1. Pharmaceutical calculation
Assist.Lecturer
Qasim Allawi Bader
MSc Pharmaceutics

2. Specific gravity of mixtures: (continued)
Example:
How many ml.s of each of two liquids with sp. Gr and should be used to prepare 1500 ml of a solution has sp. Gr ?
Sol:
parts ml
parts
relative amount 50 : 25 or 2 : 1
2+1 = 3 total parts
X x=500 ml of liquid
= 1000 ml of 0.95 liquid

3. Example: Rx Zinc oxide 1. 5 Hydrophilic petrolatum 2
Example: Rx Zinc oxide 1.5 Hydrophilic petrolatum 2.5 Purified water 5 Hydrophilic ointment ad 30 How much zinc oxide should be added to the product to make a 10% zinc oxide ointment? Sol: X 100 x= 5% 100% 5 parts of 100% 10% 5% 90 parts of 5% Relative amount 5 : 90 = 1: X 30 x = 1.667g zinc oxide

4. An other way: 10% zinc oxide mean 90% conc. of diluent 30 gm – 1
An other way: 10% zinc oxide mean 90% conc. of diluent 30 gm – 1.5 = 28.5 gm amount of diluent which is the same amount in the new 10% oint X X = gm total wt. of the oint – 30 = 1.67 gm wt. of the zinc oxide should be added.

5. Example: castor oil 5ml Resorcinol monoacetate 15ml Alcohol 85 % ad 240ml How many ml.s each of 95% v/v alcohol and water should be used in preparing the prescription? Sol.1: 5+15 = 20 ml of cast. Oil. And resorcinol = 220 ml of 85% alcohol 85% 100 X 220 x = 187 ml y y = ml of 95% alcohol. .

6. Sol.2: C1 V1 = C2 V2 85 * 220 = 95 * V2 V2 = ml of 95% alcohol Water: add water enough to make 220 ml of 85% alcohol

7. Example:
Benzalkonium chloride sol ml.
Make a solution such that 10 ml diluted to a liter equals a 1:5000 solution.
How many milliliters of a 17% stock solution of Benzalkonium chloride should be used in preparing the prescription ?
Sol:
X x = 0.2 g in liter and the same in 10 ml.s
Y y = 4.8 gm in 240 ml
Z Z = 28.2 ml stock solution.

8. Example: How many ml. s of a 2
Example: How many ml.s of a 2.5 % w/v chlorpromazine hydrochloride injection and how many ml.s of 0.9% w/v sodium chloride should be used to prepare 500 ml of a 0.3% w/v chlorpromazine hydrochloride injection? Sol: C1 V1 = C2 V2 2.5 * V1 = 0.3 * 500 V1 = 60 ml of chlorpromazine. 500 – 60 = 440 ml of sodium chloride. (H.W: answer this ex. With alligation alternate method. )

9. Isotonic solution
When a solvent passes through a semipermeable membrane from a dilute solution in to a more concentrated one, the concentrations become equalized and the phenomenon is known as osmosis.
Thus, osmotic pressure is the pressure responsible for this phenomenon which varies with the nature of the solute.
We have two type of solute:
Nonelectrolyte, its solution contains only molecules and the osmotic pressure depends on the conc. of the solute.
electrolyte solution , its solution contains ions and the osmotic pressure depends on the conc. of the solute and its degree of dissociation.
Thus , solutes that dissociate present a greater no. of particles in solution and exert a greater osmotic pressure than undissociated molecules

10. Colligative properties; such as osmotic pressure, vapor pressure, boiling point, and freezing point depend on the no. of particles in solution therefore the change in any one of them will result in a change in the other.

11. Isosmotic solutions: they are two solutions have the same osmotic pressure.
Isotonic solution: a solution have the same osmotic pressure as body fluid ( serum, lacrimal fluid). 
Hypotonic solution: a solution have lower osmotic pressure than that of a body fluid.
Hypertonic solution: a solution have higher osmotic pressure than that of a body fluid

12. Most ophthalmic preparations should be isotonic or approximately isotonic to be comfort to the patient and to reduce the irritation of the eyes.
Injections that are not isotonic should be administered slowly and in small quantities to minimize tissue irritation and pain.
Intravenous infusions which are hypotonic or hypertonic can have adverse effects because they generally are administered in large volumes.

13. We can calculate the osmotic pressure depending on colligative properties especially freezing point.
Note: freezing point of both serum and lacrimal fluid is ( ◦C)
Freezing point (or any other colligative properties) of solutions could be used for determining the tonicity of these solutions.
As we said before F.P. of body fluid is (-0.52 ◦c) so, any substance has this F.P. should be isotonic with body fluid.

14. Calculation of tonicity for nonelectrolyte substances
When one gram molecular weight of any nonelectrolyte is dissolved in gm of water, the freezing point of the solution is about ( C)
So we can calculate the weight of substance that should be dissolved in (1000 gm) of water

15. Example: Boric acid has m. wt = 61. 8 , thus 61
Example: Boric acid has m.wt = 61.8 , thus 61.8 gm when dissolved in 1000 gm of water should produce F.P. = ◦c therefore: 1.86/0.52=61.8g/x X=17.3 gm of boric acid in 1000 gm of water = 1.73% (w/v) make isotonic sol.

16. Calculation of tonicity for electrolyte substances:
Here osmotic pressure depend on the degree of dissociation.
Example: Sodium chloride in weak solution 80% dissociated ( and have m.wt. = 58.5) so, 100 molecules give 180 molecules
Dissociation factor (i) = 180/100 = 1.8 therefore: simple isotonic solution could be calculated as follow:
0.52 * m.wt. /1.86 * (i) = ( ) gm of solute in 1000 gm water.
X= 9.09 gm of NaCl in 1000 gm of water = 0.9% (w/v) make isotonic sol

17. If the no. of dissociated ions is known , (i) could be determined:
non electrolyte substances
substances that dissociate in to 2 ions
substances that dissociate in to 3 ions
substances that dissociate in to 4 ions
substances that dissociate in to 5 ions

18. Example How much Sod. Chloride should be added to 0. 5% w/v sol
Example How much Sod. Chloride should be added to 0.5% w/v sol. to make it isotonic ? 0.5% 0.5 gm /100 ml water 0.9% 0.9 gm/ 100 ml water Then 0.9 – 0.5 = 0.4 gm of NaCl should be added to 100 ml 0.5% sol. to be made isotonic with body fluid.

19. Sodium chloride equivalent (E value): is the no. of gm
Sodium chloride equivalent (E value): is the no. of gm.s of NaCl which is equivalent to 1 gm of the substance.
E value = m.wt of NaCl / (i) of NaCl * (i) of the sub / m.wt of the sub.
Ex.

20. Calculation of Sodium Chloride equivalent (E values)
Papavarine HCl (M.wt = 376) is a 2-ion electrolyte dissociating 80%. Calculate its E-value, where its dissociation factor (i) = 1.8
M.Wt of NaCl = By substituting the values in the equation: E = 58.5/1.8 X 1.8/376 = or 0.16

21. Ex.
How much NaCl should be used in preparing 100 ml of a 1% w/v solution of Atropine sulfate, which should be made isotonic?
الجواب هنا يعتمد على حساب كمية والتي تكافئ
NaCl Atropine sulfate
m.wt of NaCl = , i = 1.8
m.wt of Atropine sulfate = 695 , i = 2.6
1% w/v gm in 100 ml
X = 0.12 gm of NaCl represented by 1 gm of Atropine sulfate ( E value )
0.9 – 0.12 = 0.78 gm of NaCl should be added to 100 ml 1% w/v Atropine sulfate sol. to make it isotonic.

22. Example:
Calculate the dis. factor of zinc sulfate (2 ion electrolyte, dissociating 40 %) ?
100 particle will give : 40 zinc ions
40 sulfate ions
60 undissociated particles
140 particles
So, dissociation factor (i) = 140/100 = 1.4 , answer.

23. Example:
Calculate the dissociation factor of zinc chloride ( 3 ion electrolyte , dissociating 80 %)
100 particle will give : 80 zinc ions
80 chloride ions
20 undissociated particles
260 particles
So, dissociation factor (i) = 260/100 = 2.6 answer.

25. The amount of NaCl represented by a sub. = wt. of the sub. * E value
The procedure for calculating isotonic sol. With NaCl equivalent is:
 1. Calculate the amount of NaCl represented by the sub. In the prescription
(sub. amount * E value).
2. Calculate the amount of NaCl that would be contained in a 0.9% solution of the volume of the prescription.
3. Amount of NaCl added to make the sol. Isotonic =Amount of NaCl (step 2) – amount of NaCl (step 1)
4. If an agent other than NaCl ( Boric acid, dextrose …) is to be used to make isotonic solution , we will divide the amount of NaCl (step 3) by the E value of that agent.

26. Example:
how many grams of sodium chloride should be used in compounding the following prescription? (E value of Pilocarpine nitrate is 0.23) Rx
Pilocarpine nitrate
Sodium chloride q.s.
Purified water ad ml
Make isoton. Sol.
Sig. for the eye

28. Example:
How many grams of Boric acid should be used in compounding the following prescreptioin?
(E value of Phenacaine Hydrochloride is 0.20 and E value of Chlorobutanol is 0.24) Rx
Phenacaine Hydrochloride %
Chlorobutanol ½ %
Boric acid q.s.
Purified water ad
Make isoton. Sol.

29. 1gm 100ml 0. 5gm 100ml X 60ml X = 0. 6gm X 60ml X= 0
1gm 100ml 0.5gm 100ml X 60ml X = 0.6gm X 60ml X= 0.3gm So, the prescription calls for 0.6 g of Phenacaine Hydrochloride and 0.3 g of Chlorobutanol. Step * 0.6 g = g of NaCl represented by Phenacaine Hydrochloride 0.24* 0.3 g = g of NaCl represented by Chlorobutanol. Total g of NaCl represented by both ingredients. Step2. 60* = g of NaCl in 60 ml of an isotonic NaCl sol. Step g (from step2) g (from step1) g of NaCl required to make the sol. Isotonic. But because the prescription calls for boric acid: Step g / 0.52 (E value of Boric acid) = g of Boric acid to be used.

30. Using an isotonic sodium chloride solution to prepare other isotonic solution
0.9% w/v NaCl solution used to compound isotonic solutions of other substances as follow: Step.1: Calculate the quantity of the drug needed in the prescription. Step 2: Calculate the volume of water needed to make the solution isotonic: gm(drug) * E value(drug) = ( ) ml of water needed to make isotonic of the drug Step.3: Add 0.9% w/v NaCl solution to complete the volume of the prescription.

31. Example:
what is the volume of water and 0.9% w/v NaCl solution needed to prepare 20 ml of a 1% w/v solution of hydromorphone hydrochloride ( E- value =0.22)
Answer:
step.1:
X X= 0.2 gm hydromorphone.
Step.2:
0.2 gm * = 4.89 ml water required to make isotonic solution of hydromorphone
0.009.
Step.3:
= ml of 0.9% w/v NaCl solution required

32. Example:
 Tetracaine Hcl %
Epinephrine Bitartarate 1:
Boric acid q.s.
Purified water
Make isotonic sol.
 The solution of epinephrine is already isotonic and the E- value of tetracaine = How much boric acid used?

33. Answer: 0. 5 100 X 30 X= 0. 15 gm of tetracaine in the prescription 0
Answer: X 30 X= 0.15 gm of tetracaine in the prescription 0.15 × 0.18 = gm of NaCl represented by the sub. 30 – 10 = 20 ml (vol. of epinephrine which is isotonic) X 20 X= 0.18 gm of NaCl in 20 ml of an isotonic NaCl sol – = gm of NaCl should be added. But here we use Boric acid , so : ÷ 0.52 = 0.29 gm of boric acid used. (E-value) of Boric acid

34. Use of freezing point in calculation of isotonicity:
Substance to be isotonic should lower freezing point ∆Tf = - o.52 (which is the freezing point of blood and lacrimal fluid). Example: How many milligrams each of NaCl and dibucaine HCl are required to prepare 30 ml of 1% dibucaine HCl isotonic solution? ( ∆Tf (1% dibucaine) = , ∆Tf (1% NaCl) = )

36. Practice problems of isotonic solution

38. EX.
1. How many grams each of lidocaine hydrochloride (i) and sodium chloride (ii) are required to prepare 150ml of a 2% w/v solution of lidocaine hydrochloride isotonic with blood plasma? Freezing point depression of a 1% w/v lidocaine hydrochloride solution is 0.063◦C and that of sodium chloride is 0.576◦C.
2. A pharmacist receives a prescription for 10ml of isotonic 1% w/v atropine sulfate eye drops. What weight of sodium chloride is required to make the solution isotonic with tears?
1. (i) 3g (ii) 1.026g mg

39. Home works
1.How many grams of NaCl and Naphazoline HCl should be used in compounding the following prescription: Rx Naphazoline HCl 1% NaCl Q.S. Purified water ad 30ml Make isotonic solution. Use the freezing point depression method. Tf blood = Tf 1% NaCl = Tf 1% Naphazoline HCl = 0.16

40. 2.For agent having 0.32 sodium chloride equivalent, calculate the percentage concentration of an isotonic solution.
3.Calculate the E value for holocaine hydrochloride, if 6.7 mL of water will produce an isotonic solution from 0.3 g of drug substance.

41. 4. The freezing point of 5% solution of boric acid is -1. 55 oC
4.The freezing point of 5% solution of boric acid is oC. How many grams of boric acid should be used in preparing 1500 mL of an isotonic solution? Rx Dextrose, anhydrous 2.5% Sodium chloride q.s Sterile water for injection ad 1200mL Label: Isotonic Dextrose and Saline Solution. How many grams of sodium chloride should be used in preparing the solution?