Lec 3- Calc 2 Flashcards

1
Q

Body mass index

A
  • 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
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2
Q

Ratio strength

A
  • 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
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3
Q

Millimoles

A
  • 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
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4
Q

Millimoles- exercise

  • How many millimoles of Na+ are present in each capsule
  • Sodium Bicarbonate capsules contain 500mg of sodium bicarbornate (NaHCO3)
A
  • Na+= 23
  • H= 1
  • CO3= 60
  • 500/84= 5.95 mmol
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5
Q

Millimoles calcs exercise 2

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

A

Ca= 40

Cl2= 71

6H2O= 108

500/219 = 2.28

Cl- = 4.6

H2O= 13.8

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6
Q

Molar solutions

A
  • 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
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7
Q

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
A
  1. 5= 1000ml
  2. 5/2= 29.25= 1000ml
    (29. 25/1000) x 75ml= 2.19g
  3. 19 x 1000= 2190mg / 58.5 = 37.5 mmol
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8
Q

Milliequivalents

A
  • 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
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9
Q

Freezing point depresison

A
  • 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
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10
Q

Freezing point depression

A
  • 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
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11
Q

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
A
  • 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
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12
Q

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
A
  • 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
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13
Q

Molecular concentration

A
  • 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
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14
Q

Molecular concentration- example

A
  • 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
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15
Q

Molecular concentration- example 2

  • What mass of KCl is required to produce 500mL of a solution that is isotonic with blood
    *
A
  • 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
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16
Q

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)
A
  • 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
17
Q

Sodium chloride equivalents

A
  • 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
18
Q

Sodium chloride equivalent- examples

A
  • 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
19
Q

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
A
  • 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
20
Q

Displacement Values

A
  • 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
21
Q

Displacement values- example

A
  • 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
22
Q

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
A

*