# 17 Liquid & Solid Dilutions

Most students cringe when they think about liquid & solid dilutions. That’s because most text books make this unnecessarily complicated and frustrating. Let’s take that headache away by understanding some simple concepts.

Firstly, we can NOT change the strength, or concentration, of a preparation by simply adding more of it. Instead, we have to add some type of diluent, solvent or inactive ingredient to accomplish this (such as distilled/sterile water, lanolin, etc.). Secondly, adding this diluent results in a liquid solution or solid medication with a lower percent strength. From our last module, we learned that percent strength describes the amount of drug (in grams or mL), in exactly 100mL of solution — or similarly, grams of drug per 100g of solid. Always remember that this amount of drug is part of the total volume of solution/solid, NOT in addition to it. Lastly, these are DILUTION problems. This means we’re ADDING diluent/solvent, NOT SUBTRACTING it (at least not when it comes to the exam). In a future module, we’ll learn about a similar concept called, alligations. The formulas we’ll use for dilutions are:

1. Initial Volume · Initial Strength = Final Volume · Final Strength

$$\text{V}_1 \times \text{S}_1 = \text{V}_2 \times \text{S}_2$$

$$\frac{\text{V1}}{\text{S2}} = \frac{\text{V2}}{\text{S1}}$$

2. Initial Quantity · Initial Strength = Final Quantity · Final Strength

$$\text{Q}_1 \times \text{S}_1 = \text{Q}_2 \times \text{S}_2$$

$$\frac{\text{Q1}}{\text{S2}} = \frac{\text{Q2}}{\text{S1}}$$

The volumes/quantities need to have the same units: L, mL, g, mg, etc. Furthermore, when using these dilution formulas, we’re allowed to input the percent strengths as whole numbers. We don’t have to convert the percent strengths to decimal fractions. Now Let’s have a deeper look with some real examples.

Let’s say we have a 1.2L (1,200mL) bottle of 23% calcium gluconate solution and are out of stock of 10% calcium gluconate solution because GASP — the delivery van forgot to bring it! Well, we may be ticked off but you know what? We can fix this problem by doing some extemporaneous compounding! Assuming we use the entire bottle for the initial volume (like a boss), then we can figure out how much diluent we need to add by setting up the formula with the following variables:

$$\text{1,200 mL} \times \text{23%} = \text{(x) mL} \times \text{10%} \;\text{;}$$

$$\frac{\text{1,200 mL} \times \text{23%}}{\text{10%}} = \text{(x) mL} \;\text{;}$$

$$\frac{\text{1,200 mL} \times \text{20%}}{\text{10%}}$$

$$= \text{2,760 mL 10% calcium gluconate}$$

But wait, we’re not done yet! Since 2,760mL is the final volume this means that we need to qs ad 1,560mL sterile water, the difference between the initial and final volume:

$$\text{2,760 mL }-\text{ 1,200 mL}$$

$$= \text{1,560 mL sterile water}$$

Now let’s say we traveled back in time (because we’re cool like that), and before we decided to empty the pharmacy’s last bottle of 23% calcium gluconate, we received a prescription that calls for a 96 dram (6 oz or 355mL) vial of 10% calcium gluconate solution. In other words, instead of starting with an initial volume of 23% calcium gluconate, this time we are going to initialize our formula with a final volume of 10% calcium gluconate instead. So let’s plug in our values and solve for the initial volume of 23% calcium gluconate solution:

$$\text{(x) mL} \times \text{23%} = \text{355mL} \times \text{10%} \;\text{;}$$

$$\text{(x) mL} = \frac{\text{355mL} \times \text{10%}}{\text{23%}} \;\text{;}$$

$$\frac{\text{355mL} \times \text{10%}}{\text{23%}}$$

$$= \text{154mL 10% calcium gluconate}$$

Finally, always remember to take the difference of the two volumes to obtain the amount of solvent added (in this case, sterile water):

$$\text{355mL }-\text{ 154mL}$$

$$= \text{201mL sterile water}$$

To solidify your knowledge of liquid & solid dilutions, let’s look at 2 more examples. What is the final volume of a 70% alcohol solution produced by diluting 2 pints of 91% alcohol? Since the units must match, let’s first convert 2 pints to mL and then proceed as usual:

$$\text{2 pt} \times \frac{\text{473.176 mL}}{\text{1 pt}}$$

$$= \text{946.353 mL} \;\text{;}$$

$$\text{946.353 mL} \times \text{91%}$$

$$= \text{(x) mL} \times \text{70%} \;\text{;}$$

$$\frac{\text{946.353 mL} \times \text{91%}}{\text{70%}}$$

$$= \text{(x) mL} \;\text{;}$$

$$\frac{\text{946.353 mL} \times \text{91%}}{\text{70%}}$$

$$= \text{1,230.3 mL 70% alcohol}$$

Okay let’s do 1 more. This time we’ll work with a cream. How many g of ointment base should be mixed with a 2.5% Hydrocortisone cream to prepare 1 oz of 1% Hydrocortisone cream? Well, let’s first convert 1 oz to g and then proceed as usual:

$$\text{1 oz} \times \frac{\text{28.349 g}}{\text{1 oz}} = \text{28.349 g} \;\text{;}$$

$$\text{(x) g} \times \text{2.5%} = \text{28.349 g} \times \text{1%} \;\text{;}$$

$$\frac{\text{28.349 g} \times \text{1%}}{\text{2.5%}} = \text{(x) g} \;\text{;}$$

$$\frac{\text{28.349 g} \times \text{1%}}{\text{2.5%}}$$

$$= \text{11.3 g of 2.5% HC cream} \;\text{;}$$

$$\text{28.349 g }-\text{ 11.3 g}$$

$$= \text{17.0 g ointment base}$$

Complete the following dilution problems:
5 oz of 10% cream to g of 5% cream
7 mL of 2% solution to mL of 20% solution
48 dram of 1% solution to oz of 2% solution
5 lb of 4% cream to g of 25% cream
10 L of 7% solution to mL of 1% solution
96 dram of 8% solution to oz of 4% solution
12 g of 9% cream to g of 2% cream
21 oz of 18% cream to g of 9.5% cream
30 mL of 27% solution to mL of 5.5% solution
11 dram of 10% solution to oz of 3% solution
70 lb of 1.5% cream to g of 0.2% cream
0.5 L of 8.2% solution to mL of 5.2% solution
22 dram of 19% solution to oz of 5% solution
1.7 g of 52% cream to g of 7% cream