This presentation quotes various pharmaceutical calculations with examples. The following aspects like percentage calculations, alcoholic dilutions, Alligation method, proof spirit calculations, isotonicity adjustment, posology, temperature measurements, dialysis clearance, Pharmacokinetics calculations were covered with examples.

1. PHARMACEUTICAL CALCULATIONS
UNDER THE ESTEEMED GUIDANCE OF
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Presented By
B.LAKSHMI DURGA
11AB1R0082
Mrs.PALLAVI.K M.Pharm
0011 0010 1010 1101 0001 0100 Assistant 1011
Professor
1 VIGNAN PHARMACY COLLEGE
Approved by PCI,AICTE,New Delhi.
Affiliated to JNTU,Kakinada.
Vadlamudi-522213,Guntur,A.P 1

2. CONTENTS
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Percentage calculations
Alcoholic dilutions
Alligation method
 Proof spirit calculations
 Isotonic solutions
 Posology
 Temperature measurements
 Dialysis Clearance
 Pharmacokinetics
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3. PHARMACEUTICAL
CALCULATIONS
1 0011 Pharmacokinetic calculations
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4. INTRODUCTION
 To have a complete understanding of various types of calculations,
which are involved in dispensing, it is desirable that the
pharmacist should have a thorough knowledge regarding weights
and measures which are used in calculations.
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 There are two systems of weights and measures.
1.The Imperial system
2.The Metric system
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5. IMPERIAL SYSTEM
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• Imperial system is an old system of weight and measures.
Imperial
system
Avoirdupois
system
Apothecaries
system
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6. Avoirdupois system :
• In this system the pound is the standard unit for
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weighing and all measures of mass are derived from the
Imperial standard pound (lb).
Apothecaries system :
• This system is also known as Troy system.
• The grain is the standard unit in this system and all
other weights are derived from it.
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7. CONVERSION TABLE-1
FROM APOTHECARIES TO AVOIRDUPOIS
1 0011 1 Lb 16 oz (avoir)
1 Lb 7000 grains
1 oz (avoir) 7000/16 = 437.5 grains
20 grains (gr) 1 scruple (̶) )
60 grains 1 drachm
480 grains 1 ounce (apothe)
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12 ounces 1 pound (Lb)
(apothe)
5760 grains 1 pound
(apothe)
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8. Measurement of capacity in imperial system :
• The standard units for capacity is same for both the
avoirdupois and the apothecaries systems.
• The gallon is the standard unit and other measures of capacity
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are derived from it.
8
1gallon (c) = 160 fluid ounces
1/4th of gallon = 1 Quart
1/8th of gallon = 1 pint(O)
1/160th of gallon = 1 flounce
1 quart = 40 flounces
1 pint = 20 flounces
1 flounce = 480 minims
1 fl drachm = 60 minims
CONVERSION TABLE-2

9. METRIC SYSTEM
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• Metric system is used in the Indian pharmacopeia for the
measurement of weights and capacity.
• It was implemented in India from 1st April , 1964 in pharmacy
profession.
Measurement of weight in metric system :
• A kilogram is the standard unit for measurement of
weight and all other measures are derived from it.
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10. 0010 1010 1101 0001 01010k 1i0lo1g1 ram = 1000gms
0011 1 hectogram = 100 gms
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1 decagram = 10 gms
1 decigram = 0.1 gms
1 centigram = 0.01 gms
1 milligram = 0.001 gms
1 microgram = 0.000,001 gms
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CONVERSION TABLE - 3
FROM HIGHER TO LOWER

11. Measurement of capacity in Metric System
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A litre is the standard unit for measurement of capacity and all
measures of capacity are derived from it.
1 litre ( lt ) = 1000 milliliters (ml)
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12. Conversion Tables
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 The pharmacopeia of India uses only the metric system in
formulae, But the prescription are still written in the imperial
system by many old time physicians. So a conversion tables is
used by the pharmacists.
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13. Metric system Imperial system
1 kilogram (kg) = 2.2 lb ( pound )
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• Weight measures
• Capacity measures
30g = 1 ounce
450g = 1 pound ( avoir )
1g = 15 grains
60mg = 1 grain
1000 ml = 1Quart
500 ml = 1 pint
30 ml = 1 fluid ounce
4 ml = 1 fluid drachm
1 ml = 15 minims
0.06 ml = 1 minim

14. Conversion table for Domestic Measures
Domestic Measure Metric system Imperial system
1 drop 0.06 ml 1 minim
0011 1 tea spoonful 4.00 ml 1 fluid drachm
1 desert spoonful 8.00 ml 2 fluid drachm
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1 tablespoonful 15.00 ml 4 fluid drachm
2 tablespoonful 30.00 ml 1 fluid ounce
1 wine glassful 60.00 ml 2 fluid ounce
1 tumblerful 240.00 ml 8 fluid ounce
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15. CALCULATIONS BASED ON DENSITY :
 Density is defined as the mass of a substance per unit volume.
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 It has the units of mass over volume.
 Specific gravity is defined as the ratio of the mass of a substance
in air to that of an equal volume of water.
 In the metric system both density and specific gravity are
numerically equal.
WEIGHT = VOLUME * DENSITY
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16. Example problems for calculations based on density
1. Calculate the volume of 2kg of glycerin. The density of
glycerin is 1.25g/ml.
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= 1600 ml
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VOLUME =
=
=

17. 0010 1010 2. Calcula 1te1 0t1h 0e0 w01e 0ig10h0t 1o0f1 1200ml of alcohol whose
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density is 0.816 g/ml
WEIGHT = VOLUME X DENSITY
= 200 ml x 0.816 g/ml
= 163.2g
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18. 1 0011 42
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19. Example problem For Alcoholic Dilutions
 Calculate the amount of 95% alcohol required to prepare
400ml of 45% alcohol
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volume required = 400ml
percentage of alcohol required = 45
percentage of alcohol used = 95
Volume of strong alcohol =
to be used
=
=
= 184.47 ml
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20. 0011 Alligation method
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21. ALLIGATION METHOD
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 When the calculation involves mixing of two similar preparations
of different strenghts , to produce a preparation of intermediate
strength , the alligation method is used.
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22. Stronger Percentage Weaker percentage
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Required Percentage
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Required percentage –
Weaker Percentage
Stronger percentage –
Required Percentage

23. Example problems For Alligation method
1. Calculate the volume of 95% alcohol required to prepare 600ml of
70% alcohol.
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Volume required = 600 ml
Percentage of alcohol required = 70
Percentage of alcohol used = 95
70
0
95
70 -0=70 95-70=25
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24. (Problem Contd…..)
• 70 parts of 95 %alcohol and 25 parts of water will produce the
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required %alcohol.
Quantity of 95% alcohol required =
= 422.10ml
Quantity of water required =
=157.90ml
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25. 2. How much quantity of 60%,50%,30% and 20%
alcohol should be mixed to get 40% alcohol.
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40
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60
20
50
30
40-20=20 parts of 60%
40-30=10 parts of 50%
50-40=10 parts of 30%
60-40=20 parts of 20%

26. ( Problem contd……)
• Quantity required = 20parts of 60%+10 parts of 50%+
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10 parts of 30%+20 parts of 20% alcohol are mixed together, the
resulting solution will produce 60% alcohol.
Calculation can be checked by the following method:
= (20 X 60+10 X 50+10 X 30 +20 X 20)
=1200+500+300+400
=2400
20+10+10+20=60
i.e., 60 X 40 = 2400 26

27. PROOFSPIRIT
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CALCULATIONS
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28. Introduction
Proof spirit calculations
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• The strength of alcohol is calculated in proof degrees.The
Indian standards of 100% proof spirit is equal to 57%v/v of
ethyl alcohol.
i.e., 100% p.s = 57%v/v ethyl alcohol
• If the value is more than 57% then it is said to be as over
proof spirit.
• If the value is less than 57% then it is said to be as
under proof spirit. 28

29. (Introduction contd….)
• In India , the exercise duty is calculated in terms of rupees per
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liter of proof alcohol. So, any percentage volume in volume of
alcohol can be converted into proof strength and vice versa by
using the following method.
i. Multiply the percentage strength of alcohol by 1.753 and
deduct 100 from the product.
ii. If the result is positive , it is known as over proof
iii. If the result is negative , it is known as under proof.
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30. • 1.753 is obtained as follows:
57.1 volume of ethyl alcohol = 100 ml of proof spirit.
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1 volume of ethyl alcohol = 100 / 57.1
= 1.753 volume of proof spirit
Example problems for Proof Spirit calculations
Calculate the strength of 30 0 O.P and 400 U.P
300 OVER PROOF = 100 + 30 = 130
400 Under proof = 100 – 40 = 60
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31. (Problem Contd….)
Alcohol strength =
=
= 74.15%v/v
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Alcohol strength =
=
= 34.23%v/v
Check :
 If strength is 74.15%v/v
74.15 x 1.753 – 100 =129.9 – 100 = +29.90 or 300 O.P
If strength is 34.23 %v/v
34.23 x 1.753 – 100 = 60.00 – 100 = - 400 U.P

32. 2.What is the proof strength of 95% v/v alcohol.
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By applying the formula:
percentage strength of alcohol x 1.753 – 100
= 95 x 1.753 – 100
= 166.53 – 100
= +66.530 O.P
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33. 3. How many proof gallons are contained in 5 gallon of
70% v/v alcohol
Applying the formula:
Value in proof = % strength of alcohol x 1.753 – 100
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= 70 x 1.753 – 100
= 122.71 – 100
= 22.710 o.p
100 gallons of 70% v/v alcohol =122.71 units of p.s
1 gallon of 70% v/v alcohol = 122.71/100 = 1.2271
5 gallons of 70% alcohol = 1.2271 x 5
= 6.1355 gallons of p.s
5 gallons of 70% v/v alcohol are equivalent to 6.1355
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gallons of p.s

34. 1 0011 42
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ISOTONIC SOLUTIONS
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35. Isotonic solutions
0010 1010  1101 0001 0100 1011 Isotonic solutions are which have same osmotic pressure or
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equal solute concentrations.
 0.9% sodium chloride solution is considered to passes the
same osmotic pressure and hence it is a standard solution
which is isotonic with blood plasma.
 Any concentration above this (0.9%) is considered as
hypertonic and below this (0.9%) is considered as
hypotonic.
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36. ADJUSTMENT OF ISOTONICITY:
FP depression
/Cryoscopic
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method
ISOTONIC
ADJUSTMENT
Nacl
Equivalent
method
White
Vinscent
method
Molecular
Concentration
method

37. I. Freezing point method/ Cryoscopic method
%w/v of adjusting substance needed =
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Where
PSM = percentage strength of medicament
a = Freezing point of the unadjusted solution
b = Freezing point of a 1% w/v solution of a adjusting
substance
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38. Example Problem for Freezing point method
 Using cryoscopic method. How will you prepare 1%
solution of boric acid ,iso osmotic with blood plasma.
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[Hint : The freezing point of 1%w/v solution of boric acid is
0.2880 c
The freezing point of 1%w/v solution of sodium chloride is
0.5760 c]
Percentage of w/v of sodium chloride required =
=
= 0.402% w/v
1gm of boric acid is needed for 0.402gm of Nacl
to become isotonic.
.
38

39. 2. Nacl Equivalent method:
Quantity of Nacl required to adjust tonicity = 0.9 – ( PSM x E )
Where
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PSM = Percentage strength of medicament.
E = Nacl equivalent (quantity of Nacl i.e., equivalent to
1gm of drug)
3. White Vincet method :
V = W x E x 111.1
Where
V = volume of isotonic solution in ml that can be
prepared by dissolving drug in water
W = weight of the drug.
E = Nacl equivalent .
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40. 4 . Molar (or) Molecular concentration :
%W/V of adjusting substance required =
Where
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M = gram molecular weight.
N = no.of ions into which the substance is ionised.
Problem:
• How to prepare an iso osmotic solution using dextrose.
[Hint: mol.wgt of dextrose = 180
[dextrose is non ionising substance]
W = 0.03 M
W = 0.03 x 180 = 5.4 g/100ml
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41. POSOLOGY
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42. POSOLOGY
0011 0010 1010 1101 0001 0100 1011
Posology refers to the calculation of doses for children.
1. Proportion to Age :-
(a) Young’s formula:
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Dose for a child =
The above formula is used for calculating the
doses for children 12 years of age.
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43. • Example problem for young’s formula
What will be the dose for a child of 5 years if the adult
dose of a drug is 400 mg.
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Dose of the child =
=
= 117 mg (approximately)
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44. (b) Dilling’s formula:
Dose of a child =
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The above formula is used for calculating the doses of a children in
between 4 to 20 years of age.
Example problem:
 What will be the dose for a child of 10 years if the
adult dose of a drug is 600mg
Dose of a child =
= 300 mg 44

45. (c) Freid’s formula:
Dose for a child =
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The above formula is applicable only for infants.
Example problem:
 What is the dose for an 8months old infant if the
average adult dose of a drug is 250mg
Dose for the child =
= 13.3mg
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46. (2) Calculations based on body weight:
Catzel rule:
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Dose for the child =
The average body surface area for an adult =1.73m2
Hence
Dose for the child =
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47. Example problem for calculations based on body weight
 Calculate the dose for a child that has a body surface
area of 0.57m2 , when the adult dose of a drug is 50mg.
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Child dose =
=
= 1.65 mg
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48. Calculations based on body weight:
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• Dose =
• Dose =
Rule is applicable only when child dose is less than
150 lb or 70kg.
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49. • Example problem for calculations based on body weight
The dose of a drug is 5mg/kg body weight. How much of drug
required for a boy of 12 years weighing 21kg.
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Dose =
=1.5mg
(3) Calculations based on body surface area:
Most accurate method commonly used in onchology department
Mosteller rule:
BSA ( M2 ) =
49

50. Example problem for calculations based on body
surface area
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 Calculate the BSA of a boy of height 165 cm and
weighing 65kg.
BSA =
=
= 1.726
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51. 1 0011 42
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TEMPERATURE
MEASUREMENTS

52. TEMPERATURE
MEASUREMENTS
0010 1010 • The tem p1e1r0a1t u0r0e01i s01g0e0n 1e0r1a1lly measured in pharmacy by using
0011 either Fahrenheit or Centigrade thermometers.
1 • The relationship of Centigrade (C) and Fahrenheit (F) degrees
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is
90 (C) = 50 (F) – 160
Where
0 C is the number of degrees Centigrade.
0 F is the number of degrees Fahrenheit.
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53. Example problem for converting Fahrenheit to Centigrade
Convert 1200 F into 0 C
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90 (C) = 50 (F) -160
= 5(120) – 160
= 600 –160
= 440
0 C = 440/9
= 48.90 C
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54. Example problem for converting Centigrade to Fahrenheit
Convert 300 c into 0F
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50 F = 90 C + 160
= 9(30) + 160
= 270 + 160
= 430
0 F =
= 860 F
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55. 1 0011 42
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DIALYSIS CLEARANCE
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56. • Dialysis clearance is also called as Dialysance.
• It is used to express the ability of machine to clear the drug from
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blood.
Cld = Q(Cin – Cout ) / Cin
Where
Cld = dialysis or dialysis clearance.
Q = blood flow rate to dialyzer.
Cin = concentration of drug in blood entering the
dialyzer
Cout= concentration of drug in blood leaving the
dialyzer.
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DIALYSIS CLEARANCE

57. 1 0011 42
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.
57
Semipermeable
membrane
Blood
dialysate
Diagrammatic representation of a haemodialyser .
The blood and the dialysate flow counter-currently.

58. Example problem for Dialysis Clearance
 Calculate the dialysis clearance if the blood flow rate
to the dialyzer 50ml/min and concentration of drug
entering and leaving the dialyzer is 100 and
200microgram/ml respectively.
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Dialysis Clearance = Q[C in – Cout ] / Cin
=
=
= 40ml/min
58

59. 1 0011 42
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PHARMACOKINETICS
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60. • Pharmacokinetics is the mathematical analysis of process of
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ADME.
• The velocity with a reaction or a process occurs is called as its
rate.
• The manner in which the concentration of drug influences the
rate of reaction or process is called as the order of reaction or
order of a process.
• consider the following equation
Drug A Drug B
•
PHARMACOKINETICS
60

61. Rate of reaction can be expressed as follows
dc/dt = -k.cn
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Where
k = rate constant
n= order of reaction
• If n = 0 , it is a zero order process
• If n=1 , it is a first order process
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62. Conversion of various order of reactions
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Basic Equation Final Concentration T1/2
Zero order dc/dt = -k.c0 C = C0 – kt T1/2 = 0.5c0 /k
First order dt/ct = -k.c1 log C = log T1/2 = 0.693/k
Second order dc/dt = -k.c2 T1/2 = 1/c0 k
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63. Example problem for pharmacokinetic calculations
 A penicillin solution has a half-life of 21days.How long
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will take it for the potency to drop to 80% of the initial
potency.
t1/2 = 0.693/k
21 = 0.693/k
k = 0.693/21
= 0.033
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64. t1/2 =
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=
= 69.78 * log(1.25)
= 69.78 * 0.096
= 6.6988 ~ 7days
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65. CONCLUSION
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• Pharmaceutical calculations is a predominant area which
plays keyrole in various areas of pharmacy like
 Conversion of one measurement system to anthor.
 Dose calculations for dispensing prescriptions.
 Dilution of alcohols.
 Filling of capsules etc….
• Pharmaceutical calculations also play a crucial role
in diagnosis , Treatment and prevention of ailments.
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66. REFERENCES
Remington - science and practice of pharmacy.
Patrick J Sinko – Martins physical pharmacy and phamaceutical
sciences.
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1 M.E.Aulton - Pharmaceutics,The science of dosage form design.
Haward c Ansel / Mitchell J stoklosa Lippincott –
Pharmaceutical Calculations.
Leon Shargel , Alan Mutnic , Paulf Souney , Larry N Swa –
Comprehensive Pharmacy Review. 66

67. REFERENCES
0011  0010 1010 1101 0001 0100 1011 R.M.Mehta – Dispensing Pharmacy. Pg.no:77-103
 B.M.Brahmankar Sunil B.Jaiswal – Pharmaceutics and
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pharmacokinetics. Pg.no: 211-213,239-243.
 R.M.Mehta – Pharmaceutics-II. Pg.no:14-38,65.
 CVS.Subrahmanyam – Essentials of Physical Pharmacy.
pg.no:29,195-198
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68. ACKNOWLEDGEMENT
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 I Sincerely thank my guide Mrs.K.Pallavi garu for
her constant guidance and support.
 I’m very thankful to our beloved principal sir
Dr.P.Srinivas Babu garu and also the seminar
committee.
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69. 1 0011 42
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