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