Dana Zimmer, Rhena Schumann

Very high analyte concentrations can be above the calibration line or even the linear range of the calibration line. If the element concentrations are higher than the linear range of the method, the extract has to be diluted. If the element concentration is "only" above the highest standard of the calibration line but within the linear range of the method, it has to be decided case by case. For element determination by ICP-OES, exceedances of the highest standard by a factor of four or five are normally unproblematic, as long as the concentrations are in the linear range. For higher exceedances it has to be decided if either an additional higher standard is set and the calibration line is extended or if the samples should be diluted. The setting of an extended calibration line is normally useful if a higher percentage of the samples is higher than the highest standard. A dilution of samples is normally useful if only some samples are higher. This decision should be made together with the technical assistant of the ICP-OES.

If samples have to be diluted, it has to be considered that it has to be differentiated between dilution and mixtures for specification of dilution factor and converting in "real" element concentrations.

► Definition of mixture:

► If the specification of a dilution is referred to the single components of a solution, this has to be called a mixture and the so-called dilution factor for calculation of the "real" concentration is determined by addition of single components.

Example for the calculation of the dilution factor in a mixture:

It is necessary to dilute a sample for P determination. For this reason, a sample solution is mixed with ultra-pure water (RW) as followed: to 1 part sample 9 parts of water are added, that means 1 part sample + 9 parts water. This is called a mixture with 1 + 9. If both is added (1 + 9), the dilution factor is 10. With this dilution factor all measured element concentrations of the samples have to be multiplied to get the "real" element concentration in the sample.

If the sample solution is not only diluted by RW but also other solutions, all parts have to be added analogously. For example: 1 part of sample is mixed with 3 parts of a solution X and 5 parts of a solution Y, so the dilution factor is calculated as follows by addition: 1 + 3 + 5. This means a dilution factor of 9; therefore, the measurement value has to be multiplied by 9.  

► Definition of a dilution:

► A dilution in the narrow sense of the word means the ratio of the initial concentration to the end concentration. It is related to the percentage of a component in the total solution.

Example for the calculation of the dilution factor in a dilution:

At a dilution of 1 : 10, the 1 is the proportion in the total solution (10). The dilution factor for calculation of concentration is therefore 10 and not additive as in a mixture (1 + 10 = 11). That means, for example, in 10 ml of the total solution is 1 ml of the component. The tenth of the total solution is the sample. As a mixture it would be specified as 1 + 9.

Calculation of the dilutions of calibration standards for calibration lines

The calibration standards for a calibration line are made from so-called P stock solution by dilution. For calculation of necessary volumes of the P stock solution (in ml) the concentration of the P stock solution (in mg P L^-1), the end concentration of the calibration standards (in mg P L^-1) and the necessary volume of the calibration standards (in ml) is necessary.

given:
X mg P per litre in the P stock solution           e.g. 1 g P per litre

searched:
Necessary ml of stock solution for 100 ml calibration standard for y mg P per litre    e.g. 2 mg P; 4 mg P and 6 mg P per litre

solution steps (with examples):
1. question: How many mg of P are in 1 ml of P stock solution?
For this calculation 1 litre is presented as 1000 ml and 1 g P as 1000 mg P for uniform units. This question can be solved by a ratio equation.
_{Abbreviations: Stock solution= stock sol.        volume = vol}

Converting to x mg of P means: 1 mg P per 1 ml stock solution

2. question: How many mg of P have to be in 100 ml calibration standard at a given concentration? This question can be solved by a ratio equation.

Converting to y

That means 0.2 mg P are necessary for 100 ml of the standard.

3. question: If I have x mg P in 1 ml of stock solution and y mg P are necessary in the 100 ml standard, how many ml of stock solution are necessary?

Converting to z means that 0.2 ml stock solution are necessary to set 100 ml standard with a concentration of 2 mg P per litre from a stock solution with 1 g P per litre.

Formulas can be brought together and so the calculation can be shortened.

For x mg P per 1 ml in formula 3 the stock solution concentration in mg P per 1000 ml (1 litre) of formula 1 can be inserted

That means:

Converting to z means:

For y mg P in the standard, formula 2, which is converted to y, is inserted, that means:

Both “per 1000 ml” can be cancelled, that means:

Thus:

For the calculation of the necessary volume of the stock solution the necessary standard volume (in ml) is multiplied with the searched P concentration in the standard (in mg P per litre) and divided by P concentration in the stock solution (in mg P per litre).

For our example this means:

For both of the other standards 0.4 and 0.6 ml of the stock solution are necessary. After the pipetting of the stock solution in the volumetric flask the standards are filled by the necessary solution. It has to be paid attention that the stock solution has room temperature for the setting of standards, since the temperature changes the density and therefore, the volumes would be pipetted but not the necessary mass of the P!

For the dilution of acids or other chemicals the procedure is generally the same. Room temperature of the used chemicals is a prerequisite as well. Additionally, it has to be considered that first the water and subsequently the acid or base is filled into the flask! This means: after calculation of the necessary acid or base volume, the necessary water volume has to be calculated. A little bit less than the calculated water volume is filled into the volumetric flask. For a 1 litre volumetric flask that means for example 50 to 100 ml less water. After the water is filled the acid or base is added carefully. After cooling down of the solution the water is filled to the calibration mark. On bottles with concentrated acid the concentration is specified in % as well as in mol per litre. Therefore, it can be calculated for searched concentrations in % as well as in mol per litre.

The following calculation is possible:

General formula for dilution of solutions:

V_A = searched volume of initial solution (e.g. in ml)
V_B = necessary volume of diluted solution (e.g. in ml)
C_A = concentration of initial solution (e.g. in mg element per litre)
C_B = concentration of diluted solution (e.g. in mg element per litre)

Example for dilution of an acid:
Initial acid conc. HCl: 37 % or 12 M per litre
searched: 1 litre 10 % HCl or 1 M HCl

solution for 10 % HCl:

For a 1 litre volumetric flask 270.27 ml conc. HCl are necessary. First, 600 to 700 ml water are filled into the volumetric flask. Subsequently, 270.72 ml of conc. HCl are added. After cooling down it has to be filled with water to the calibration mark.

Solution for 1 M HCl:

For a 1 litre volumetric flask 83.33 ml conc. HCl are necessary. First, 750 to 850 ml water are filled into the volumetric flask. Subsequently, 83.33 ml of conc. HCl are added. After cooling down it has to be filled with water to the calibration mark.

Detailed example: 1 litre of a 10 % HCl solution has to be made from conc. HCl (37 %). How many ml of the conc. HCl have to be pipetted into a 1 litre volumetric flask?

Given: 37 % HCl                              searched: 1 litre of 10 % HCl

Shortened steps of solution analogous to setting of calibration standards:

► In a 37 % HCl are 37 ml HCl ions in 100 ml of the solution. That means that in 1 ml of the conc. HCl solution are 0.37 ml of HCl ions.
► If I want to make a 10 % HCl, there have to be 100 ml of HCl ions in 1 litre solution.

Question: How many ml of the conc. HCl are necessary to add 100 ml HCl ions into a 1 litre volumetric flask? This can be calculated by a ratio equation:

Conversion to search x ml of the conc. HCl means:

Therefore, 270.27 ml of conc. HCl are necessary and have to be given into the 1 litre volumetric flask and filled with water to get a 10 % HCl. However, first the water has to be added and subsequently the acid. First, estimate how many ml of water are approximately necessary to fill to 1 litre. 1000 ml necessary solution minus around 271 ml acid means 729 ml water. Fill 600 to 700 ml water and then the acid into the flask. After cooling down fill with water to calibration mark.

For citation: Zimmer D, Schumann R (year of download) Chapter 7.1 Dilution and Mixtures (Version 1.0) in Zimmer D, Baumann K, Berthold M, Schumann R: Handbook on the Selection of Methods for Digestion and Determination of Total Phosphorus in Environmental Samples. DOI: 10.12754/misc-2020-0001

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Dana Zimmer, Rhena Schumann

At the ICP-OES, the P concentration per litre in the sample solution is automatically calculated from the measured element-typical light intensity with the help of the previously measured and automatically calculated calibration line. Regarding the molybdenum blue method concentrations of P per litre solutions are calculated as well. Normally, concentrations of P are presented in µg or mg per litre. In the following example for calculation the P concentration in mg l^-1 is converted into mg kg^-1. For all other units the procedure is analogous. In any case the units of the weigh-in and of the extract volume also have to be considered. Before converting the P concentration of sample per litre in P concentration per kg, the blanks have to be subtracted. The extract volume is either the added extracting agent volume (if not filled to defined volume after extraction) or that volume, to which the extract was filled after extraction. It is not the subvolume, which is taken for measurement of the analyte.

List of given and searched variables

Example

given:

digestion of a soil sample with aqua regia
weigh-in: 0.500 g
Extraction agent: 6 ml conc. HCl + 2 ml conc. HNO₃
Extract volume (filled after digestion): 100 ml
Concentration of analyte in mg P l^-1            e.g. 2.635 mg P L^-1 

searched:

concentration of analyte in mg P per kg material

Approach to find the equation

1. question: Which amount of the analyte is in the extract volume?

This is calculated by a ratio equation from concentration of the analyte in the extract and the extract volume. Since the extract volume is given in ml, the concentration is also calculated in ml.

given:

concentration: x mg P per 10³ ml   e.g. 2,635 mg P 10^-3 ml^-1
extract volume: a ml                           100 ml

searched:

mg P per extract volume: y₁ 

ratio equation:

Conversion of equation to y₁ mg means:

In our example 0.2635 mg P are in 100 ml. Since the complete P of the weigh-in material is in this extract volume (100 ml), the P in the extract volume is equivalent to the P of the weigh-in. Therefore, 0.2635 mg P are in 0.5 g soil.

2. question: If y₁ mg P are in the weigh-in of the materials, how many mg P are in 1 kg of the material?

This is calculated by a ratio equation as well. It has to considered that the weigh-in is in g and the mass reference of the material is in kg. For this reason, it is calculated with 103 g (instead of 1 kg) for the material.

given:

        y₁ mg P in b g weigh-in               e.g. 0.2635 mg P in 0.5 g soil

searched:

        y₂ mg P in 10³ g material

ratio equation:

Converting the equation to y₂ mg means:

In our example the soil has a P concentration of 527 mg kg^-1.

To calculate not always stepwise, the formula converted to y₁ can be inserted for y1 in the formula converted to y2.

The formula   

is inserted in the formula  

instead of y₁. This results in the following formula:

Both 10³ cancel each other. If P concentration is measured in (mg) per litre, the weigh-in is given in g and the extract volume in ml, the P concentration in the environmental sample in (mg) per kg material can be calculated by the following formula:

This means for our example:

If the extract was diluted, the dilution factor (DF) has to be considered. This factor has to be inserted as multiplier in the formula, since without dilution the concentration in the extract would have been higher.

If in our example one part of the soil extract would have been diluted     (= mixed) with one part of water, the DF would be 2. If one part of the soil extract would have been diluted (= mixed) with two parts of water, the DF would be 3.
The P concentration of 527 mg P per kg soil would be therefore 1054 mg P with DF 2 and for DF 3 1581 mg per kg soil. More information about dilution and mixture can be found in the chapter 7.1 Dilution and mixtures.

In an excel sheet the order of columns could be as follows. If further elements were determined by ICP-OES, the columns for other elements have to be added.

For citation: Zimmer D, Schumann R (year of download) Chapter 7.2 Conversion of Measurement Values (per Litre) into the Concentration in the Environmental Sample (per kg) (Version 1.0) in Zimmer D, Baumann K, Berthold M, Schumann R: Handbook on the Selection of Methods for Digestion and Determination of Total Phosphorus in Environmental Samples. DOI: 10.12754/misc-2020-0001

download chapter (PDF)