dyes have different relative responses at the two wavelengths. The second wavelength provides information that the first cannot.

Part (c): When even two wavelengths are not enough

If the response coefficients at \(\lambda_2\) were exactly twice those at \(\lambda_1\) for both dyes, the second measurement would not provide an independent equation.

For example, the two calibrations would have the form:

\[ A_1=0.050c_X+0.020c_Y \]

\[ A_2=0.100c_X+0.040c_Y \]

The second equation is simply twice the first:

\[ A_2=2A_1 \]

Every possible pair of concentrations that fits the first equation would also fit the second.

There would be infinitely many possible combinations of X and Y consistent with the measurements, rather than one unique answer.

Repeating the analysis at more dilution factors would not solve this. Both substances would keep changing by the same proportion.

To separate the concentrations, the laboratory would need another independent measurement. For example, it could use a wavelength where the two dyes have a different response ratio, apply a selective chemical reaction, or physically separate the substances before measuring them.

This is the deeper limitation of quantitative analysis: a consistent signal is not necessarily a selective signal.

The original sample concentration can only be recovered reliably when the calibration genuinely measures the intended analyte and the dilution history is known.

08Choosing the next analytical step

Once you have calculated the original concentration, decide whether the result is actually supported by the experiment.

Check that the analysed solution was within the validated quantitative range, that every dilution and reagent-volume convention was handled correctly, and that the blank and calibration dealt adequately with background signals.

If those conditions are satisfied, the concentration can be used to calculate the amount of analyte in a larger sample using \(n=cV\) for molar concentration or \(m=cV\) for mass concentration, with consistent units.

If the instrument is measuring more than one substance, improving the dilution calculation will not fix that selectivity problem. You need a better chemical measurement.

The same distinction appears when the signal is a titration volume instead of absorbance. In precipitation titrations, the titre is converted into analyte amount using a reaction ratio before the original sample concentration is recovered.

The experimental signal changes, but the essential reasoning stays the same: establish what the measurement represents, calculate the concentration of the material actually analysed, and only then work backwards to the original sample.