1. An important equation for us in this chapter is
Calculate the amount of drug present in 300 mL of a 4.0% w/v solution of that drug.
Solution. 0.040 g/mL × 300 mL = 12 g
2. Using the same equation, calculate the amount of alcohol present in 35 mL of a 60% solution of alcohol in water.
Solution. 0.60 × 35 mL = 21 mL
3. Calculate the amount of calamine in 13 g of a 10% w/w calamine ointment.
Solution. (0.10)(13 g)= 1.3 g
4. Calculate the number of milligrams of potassium permanganate in 90 mL of a 1:500 w/v potassium permanganate solution.
Solution. 180 mg
CALCULATIONS
90 mL × = 0.18 g = 180 mg
5. Try these problems.
A. How many milliliters of solute are there in,
(1) 350 mL of 5% v/v solution?
(2) 4.20 L of a 1:2000 v/v solution?
B. How many grams of solute are there in
(1) 220 g of a 1:40 w/w solution?
(2) 170 mL of a 15.0% w/v solution?
C. How many milligrams of drug are there in
(1) 25.0 mL of a 3.25% w/v solution?
(2) 900 g of a 1:150 w/w mixture?
Solutions.
A. (1) 17.5 mL
(2) 2.1 mL
B. (1) 5.5 g
(2) 25.5 g
C. (1) 813 mg
(2) 6000 mg
PROBLEM SOLVING APPROACHES
6. You will be presented now with a simple and clear-cut way to deal with dilution and concentration of pharmaceutical preparations. In case your learning style does not correlate with algebraic methods, you may “connect” to one of the alternative techniques that we will provide later in this chapter. Some of these mathematical methods will require a few more steps during calculations but if you feel more comfortable and confident when using them, there is no problem. We will present the basics of each approach and follow with some examples solved by each one of the methods proposed.
ALGEBRAIC METHODS
Mass Balance Equation
7. It is sometimes necessary for the pharmacist to dilute preparations to make a product of lower strength. Here is an example:
Ichthammol ointment 8% w/w 90.0 g
Ft. ung.
The pharmacist has some 20% w/w ichthammol ointment. He wishes to dilute it with petrolatum, an inert semisolid that contains no drug of any kind, to the proper strength. The quantity of the 20% ointment and of petrolatum to be mixed must be determined.
The key to solving problems of this type is the realization that the amount of an ingredient in the finished product must be equal to the contributions of the components of the formula. In terms of our example, we know that the ichthammol in our final product, the 8% ointment, must be the sum of the ichthammol contributed by the 20% ichthammol ointment and by the petrolatum. We may therefore write
g ichthammol from 20% ointment + g ichthammol from petrolatum = g ichthammol in 8% ointment
To conserve space, we are going to rewrite this equation, which is called a mass balance equation, using a bit of shorthand. The preparation that contains the ingredient will be written in parentheses and “ichthammol” will be abbreviated as “ich.”
Of course, there is no ichthammol in petrolatum, so our equation becomes
Recall that,
Let j equal the quantity of 20% ointment to be used:
Substituting these quantities in our mass balance equation yields
Use 36 g of the 20% ichthammol ointment; the amount of petrolatum needed is 90 g − 36 g = 54 g.
8. In summary, to solve problems involving dilution or concentration:
(1) Write the mass balance equation.
(2) Substitute in the mass balance equation.
(3) Solve the resulting algebraic expression.
Now try this problem.
We wish to dilute an ointment containing 14% sulfur with petrolatum to make 60 g of an ointment containing 10% sulfur. How many grams of 14% sulfur ointment and how many grams of petrolatum will be necessary to make the dilution?
(1) Write a mass balance equation that shows where the sulfur in the finished product (the 10% ointment) comes from.
Solution. g S (14% oint) = g S (10% oint) (petrolatum contains no sulfur).
9. Calculate the amount of sulfur in the 10% ointment using the equation quantity of solute = concentration × quantity of preparation
Solution. g S (10% oint) = (0.100)(60 g) = 6 g
CALCULATIONS
Let j equal the quantity of 14% ointment to be used. Calculate the number of grams of sulfur in the 14% ointment.
g S (14% oint) = 0.14j
(2) Substitute in the mass balance equation and (3) solve.
0.14j = 6 g
j = 42.9 g
Use 42.9 g of the 14% sulfur ointment. (The amount of petrolatum is 60.0 g − 42.9 g = 17.1 g)
10. It is necessary to prepare 180 mL of a 1:200 solution of potassium permanganate (KMnO4). What quantity of a 5% stock solution of KMnO4 should be diluted with water?
Fill in the blank in the mass balance equation:
_______________ = g KMnO4 (1:200 sol)
Solution. g KMnO4 (5% sol)
11. Let j equal the number of milliliters of 5% KMnO4, solution to be diluted. Fill in the blanks:
A. g KMnO4 (5% sol) = __________
B. g KMnO4 (1:200 sol) = __________
Solutions.
A. (0.05 g/mL)j
B. 180 mL = 0.9 g
12. Now, substitute in the mass balance equation and solve.
Solution. Use 18 mL of the 5% solution.
CALCULATIONS
(0.05 g/mL)j = 0.9 g
j = = 18 mL
13. How many milliliters of a 10% w/v Merthiolate solution should be diluted with water to make 440 mL of a 0.25% w/v Merthiolate solution?
Solution. 11 mL
CALCULATIONS
Let j equal the milliliters of 10% Merthiolate solution:
g merth (10%) = g merth (0.25%)
(0.10 g/mL)j = (440 mL)(0.0025 g/mL) = 1.1 g
j = 11 mL
14. How many milliliters of water must be added to 180 mL of 36% w/v acetic acid solution in order to make up a solution of 10% w/v strength? (Assume that no shrinkage or expansion of volume occurs on mixing.)
Solution. 468 mL
CALCULATIONS
Let j equal the quantity of water to be added. The volume of the 10% solution will then be 180 mL + j.
g acetic acid (36% sol) = g acetic acid (10% sol)
(0.36 g/mL)(180 mL) = (0.1 g/mL)(180 mL +j)
64.8 g = 18.0 g + (0.1 g/mL)j
15. How many milliliters of Zephiran concentrate (17% w/v of Zephiran) are required to prepare 2L of a 1:1500 solution of Zephiran?
Solution. 7.84 mL
CALCULATIONS
Let j equal the milliliters of Zephiran concentrate needed:
g Zephiran (17% sol) = g Zephiran (1:1500 sol)
(0.17 g/mL)j = (1/1500 g/mL)(2000 mL)
j = 7.84 mL
16. If 12.5 g of a 10% zinc oxide ointment are diluted with 17.5 g of white petrolatum, what is the percentage strength of the resulting product?
Solution. 4.17%
CALCULATIONS
Let j equal the concentration of the diluted ointment:
g zinc oxide (10% oint) = g zinc oxide (dil oint) (0.1)(12.5 g) = (j)(30.0 g)
17. Hydrochloric acid USP is a 36% w/w solution of HCl in water with a specific gravity of 1.18. How many milliliters should be used to prepare 200 mL of a 5% w/v solution of HCl?
Solution. 23.6 mL
(If you ran into trouble, see the hint and solution presented below. If you were successful, continue after the solution.)
Hint. The mass balance equation is
g HCl (36% w/w sol) = g HCl (5% w/v sol)
We cannot set j equal to volume of 36% w/w HCl solution and multiply 0.36 by j to get the amount of HCl. j is a volume; 0.36 refers to concentration on a weight basis. The two quantities are just not compatible. However, it is possible to calculate the weight of the 36% solution and then convert that to volume using its density. Try the problem again. If you still have difficulty, consult the solution.
CALCULATIONS
Let j equal the weight of hydrochloric acid needed:
0.36j = (0.05 g/mL)(200 mL)
Use 23.6 mL of hydrochloric acid USP.
18. A pharmacist wishes to prepare 150 mL of ammonia (NH3) solution, 10% w/v. How many milliliters of strong ammonia solution (28.5% w/w) should be used? (Sp g of strong ammonia solution is 0.900.)
Solution. 58.4 mL
CALCULATIONS
Let j equal the weight of 28.5% solution to be used:
g NH3 (28.5% w/w sol) = g NH3 (10% w/v sol)
0.285j = (0.1 g/mL)(150 mL)
j = 52.6 g
19. A crude drug is required to contain 0.28% w/w of an alkaloid, the active compound. How many kilograms of a batch of crude drug containing 0.30% w/w alkaloid must be combined with 500 g of crude drug containing 0.20% w/w alkaloid in order that the resulting mixture will meet the required standard? (Hint Use the same procedure as for previous problems in this chapter; write the mass balance equation and then substitute in it.)
Solution. 2.0 kg
CALCULATIONS
The mass balance equation is
g alkaloid (0.20%) + g alkaloid (0.30%) = g alkaloid (0.28%)
Let j equal the quantity of 0.30% crude drug required. Can you go on from here? Try the problem again. If you still have trouble, go back to frame 4.
Let j equal the quantity of 0.30% crude drug necessary:
g alkaloid (0.20%) = (500 g)(0.0020)
g alkaloid (0.30%) = 0.0030j
g alkaloid (0.28%) = (500 g + j)(0.0028)
Now complete the solution.
The solution is as follows:
g alkaloid (0.20%) + g alkaloid (0.30%) = g alkaloid (0.28%)
(500 g)(0.0020) + 0.0030j = (500 g + j)(0.0028)
1.0 g + 0.0030j = 1.4 g + 0.0028j
j = 2000 g = 2.0 kg
20. How many milliliters of syrup containing 85.0% w/v sucrose should be mixed with 115 mL of syrup containing 60.0% w/v sucrose to prepare a syrup containing 76.0% w/v sucrose? (Assume that there is no expansion or shrinkage of volume when the two liquids are mixed.)
Solution. 204 mL
CALCULATIONS
Let j equal the volume of 85% sucrose:
g sucrose (85%) + g sucrose (60%) = g sucrose (76%)
How much alcohol USP (95% C2H5OH) should be used in this prescription in order that the concentration of ethanol (C2H5OH) in the finished product be 20% v/v?
Solution. 12.7 mL
CALCULATIONS
Let j equal the milliliters of alcohol USP needed:
mL eth (67%) + mL eth (15%) + mL eth (95%) = mL eth (product)
22. If 300 g of an ointment containing 4% (w/w) sulfur is combined with 220 g of a 10% (w/w) sulfur ointment, what is the strength of the mixture that results?
Solution. 6.54% w/w
CALCULATIONS
Let j equal the strength of finished product:
g S (4% oint) + g S (10% oint) = g S (mixture)
(300 g)(0.04) + (220 g)(0.1) = (520 g)(j)
j = 0.0654 = 6.54%
23. A pharmacist combines 140 mL of a 0.90% sodium chloride solution with 250 mL of 3.4% sodium chloride solution. Assuming no expansion or contraction in volume, calculate the percentage strength of the mixture.
Solution. 2.5%
CALCULATIONS
Let j equal the concentration of the mixture:
g NaCl (0.90%) + g NaCl (3.4%) = g NaCl (mixture) (140 mL)(0.0090 g/mL) + (250 mL)(0.034 g/mL) = (390 mL)(j)
j = 0.025 g/mL = 2.5%
24. Try these practice problems that review the material covered so far. Check your answers at the end.
A. If 4.0 mL of a 5.0% benzethonium chloride solution are diluted with water to 500 mL, what will the ratio strength of the resulting solution be?
B. How much 14% sodium solution should be diluted with water to make 350 mL of 6% sodium acetate solution?
C. How much lanolin (an inert ointment base) should be added to 300 g of cortisone ointment, 3%, to make an ointment containing cortisone, 1:250?
D. A pharmacist needs 60 mL of diluted acetic acid (a 6% w/v solution of C2H4O2 in water) for a prescription. The pharmacist has only acetic acid USP on hand. This is a 36% w/w solution of C2H4O2 in water with a specific gravity of 1.045. How many milliliters of acetic acid USP should the pharmacist use?
E. How much pure zinc oxide should be mixed with a 10.0% zinc oxide ointment to make 3.75 kg of a 12.0% zinc oxide ointment?
F. Three samples of a plant have the following potencies:
Sample 1 (220 g) contains 2.40% alkaloids
Sample 2 (50.0 g) contains 1.97% alkaloids
Sample 3 (450 g) contains 3.85% alkaloids
If the three samples are combined, what is the alkaloid content, expressed as a percent, of the mixture?
G. What is the percentage of benzalkonium chloride in a solution made by mixing 350 mL of a 1:100 benzalkonium chloride solution with 250 mL of a 1:300 benzalkonium chloride solution?
H.
Phenobarbital elixir (15% C2H5OH)
30 mL
High-alcoholic elixir (78% C2H5OH)
qs
Elixir terpin hydrate (42% C2H5OH)
45 mL
Syrup qs ad 120 mL
How many milliliters of high-alcoholic elixir should be used to make the C2H5OH content of the final solution 35%?
= 0.18 g = 180 mg
180 mL = 0.9 g
= 18 mL
