1. PH 101.91 METROLOGY

2. Definition  Metrology may be defined as the study of the various systems of weights and measures, their relationships and the mathematical principles involved in their inter conversion. PH 101.92

3. Introduction  The knowledge of weights and measures is the important tool for a Pharmacist to dispense dosage forms  In the practice of Pharmacy Weighing – used for both solids and liquids Measuring – used for liquids only PH 101.93

4.  Weight is the measure of gravitational force acting on a body and is directly proportional to its mass.  Mass remains constant and never varies because it is based on inertia.  Weight varies slightly with change in latitude, altitude, temperature, pressure and humidity. PH 101.94

5. Measure is the determination of definite volume of any substance or capacity of the liquid.  Temperature and pressure exert their effect specially on liquids and gases.  Standard weights and measures will be issued by NIST(National Institute of Standards Technology) PH 101.95

6. PH 101.136 Introduction Generally official preparations and most often prescriptions are Given for the quantity of 1000 ml or 1000 gms of product. The pharmacist has to prepare either smaller or greater quantities than the original quantity. Therefore the prescription is either reduced or enlarged to maintain the correct proportion of each ingredient.

7. PH 101.137 General formula for reducing and enlarging the prescription is Quantity of each ingredient in amount desired = Quantity of each ingredient given Total amount specified in formula x Total amount desired

8. PH 101.138 Eg : Drug 100gms Sucrose 450 gms Purified water to make 1000 ml What quantities are required to make 50 ml of the syrup ?

9. PH 101.139 Quantity of each ingredient given Total amount specified in formula Quantity of each ingredient in amount desired = x Total amount Desired Using the formula Quantity of drug = 100 x 50 = 5 gms Desired 1000 Quantity of sucrose = 450 x 50 = 22.5 gms Desired 1000 Purified water to make 50 ml

10. PH 101.1410 Stock solutions Are of known concentration that we prepare for convenience in dispensing. Diluted by diluents to obtain a product of desired strength. Stock solutions facilitate the dispensing of the soluble substance. This technique is adopted for substances stable in solution.

11. PH 101.1511 Pharmaceutical dilutions - Allegation method Proof spirit and proof strength Solved problems and exercises. Objectives

12. PH 101.1512 Introduction Alcohol dilutions : Usually made from 95% alcohol. Means 95 parts of ethyl alcohol and 5 parts of water. All other dilutions are prepared by mixing with water. On mixing alcohol with water, - contraction in volume - rise in temperature

13. PH 101.1513 - Mixture becomes turbid due to evolution of air bubbles - Because of less solubility of air in water than in alcohol. - Cooled to 20ºC before adjusting volume. The general formula for alcohol dilution is Volume of stronger Alcohol to be used Volume required x percentage required Percentage used =

14. PH 101.1514 Solved problem:1 Prepare 600 ml of 60 % alcohol According to formula Volume of stronger Alcohol to be used Volume required x percentage required Percentage used = = 600 x 60 95 = 378.95 ml Therefore dilute 379 ml of 95% alcohol upto 600 ml with water. The resulting dilution will contain 60% alcohol.

15. PH 101.1515 Exercise :1 Prepare 500 ml of 40% alcohol from 95% alcohol Ans:210 ml

16. PH 101.1516 Weight in weight (w/w) solutions: The general formula is Weight of stronger acid to be used Weight required X percentage required percentage used =

17. PH 101.1517 Solved problem:2 Send 200 ml of solution of acetic acid containing 4 % of real acetic acid. The strength of real acetic acid is 33 % Using formula Weight of stronger acid to be used Weight required x percentage required percentage used = = 200 x 4 = 24.2 gm 33 Therefore weigh 24.2 gms of real acetic acid and dilute it upto 200 ml with water, the resulting solution will contain 4% acetic acid.

18. PH 101.1518 Exercise: 2 Send 200 ml of a solution of ammonia containing 4% by weight of ammonia.The strong solution of ammonia contains 32.5% w/w of ammonia. Ans:- 24.615 gms

19. PH 101.1519 Allegation method The name allegation is derived from the Latin word “allegatio” meaning the act of attaching. It refers to lines drawn during calculation to bind quantities together. Also called method of Rectangles. Rapid means of calculation when a preparation of intermediate strength is desired by mixing two preparations of lower and higher strengths respectively.

20. PH 101.1520 Higher strengths Lower strength Required strength Lower strength Higher strength Required strength 1 2 o 4 3 Allegation method

21. PH 101.1521 The known concentrations of the preparation are written at positions 1 and 2. Required strength is written in the middle (position 0). The required strength is subtracted from the higher strength and put in position 3. (contd..) Allegation Method

22. PH 101.1522 The results obtained at positions 3 and 4 gives the parts respectively of the lower and higher strengths to be mixed to give a preparation of the required strength. The lower strength is subtracted from the required strength and the result is written at position 4. Allegation Method

23. PH 101.1523 Solved problem: 3 Prepare 800 gms of dilute acetic acid from acetic acid I.P. the strength of dilute acetic acid is 6 % w/w, while that of acetic acid I.P is 33 % w/w. Solution :- Strength of acetic acid I.P = 33%w/w Required strength of dilute acetic acid = 6%w/w Dilute acetic acid is prepared by diluting acetic acid with water. Hence strength of the second preparation to be mixed = 0% (contd..)

24. PH 101.1524 By the method of allegation 33 6 6 0 27 27parts of water mixed with 6 parts of acetic acid I.P to prepare 33 parts of dilute acetic acid. Quantity of water required to prepare 800 gms of dilute acetic acid = 27 x 800 = 654.55gms 33 (contd..)

25. PH 101.1525 Quantity of acetic acid required to prepare 800 gms of dilute acetic acid = 6 x 800 = 145.45 gms 33 Therefore 654.55 gms of water have to be mixed with 145.45 gms of acetic acid I.P to prepare 800 gms of dilute acetic acid.

26. PH 101.1526 Exercise :3 Prepare 300 ml of 45 % alcohol from 90 % alcohol by Alligation method. Ans :150 ml

27. PH 101.1527 Exercise: 4 Calculate the volume of each of 80%,70%,30% and water required to produce 500 ml of 50% alcohol. Ans:- 208.33 ml,83.33 ml, 41.67 ml and water 166.67 ml

28. PH 101.1528 Proof spirit The strength of alcohol is expressed in terms of proof spirit and proof strength. mixture of absolute alcohol and water having specific gravity equal to 0.91976. Means a mixture of absolute alcohol and water contains 57.1%v/v pure alcohol, which is 100 proof spirit. Proof spirit = % strength of alcohol x 1.7513 Since 57.1%v/v volume of alcohol = 100 volumes of proof spirit 1 volume of alcohol = (1 x 100) /57.1 = 1.7513 volumes of proof spirit

29. PH 101.1529 Proof strength % strength v/v alcohol is multiplied by 1.7513 will give the corresponding Proof strength If proof strength is divided by 1.7513 will give % strength of an Absolute alcohol. The strength above proof strength is expressed as “over proof” Any strength below proof strength is expressed as “under proof” Proof strength = (% strength of alcohol x 1.7513) – 100 = Proof spirit - 100

30. PH 101.1530 Solved problem: 5 What will be the proof spirit of an elixir containing 40% v/v alcohol ? Proof spirit = % strength of alcohol x 1.7513 = 40 x 1.7513 Proof spirit = 70.052 Exercise : 5 Find the strength of 80% v/v alcohol in terms of proof spirit Ans:- 40.24º over proof

31. PH 101.1531 Solved problem: 6 Calculate the real strength of 20º Over Proof and 30º Under Proof mixed to get 50% alcohol. 20º OP means 100 + 20 = 120 Alcohol strength = 120 / 1.7513 = 48.45%v/v 30º UP means 100 – 30 = 70 Alcohol strength = 70 / 1.7513 = 39.93%v/v

32. PH 101.1532 What will be the % strength corresponding to 60 º OP and 20 º UP. Exercise: 6 Ans:- 91.4% v/v & 46.6% v/v

33. PH 101.1633 Isotonicity and isotonic solutions. Reasons for maintenance of isotonicity. Applications of isotonicity 33 Objectives

34. PH 101.1634 Introduction Osmosis Definition: The flow of the solvent through a semi permeable membrane from pure solvent or from a dilute solution to concentrated solution, until the concentrations of both the solutions become equal. 34

35. PH 101.1635 It is the hydrostatic pressure built up on the solution which just stops the osmosis of pure solvent into the solution through a semi permeable membrane. Two solutions that have the same osmotic pressure are termed iso - osmotic to each other. 35 Osmotic pressure

36. PH 101.1636 Isotonic solution Means solutions have same osmotic pressure equal to physiological fluids. Ex: plasma, tears, extra and intra cellular fluids Isotonic solutions have equimolar concentrations.. 36 The solutions which are not isotonic with plasma may be harmful to use. For the adjustment of isotonicity of injectable solutions, substances like Nacl, dextrose etc. are used.

37. PH 101.1637 Hypertonic solution : Solutions having higher osmotic pressure than physiological fluids. Hypotonic solution: Solutions having lower osmotic pressure than physiological fluids Hypo/Hypertonic solutions together called as paratonic solutions

38. PH 101.1638 Reasons for maintenance of isotonicity The formulations of the parenteral product should be isotonic with invivo body fluids. If any isotonicity and pH differences may cause irritation, haemolysis, necrosis and tissue toxicity. Ophthalmic solutions should be isotonic with lachrymal fluid (tears) to prevent irritation and pain. Similarly injectable solutions should be isotonic with blood plasma 38

39. PH 101.1639 On injecting the hypotonic solution into blood stream, It may enter the red blood cells in an attempt to produce equilibrium. The cells swells rapidly until they burst leading to haemolysis. As this damage is irreversible may lead to serious danger to red blood cells. 39

40. PH 101.1640 When hypertonic solution is injected into the blood stream, The water comes out of the membrane of red blood cells in order to reach the equilibrium. The cells shrink leading to crenulations which is only a temporary damage When the osmotic pressure of two solutions becomes equal, the shrinken cells will come to its original position. 40

41. PH 101.1641 Hence hypertonic solutions may therefore be administered without permanent damage to the blood cells. Should be injected slowly to ensure rapid dilution into the bloods stream and To minimize the crenulations of blood cells.

42. PH 101.1642 This can be showed diagrammatically as shown below Red blood cells are taken and are placed in 3 different solutions like hypotonic, isotonic and hypertonic solutions. 42 FIG : 16.1 hypotonic solution hypertonic solution Isotonic solution

43. PH 101.1643 Diagram I: The RBC will absorbs water and the cell will be bursted due to hypotonicity. Diagram II: The medium is isotonic with blood. So no entering or outcoming of water molecules. Diagram III :Due to higher concentration the water will be released from RBC to the medium and cell shrinkage may be observed 43

44. PH 101.1644 Applications of isotonicity The following preparations need to be isotonic 1) All the parenteral preparations 2) Intravenous infusions, irrigating solutions, lotions for wounds and subcutaneous injections 3) Parenteral preparations used for diagnostic purpose 4) Solution meant for intra thecal injections 5) Aqueous solution used for nasal applications 6) Ophthalmic preparations 44

45. PH 101.1645 Calculations for preparing Isotonic solutions For making isotonic solutions the quantities of substances to be added may be calculated by following methods. 1.Based on freezing point data. 2. Based on molecular concentration. 3. Based on sodium chloride equivalents.

46. PH 101.1746 Introduction Freezing point :- The temperature at which the solution freezes is known as freezing point. Freezing point of pure water is 0ºC Freezing point of blood plasma is – 0.52º C 46

47. PH 101.1747 Calculations for preparing Isotonic solutions For making isotonic solutions the quantities of substances to be added may be calculated by following methods. 1.Based on freezing point data. 2. Based on molecular concentration. 3. Graphical method based on vapour pressure and freezing point determinations. 4. Based on sodium chloride equivalents. 47

48. PH 101.1748 Based on Freezing point data Freezing point is a physical property of solutions. Most often used in the calculations of isotonic solutions. Because it can be measured easily and accurately. Body fluids such as blood plasma and lachrymal secretions have a freezing point of – 0.52ºC. Which is same value of a 0.9% solution of sodium chloride. All solutions which freeze at -0.52 º C will be isotonic with body fluids 48

49. PH 101.1749 The quantity of adjusting substance required for making solution isotonic with physiological fluid may be calculated from the formula. % w/v of adjusting substance required = 0.52 – a b Where ‘a’ is the figure representing the freezing point of 1% w/v solution of unadjusted substance 49 Based on Freezing point data

50. PH 101.1750 Solved problem: 1 Find the concentration of procaine hydrochloride required to make a solution isotonic with blood plasma. The freezing point of 1% w/v solution of procaine hydrochloride is -0.122ºC. By applying the general formula 0.52 – a b 50

51. PH 101.1751 ‘a’ the freezing point of unadjusted solution = 0.00ºC (water) ‘b’the freezing point of 1%w/v solution of the adjusting substance (procaine hydrochloride) = - 0.122ºC % w/v of procaine hydrochloride required = 0.52 – 0.00 0.122 = 4.26%w/v 51

52. PH 101.1752 Exercise: 1 Find the concentration of sodium chloride required to render 1%w/v solution of cocaine hydrochloride isotonic with blood plasma. The freezing point of 1% w/v cocaine hydrochloride is -0.090ºC and that of sodium chloride is -0.576ºC. Ans : 0.746% w/v 52

53. PH 101.1753 Based on molecular concentration Freezing point of a solution depends on the concentration of the solute dissolved in it Greater the concentration of solute, lower will be the freezing point. If 1 gram molecular weight of a substance is dissolved in 100 gm of water, the resulting solution is said to have a gram molecular concentration of 1 %. 53

54. PH 101.1754 An aqueous solution having 1% molecular concentration depresses the freezing point to -18.6ºC. The freezing point of blood plasma is -0.52ºC. Therefore the molecular concentration of blood plasma and lachrymal fluid is A depression of 18.6º C is given by a mol. Conc. = 1% A depression of 0.52ºC is given by a mol. Conc.= 1 x 0.52 = @ 0.03% 18.6 54 Based on molecular concentration

55. PH 101.1755 Hence any solution which have a molecular concentration of 0.03 % will be isotonic with blood plasma and tears. The formula for calculating the w/v % of non- ionizing substance required to make the solution isotonic with blood plasma is Based on molecular concentration

56. PH 101.1756 w/v % of substance required = 0.03 x gram molecular weight of the substance For ionizing substances : w/v % of substance required = 0.03 x gram molecular weight of the substance number of ions yielded by the substance Based on molecular concentration

57. PH 101.1757 Solved problem: 2 Calculate the quantity of dextrose required to make a solution isotonic with blood plasma. The molecular weight of dextrose is 180 and is non-ionizing. By applying the formula %w/v of dextrose required = 0.03 x gm. Mol. Weight of dextrose = 0.03 x 180 = 5.4% Therefore 5.4 gm of dextrose is required for making 100 ml solution isotonic with blood plasma. 57

58. PH 101.1758 Exercise: 2 Determine the quantity of sodium chloride required to prepare a solution isotonic with blood plasma. Molecular weight of sodium chloride is 58.5 and it dissociates into 2 ions. Ans :- 0.88% w/v 58

59. PH 101.1759 Based on sodium chloride equivalents This is the value by taking the amount of sodium chloride equivalent to 1 gm of drug. The value of sodium chloride equivalent is noted from the table provided, to the % of medicament in solution. This value is multiplied by the % of medicament. 59

60. PH 101.1760 The result so obtained is subtracted from 0.9% (the concentration of sodium chloride which will make the solution isotonic with blood plasma). The difference in values is the % of sodium chloride required to adjust the tonicity of the solutions. Amount of sodium chloride required = 0.9 – (% of medicament x sodiumchloride equivalent to that %) 60 Based on sodium chloride equivalents

61. PH 101.1761 Solved problem: 3 Find the proportion of sodium chloride required to make a 0.5% solution of potassium chloride isotonic with blood plasma. Sodium chloride equivalent of 0.5% is 0.76. By applying the formula Amount of sodium chloride required = 0.9 – ( 0.5 x 0.76) = 0.9 – 0.38 = 0.52% Therefore 0.52 gms of sodium chloride is required to make 100 ml solution isotonic with blood plasma. 61