Equations Flashcards
Density
= mass (or weight) / volume
Specific Gravity
= g/ml (density in these units)
% Error
= (error x 100%) / quantity desired
= (SR x 100%) / Minimum Weighable Quantity
Error is max error of measurement (larger - smaller)
Quantity Desired is total amount measured
USP % Error (for pharmD measuring)
5%
Sensitivity Requirement (SR)
weight of material that will move the indicator 1 marked unit on index plate
Class A Balance SR
= 6 mg
Minimum Weighable Quantity
= (SR x 100%) / 5% (or whatever % error)
IBW
= (45.5 or 50) + (2.3 x inches over 5 ft)
Clark’s Rule
in absence of specific info, adjust dose for really big or small person by ratio of pt wt to 70kg
BSA (m^2)
{[ht (cm) x wt (kg)]/3600} ^1/2
Average Adult BSA
1.73 m^2
CrCl
= {[(140 - age) x wt(kg)] / (72 x SCr (mg/dl))} x 0.85maybe
Schwartz
=Kids CrCl = k x ht(cm) / SCr (mg/dl) Premature <1y k=0.45 Child/Teen Girl k=0.55 Teen Boy k=0.7
% w/w (% strength)
= g (ingredient) / 100 g
% v/v (% strength)
= ml (ingredient) / 100 ml
% w/v (% strength)
= g (ingredient) / 100 ml
1 Molar Solution
= MW in g / L
1 Millimolar Solution
= (1000 x MW in g) / L
Equivalent Weight
= MW / Valence
mEq
= Eq Wt / 1000
Milliosmoles (mOsm) Def
particles in solution and is based on total cation and anion #s
Milliosmolarity
= mOsm / L
= (mols/L) x # species x (1000 mOsm/mols)
# species = # ionic species upon complete dissociation
(NaCl = 2, CaCl2 = 3)
Milliosmolality
= mOsm / kg of solution
mg %
mg/100 ml
convert 230 mg/dl to % w/v
230 mg/dl
1 dl = 100 ml
230 mg = 0.23g
% w/v = 0.23 g / 100 mL = 0.23%
Find Na+ mg in 5 million units. PenG has 2.2 mEq Na+ (MW=23, val=1) per 1 million units
5 million units has 5 x 2.2 = 1 mEq Na+
Eq Wt = 23/1 = 23 g(?) so 1 mEq = 23g/1000 = 23 mg
Na+ (mg) = (23 mg/mEq) x 11 mEq = 253 mg
Parts Per Million (ppm)
Find 1:25,000 in ppm
1/x = 25,000/1,000,000
x=40 ppm
Simple Dilution Equations
C1 x Q1 = C2 x Q2 [C=concentration, Q=quantity)]
OR C1/C2 = Q1/Q2
How much h2o added to 250 ml of 0.2% w/v to make 0.05% solution?
C1=-0.2%, Q1=250 ml, C2=0.05%, x=Q2-250ml
Q2=(C1 x Q1)/C2
Q2 = 1000 ml, x = 750 ml
Alcohol Solutions Dilutions
Ethyl alcohol and water solutions is not volume additive because contraction on mixing.
Etoh Solution Example: Find amount h2o to be added to 100 ml of 95% ethanol to make 50% ethanol.
Q2=(C1xQ1)/C2 = (95% x 100 ml)/50% Q2 = 190 ml so add 90 ml EXCEPT it will contract so it will be more than 90 ml.
Concentrated Acids
C given in %w/w but pharmD preparing must express in % w/v using SpGr
Acid Example: What volume of 35% HCl (w/w) SpGr=1.2 is needed to make 500 ml of 5% w/v?
Step 1. Find wt HCl needed for 5% 500 ml so 5% = 5g/100ml so 25 g in 500 ml.
Step 2. What weight of 35% has 25 g HCl (35%=35 g per 100 g). So x/100g = 25 g/35g and x = 71.4 g
Step 3. Find 71.4g in ml. So 1.2 g/mL = 71.4 g/x ml so x=59.5ml
Triturations
are 10% w/w dilutions of finely powdered (triturated) mixtures of drug in an inert substance
Trituration Example: Find wt of drug to make 30 doses of 0.25 mg ea?
30 x 0.25 mg = 7.5 mg drug needed
Trituration is 10% so 10g drug:100gmix = 1g:10g = 1 mg:10mg.
1mg drug:10mg mix = 7.5mg drug: x mix So x=75 mg trituration.
Simple Concentrations (Making More Concentrated) 3200 g of 5% oint. How many g drug for 20% oint?
Opposite of dilution so find conc of base or diluent
95% base x 3200 g = 80% base x Q2
Need to add 600 g to make 3800=Q2
Alligation two stock solutions A% > B% combined to make C% (inbetween). What proportion 20% w/v added to 5% w/v to make 15% w/v? How much of each to make 75 ml of 15%?
A-C = parts of B, C-B = parts of A
20-15 = 5 parts B, 15-5 = 10 parts A, Total 15 parts
10 of 15 parts 75 ml is A = 50 ml and B = 25ml
Alligation Medial: 100 ml of 10% solution added to 200 ml of 20% solution and 300 ml of 30% solution. Final Conc?
0.1 x 100 ml = 10 ml
0.2 x 200 ml = 40 ml
0.3 x 300 ml = 90 ml
=600 ml total and 140 ml drug
(140/600) x 100=23.3%
Dissociation factor (i) of 100 molecules of NaCl
NaCl dissociates at 80% so 80 particles Na, 80 particles Cl and 20 particles NaCl. Total Dissociated is 180.
i = total dissociated/start =180/100=1.8
What is the i for 80% dissociation with 2 ions? 3 ions? 4 ions?
2 ions 80%: i = 1.8
3 ions 80%: i = 2.6
4 ions 80%: i = 3.4
i for 60% dissociation into 3 ions
60 particles x 3 + 40 undissociated = 220
220/100 = 2.2
NaCl Equivalent (E value)
number of grams of NaCl that would produce the same tonicity effect as 1 g drug
NaCl Equivalent =
(MW of NaCl x drug dissociation factor) /
(MW of drug x NaCl dissociation factor)
=(58.5 x drug i) / (MW drug x 1.8)
NaCl Equivalents (E) Ex: Find wt needed of NaCl to make 50 ml of isotonic solution with 1000 mg drug (E=0.23)
Isotonic saline 0.9% = 0.9 g/100ml so 50 ml has 0.45g
1000 mg drug = 1 g drug with E=0.23
So 1 g drug x 0.23 = 0.23 g NaCl
0.45 g - 0.23 g = 0.22 g NaCl needed
NaCl Equiv Ex2: Drug MW 376, i=2.6 is provided as 500 mg in 60 ml made to be isotonic. Wt of NaCl?
(58.5 g x 2.6) / (376 x 1.8) = 0.22 = E
0.9g/100 ml = x g/60 ml = 0.54 g
1 g drug = 0.22 g NaCl so 0.5 g drug = 0.11 g NaCl
0.54 - 0.11 = 0.43 g NaCl
IV Flow Rate Ex: 250 mg drug in 500 ml d5w, what should flow rate be for 50 mg/h
250 mg/500 ml = 0.5 mg/ml
50 mg/h x 1ml/0.5 mg = 100 mg/h
Henderson-Hasselbalch Eqns (Buffers)
Weak Acid: pH = pKa + log (salt/acid)
Weak Base: pH = pKw - pKb + log (base/salt), pKw=14
Buffer Ex: Find pH of buffer solution with 0.5 mol (M) sodium acetate and 0.05 M acetic acid. pKa acetic acid = 4.76.
Weak Acid: pH = pKa + log (salt/acid)
pH = 4.76 + log (0.5/0.05) = 5.76
Buffer Ex 2: Find pH 0.5 M ammonia pKb=4.74 and 0.05 M ammonium Cl
Weak Base: pH = pKw - pKb + log (base/salt), pKw=14
pH=14-4.74 + log(0.5/0.05) = 10.26
Temperature Conversion
9 C = 5F - 160
Temp Ex: 100 F = what in C?
9C = 5F - 160 C = ((5x100) - 160) / 9 = 37.8
