This is the second of our seven-part series on pharmaceutical calculations - this time dealing with the basic concept of dilution. It aims to cover simple dilutions, serial dilutions, mixing dilutions and triturations.

**Concentrations***Dilutions***Formulations****Calculation of Doses****Density & Displacement Values****Molecular Weights & Moles****Parenteral Solutions & Isotonicity**

#### Section 1 – Simple Dilutions

When a product is diluted there is a change in the amount of product even though the quantity of ingredient remains unchanged.

For example, if we have a solution which contains 100mg of an ingredient in 200mL of water and if this is diluted to 400mL then the strength of the product is less as the 100mg is dispersed through a greater volume.

##### Example 1

*A product consists of a solid ingredient in a solid vehicle; 2g of the solid ingredient is contained in 500g of product. A further 250g of vehicle is added to the product. Find the concentration of the diluted mixture as a percentage strength, a ratio strength and an amount strength in mg/g*.

The amount of ingredient will remain the same i.e. 2g. The total amount of product changes to 750g.

Amount | Percentage | Ratio | A.S (g/g) | |
---|---|---|---|---|

Ingredient (g) | 2 | p | 1 | a |

Product (g) | 750 | 100 | r | 1 |

By using proportional sets we find that:

- P = 0.27
- R = 375
- A = 0.0027

However the question asks us for mg thus we convert the g into mg.

Thus, the diluted product has a percentage strength of 0.27%w/w, a ratio strength of 1 in 375w/w and an amount strength of 2.7mg/g.

##### Example 2

*100mL of a 1 in 50 w/v solution is diluted to 1000mL. Find the concentration of the diluted product as a percentage strength, a ratio strength and an amount strength expressed as mg/mL.*

By convention, 1 in 50 means 1g in 50mL.

If there is 1g in 50mL, then obviously we have 2g in 100mL.

After dilution, the amount of ingredient remains the same (2g) and the total amount of vehicle becomes 1000mL. We can set up the following proportional set:

Amount | Percentage | Ratio | A.S (g/g) | |
---|---|---|---|---|

Ingredient (g) | 2 | p | 1 | a |

Product (g) | 1000 | 100 | r | 1 |

By ‘spotting’:

- P = 0.2
- R = 500
- A = 0.002

We need to convert mg to g and hence ‘a’ = 2mg.

The final solution can therefore be expressed as a percentage strength of 0.2%w/v, a ratio strength of 1 in 500w/v or an amount strength of 2mg/mL.

#### Section 2 – Serial Dilutions

Rather than keeping large amounts of products in the dispensary it is usual to keep concentrated products. These stock solutions can then be diluted to the desired concentration for the final product.

##### Example 3

*What volume of a 1 in 400v/v solution is needed to product 5L of a 1 in 2000v/v solution?*

Let y mL be the volume of 1 in 400 solution required.

The amount of the ingredient is the same in both y mL of 1 in 400 solution and 5L of 1 in 2,000 solution. Let this amount be x mL. Set up proportional sets:

For 1 in 400v/v:

Ingredient (mL) | 1 | x |

Product (mL) | 400 | y |

For 5L of 1 in 2000v/v:

Ingredient (mL) | 1 | x |

Product (mL) | 2000 | 5000 |

From the second pair, we can determine x to be 2.5

By inserting 2.5 into x for the first equation, we find y to be 1,000

So 1,000mL of 1L of the 1 in 400 mixture is diluted to 5L to product a 1 in 2000 product.

##### Example 4

*What volume of a 40% v/v solution needs to be used to product 500mL of 5% v/v solution?*

Let y mL equal the volume of the 40% v/v solution required and let x mL equal the volume of ingredient in y mL of 40% v/v solution. There is also x mL of ingredient in 500mL of the 5% v/v solution. We can therefore set up the following proportional sets:

For 40% v/v:

Ingredient (mL) | 40 | x |

Product (mL) | 100 | y |

For 500mL of 5% v/v:

Ingredient (mL) | 5 | x |

Product (mL) | 100 | 500 |

From the 2nd set, we can determine x to be 2.5. We can insert 2.5 into the first proportional set and determine y to be 62.5

Therefore, 62.5mL of the 40% v/v solution is required to produce 500mL of 5% v/v solution.

##### Example 5

*To what volume must 250mL of a 25%w/v solution be diluted to product a 10% solution?*

First calculate the amount of ingredient in 250mL of a 25% solution.

Let the number of grams of ingredient in 250mL of 25%w/v solution be x.

By convention, 20%w/v solution has 25g of ingredient in 100mL of solution.

We can use the following proportional set:

From 250mL of a 25%w/v:

Ingredient (g) | 25 | x |

Product (mL) | 100 | 250 |

We can work out x to be 62.5

After dilution, the amount of the ingredient will stay the same i.e. 62.5g.

Let y mL be the final volume of the 10%w/v solution. We can set up the following proportional set:

For 10%w/v:

Ingredient (g) | 10 | 62.5 |

Product (mL) | 100 | y |

By ‘spotting’, we can determine y to be 625mL

Thus, we need to dilute 250mL of a 25%w/v solution to 625mL to produce a 10%w/v solution.

#### Section 3 – Concentrated Waters

Concentrated waters, such as rose water, peppermint water and chloroform water, are used to produce single-strength solutions. They are intended for dilution in the ratio 1 part of concentrated water with 39 parts of water. To produce a single strength product, we take one part of the concentrate and dilute it to 40 parts of water.

Suppose that we have to make volumes of single-strength chloroform water (50mL, 100mL, 200mL, 300mL, and 500mL) from chloroform water concentrate. Since chloroform water concentrate is 1 part in 40, we have to take 1mL and dilute it to 40mL with water in order to obtain single-strength chloroform water.

Setting up the following proportional sets:

Chloroform Conc. (mL) |
1 | a | b | c | d | e |

Water (mL) to… |
40 | 50 | 100 | 200 | 300 | 500 |

For simplicity sake, we’ll try to calculate ‘c’ merely because we can easily do it from ‘Spotting’.

If there is 1mL of Chloroform conc in 40mL, then obviously we have 5mL for 200mL of water. (c = 5mL)

Thus, we need 5mL of chloroform water concentrate made up to 200mL with water to produce 200mL of single strength chloroform water. We can repeat the calculation for a, b, d and e.

In most formulae, we’ll find the chloroform expressed as a double strength for half the total volume. To form double-strength chloroform water we have to take twice the volume of the chloroform water concentrate. Hence:

and so on...

#### Section 4 – Triturations

One of the many problems when weighing ingredients for preparations is that amounts of less than 100mg cannot be weighed with sufficient accuracy. We need to use a form of calculation called a trituration to find the required amount.

##### Example 6

*How would you prepare 100mL of a preparation to the following formulae?*

Hyoscine Hydrobromide (micrograms) |
500 |

Chloroform Water (mL)to... |
5 |

Let ‘y’ be the number of milligrams of hyoscine hydrobromide in 100mL. We can now set up the following proportional set:

Hyoscine Hydrobromide (micrograms) |
500 | y |

Chloroform Water (mL) to... |
5 | 100 |

By ‘spotting’, we can find y to be 10,000 micrograms or 10mg.

This is all well and good but we can’t accurately weigh less than 100mg with sufficient accuracy. So what do we do now? Leave the lab? No.

Let the number of millilitres of product that contains 100mg of hyoscine hydrobromide be x. Now we can set up the following proportional set:

Hyoscine Hydrobromide (mg) |
10 | 100 |

Chloroform Water (mL) to… |
100 | x |

By ‘spotting’, we find x to be 1,000.

Therefore, if we can only weigh 100mg of hyoscine hydrobromide we would have to make this up to 1,000mL with chloroform water to get the required strength. It’s also important to stress that there are multiple ways to perform this calculation, but this is one of the simpler explanations for beginners.

#### Section 5 – Powder Calculations

The Pharmaceutical Codex states that powders must weigh a minimum of 120mg. No maximum weight is stated.

If the amount of drug in the powder is less than 120mg, it is necessary to include an inert powder to bulk up the powder to the minimum weight.

##### Example 7

*Prepare 5 powders each containing 100mg of Paracetamol.*

Setting up proportional sets:

Powders | 1 | 5 |
---|---|---|

Paracetamol (mg) | 100 | a |

Diluent (mg) | y | b |

Total Weight (mg) | 120 | c |

Y is the amount of diluents required to increase the final weight of one powder to 120mg, therefore:

A is the amount of paracetamol in 5 powders, b is the amount of diluents and c is the total weight of the 5 powders. The ratio of the powders is 1 to 5 so we can calculate these values by multiplying by 5:

Powders | 1 | 5 |
---|---|---|

Paracetamol (mg) | 100 | 500 |

Diluent (mg) | 20 | 100 |

Total Weight (mg) | 120 | 600 |

To prepare 5 powders each containing 100mg of paracetamol, we need to weigh 500mg of paracetamol and add it to 100mg of diluents. This mixture would then be divided into 5 powders of 120mg.

##### Example 8

*Prepare 7 powders each containing 0.37mg of drug.*

Setting up proportional sets for one and seven powders:

Powders | 1 | 7 |
---|---|---|

Drug (mg) | 0.37 | a |

Diluent (mg) | y | b |

Total Weight (mg) | 120 | c |

Therefore; 120 – 0.37 = 119.63

The figures are in the ratio 1:7 so the proportional sets become:

Powders | 1 | 7 |
---|---|---|

Drug (mg) | 0.37 | 2.59 |

Diluent (mg) | 119.63 | 837.41 |

Total Weight (mg) | 120 | 840 |

Again, we cannot weigh less than 100mg of drug.

Let ‘z’ equal the number of powders containing a total of 100mg of drug:

Powders | 1 | z |
---|---|---|

Drug (mg) | 0.37 | 100 |

Diluent (mg) | 119.63 | d |

Total Weight (mg) | 120 | e |

By simple calculation we find z to be 270.3

All we need do now is multiply the diluents and total weight by z to find values for d and e.

We would have to make enough for 270 powders and try to mix 100mg of drug with 32,336mg of diluent.

Alternatively you can try and this calculation using trituration. Your choice!

#### Section 6 – Multiple Dilutions

Now we’re equipped to move onto the next section which involves trying to calculate the amount of ingredient required in the initial product when given the final concentration and the degree of dilution.

##### Example 9

*What weight of ingredient is required to produce 1,000mL of a solution such that, when 2.5mL of it is diluted to 50mL of water, it gives a 0.25% w/v solution?*

First and foremost, consider the 0.25%w/v solution and let y be the weight of ingredient in 100mL. By convention, 0.25%w/v means 0.25g in 100mL. For 100mL of 0.25%w/v:

Ingredient (g) | 0.25 | y |

Water (mL) to | 100 | 50 |

We can see that y is 0.125

Now we take 2.5mL of the original solution and increase the volume to 50mL. The amount of ingredient will stay the same i.e. y, so we get proportional sets that relate the weight of ingredient in 50mL to the weight in 2.5mL.

Let z be the number of grams of ingredient in 1,000mL of solution:

Ingredient (g) | z | y |

Water (mL) to | 1000 | 2.5 |

By inserting y and solving for z, we get 50.

Therefore, the original solution is 50g of ingredient made up to 1,000mL with water.

##### Example 10

*What weight of malachite green oxalate is required to produce 500mL of solution such that 25mL of this solution diluted to 4000L gives a 1 in 2,000,000 solution?*

This is easy. Start by converting the litres into millilitres so that all the units are the same. This makes for 4,000,000mL.

Let the amount of malachite green oxalate in 4,000,000mL by y. We can now set up the following proportional set:

Malachite Green Oxalate (g) | 1 | y |

Water (mL) to | 2,000,000 | 4,000,000 |

By ‘spotting’, we can find y to be 2.

As we took 25mL of the original and diluted it to 4,000,000mL, we know that 25mL contains 1g of Malachite because we have only increased the volume and the amount has stayed the same.

Let z be the amount of malachite green oxalate in the 500mL of original solution. We can set up the following proportional set:

Malachite Green Oxalate (g) | z | 2 |

Water (mL) to | 500 | 25 |

By ‘spotting’, we can find z to be 40.

Therefore, 40g of malachite green oxalate is dissolved in water to produce 500mL of solution.

##### Example 11

*How many millilitres of a 1 in 80 w/v solution are required to make 500mL of a 0.02% solution?*

This probably looks familiar! By convention, 1 in 80 means 1g in 80mL and 0.02w/v means 0.02g in 100mL.

Let the number of millilitres of the 1 in 80 solution be y and let the amount of ingredient in grams in 500mL of 0.02% solution be x. The amount of ingredient in grams in y mL of 1 in 80 solution will also be x.

We can now set up the following proportional set:

For 1 in 80… | ||
---|---|---|

Ingredient (g) | 1 | x |

Product (mL) | 80 | y |

For 500mL of 0.02%... | ||
---|---|---|

Ingredient (g) | 0.02 | x |

Product (mL) | 100 | 500 |

By spotting, x is 0.1

Substituting into the first pair of proportional sets, y can be found to be 8.

Therefore, 8mL of a 1 in 80w/v solution is required to make 500mL of a 0.02% w/v solution.

#### Section 7 – Mixing Concentrations

There are situations when two or more strengths of product are mixed at stated volumes and the final concentration must be calculated.

##### Example 12

*What is the final %v/v of a solution if 200mL of 40%v/v solution is added to 300mL of 70%v/v solution?*

The two volumes will be added together to make the final volume of 500mL. The volumes of ingredients present in the two volumes will therefore need to be added together. Let x mL be the volume of ingredient in 200mL of 40%v/v solution. Setting up proportional sets:

Ingredient (mL) | 40 | x |

Product (mL) | 100 | 200 |

By ‘spotting’, we can determine x to be 80.

Let y be the number of millilitres of ingredient in 300mL of 70%v/v solution. Setting up proportional sets:

Ingredient (mL) | 70 | y |

Product (mL) | 100 | 300 |

By ‘spotting’, y is 210.

The total volume of ingredient in 500mL of final solution is therefore 290mL.

Let p be the percentage strength. Setting up proportional sets:

Ingredient (mL) | 290 | p |

Product (mL) | 500 | 100 |

By ‘spotting’, p equals 58.

Therefore, the final solution is 58% v/v.

**Note:**

Sometimes we are required to calculate the ratio in which two different products of given strengths must be mixed to produce a product of a given concentration.

Let us consider what happens when two products of different concentrations are mixed. The final volume is the sum of the two volumes and the final amount is the sum of the two amounts present in the volumes used.

For product 1 let the concentration be c1, the volume of product used be v1 and the amount of ingredient in v1 be a1. These values are c2, v2, and a2 for product 2 and c3, v3 and a3 for the final combined product.

The volume of product 1 (v1) + volume of product 2 (v2) = volume of final product (v3), i.e.

The amount of ingredient in v1 (a1) + amount of ingredient in v2 (a2) = amount of ingredient in v3 (a3), i.e.

If c1 if the concentration of product 1 (expressed as a fraction or decimal):

*Therefore:*

The derivation is included for completion, but it’s the equation above which we’ll need for the final few problems.

##### Example 13

*A suspension used to be provided as two strengths, both 100mg/5mL and 25mg/5mL solutions, which could be mixed to produce intermediate doses. If we want to produce a product that has a strength of 75mg/mL, what proportion of the two mixtures do we need?*

We therefore need 2 parts 100mg/5mL solution and 1 part 25mg/5mL solution to produce a 75mg/5mL solution.

##### Example 14

*What proportions of 90%v/v and 50%v/v ethanol mixtures would produce a 70%v/v mixture. Assume no contractions.*

Once we use the formula above V1/V2, we find the answer to be *straightforwardly* 1/1.

We therefore need to mix 1 part 90% solution with 1 part 50% solution.

GeraldJanuary 13, 2016 at 08:53Hallo, i've liked your lessons. Is it possible for me to get all the seven part lessons downloads for my private studies purpose