Pharmaceutical Calculations (5)

1. Pharmaceutical Calculations (5) Dilutions Phil Rowe Liverpool School of Pharmacy

2. Dilutions When we dilute a solution, we can inter-relate 4 things: C1 The initial concentration (i.e. prior to dilution)C2 The final concentration (i.e. after dilution) V1 The initial volume of solution which is to be dilutedV2 The final volume of solution after dilution The equation linking these is: C1.V1 = C2.V2

3. Dilution factor When we use the equation, it is usually re-arranged with one term on the left and three on the right. Two of the terms on the right hand side will form a dilution factor. e.g. If we want to calculate C2, knowing that 10mL of a 20mg/mL solution has been diluted to 200mL, we would re-arrange the equation to: C2 = C1 x V1 V2 C2 = 20mg/mL x 10mL 200mL C2 = 20mg/mL x 0.05 = 1mg/mL The ratio of the two volumes on the right hand side indicates the extent of dilution. In this case it is 10:200, indicating a 20 fold dilution. As they have to form a ratio, it is essential that the two volumes are in the same units. There is no restriction on the units of C1. Dilution factor of 1/20 = 0.05

4. Re-arrangements of C1.V1 = C2.V2 This equation can be re-arranged to allow us to calculate any one of the terms, so long as we know the values for the other three. C1 = C2 x V2 V1 C2 = C1 x V1 V2 V1 = V2 xC2 C1 V2 = V1 x C1 C2 Dilution factors. Units of the two volumes or concentrations must match.

5. C1 = C2 x V2 / V1 Example question that would require equation above: What concentration of stock solution should we prepare, so that we can take 10mL of that stock and dilute it to 2.5L and the result will be a 200mg/L solution? Note that we are provided with values for all the terms on the right hand side (C2 ,V2 & V1) and so we can calculate an appropriate initial concentration (C1) C1 = C2 x V2 / V1 = 200mg/L x 2.5L / 10ml (Use mL as common unit for dilution) = 200mg/L x 2,500mL / 10mL = 200mg/L x 250 = 50,000mg/L = 50g/L

6. C2 = C1 x V1 / V2 Example question that would require equation above: If we diluted 50ml of 0.45M phenol to a final volume of 15L, what would be the resultant concentration? C2 = C1 x V1 / V2 = 0.45M x 50ml / 15L (Use mL as common unit for dilution) = 0.45M x 50mL / 15,000mL = 0.45M x 0.00333 = 0.0015M = 1.5mM

7. V1 = V2 x C2 / C1 Example question that would require equation above: If we want to dilute a solution of sodium dodecyl sulphate (5g/L) to produce 2L of a 50mg/100mL solution, what volume of the original solution should we use? V1 = V2 x C2 / C1 = 2L x 50mg/100mL / 5g/L (Use mg/L as common unit for dilution) = 2L x 500mg/L / 5,000mg/L = 2L x 0.1 = 0.2L = 200mL

8. V2 = V1 x C1 / C2 Example question that would require equation above: If we are starting with 50ml of a solution contaminated with 12ppm of aluminium, to what volume would we need to dilute it, to reduce the aluminium concentration to 0.1ppm? (Assume the diluent is aluminium free.) V2 = V1 x C1 / C2 = 50ml x 12ppm / 0.1ppm (Both parts of dilution= 50mL x 120 already in same units) = 6,000ml = 6L

9. Caution!!!What exactly is V2? Some dilution questions can be deceptive. For example: If we diluted 1L of sulphuric acid (1M) with 5L of water, what is the resultant concentration? C2 is the unknown and It is all too easy to think that we know the values of the other 3 parameters to be: C1 = 1MV1 = 1LV2 = 5L No!!! V2 Is supposed to be the total final volume. 5L is the volume of diluent, but the total final volume is 6L and this is the true value of V2

10. Assumption that volumes are additive In order to be able to calculate the concentration in the previous slide, we would have to assume that mixing 1L of dilute sulphuric acid and 5L of water will result in a total volume of 6L. This assumption does not necessarily hold for all liquids, but if all the materials are reasonably dilute aqueous solutions, there will be no significant error.

11. Being aware that the volume of diluent, is NOT V2 (Example 1) What concentration will arise if we mix together the following:- 250ml carbamazepine solution (1g/L)750ml phosphate buffer (pH 7.4) C1 = 1g/LV1 = 50mlV2 = 250ml + 750ml = 1,000ml C2 = C1 x V1 / V2 = 1g/L x 250ml / 1,000ml = 1g/L x 0.25 = 0.25g/L

12. Being aware that the volume of diluent, is NOT V2 (Example 2) We start with 2L of 0.2M sodium carbonate and we need to dilute it to 0.01M. How much water should we add? C1 = 0.2MC2 = 0.01MV1 = 2L V2 = V1 x C1 / C2 = 2L x 0.2M / 0.01M = 2L x 20 = 40L Beware: The figure of 40L is the final volume required. That does not mean that we should dilute with 40L of water. What we should do is to add 38L of water to bring the final total volume up to 40L

13. Examples of dilution calculations Question 1 What concentration would arise if 350ml of potassium chloride solution (1.8g/100ml) was diluted to a volume of 7L?

14. Examples of dilution calculations Question 2 We want to be able to take 500ml of a stock solution of hydrochloric acid and dilute it by the addition of 3L of water, producing a final concentration of 0.1M. What should be the strength of the stock solution?

15. Examples of dilution calculations Question 3 What volume of a 400ppm solution of fluoride should be diluted to 200L to produce a 0.8ppm solution?

16. Examples of dilution calculations Question 4 If we start with 100ml of a 5mM solution of human serum albumin and need to dilute this to 12.5microM, to what volume should it be diluted?

17. Examples of dilution calculations Question 5 How much water should we add to 2ml of theophylline solution (60mg per 5ml) to produce a concentration of 2mg/ml?

18. Examples of dilution calculations Question 6 We want to be able to take 200microlitres of ethanol solution and dilute it to 4ml thereby achieving a concentration of 0.5%(v/v). What should be the strength of the original solution?

19. Examples of dilution calculations Answer 1 What concentration would arise if 350ml of potassium chloride solution (1.8g/100ml) was diluted to a volume of 7L? (Answer in units of g/L) C2 = C1 x V1 / V2 = 1.8g/100ml x 350ml / 7L (Use mL as common units = 1.8g/100ml x 350ml / 7,000mL for dilution) = 1.8g/100mL x 0.05= 0.9g/L

20. Examples of dilution calculations Answer 2 We want to be able to take 500ml of a stock solution of hydrochloric acid and dilute it by the addition of 3L of water, producing a final concentration of 0.1M. What should be the strength of the stock solution? Caution: Final volume produced (V2) is 3.5L (Not 3L). C1 = C2 X V2 / V1 = 0.1M x 3.5L / 500ml (Use L as common units for dilution) = 0.1M x 3.5L / 0.5L = 0.1M x 7 = 0.7M

21. Examples of dilution calculations Answer 3 What volume of a 400ppm solution of fluoride should be diluted to 200L to produce a 0.8ppm solution? V1 = V2 x C2 / C1 = 200L x 0.8ppm / 400ppm (Both concs already in same units) = 200L x 0.002 = 0.4L = 400mL

22. Examples of dilution calculations Answer 4 If we start with 100ml of a 5mM solution of human serum albumin and need to dilute this to 12.5 microM, to what volume should it be diluted? V2 = V1 x C1 / C2 = 100ml x 5mM / 12.5microM (Concentration units need to match) = 100ml x 5,000microM/12.5microM = 100mL x 400 = 40,000ml = 40L

23. Examples of dilution calculations Answer 5 How much water should we add to 2mL of theophylline solution (60mg per 5mL) to produce a concentration of 2mg/mL? V2 = V1 x C1 / C2 = 2mL x 60mg/5mL / 2mg/mL (Use mg/mL as common units) = 2mL x 12mg/mL / 2mg/mL = 2ml x 6 = 12mL BUT this is the final volume and we were asked for the volume of diluent. Diluent vol = 12ml – 2ml = 10ml

24. Examples of dilution calculations Answer 6 We want to be able to take 200microlitres of ethanol solution and dilute it to 4ml thereby achieving a concentration of 0.5%(v/v). What should be the strength of the original solution? C1 = C2 X V2 / V1 = 0.5% x 4ml / 200microlitres(Use mL as common units for dilution) = 0.5% x 4mL / 0.2mL = 0.5% x 20 = 10%(v/v)

25. What you should be able to do • Calculate … • An initial concentration (C1), so that when a volume (V1) is diluted to a volume (V2) it will yield a concentration (C2). • A volume (V1) of solution of concentration (C1) that when diluted to volume (V2) will yield concentration (C2). • The concentration (C2) that will arise if volume (V1) of a solution of concentration (C1) is diluted to volume (V2). • The volume (V2) to which volume (V1) of a solution of concentration (C1) must be diluted in order to yield concentration (C2) • Deal properly with those cases that either provide, or ask you to calculate, the volume of diluent rather than the final total volume (V2).