Alligation Methods Workbook

ALLIGATION Introduction Alligation is an arithmetical method for solving calculations involving mixtures of an active pharmaceutical ingredient (API) of different strengths. There are two types of alligation: alligation medial and alligation alternate. 1) Alligation medial may be used to determine the strength of a common API in a mixture of two or more solutions, where the amounts of that API in each solution is known. The quantities must be expressed in a common denomination, such as %w/v, whether by weight or volume. Example of the alligation medial method What is the final concentration of glucose (expressed as %w/v) if you mix together the following glucose solutions: 50 mL of 5% w/v + 100 mL of 10% w/v + 200 mL of 20 % w/v? Remember: ‘of ’ as a mathematical term means ‘multiplied by’ Totals 50 + 100 + 200 350 (mL) (mL) (mL) of (multiplied by) of (multiplied by) of (multiplied by) 5 % 10 % 20 % = = = Take the totals and divide as follows: 5250÷350 = 15. concentration! So the resulting solution contains glucose 15 %w/v 250 1000 4000 5250 + + This is the final % In future working out for alligation medial a simplified version of this table will be used, as shown below: + + Totals 50 (mL) × 100 (mL) × 200 (mL) × 350 5 % = 10 % = 20 % = 250 1000 + 4000 + 5250 2) Alligation alternate may be used to determine the proportion (or quantities) of two [or more] preparations containing different strengths of an API which are combined in order to prepare a mixture of a desired, intermediate strength. Again, the quantities must be expressed in a common denomination, such as %w/v, whether by weight or volume. Example 1 (of the alligation alternate method) Imagine the following scenario A prescription arrives in your dispensary for 30g of 6.5% w/w sulfur ointment. The only items available to you are stocks of 3% w/w sulfur ointment and 8% w/w sulfur ointment. How much of the 3% w/w sulfur ointment must be mixed with 8% w/w sulfur ointment to fulfil this prescription? The ‘C1M1 = C2M2’ method cannot be used for this calculation as there are three different concentrations involved (3%, 6.5% and 8%). Various textbooks illustrate the alligation alternate method in a number of different ways but the ‘noughts and crosses’ grid method works well for this purpose. If you have played this game you will know that you can move diagonally, horizontally and vertically when filling the grid. This principle is used in alligation alternate. Using the example above, the following steps should be taken to work out the relative proportions of 3% w/w and 8% w/w sulfur ointment that should be mixed together to produce an intermediate strength of 6.5% w/w: 1. Draw your ‘noughts and crosses’ grid 2. Working HORIZONTALLY enter the values of the two strengths being mixed together in the top corners of the grid 3 8 3. Enter the value of the final concentration required in the centre of the grid 4. Working DIAGONALLY from each top corner to the opposite bottom corner, calculate the difference between the two numerals as an absolute number (disregard any negative signs) and enter that number into the bottom corner of the grid. 6.5 5. Now you must work VERTICALLY. The 1.5 represents the number of parts of 3%w/w ointment required and the 3.5 represents the number of parts of 8%w/w ointment needed. This means that in order to produce any quantity of a preparation containing 6.5% w/w API using a combination of 3% w/w and 8% w/w ointments, they must be combined in the proportions of 1.5 parts of 3% w/w ointment and 3.5 parts of 8%w/w ointment 6. Add together the total number of parts: 1.5 (of 3%) + 3.5 (of 8%) = 5 parts total These 5 parts represent the total quantity of ointment. In the above scenario this quantity is 30 g. If 5 parts = 30 g, then 1 part = 30/5 = 6 g. If you know the value for 1 part then you can calculate the values for 1.5 parts and 3.5 parts If 1 part = 6 grams then: 1.5 parts = 1.5 x 6(g) = 9 g and 3.5 parts = 3.5 x 6 (g) = 21 g So you need to combine 9 g of 3% w/w ointment with 21 g of 8% w/w ointment to make 30 g of 6.5% w/w ointment (9 + 21 = 30) 7. You can double check that what you have done is correct as follows: Calculate the weight of sulfur in 9 g of 3% w/w ointment (0.27 g) Calculate the weight of sulfur in 21 g of 8% w/w ointment (1.68 g) Adding these quantities together there is 1.95 g of sulfur in the finished ointment. 30 g of 6.5% ointment will contain 1.95 g sulfur, confirming that the alligation calculation is correct. Alternatively you can check it using the alligation medial method! + Totals 9 (g) 21 (g) 30 × × 3 % = 8 % = 27 168 + 195 195/30 = 6.5 (%) Key points regarding both alligation medial and alligation alternate: As with other pharmaceutical calculations you must be comparing ‘like with like’ so ensure that all expressions of concentration are in the same units before using an alligation method Regarding alligation alternate: When inserting the numbers in the grid remember that the concentration of the preparation you are making must be in the centre of the grid Practice this method until you are confident about what each number in the alligation grid represents for a particular question. Once you have worked out the relative proportions needed of each strength that you are combining, you may wish to manipulate these numbers to make them easier to use in the calculation (this is not essential but can often make the maths a bit easier). Examples of manipulating proportions will be demonstrated in the workshop video associated in this workbook. Only do this if you are confident to do so. Other applications of Alligation Alternate a) Using alligation alternate when diluting a preparation with a ‘true diluent’ (which does not contain any API). Alligation alternate may also be used in calculations involving a ‘standard diluent’ (which does not contain any API) and which could also be solved using the C1M1 = C2M2 equation. Example 2 How many grams of an ointment containing 12 %w/w API should be mixed with a suitable diluent to prepare 150 g of an ointment containing 3 %w/w? Method a) Using the C1M1 = C2M2 equation (as in Workbook 1:3 from last year) If C1 is the initial strength of the ointment (12% w/w) then M1 will be the mass associated with that strength C1 = 12 (%), M1 = unknown, C2 = 3 (%), M2 = 150 (g) – so this is : 12(%) x M1 (g) = 3(%) x 150 (g) so by re-arranging the equation: M1 = 3 x 150 = 37.5 12  37.5 g of 12% w/w ointment, when combined with a suitable diluent, will produce 150 g of a 3 % w/w ointment. Method b) Using the alligation alternate method: Consider the amounts of API in the two products that you are mixing together in order to prepare an ointment containing 3 %w/w. The 12 %w/w product can be entered into one of the top corners of the alligation grid as 12 (its %w/w strength). The other product you are mixing with it is an unspecified diluent which does not contain any API as part of its formulation. Therefore it has 0 (zero) %w/w API. Enter a value of 0 in the opposite top corner of the alligation alternate grid. Then follow steps 4 and 5 as in Example 1. 0 12 3 3 Parts of 12% 9 Parts of diluent Add together the total number of parts: 3 (of 12% w/w) + 9 (of diluent) = 12 parts total These 12 parts represent the total quantity of ointment. In this particular question this quantity is 150 g. If 12 parts = 150 g, then 1 part = 150/12 = 12.5 g. If you know the value for 1 part then you can calculate the values for 3 parts and 9 parts respectively 3 parts = 3 x 12.5 g = 37.5 g of 12% w/w ointment (as expected from method a) 9 parts = 9 x 12.5 g = 112.5 g of diluent Now you have worked out the quantities required for both components of the mixture. You can check your answer by confirming that the total quantity (37.5 + 112.5 = 150 g) matches that specified in the question, providing reassurance that you have the correct answers. The advantage of using the alligation alternate method for this type of question is that you also calculate the amount of diluent required so if the question asks for ‘how many grams of a suitable diluent ……..’ then you do not have to remember the final step in the C1M1 = C2M2 method to subtract the quantity of the original ointment (that gets diluted) from the total quantity prepared. b) An alternative question in which the C1M1 = C2M2 method can be used Example 3 You have 50 g of salicylic acid ointment 2% w/w. What weight of emulsifying ointment should be added to this quantity of 2% w/w ointment to reduce the concentration of salicylic acid to 0.5% w/w? Method a) Using the C1M1 = C2M2 equation C1 = 2(%), M1 = 50 g, C2 = 0.5 (%), M2 = unknown, so By re-arranging the equation: M2 = 2 x 50 = 200 g 0.5 2(%) x 50(g) = 0.5(%) x M2 (g) THIS IS THE AMOUNT OF 0.5% OINTMENT PREPARED BY DILUTING 50 g OF 2% therefore the amount of diluent required is the difference in mass between the final amount of 0.5% w/w strength and the initial mass of the 2% w/w ointment = 200g – 50g = 150 g Method b) Using the alligation method: As in Example 2, consider the amounts of API (salicylic acid in this instance) in the two products that you are mixing together in order to prepare an ointment containing 0.5%w/w. The 2%w/w product can be entered into one of the top corners of the alligation grid as 2 (its %w/w strength). The other product you are mixing it with is emulsifying ointment (the diluent) which does not contain salicylic acid as part of its formulation. Therefore it has 0 (zero) %w/w salicylic acid. Enter a value of 0 in the opposite top corner of the alligation grid, and then follow steps 4 and 5 as in Example 1. 2 0 0.5 0.5 Parts of 2% 1.5 Parts of emulsifying ointment For this particular question we do not know the total quantity that is being prepared, but knowing that we start with 50 g of 2% w/w salicylic acid ointment, 0.5 parts is actually 50 g in this instance. Therefore if 0.5 parts = 50 g then 1 part will be 50÷0.5 = 100 gso 1.5 parts = 100 (1 part) x 1.5 = 150 g of emulsifying ointment. Once again, we can simply check the answers by adding the two constituents together (50g + 150g = 200g) – which tallies with the question. c) Using alligation alternate to calculate how much pure ingredient to add to a preparation to achieve a higher desired concentration. This method works on exactly the same principle as in Example 2 (above) but instead of using a ‘zero’ concentration diluent, the preparation is mixed with pure API (i.e. it is 100% API) rather than 0%, in order to increase the concentration of the preparation. Consider the following example: Example 4 How many mLs of pure ingredient A (the API) should you add to 100mL of a 10% v/v solution of A to increase it in strength to a 20% v/v solution? If you are adding the pure API to a solution which contains that API then you are adding 100% API to a known strength (10%v/v in this example). Therefore enter the values into the alligation alternate grid, using 10 (as it is a 10%v/v solution) and 100 (the pure API, which is a liquid as it is a v/v solution) on the top corners of the grid. Enter the value of the desired concentration in the centre of the grid (20%v/v in this example). Then follow steps 4 and 5 as in Example 1 (alligation alternate). 10 100 20 80 10 v Parts of 10% /v Parts of pure API (100%) For this particular question we do not know the total quantity that is being prepared, but we do know that we start with 100mL of 10% v/v solution, so 80 parts is actually 100 mL in this instance. Therefore if 80 parts = 100 mL then 1 part will be 100/80 = 1.25 mL so 10 parts = 1.25 (1 part) x 10 = 12.5 mL of ingredient A (the pure API) So 12.5 mL of ingredient A should be added to 100 mL of 10% v/v solution to increase the strength to 20% v/v To check that this is correct, using alligation medial: + Totals 100 (mL) 12.5 (mL) 112.5 × × 10 % = 100 % = 1000 1250 + 2250 2250 ÷ 112.5 = 20, so resulting strength is 20% v/v This method of increasing concentration will only work for % v/v or w/w preparations The following questions all require the use of the alligation methods. Make sure you follow the steps involved in each process and that you have read and understood the key points about alligation. Exercise 1 Question 1 What is the %v/v strength of alcohol in a mixture of 3 L (litres) of 40 %v/v alcohol, 1L of 60%v/v and 1 L of 70 %v/v alcohol? Answer: ____________ % v/v Question 2 How many mLs of 96% v/v alcohol should be mixed with an appropriate volume of 36% v/v alcohol to produce 175 mL of a solution containing 60% v/v alcohol? Answer: ____________ mL of 96% v/v Question 3 How many mL of Stock Solution A (containing 100 mg/5 mL) should be mixed with an appropriate volume of Stock Solution B (containing 60 mg/5 mL) to produce 600 mL of a 1.8 %w/v solution? Answer: ____________ mL of Solution A Question 4 You have a prescription from a local dermatologist requesting 100 g of a 10% w/w sulfur cream. You have a 50 g tube of 25% w/w sulfur and 100 g tube of 5% w/w in stock in the dispensary. How many grams of the 25% w/w sulfur should be added to an appropriate quantity of the 5% w/w strength to make 100 g of 10% w/w sulfur cream? Answer: ____________ g of 25% w/w Question 5 How many mg of ingredient B should be mixed with 2500 mg of an 8% w/w concentration of ingredient B to increase it in strength to 20% w/w? Answer: ____________ mg of Ingredient B