This is the first of the seven-part series on pharmaceutical calculations – this time dealing with concentrations. It examines the concepts behind concentration before going on to look at how these concepts are expressed in basic calculations.

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

Pharmaceutical preparations are often composed of many and varying ingredients – or excipients – which are contained within a defined vehicle. These excipients can take various forms – some solid and some liquid, for example – and, as such, these differing concentrations need to be taken into consideration. A medicinal product needs to be defined by its concentration as it is this concentration that lets someone know how much of the drug is present. This is why understanding concentrations is important, but it’s also important to know how to convert from one type of concentration to another. This section goes some way toward understanding these ideas.

**Concentration** is the expression of the ratio of the amount of an ingredient to the amount of product. Concentration can be measured in many different ways:

- In the case of a
**solid**ingredient in a**liquid**vehicle, the ratio is expressed as**weight**in**volume**or**w/v**. - If a
**liquid**ingredient is formulated into a**solid**vehicle, the ratio is expressed as**v/w**. - If both drug and vehicle are
**liquids**, it is expressed as**v/v**. - If both the drug and the vehicle are
**solid**in nature, the ratio is expressed as**w/w.**

For ratio strengths, parts per million and percentage strengths in w/w or v/v – the same units must be used.

##### Example 1

- A ratio of 5mL to 12mL is the ratio 5:12 v/v
- A ratio of 3mg to 5mg is the ratio 3:5 w/w

For a concentration of 3mg to 5g, one of these values must be converted so the units are equal:

- 5g = 5,000mg – meaning the ratio would be 3:5,000 w/w, for example.

#### Section 1 – Amount Strengths

##### Example 2

*Your colleague gives you a preparation which contains 1,200mg of Sodium Chloride dissolved in water to produce 120mL of solution. Express the concentration of the solution as amount strength.*

**Answer**

The concentration of this solution can be expressed as an amount strength in units of mg/100mL, mg/mL, g/100mL, g/L and so on.

We can use Proportional Sets to find out what the amount strength would be in terms of mg/mL or g/mL.

Sodium Chloride: | 1,200mg | z |

Water (mL) to: | 100mL | 1 |

Z represents the number of milligrams of sodium chloride in 1ml of water.

** Why did we represent the water as ‘to 100mL’ above and not ‘in 100mL’?** This is because the sodium chloride is dissolved in water and made up to 100mL. 1,200mg of NaCl + 100mL of water produce more than 100mL of solution itself. Thus, the amount of water required is actually less than 100mL to make the 100mL of solution. Sodium Chloride essentially causes displacement but we’ll look at that in more detail later. Another factor, such as the strong concentration of some drugs such as alcohol, causes a contraction in volume when dissolved in water. Hence, in pharmacy it’s said that you ‘make up to volume’.

It’s pretty straightforward from the proportional sets above that **z=12mg**.

Thus, given that 12mg of NaCl is in 1mL of solution, we can represent this as **12mg/mL**. This is the amount strength we were looking for.

We also know that there are 1,000mg in a gram, hence we can also represent this figure at **0.012g/L**.

#### Section 2 – Ratio Strengths

Ratio Strengths is expressed as a ratio in the form of 1 in r. The corresponding fraction would have the numerator of 1. The current agreed convention would be that when a ratio strength represents a solid in a liquid, we use grams for the solid and millilitres for the liquid.

1 in 500 KMnO_{4} in water is a solid in a liquid such that its amount strength is represented as the weight in volume (w/v) ratio strength. This means that the solution contains 1g of potassium permanganate made up to 500mL with water.

##### Example 3

*2L of an aqueous solution contains 50mL of ethanol. Express this as a ratio strength.*

**Answer**

As this solution is a v/v we need to convert to the same units before we can express this as a ratio.

\[2L = 2,000mL\]

Let the volume of product in millilitres containing 1mL of ethanol be r.

Using Proportional Sets:

Ethanol (mL) – | 50mL | 1 |

Product (mL) – | 2,000mL | r |

We can easily deduce that r=40, so the ratio strength is **1 in 40v/v**

##### Example 4

*5g of product contains 250mg of sulphur in yellow soft paraffin. Express this as a ratio strength.*

Given the g and mg, this is going to be a **w/w ratio**.

As with the previous example, they must be the same unit before continuing.

250mg = 0.25g – Then we use Proportional Sets:

Sulphur (g) – | 0.25 | 1 |

Product (g) – | 5 | r |

\[\frac{5}{0.25} = \frac{r}{1}\]

\[r = 20\]

The ratio strength is **1 in 20 w/w**.

#### Section 3 – Parts Per Million

Parts Per Million (ppm) is used to denote concentrations in cases when the ratio of ingredient to product is very small. It is the equivalent to a ratio in the form of p in 1,000,000 or a fraction in which the denominator is 1,000,000.

By agreed convention:

- 1ppm weight in volume is 1g in 1,000,000mL
- 1ppm weight in weight is 1mg in 1,000,000mg/1g in 1,000,000g
- 1ppm volume in volume is 1mL in 1,000,000mL/1L in 1,000,000L

##### Example 5

*Fluoride in a water supply is expressed as parts per million w/v. Fluoride supplements should not be taken if the amount of fluoride taken in the water supply exceeds 0.7 parts per million w/v according to the BNF. Express this as a ratio in mg/L.*

By agreed convention, 0.7ppm w/v can be represented as 0.7g in 1,000,000mL.

Thus, the question required us to make the ratio mg/L.

Convert 0.7g into mg and we get 700mg.

Converting 1,000,000mL into L and we get 1,000L.

0.7ppm w/v = 700mg per 1,000L = **0.7mg/L**

Thus, we can see that ppm is the same as mg/L. These representations of the concentration of fluoride are used interchangeably.

##### Example 6

*If the concentration of fluoride is 0.25ppm w/v, how many litres would contain 1mg of fluoride?*

We can see from the above example that 0.25ppm w/v = 0.25mg/L

Using Proportional Sets, we can find that we’d need 4L to contain 1mg.

Answer: 4 Litres.

#### Section 4 – Percentage Concentration

##### Example 7

*A cream contains 12g of drug X made up to 100g with cream base. What is the percentage concentration?*

As we have 12g in 100g, it’s 12% w/w.

##### Example 8

*Express 1 in 500w/v solution of potassium permanganate as a percentage.*

Let the number of grams of potassium permanganate in 100mL of product be x. Setting up proportional sets:

Potassium Permanganate (g) | 1 | x |

Product (mL) | 500 | 100 |

We can see that we can divide 500 by 5 to give 100 and proportionally we can divide the 1 by 5 to give us a fifth or 0.2.

Hence x = 0.2 giving the percentage of potassium permanganate as 0.2% w/v.

##### Example 9

*Express 900mg of sodium chloride made up to 100mL with water as a percentage.*

To express the value as a percentage, we need to convert the number of milligrams in 100mL to grams in 100mL:

\[900mg = 0.9g\]

There is 0.9g in 100mL of solution, thus the percentage is 0.9% w/v.

##### Example 10

*A morphine sulphate injection contains 10mg/ml. What is the percentage concentration?*

As usual, to express the value as a percentage, we need to convert the number of milligrams in 1mL to grams in 100mL.

\[10mg = 0.01g\]

There is 0.01g of morphine sulphate in 1mL of solution which means there is 1g in 100mL of solution or 1% w/v.

#### Section 5 – Converting Expressions of Concentration

Let’s consider a general case – Let the amount of the ingredient be ‘a’ and the amount of product be ‘b’. Let ‘p’ be the amount in 100 parts (the percentage concentration), 1 in r to be the ratio strength and ‘m’ to be the number of parts per million.

We can set up the following proportional sets:

Amount | Percentage | Ratio Strength | ppm | |
---|---|---|---|---|

Ingredient | a | 100 | 1 | m |

Product | b | p | r | 1,000,000 |

This table shows the relationship between the different expressions of concentration. Using these relationships, it’s possible to convert from one form of expression to another.

##### Example 11

*A solution contains 20mL of ethanol in 500mL of product. Express the concentration as a ratio strength and as a percentage strength.*

Let p be the percentage strength and let the ratio strength be 1 in r.

We can set up the following proportional sets:

Ratio Strength | Percentage | Amount Strength | |
---|---|---|---|

Ingredient (g) | 1 | p | 8 |

Product (g) | 40 | 100 | 1 |

\[\frac{500}{20} = \frac{r}{1} \dots r = 25\]

If r = 25, then by ‘spotting’, we can see that p would have to be 4.

Thus, the Ratio Strength would be 1 in 25v/v.

The Percentage Strength would be 4%v/v.

##### Example 12

*A solid ingredient mixed with a solid vehicle has a ratio strength of 1 in 40. Find the percentage strength and the amount strength expressed as grams per gram.*

Let p represent the percentage strength and let ‘a’ grams be the weight of ingredient in 1g of product. We can then set up the following proportional sets:

Ratio Strength | Percentage | Amount Strength | |
---|---|---|---|

Ingredient (g) | 1 | p | a |

Product (g) | 40 | 100 | 1 |

Corresponding pairs of values are in the same ratio so…

\[\frac{1}{40} = \frac{p}{100} \dots p = 100/40\hspace{1 mm}or\hspace{1 mm}2.5\]

Corresponding pairs of values are in the same ratio so…

\[\frac{2.5}{100} = \frac{a}{1}\]

Solving for the unknown in the proportional sets and we get a = 0.025

Thus, the concentration of the mixture can be expressed as either:

- 2.5% w/w
- 0.025 g/g

##### Example 13

*A solution contains a solid dissolved in a liquid. The ratio strength is 1 in 2000w/v. What are the percentage strength and the amount concentration expressed as mg/mL?*

By convention, a ratio strength 1:2,000 w/v means 1g in 2,000mL and the percentage strength is the number of grams of ingredient in 100mL of product.

Let the percentage strength be p and the amount of solid in grams in 1mL of product be ‘a’. We can now set up the following proportional set:

Ratio | Percentage | Amount Strength (g/mL) | |
---|---|---|---|

Solid (g) | 1 | p | a |

Product (mL) | 2,000 | 100 | 1 |

Without going through the proportional sets, we can find that:

\[P = 0.05\]

We now have enough information for the 2^{nd} set from which we discover that:

\[A = 0.0005g\]

As the question asked for the amount concentration to be expressed in mg, we now convert 0.0005g into 0.5mg.

Answer:

- Concentration of 1 in 2000w/v can be expressed as 0.05%w/v or 0.5mg/mL.

##### Example 14

*A liquid ingredient mixed with another liquid vehicle has a concentration of 5% v/v. Find the ratio strength and the amount strength expressed as mL/mL.*

5% v/v can be expressed as 5mL of ingredient in 100mL of product.

Let the ratio strength be 1 in r and the amount of ingredient in millilitres in 1mL of product be a. We can set up the following proportional set.

Ratio | Percentage | Amount Strength (mL/mL) | |
---|---|---|---|

Ingredient (g) | 5 | 1 | a |

Product (g) | 100 | r | 1 |

From ‘spotting’, we can find that r=20 and ‘a’ corresponds to 0.05

The ratio is 1 in 20v/v and the amount strength in mL/mL is 0.05mL/mL.

##### Example 15

*5g of solid ingredient is added to 45g of a base. Find the percentage strength, the ratio strength and the amount strength expressed as g/g.*

Remember that for weight in weight and volume in volume the product is equal to the sum of the vehicle and the ingredient, in this case 5+45=50g of product.

Let ‘p’ be the percentage strength, 1 in ‘r’ be the ratio strength and ‘a’ grams be the amount in 1g of product. We can now set up the following proportional set: [Note: A.S = Amount Strength]

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

Ingredient (g) | 5 | p | 1 | a |

Product (g) | 50 | 100 | r | 1 |

Percentage is pretty straightforward as it’s 5/50 x 100 = 10%

\[Ratio\hspace{1 mm}is\hspace{1 mm}(50/5\times1) = 10\]

\[a = 5/50\times1 = 0.1\]

Thus, the concentration can be expressed as 10%w/w or 1 in 10w/w or 0.1g/g

This is a simple example but highlights how to use these tables even for more difficult examples.

#### Section 6 – Calculating the amount of ingredient required to make up a % solution

In the same way as converting from one expressed of concentration to another. It is also possible to use the proportional sets to calculate the amount of ingredient required to product a known amount of a known percentage product. This can be achieved by using the following proportional set:

Amount | Percentage | |
---|---|---|

Ingredient | a | p |

Product | b | 100 |

Values of p and b will be known allowing ‘a’ to be calculated.

##### Example 16

*How many milligrams of aluminium acetate are required to prepare 500mL of a 0.03% w/v solution?*

Aluminium acetate is a solid and is expressed as a weight, in this case milligrams. The vehicle is a liquid and is expressed in millilitres. By convention, 0.03% w/v means 0.03g per 100mL so each 100mL contains 0.03g of aluminium acetate.

Converting 0.03g to milligrams = 30mg.

Let ‘x’ be the number of milligrams of aluminium acetate in 500mL.

Aluminium Acetate (mg) | x | 30 |

Product (mL) | 500 | 100 |

By ‘spotting’ we can find that x = 150mg.

Alternatively, by now you can probably do it in your head logically rather than needing sets.

For example, in the solution of 100mL we discovered we have 30mg of aluminium. So obviously, if we have 500mL, then we need 5 times as much mg of aluminium making it 150mg. There’ll come a point when using sets for certain problems is actually more tedious than working it out logically.

#### Section 7 – Calculating the amount of ingredient required to prepare a ratio strength solution

When the final concentration of the product is expressed as a ratio strength, the following proportional sets can be used to calculate the amount of ingredient to produce a known amount of product.

Ingredient | a | 1 |

Product | b | r |

In this situation, r and b will be known and allows for ‘a’ to be calculated.

##### Example 17

*What is the amount of potassium permanganate in 300mL of a 1in 25 solution and what is the percentage strength of the solution?*

We need to know what ‘1 in 25’ means; and by convention it’s 1g in 25mL as we learned earlier.

There is 1g in 25mL, thus all we need to do is find a common divisor.

300mL/25mL = 12, thus we have 12g of Potassium Permanganate in solution.

In other words:

- We have 1g of Potassium Permanganate in 25mL
- We have 300mL of solution
- For every 25mL of that 300mL, we have 1g of KMnO4
- Thus, the 300mL must have 12g of Potassium Permanganate (300ml/25mL)

Alternatively, we could simply use sets, which is a formal set up of this process.

We were also asked for the percentage strength of the solution. Well, now that we know that the solution has 12g of potassium permanganate in 300mL, the percentage must be 12g/300mL or 4% w/v.