Lec 3- Calc 2

Question 1

Body mass index

Answer 1

• BMI= weight (Kg) / Height (m²)
• Weight= 75Kg; Height= 1.72m
• BMI= 75/ 1.72² = 25.351541
• Should be quoted at 2 s.f. e.g. 25

Question 2

Ratio strength

Answer 2

• E.g. prepare 250ml of a 1 in 500 potassium permanganate solution
• Solid in a liquid i.e. 1g drug in 500mL product
• Potassium permanganate- 500mg
• Water- 250mg

Question 3

Millimoles

Answer 3

• One MOLE is the ionic/Molecular weight of an ion expressed in grams
  
  • ONE mole of Na⁺ = 23g
  • ONE mole of NaCl = 58.5g
• One MILLIMOLE is this weight expressed in milligram
  
  • One millimole of Cl^- = 35.5mg
  • One millimole of CaCl₂.2H₂O = 147mg

Question 4

Millimoles- exercise

• How many millimoles of Na⁺ are present in each capsule
• Sodium Bicarbonate capsules contain 500mg of sodium bicarbornate (NaHCO₃)

Answer 4

• Na+= 23
• H= 1
• CO₃= 60
• 500/84= 5.95 mmol

Question 5

Millimoles calcs exercise 2

How much Cl and H₂O in 500mg CaCl₂.6H₂O

Answer 5

Ca= 40

Cl₂= 71

6H₂O= 108

500/219 = 2.28

Cl^- = 4.6

H₂O= 13.8

Question 6

Molar solutions

Answer 6

• A molar solution contains ONE MOLE PER LITRE
• example: Mr of NaCl is 58.5
  
  • i.e. one mole NaCl= 58.5g
  • If 58.5g NaCl is dissolved in 1L concentration = 1mol/L

Question 7

Molar solution example

• What mass of sodium chloride is required to prepare 75mL of a 0.5M solution
• How many millimoles of Cl^- ions will there be in the final solution

Answer 7

58. 5= 1000ml
59. 5/2= 29.25= 1000ml
    (29. 25/1000) x 75ml= 2.19g
60. 19 x 1000= 2190mg / 58.5 = 37.5 mmol

Question 8

Milliequivalents

Answer 8

• One EQUIVALENT is the ionic weight divided by the valency, expressed in grams
  
  • 1 Eq of Na+ = 23/1 = 23g
  • 1 Eq of Ca²⁺= 40/2 = 20g
• One Milliequivalent is the Eq expressed in Mg
  
  • 1mEq of Ca²⁺ = 40/2 = 20mg
• mg= mEq x (MW/valency)
• mEq = mg x (Valency/ MW)
• mEq = mmol x valency

Question 9

Freezing point depresison

Answer 9

• Salts present in a solution affect the freezing point of the solution
• Salts present in blood and tears reduce the freezing point to -0.52’C
• Hence if the formulation also freezezat -0.52’C it will be isotonic with blood and tears

Question 10

Freezing point depression

Answer 10

• 1% w/v NaCl solution freezes at -0.576’C
• Therefore, the percentage of NaCl required to make an isotonic saline solutions
• 1%/-0.576’C = x%/-0.52’C
• x= 0.9% w/v (tonicity of blood and tears)
• Cannot alter the concentration of a drug solution to make isotonic
  
  • Can, however, lower the freezing point of a hypotonic solution by adding NaCl

Question 11

Freezing point depression- example

• 1% w/v apomorphine hydrochloride solution freezes at -0.08’C
• What % w/v NaCl is required to make the solution isotonic with blood

Answer 11

• 1%/-0.576’C
• 1% w/v (apomorphine) = -0.08
• Blood freezes at -0.52: so we do -0.52-0.08= 0.44
• 1% w/v NaCl freezes at -0.576’C
• x% freezes at -0.44’C
• 1%/ -0.576’C = X%/ -0.44’C
• 1%/ -0.576’C x -0.44’C = X = 0.76% w/v NaCl

Question 12

Freezing point depression- exercise

• 1% lidocaine freezes -0.13’C
• What mass of NaCl is required to make 300ml of a 1% w/v lidocaine HCL solution isatanic with blood

Answer 12

• Lidocaine freezes at -0.13’C
• Blood freezes at -0.52
• -0.52- 0.13= -0.39’C
• 1%/-0.576 = X%/-0.52
• 1%/ -0.576 x -0.39’C = 0.68% w/v NaCl
• 0.68 x 3= 2.04g

Question 13

Molecular concentration

Answer 13

• Uses osmotic theory to calculate the tonicity of a solution
• Osmotic pressure depends only on the number of osmotically active particles in the system
• Osmotic pressure of blood plasma is 6.7 atmospheres
• Osmolarity of blood is 0.3M
  
  • i.e. 0.3 moles of osmotically active particles per L of blood

Question 14

Molecular concentration- example

Answer 14

• What mass of dextrose is required to produce 100mL of a solution that is isotonic with the blood
• Osmolarity of blood = 0.3M
• Dextrose contains only 1 osmotically active particle therefore conc of dextrose =0.3M
• MW dextrose= 180 = 1M = 180 g/L
• 0.3M = 0.3 x 180 g/L = 54 g/L
• In 100mL, require 5.4g dextrose

Question 15

Molecular concentration- example 2

• What mass of KCl is required to produce 500mL of a solution that is isotonic with blood
  *

Answer 15

• Osmolarity of blood= 0.3M
• KCl has 2 osmotically active species K⁺ Cl^-
• KCl = 74.5 g/L = 1M
• 0.3 x 74.5g/L = 22.35/2 = 11.175g= 1000mL
• 11.175= 1000mL 5.59g = 500mL

Question 16

Molecular concentration- exercise 3

• What molar conc of NaCl is required to make a 1.8% w/v dextrose solution isotonic with the blood
• Dextrose= 180 (non-ionising species)

Answer 16

• Dextrose= 180MW = 1M = 180g/L
• 1.8g = 100mL or 18g/L
• Molar concentration= 18/180= 0.1M
• dextrose has 1 osmotically active particle, therefore dextrose solution supplies 0.1M
• Require an additional 0.2M of osmotically active particles
• NaCl dissociates to 2 active species, therefore 0.1M NaCl required

Question 17

Sodium chloride equivalents

Answer 17

• Sodium chloride is the most common agent used to adjust the tonicity of solutions
• Amount of NaCl required can be calculated from the previous methods
• Sodium chloride equivalents show the concentration of NaCl that would have the same effect on tonicity as 1% of an alternative agent

Question 18

Sodium chloride equivalent- examples

Answer 18

• What mass of NaCl should be added to 50mL of a 1% w/v procaine HCL solution to make it isotonic with blood
• (NaCl equivalent of procaine HCl is 0.21)
• 1% procaine HCl is equivalent of 0.21% NaCl
• From before 0.9% w/v NaCl is isotonic with blood therefore 0.9%-0.21%= 0.69% w/v NaCl required
• For 50mL, require 0.345g NaCl

Question 19

Sodium Chloride equivalents- exercise

• What mass of dextrose is required to make 50mL of a 1% w/v ephedrine sulfate solution to make it isotonic with blood
• NaCl of ephedrine is 0.23
• NaCl of dextrose is 0.16

Answer 19

• 0.9-0.23= 0.67% w/v NaCl
• However this is dextrose
• 1% dextrose has the same effect as 0.16% NaCl
• 0.67%/0.16% = 4.2% w/v 4.2g = 100mL
• For 50mL= 2.1g

Question 20

Displacement Values

Answer 20

• Displacement value= the number of parts by weight of the drug which displaces one part by weight of theobroma oil
• Displacement values of other fatty bases e.g. witepsolH15 are taken to be the same as theobroma oil
• For glycerol suppository base, these displacement values are 1/1.2 of the values for theobroma oil

Question 21

Displacement values- example

Answer 21

• Rx Paracetamol 125mg; witepsol H15 q.s
• Supply 4 suppositories (using 2g nominal moulds)
• Calculate for 6 suppositories
• Mass of paracetamol required= mass per suppository multiplied by a number of suppositories
• 6 x 125mg = 750mg = 0.75g
• Displacement value of paracetamol is 1.5
• i.e. 1.5g paracetamol will displace 1g base
• Hence 0.75g paracetamol will displace 0.5g base
• A nominal 2g mould actually holds 2.12g of witepsol H15
• For drug-free suppositories, mass Witepsol H15 required
• 6 x 2.12g = 12.72g
• BUT we have already calculated that adding 0.75g paracetamol will displace 0.5g base
• 12.72g - 0.5g= 12.22g

Question 22

Pessary/ suppository displacement

• Rx Drug X= 100mg, Glycerol suppository base q.s., Send 6 pessaries
• Glycerol suppositories= 1/1.2
• Displacement of drug X= 2.5

Answer 22

*