Equations

Question 1

Density

Answer 1

= mass (or weight) / volume

Question 2

Specific Gravity

Answer 2

= g/ml (density in these units)

Question 3

% Error

Answer 3

= (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

Question 4

USP % Error (for pharmD measuring)

Answer 4

5%

Question 5

Sensitivity Requirement (SR)

Answer 5

weight of material that will move the indicator 1 marked unit on index plate

Question 6

Class A Balance SR

Answer 6

= 6 mg

Question 7

Minimum Weighable Quantity

Answer 7

= (SR x 100%) / 5% (or whatever % error)

Question 8

IBW

Answer 8

= (45.5 or 50) + (2.3 x inches over 5 ft)

Question 9

Clark’s Rule

Answer 9

in absence of specific info, adjust dose for really big or small person by ratio of pt wt to 70kg

Question 10

BSA (m^2)

Answer 10

{[ht (cm) x wt (kg)]/3600} ^1/2

Question 11

Average Adult BSA

Answer 11

1.73 m^2

Question 12

CrCl

Answer 12

= {[(140 - age) x wt(kg)] / (72 x SCr (mg/dl))} x 0.85maybe

Question 13

Schwartz

Answer 13

=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

Question 14

% w/w (% strength)

Answer 14

= g (ingredient) / 100 g

Question 15

% v/v (% strength)

Answer 15

= ml (ingredient) / 100 ml

Question 16

% w/v (% strength)

Answer 16

= g (ingredient) / 100 ml

Question 17

1 Molar Solution

Answer 17

= MW in g / L

Question 18

1 Millimolar Solution

Answer 18

= (1000 x MW in g) / L

Question 19

Equivalent Weight

Answer 19

= MW / Valence

Question 20

mEq

Answer 20

= Eq Wt / 1000

Question 21

Milliosmoles (mOsm) Def

Answer 21

particles in solution and is based on total cation and anion #s

Question 22

Milliosmolarity

Answer 22

= mOsm / L
= (mols/L) x # species x (1000 mOsm/mols)
# species = # ionic species upon complete dissociation
(NaCl = 2, CaCl2 = 3)

Question 23

Milliosmolality

Answer 23

= mOsm / kg of solution

Question 24

mg %

Answer 24

mg/100 ml

Question 25

convert 230 mg/dl to % w/v

Answer 25

230 mg/dl 1 dl = 100 ml 230 mg = 0.23g % w/v = 0.23 g / 100 mL = 0.23%

Question 26

Find Na+ mg in 5 million units. PenG has 2.2 mEq Na+ (MW=23, val=1) per 1 million units

Answer 26

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

Question 27

Parts Per Million (ppm) | Find 1:25,000 in ppm

Answer 27

1/x = 25,000/1,000,000 | x=40 ppm

Question 28

Simple Dilution Equations

Answer 28

C1 x Q1 = C2 x Q2 [C=concentration, Q=quantity)] | OR C1/C2 = Q1/Q2

Question 29

How much h2o added to 250 ml of 0.2% w/v to make 0.05% solution?

Answer 29

C1=-0.2%, Q1=250 ml, C2=0.05%, x=Q2-250ml Q2=(C1 x Q1)/C2 Q2 = 1000 ml, x = 750 ml

Question 30

Alcohol Solutions Dilutions

Answer 30

Ethyl alcohol and water solutions is not volume additive because contraction on mixing.

Question 31

Etoh Solution Example: Find amount h2o to be added to 100 ml of 95% ethanol to make 50% ethanol.

Answer 31

``` 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. ```

Question 32

Concentrated Acids

Answer 32

C given in %w/w but pharmD preparing must express in % w/v using SpGr

Question 33

Acid Example: What volume of 35% HCl (w/w) SpGr=1.2 is needed to make 500 ml of 5% w/v?

Answer 33

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

Question 34

Triturations

Answer 34

are 10% w/w dilutions of finely powdered (triturated) mixtures of drug in an inert substance

Question 35

Trituration Example: Find wt of drug to make 30 doses of 0.25 mg ea?

Answer 35

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.

Question 36

``` Simple Concentrations (Making More Concentrated) 3200 g of 5% oint. How many g drug for 20% oint? ```

Answer 36

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

Question 37

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%?

Answer 37

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

Question 38

Alligation Medial: 100 ml of 10% solution added to 200 ml of 20% solution and 300 ml of 30% solution. Final Conc?

Answer 38

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%

Question 39

Dissociation factor (i) of 100 molecules of NaCl

Answer 39

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

Question 40

What is the i for 80% dissociation with 2 ions? 3 ions? 4 ions?

Answer 40

2 ions 80%: i = 1.8 3 ions 80%: i = 2.6 4 ions 80%: i = 3.4

Question 41

i for 60% dissociation into 3 ions

Answer 41

60 particles x 3 + 40 undissociated = 220 | 220/100 = 2.2

Question 42

NaCl Equivalent (E value)

Answer 42

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)

Question 43

NaCl Equivalents (E) Ex: Find wt needed of NaCl to make 50 ml of isotonic solution with 1000 mg drug (E=0.23)

Answer 43

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

Question 44

NaCl Equiv Ex2: Drug MW 376, i=2.6 is provided as 500 mg in 60 ml made to be isotonic. Wt of NaCl?

Answer 44

(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

Question 45

IV Flow Rate Ex: 250 mg drug in 500 ml d5w, what should flow rate be for 50 mg/h

Answer 45

250 mg/500 ml = 0.5 mg/ml | 50 mg/h x 1ml/0.5 mg = 100 mg/h

Question 46

Henderson-Hasselbalch Eqns (Buffers)

Answer 46

Weak Acid: pH = pKa + log (salt/acid) | Weak Base: pH = pKw - pKb + log (base/salt), pKw=14

Question 47

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.

Answer 47

Weak Acid: pH = pKa + log (salt/acid) | pH = 4.76 + log (0.5/0.05) = 5.76

Question 48

Buffer Ex 2: Find pH 0.5 M ammonia pKb=4.74 and 0.05 M ammonium Cl

Answer 48

Weak Base: pH = pKw - pKb + log (base/salt), pKw=14 | pH=14-4.74 + log(0.5/0.05) = 10.26

Question 49

Temperature Conversion

Answer 49

9 C = 5F - 160

Question 50

Temp Ex: 100 F = what in C?

Answer 50

``` 9C = 5F - 160 C = ((5x100) - 160) / 9 = 37.8 ```