SC VI A. Pharmacopoeial Calculations

Introduction

This Chapter provides information on methods of calculating numerical limits specified in monographs of the British Pharmacopoeia. To facilitate calculations, information on Weights and Measures is included in Appendix XXIII.

The presentation of this Chapter follows the order in which the tests appear in monographs for medicinal substances and formulated preparations. The calculations will apply where techniques are common for tests and assays. Examples provided are those specified in the British Pharmacopoeia, with varying degrees of complexity.

The majority of pharmacopoeial calculations use the ratio of the measured sample response (that is, peak area, titration volume, UVabsorbance, etc) to the standard response (measured, stoichiometric or a fixed value).

Limit Tests for Anions

Limit tests for anions are usually specified in parts per million(ppm) in British Pharmacopoeia monographs unless the limit exceeds 500ppm in which case the limits are expressed in per cent weight in weight (%w/w).

The key to determining the nominal limit set in the test is to calculate the weight of the sample in the same volume as the standard solution and convert this to µg perg. It is also helpful to associate parts per million in terms of the number ofµg of the impurity being limited per g of substance being examined.

The standard solutions used as reference standards in limit tests for anions are published in Appendix I C and their concentrations are expressed inppm. These concentrations may more conveniently be considered in terms of µg per mL so that the calculations are made in the same numerical value or units and are based on the assumption that 1 g is approximately equivalent to 1mL in dilute aqueous solutions since the density of water is about 1.0 gper mL.

For example, in the Limit Test for Chlorides (AppendixVII), the standard solution contains 50 µg ofCl (10mL of 5 ppmCl). Thus, if the sample solution contains 1g of the substance being examined in the same volume as the standard solution, the limit being set is 50 µg perg, that is, 50ppm.

As an illustration of the above, in the monograph for Aluminium Glycinate, the test for Chlorides is as follows.

Dissolve 1.0 g in 10mL of 2M nitric acid and dilute to 100 mL with water. 15mL of the resulting solution complies with the limit test for chlorides, Appendix VII(330 ppm).

To calculate the limit, determine the actual weight of the substance being examined. In the above example, it is:

(15 ÷ 100) × 1.0 (the original weight taken for the test) = 0.15 g.

This gives a sample weight of 0.15g.

In order to compare like with like, 15 mL of the standard solution in the limit test for Chlorides contains 10 mL of a 5 ppm chloride solution equivalent to 50 µg of Cl. Therefore 1g of the substance being examined contains:

(1 ÷ 0.15) × 50 = 333.33µg (nominally 330ppm).

An illustration of how to calculate the limit in tests where no value is specified is given below using the test for Inorganic phosphates in the monograph for Hydrocortisone Sodium Phosphate.

Inorganic phosphate Dissolve 25 mg in 10mL of water, add 4mL of 1M sulfuric acid, 1mL of a 10%w/v solution of ammonium molybdate and 2mL of methylaminophenol-sulfite reagent and allow to stand for 15 minutes. Add sufficient water to produce 25mL and allow to stand for a further 15 minutes. The absorbance of a 4-cmlayer of the resulting solution at 730 nm, AppendixIIB, is not more than that of a 4-cmlayer of a solution prepared by treating 10 mL of a 0.0036% w/v solution of potassium dihydrogen orthophosphate in the same manner, beginning at the words ‘add 4 mL …’.

As a first step, determine the percentage content of phosphate (PO₄) in potassium dihydrogen orthophosphate (KH₂PO₄) using the atomic weight of phosphate (PO₄) (P + O₄; i.e. 30.97 + 64.00 = 94.97) and the molecular weight of KH₂PO₄(136.10). Thus:

0.0036%w/v of potassium dihydrogen orthophosphate is equivalent to 0.0025% of phosphate. Therefore 10 mL of a 0.0036% w/v solution contains 0.00025 g of phosphate.

Ensure that the weights for comparison purposes are in the same units. Therefore, convert the sample weight (25 mg) to grams. Thus:

The sample solution contains 0.025 g in 10 mL.

To calculate the percentage limit, use the values obtained above, compare the weight of the standard solution with the weight of the sample and multiply by 100. Thus:

Limit Tests for Cations

The principles for determining the limit for anions apply when determining the limit for cations. The following example is taken from the European Pharmacopoeia monograph for Bismuth Subcarbonate.

Silver To 2.0 g add 1mL of water R and 4mL of nitric acidR. Heat gently until dissolved and dilute to 11 mL with water R. Cool and add 2 mL of 1M hydrochloric acid. Allow to stand for 5 min, protected from light. Any opalescence in the solution is not more intense than that in a standard prepared at the same time in the same manner using a mixture of 10 mL of silver standard solution (5 ppm Ag)R, 1mL of nitric acid R and 2 mL of 1M hydrochloric acid (25ppm).

In the reference solution, 10mL of 5 ppmAg is specified. This is equivalent to 10 mL of a 5 ppm solution:

50 µg in 2 g is 25 µg per gram which is 25 ppm.

The limit is calculated by dividing the maximum amount of Ag permitted inmg by the weight of the sample taken in mg and multiplying this by 1 000000. Thus:

Note that, as a check, the calculated ‘weight equivalent’ of the limit in ppm should be the same as the calculated weight of impurity present in the reference solution.

Related substances

The key to determining the nominal limit set in the test is to compare the absolute concentrations (e.g. in mg/mL or %w/v) of the test solution and the limiting solution. The relative concentration may then be calculated as a percentage and this gives the limit.

Examples below are based on the various chromatographic techniques used in British Pharmacopoeia monographs.

1. The following test is included in the monograph for Choline Theophyllinate.

Carry out the method for thin-layer chromatography, Appendix III A, using silica gelHF₂₅₄ as the coating substance and a mixture of 95 volumes of chloroform and 5volumes of ethanol (96%) as the mobile phase. Apply separately to the plate 5µL of each of two solutions of the substance being examined in ethanol(96%) containing (1)1.0%w/v and (2)0.010%w/v. After removal of the plate, allow it to dry in air and examine under ultraviolet light (254 nm). Any secondary spot in the chromatogram obtained with solution(1) is not more intense than the spot in the chromatogram obtained with solution (2)(1%).

To calculate the nominal limit, compare the concentration of the limiting solution with the test solution prescribed and multiply by100 to obtain the percentage content of related substances permitted. Thus, for the above example, the calculation is as follows:

[Concentration of solution (2) ÷ Concentration of solution (1)] × 100

2. The example below is an extract from the British Pharmacopoeia monograph for Ethanolamine. The chromatographic conditions included in the monograph are not reproduced below.

Prepare a 0.1% w/v solution of 3-aminopropan-1-ol (internal standard) in dichloromethane (solution A). Carry out the method for gas chromatography, Appendix III B, using the following solutions prepared in suitable sealed reaction vials. For solution (1) prepare a solution containing 0.05% w/v of ethanolamine and 0.1% w/v each of diethanolamine and triethanolamine in solution A. To 0.5 mL of this solution add 0.5 mL of trifluoroacetic anhydride, mix and allow to stand for 10 minutes. For solution (2) prepare a 10%w/v solution of the substance being examined in solution A. To 0.5 mL of this solution add 0.5 mL of trifluoroacetic anhydride, mix and allow to stand for 10 minutes.

In the chromatogram obtained with solution (1) the peaks eluting after the solvent peak in order of emergence are due to (a) ethanolamine, (b) 3-aminopropan-1-ol, (c) diethanolamine and (d)triethanolamine. In the chromatogram obtained with solution (2) calculate the content of diethanolamine and triethanolamine by reference to the corresponding peaks in the chromatogram obtained with solution (1). Calculate the content of any other impurity by reference to the peak due to ethanolamine in the chromatogram obtained with solution (1). The content of diethanolamine and triethanolamine is not more than 1.0%w/w of each, the content of any other impurity is not more than 0.5%w/w and the sum of the contents of all the impurities is not more than 2.0% w/w.

The limits are expressed in exactly the same way as previously, by comparing the absolute concentrations of the impurities in solution (1) to the concentration of solution (2). Thus:

For ethanolamine:

For diethanolamine and triethanolamine:

However, as this method includes an internal standard, the ratios of the peak areas of the impurity and the internal standard are determined before the results are calculated.

3. The following example is taken from the British Pharmacopoeia monograph for Amantadine Capsules. The chromatographic conditions included in the monograph are not reproduced below.

Carry out the method for gas chromatography, AppendixIIIB, using 1µL or other suitable volume of the following solution. Dissolve a quantity of the contents of the capsules containing 0.1 g of Amantadine Hydrochloride in 2 mL of water, add 2mL of a 20%w/v solution of sodium hydroxide and 2mL of chloroform and shake for 10minutes. Separate the chloroform layer, dry over anhydrous sodium sulfate and filter.

The area of any secondary peak is not greater than0.3% and the sum of the areas of any secondary peaks is not greater than 1% by normalisation.

The impurity content in the above example is determined by normalisation. The percentage content of any secondary peak is calculated by determining the area of the peak as a percentage of the total area of all the peaks, excluding those due to solvents or any added reagents.

4. The following example is taken from the British Pharmacopoeia monograph for Lormetazepam. The chromatographic conditions included in the monograph are not reproduced below.

Carry out the method for liquid chromatography, Appendix III D, using five solutions in methanol (70%) containing (1)0.25%w/v of the substance being examined, (2) 0.0005%w/v of the substance being examined, (3) 0.00025%w/v of the substance being examined, (4) 0.0005% w/v of lormetazepam BPCRS and (5) 0.00025% w/v each of lormetazepam BPCRS and lorazepam BPCRS.

In the chromatogram obtained with solution (1) the area of any secondary peak is not greater than that of the principal peak in the chromatogram obtained with solution (2)(0.2%) and not more than two such peaks have an area greater than the area of the principal peak in the chromatogram obtained with solution (3)(0.1%). The sum of the areas of all such peaks is not greater than 2.5times the area of the principal peak obtained with solution (2)(0.5%).

To calculate the first limit, compare the concentrations of the two solutions specified, that is, solutions (1) and (2) and multiply by100 to obtain the percentage limit. Thus:

[Concentration of solution (2) ÷ Concentration of solution (1)] × 100%

For the second limit, compare the concentrations of the two solutions specified, that is, solution (1) and (3) and multiply by 100 to obtain the percentage limit. Thus:

[Concentration of solution (3) ÷ Concentration of solution (1)] × 100%

For the third limit, the value obtained for the first limit multiplied by 2.5 will give the sum. Thus:

Note that certain solutions specified within the test are not used to calculate the limits but are used for system suitability purposes.

Common Solvents and other Residues

The following example is taken from the British Pharmacopoeia monograph for Azapropazone.

Acetic acid Not more than 0.2%, determined by the following method. Dissolve 10 g in 25mL of methanol, add 75mL of water and carry out a potentiometric titration, AppendixVIIIB, using 0.1M sodium hydroxide VS as titrant to a pHof5.9. Each mL of 0.1M sodium hydroxide VS is equivalent to 6.005mg of acetic acid,C₂H₄O₂.

The formula to calculate the result is as follows:

V = the volume used (mL)

C = the exact molarity of sodium hydroxide VS

w = weight in mg of the substance being examined

Residue on ignition

The following example is taken from the British Pharmacopoeia monograph for Calamine.

68.0 to 74.0%, when ignited at a temperature not lower than900° until, after further ignition, two successive weighings do not differ by more than0.2% of the weight of the residue. Use 1 g.

To calculate the percentage Residue on ignition in the example above, the formula to use is as follows.

w₁ = weight in g of the ignited crucible and residue

w₂ = weight in g of the crucible

w₃ = weight in g of the crucible and sample