Concentration Expressions

D5w = Dextrose 5%

1 DL = means 1 Deciliter = 1/10 of the Liter = 100 ml

1 PPM = 1 gm /1,000,000 ml

5% W/V = 5 gm /100 ml

5% W/W = 5 gm /100g

1:50 W/V = 1 gm /50 ml

1:50 W/W= 1 gm/50 gm

1:50 V/V= 1 ml/50 ml

1 part in 500 part Phenol in Water = 1 gm / 500 ml (called Ratio strength)

1 teaspoonful = 5 ml1 dessertspoonful =10 ml1 tablespoonful =15 ml

1 teacupful =120 ml1 measuring cup =glassful = 240 ml

1 Kg = 2.2 Lb1 Gallon = 3784 ml1 Fz (Fluid Ounce) = 29.57 ml1 Pint = 473 ml1 Inch = 2.54 cm

1 Gallon = 4 quarters / 8 pints / 16 Fluid Ounce

9 C = 5F-160 Ex. Convert 212 C to F (9x 212 = 5 F -160, F= 212

Molarity = Moles of solute /Liters of Solution Molality = Moles of solute/Kg of Solvent Normality = Weight of solute (gms) /gm equivalent weight x volume (liters)

N.B. Normality + Molarity changes with Temp. Molality doesn’t change by temp

Dilution from Stock

I think it’s better to name the quantity required by Volume required since it’s in mL

Also the expression Amount/Unit dose is simply the Final Concentration mg/ml

This type of equations has the expression (When X ml is diluted to Y ml a final conc. is ….) Or (when diluted X in Y using only numbers ex. Diluted 1 in 10)The X is a Low volume while the Y is the higher volume

We use this expression to get the Dilution Factor and the needed outcome is usually the Weight of the stock solution.

Ex.1 How many grams of Terbutaline Sulphate would be required to prepare a 100 ml solution such that when 1 ml is diluted to 5 ml, a final concentration of 500 mcg/ml of Terbutaline Sulphate would be produced

X mg/100 ml x 1ml /5ml = 0.5 mg/1 ml x = 250 mg Terbutaline Sulphate

Ex.2 what is the amount of magnesium chloride (Mgcl2.6 H20 ; MW= 203.3) required to prepare 100 ml of a solution such that 10 ml is diluted to 1 liter yields a solution containing 0.3 Meq Mg+2 /ml

Given:

Volume required = 100 ml, Dilution Factor = 10/1000, Final Conc. = 0.3 Meq Mg+2/ml (has to be mg/ml)

Solution

Mgcl2.6 H20 Mg+2 + 2 Cl- + 6 H201 millimole Mgcl2.6 H20 2 Meq Mg+2 = Mgcl2.6 H20 m.wt (203.3 mg)

0.3 Meq Mg+2 = X, X = 30.495 mg

Amount of Drug /required Volume x Dilution Factor = Final Conc.Amount of Drug/ 100 x 10/1000 = 30.495 mg/ 1 mlAmount of Drug /10,000 = 30.495 / 1 , Amount of Drug = 304950 mg = 304.95 gm

Ex. A Hospital Pharmacy Technician adds by a syringe 20 ml of a concentrated sterile 2% w/v dye solution to a 250 ml commercial bag of saline what’s the concentration of the dye in Final solution ?

2% w/v = 2 Gm 100 ml X Gm 20 ml, X = 40/100 = 0.4 gm

The 4 gm in the 20 ml have an additional 250 saline = 0.4 gm 270 ml (Don’t use 250) Y 100 ml

Y = 400/270 = 0.15% W/VAn important note when solving this don’t do 0.4 gm 20 ml Y gm 100 ml, Y = 2% (wrong)

This is because The 0.4 gm Dye is the Total amount in 20 ml while if we use 0.4 Gm /20 ml we will get the amount of Dye in only 1 ml

Ex. Codeine Phosphate 200 mg Dimetab Elixir qs 120 ml

Sig: 1 tsp t.i.d. P.C. and H.S.What’s the amount of daily Codeine consumed

1 tea spoonful 3 times daily post cibum + Hora Somni5 ml x 3 after Meals + 5 ml at bed time = 20 ml

Take care not to be mixed up Tsp (tea spoonful) with Tbsp (table spoonful)200 mg 120 ml X mg 20 ml, X = 200x 20 /120 = 33 mg / dy , 8.25 mg / dose

EX.How many mg of the codeine base in each dose of the cough product used in question (mwt codeine = 299, mwt Codeine phosphate = 406)

Codeine phosphate Codeine base + Po4 2-1 millimole = 1 millimole Codeine base = 406 mg X = 8.25 mg , X = 0.02 millimole

N = Mass/mwt , Mass = 0.02 x 299 = 5.98 mg

Alligation Technique: (Liquids, Semisolids)

X Parts = C-B , Y Parts = A-C

This types of equations the Conc. Is given in % and the wanted is in Parts or Volume if C’s volume is given

It could be a Solution and a Diluent solution (Ex. Distilled Water) or 2 of the Same Solution with different Concentrations

Types:

1-What Volumes of Ethanol 95% (A) and distilled Water (B, Its Conc. Will be 0% ) should be mixed to produce 1000 ml 70% Ethanol (C) , The Answer will be in Volume as C volume is given

Solution:

95% X 70%0% Y

X= 70 parts, Y = 25 parts, total parts = 95 parts

Volume of X = 70/95 x 1000 ml (C Volume) = 736.84 ml EthanolVolume of Y = 25/95 x 1000 ml = 263.2 ml Distilled water

2-In what Proportions should Ethanol 95 % (A) and Ethanol 50% (B) be mixed to make Ethanol 70% (C), The Answer will be in parts as C volume not given

95% X 70%50% Y

X = 20, Y = 25

20 parts Ethanol 95% to 25 parts Ethanol 50%No definite volume could be obtained as the Final Volume is unknown

3-If we are given more than 2 conc. and the Final conc. Is needed

What is the percentage of Alcohol in the following mix

300 ml 95% V/V alcohol1000 ml 70% V/V alcohol200 ml 50% V/V alcohol

Solution

300 x 95% = 285 ml absolute alcohol1000 x 70% = 700 ml absolute alcohol200 x 50% = 100 ml absolute alcohol--------------- ------------------------------1500 ml (Total Volume) 1085 absolute alcohol

1085/1500 = 0.72 = 72% V/V

4- If we are given more than 2 conc. and the parts are asked which are need to make a certain new conc.

In what proportion should the following strengths be mixed to get a 500 ml 10 % mixture? 20 %, 5%, 3 %, 15 %

20 \ / 7 parts 20 % 7/27 x 500 ml =129.6 ml 20 % 15 - - 10 -- 5 parts 15 % 5/27 x 500 ml = 92.59 ml 15 % 5 / \ 5 parts 5% 5/27 x 500 ml = 92.59 ml 5% 3 / \ 10 parts 3 % 10/27 x 500 ml =185.1 ml 3 % ----------------- 27 parts

When the Alligation problem has no dilution factor the problem could be solved in another way

Ex 1 .Given

Stock HCL = 36.8% W/W, SP. Gravity = 1.19

Needed HCL = 10% W/V, needed HCL vol. = 4 litres

Problem

How many ml of stock HCL is needed?

Calculate the weight in gm of pure HCL in 10% HCL W/V

4000 ml x 10% = 400 gm of HCL (take care not ml)

Since we are not using pure HCL but 36.8 W/W HCL Calculate

How many gm we need from stock to provide 400 gm HCL

36.8 gm 100 gm400 gm X , X = 400x 100/36.8 = 1086.9 gm

Now we convert weight to volume Density = mass / volume

Volume = 1086.9 gm/ 1.19 = 913.36

Another fast solution

Stock Hcl = 36.8 gm 100 gm Density = Mass / Volume Volume = Mass / Density = 100 / 1.19 = 84 mlStock Hcl = 36.8 gm 84 ml X 100 ml = 43.8 %

C x V = C1 x V1 43.8 x V = 10 x 4000 (ml) V = 10x 4000/43.8 = 913.24

Ex. A pharmacist adds 2 ml of Tobramycin injection 40 mg/ml to 4 ml of Tobramycin ophthalmic solution 0.3% the concentration in the final mixture will be in gm/ml ?

Calculating Total Amount of Tobramycin

40 mg / ml x 4 = 80 mg = 0.08 gm0.3 gm 100 ml

X gm 4 ml , X = 4 x 0.3/100 = 0.012 gm

0.16 gm + 0.012 gm = 0.092 gm

Calculating Total Volume = 2+ 4 = 6 ml

0.092 gm 6ml Y gm 1 ml , Y = 0.092 /6 = 0.0153 mg/ml

Ex 2 .Given

1 gm chemical solution in 10 ml alcohol Alcohol sp gravity = 0.8

Problem % W/W strength of this solution?

Convert Alcohol’s Volume to Weight

Density = M / VM= 0.8 x 10 = 8 gm

Calculate the % of the chemical in Alcohol1gm 8 gm + 1 gm (don’t forget to add the weight of the chemical to total weightX 100 gmX= 11.11 % w/w

Ex.3

Calculate the amount of histamine acid phosphate required to prepare 10 ml of a solution containing 0.03% histamine baseMartindale quotes: 2.76 mg of histamine acid phosphate is equivalent to 1 mg histamine base

0.03 gm Histamine base 100 mlX 10 ml, X = 0.003 gm = 3 mg

2.76 mg Histamine 1 mg histamine baseY 3 mg histamine base, Y = 8.28 mg

Formula for Magnesium Citrate solution requires 27.4 gm of Anhydrous Citric acid (m.wt = 192) how many gm of Citric Acid monohydrate could be used as a replacement m.wt H20 = 18

Solution

The replacement to the An. Citric Acid is Citric Acid Monohydrate which has 1 H20 molecule more

192 27.4 g192+18 X , X= = 30 gms

Another solution

Anhydrous Citric acid = 27.4 gm millimmoles = mass /m.wt. 27400 /192 = 142.7 millimoles

Citric Acid monohydrate Anhydrous Citric acid + H201 millimole 1 millimole = (192 +18) 142.7 millimole X , X = 29967 mg = 29.967 g

How much saline should be mixed in a syringe with 1 ml of 1:1000 strength solution in order to obtain a 1:2500 dilution

Q1 X C1 = Q2 X C2 (this could be used in semisolid dilutions too)

1 ml x 1/100 = X x 1/2500, X = 2.5 ml, 2.5 ml - 1ml (original volume) = 1.5 ml

Ex .How much dil. HCL (10% w/w, HCL specific gravity = 1.045)Would be required to prepare 2000 ml of a solution containing 0.05 molar HCL? HCL m.wt. = 36.5

0.05 Mole 1000 mlX 100 ml

X = 0.005 Mole

N = Mass/M.wt. , Mass = 0.005 x 36.5 = 0.1825 GmThe wanted Hcl conc. Is 0.1825 % w/v

10% w/w = 10 Gm 100 gm Density = Mass / volume Volume = Mass/Density = 100 gm / 1.045 = 95.69 10 Gm 95.69 ml Y 100 ml , Y = 10.45 gm % w/vQ1 X C1 =Q2 X C22000 x 0.1825 = Q2 x 10.45Q2 = 365/10.425 = 34.9 ml

Another Easy Solution:

Whenever I see a w/w conc. I like to convert to w/v as it’s more reasonable

10% w/w HCL = 10 gm 100 gm 10 gm 100 gm/1.045 (Specific Gravity) 10 gm 95.69 ml X gm 100 ml, X = 10 x 100 /95.69 = 10.45 % W/V

0.05 Molar = 0.05 Mole 1000 ml Y 2000 ml , Y = 0.1 Mole

Converting the Mole into mg , No. of moles = Mass/mwt0.1 = Mass /36.5Mass = 3.65 gm (take care not mg since it is mole)

Calculating how much the mass is equivalent to Volume10.45 % W/V HCL = 10.45 gm 100 ml 3.65 gm Z ml, Z = 100 x 3.65 / 10.45 = 34.928 ml Ex.How much Chlorohexidene base is contained in 20 gm of Chlorohexidine Gluconate?

Chlorohexidene Gluconate C12 H30 C12 N10. 2 C6 H12 O7, M.wt. = 897.8Gluconic Acid C6H 12 O7 m.wt. = 196.19

2 x 196.19 = 392.38 subtract from 897.8 = 505.42 (mwt of Cholorohexidene )Moles of Chlo.Gl. = M/m.wt. = 20/897.8 = 0.222 mole

Chlorohexidene Gluconate Cholorohexidene + 2 Gluconic Acid 1 Mole 1 mole + 2 Mole = (m.wt. Chlo.Gl.= 897.8 )0.222 mole X , X = 0.222 mole

Mass of Chlorohexidene = Moles x M.wt. = 0.222 x 505.42 = 11.25 gm

Ex .the package information enclosed with the vial containing 5,000,000 units of penicillin G specifies that when 23 ml of solvent are mixed to dry powder the resultant concentration is 200,000 units/ml.

On the basis of this information, how many ml of sterile water injection should be used to prepare a solution containing penicillin G potassium 500,000 unit/ml?

Solution

Getting the Displacement Value:

To make a concentration of 200,000 units/ml although we add 23 ml the total volume of the final solution must have actually been 25ml.

5,000,000 X ml200,000 1 ml , X = 25 ml

(5,000,000 units / 25ml = 200,000units/ml) Ie the Displacement Value of the Penicillin G adds 2 ml to the final volume.

Knowing that once dissolved 5,000,000 unit powder takes up 2 ml of volume we can use the following equation

Conc. required =500,000 units 1 ml5,000,000 units (2 ml + Volume) if there was no D.V it would have been only volume

2 ml + Volume = 5,000,000 /500,000 Volume = 10-2 = 8 ml

Ex. Hospital Formula

Cocaine HCL 4%Tetracaine HCL 2% 0.5 mlEpinephrine HCL 1/2000

Sodium chloride qs 4 ml

1-How many mg cocaine HCL in the final solution?

2-How many ml of epinephrine HCL solution (0.1%) maybe used to prepare the solution?

N.B. 0.1 % = 0.1 gm 100 ml = 1 gm 1000 ml = 1/1000

Solution

1-

4 gm 100 mlX gm 4 ml, X = 4 x 4 /100 = 16/100 = 0.16 gm = 160 mg

2-

Q1 x C1 = Q2 x C24ml x 1/2000 = Q2 X 1/1000Q2 = 0.002 / 0.001 = 2 ml

Or

1 gm 1000 mlX 4 ml, X = 4/1000 = 0.002 gm

0.1 gm 100 ml0.002 gm Y , Y = 100 x 0.002 / 0.1 = 2 ml

Ex. Parentral Admixture OrderFor Alex Sanders room: M 704Cefazolin Sodium 400 mg in 100 ml normal saline

Infuse over 20 minutes q6h ATC for 3 days

Available in the pharmacy are cefazolin sodium 1 gm vials with reconstitution directions of addition of 2.5 ml SWFI will give 3 ml of solution

1-How many ml of reconstituted solution are required for each day of therapy?

2-What infusion rate in ml/min should the nurse establish for each bottle?

Solution

1-Infuse over 20 minutes q6h (every 6 hours) ATC (around the clock) for 3 days400 mg /100 ml x 4 = 1600 mg /400 ml so we need 1600 mg daily

Our stock Cefazolin Na = 1000 mg / 3 ml 1600 mg X ml, X = 3x 1600/1000 = 4.8 ml2-

100 ml 20 minutesY ml 1 minute = Y = 100 /20 = 5 ml (5 ml/min)

Ex. How many ml of normal saline would be mixed in a syringe with 1 ml of 1:1000 strength solutions I order to obtain a 1:2500 dilution?

Solution

Q1 x C1 = Q2 x C21 x 1/1000 = Q2 x 1/2500Q2 = 0.001/0.0004 = 2.5 ml (Total Volume)X = 2.5 – 1 = 1.5 ml

Dopamine 200 mg in 500 ml of normal saline at 5 mg/Kg/min ordered for a 155 lb patient what is the final conc. of solution in Micro gram /ml?

Solution

Sometimes the example has loads of unneeded information which could lead to confusion.So there is no need for the patient weight, the Infusion rate

200 mg / 500 ml = 200,000 microgram / 500 ml X 1 ml, X = 200,000/500 = 400 micro gm / ml

Ex. Morphine sulphate 500 mg in 100 ml Pca unit to deliver 0.05 mg/min

Dosing to start at 8:00 am tomorrow

How many ml of commercial morphine sulphate vial (25 mg/ml) is needed to fill this order?

What flow rate must be programmed into the Pca unit to obtain the desired amount of morphine / minute?

Upon consultation the prescriber decides to add bolus PRN dosing of 2 mg with a lockout period of 1 h Intervals assuming the patient uses all bolus dosing intervals ,approximately how long will the PCA last ?

Solution

1-We need to deliver 500 mg total, 25 mg ml 500 mg X, X = 500/25 = 20 ml2- 500 mg 100 0.05 mg Y , Y = 100 x 0.05 / 500 = 0.01 ml/min

3-The maximum volume /Hr =

0.01 ml /min x 60 = 0.6 ml

N.B. 2 mg with a lockout period of 1 h Intervals = 2 mg/Hr

500 mg 100 ml2mg Z , Z = 200/500 = 0.4 ml

0.6 ml + 0.4 ml Total = 1 ml and since we have 100 ml so we have 100 hrs / 24 = 4.2 days

Ex.How many mg of Iron is present in every 300 g Ferrous Sulphate tablet? Ferrous Sulphate FeSO4.7 H20 = 278, Fe= 56, S= 32, O = 16 , H = 1

The resident changes the above order of 300 mg ferrous Sulphate to Ferrous Gluconate, which maybe less irritating what strength tablet should be ordered? Ferrous Gluconate C 12 H 22 FeO 14 .2 H20 = 482?

Solution

Mg of Fe = 300 mg x Atomic wt. Iron /Formula wt. Ferrous Sulphate = = 300 mg x 56/278 = 60.4 m

60 mg elemental iron was present in 30 mg Ferrous Sulphate56 (Fe M.wt) 482 (ferrous gluconate)60 (Iron required) X , X = 526 mg

Moles and Millimoles:

Density D = M (mass) /V (volume) gm / ml

Water Density = 1Specific Gravity = Density of Substance /density of Water gm /ml

Mole Definition

One Mole or Gram Molecular Weight of a Substance is the Molecular Weight of That substance expressed in Grams If in Milligrams it will be Millimoles

N (Number of Moles) = M (mass in gm)/ Molecular Weight

Comments

1. The word Dissociation accompanied by Millimoles :Should remind us of writing an Equation and calculate how many Millimoles

2. how many Gms of Salt in X moles :Should remind us of writing an Equation and using N = mass / m.wt.

3. How many Moles are Equivalent to X Gm of Salt :Should remind us of writing an Equation and using N = mass / m.wt.

4. How to Express a certain Ion as Mole given the Ion’s weight Ex. 1000 mg NA+, NA2CO3.10 H2O mwt= 286, NA+ m.wt. = 23

N = M / m.wt. = 1 Gm (wt in Gm of the Ion not salt) / m.wt. of the IonN = 0.043 mole

The 2nd demand is to find how many Gms Salt has 1000 mg NA + Since we already know how much 1000 Gm Na+ in Moles we use the Equation1 mole of salt 2 mol NA+X 0.043

X= 0.0215 mole of the salt (the amount of moles of salt that contain 1000 mg NA+)Now we convert the moles into weightN = M / m.wt M = 0.0215 x 286 = 6.149 gm

If the question has a m.wt or A.wt. And a Conc. of an Ion its probably a Millimole Calculation

if we have (a mass of an Ion) and we want to convert it to milllimole to be Equivalent to the salt its part of we have to divide the Millimole by the number of Atoms of that ion in the salt

Ex.We have 10 gm F- and the salt is CAF2 , F- At.Wt. = 19To get CAF2 millimole we doAmount of F- (10 gm)Millimole F- = 10000/19 = 526 millimole To get CAF2 millimole divide by 2 = 263

Important if you get Number of Millimoles of ex. Na in a Salt ex. NA2SO4 and you want to get Milliequivalent of the salt divide the Millimole by 2And vice versa if we get Millimole NA2SO4 and want NA+ multiply by 2

Important Rule if we have a Salt and we know that we have a certain number of Moles of that salt and we need to get the Amount of Grams of any Ion accompanying the salt we multiply (The number of Moles X the Atomic Weight of that Ion X Number of Atoms of that Ion)

Ex. we have a salt NAF, NA+ = 23The NAF m.wt. is not given

We have 10 moles of NAFWe need the amount of NA in 10 moles NAFN = M / m.wt.M = 10 x 23 = 230 Gms

N.B.When using the N = M / m.wt. In Millimoles equations the M is in Millgrams not grams

1 Mmole NACL = 58.5 Mg NACLN = M / m.wt The 58.5 Mg is the m.wt. Of the salt

If a question comes that has a Prescription and its asked to calculate Millimoles for an Ion calculate the Ion in each salt and add them together ex.

Na Acetate 1 gmNAHCO3 0.75 gmOrange Syrup to 1 mlPurified Water to 10 ml

N.B. Millimole is a measure of Quantity not concentration

Milliequivalent

Multiply the number of Millimoles by the absolute value of the ValenceEx. NaCl NA+ + CL-

1 mole of NA+ and CL- producedEquivalent = Moles x Valence= 1x 1 = 1 Equivaelnt of CL- + 1 Equivaelnt of Na+

We can get milliequivalents without using Millimoles by using m.wt /valency

Ex.m.wt. of KCL = 74.51 gm quivalent = mwt/Valency = 74.5/1 = 74.5 gm/ml1 milliequivalent = mwt/Valency = 74.5/1 = 74.5 mgm/ml

Ex.what’s the conc. Of solution containing 4 meq/ml of CaCL2.2h2O in mg/ml ?m.wt. CaCL2.2H2O = 147

1 milliequivalent = mwt/Valency = 147/2 = 73.5 mg/ml4 milliequivalent = 4x 73.5 = 294 mg/ml

Ex.2 a solution labeled as containing (10 mg % K+ w/v) express it in meq/Litrem.wt. of K+ = 39

Get the weight of K+ in / Litre10 % k = 10 mg 100 ml X mg 1000 ml, X = 100 mg

1 milliequivalent = m.wt /valency = 39/1 = 39 mg/mlNo of milliequivaelnt in 100 mg = wt/meq = 100 /39 = 2.564 meq /Litre

Ex.3 A solution contains 15 mg % Ca+2 w/v express it in meq/litreCa+2 m.wt. = 40

Get the weight of ca+2+ in / Litre

15 mg % = 15 mg 100 ml X mg 1000 ml , X = 150 mg

1 milliequivalent = m.wt /valency = 40/2 = 20 meq/mlNo of milliequivaelnt in 150 mg = wt/meq = 150/20 = 7.5 meq/litre

Ex.4 whats the % Conc. w/v of KCL containing 20 meq/10 ml?KCL mwt =74.51 milliequivalent = m.wt /valency = 74.5 /1 = 74.5 meq/ml20 meq/10 ml = 2 meq/mlNo of milliequivaelnt in X mg = wt/meq 2 = wt /74.5 , wt = 149 mg

149 mg ml Y mg 100 ml , Y = 14900 mg = 14.9 gm %

Ex.5 a patient is required to receive 4 meq/kg of Nacl if the person’s weight is 70 kg what is the Volumet of Normal Saline that should be used to provide that?M.wt NA = 23, M.wt. Cl = 35.5

Mg/L = Meq x M.Wt. / Valency = (4 x 70) x (23+35.5) / 1 = 280 x 58.5 = 16380 mg / L

0.9 Gm Nacl 100 ml16.38 Gm Nacl X , X = 1820 ml

OR another easy solution

4 meq kgX 70 kg , x = 280 meq

NaCl Na + Cl1 mmole = 1 meq = 58.5 mg 280 meq = Y , Y = 163,800 mg = 16.38 gm

0.9 gm 100 ml 16.38 gm Z, Z = 1820 ml

If we are given 2 Salts and the weight of one of them and it’s asked to get the Weight of the other since (the quantity of the Ion in the 2 is equal) ex.

K2CO3 and KCL,K2CO3 weight = 7.5 mgK2CO3 m.wt. = 138 mgKCL m.wt. = 74.6 mg

We get the no of moles in K2CO3 M/m.wt. = 7.5/138 = 0.054 millimole

Now we need to multiply the number of millimoles by 2 to get the no. of millimoles of KCL

(the question says that the amount of k+ in both is equal but K2CO3 contains 2 k while KCL has only 1 )

So we have to multiply the amount of moles by 2 0.054 x 2 = 0.108 millimole (Very Important)

We get the no of gms in the moles using Moles = Mass/mwt Mass = Moles x mwt =0.108 x 74.6 = 8.05 mg

Ex. If we get a calculation with Meq in the form of Multi salt solution

What’s the weight of each salt needed to prepare the following prescription?

Na+ 67 Meq m.wt. NACL =58.5K+ 6 Meq m.wt. KCL = 74.6Ca+2 4 Meq m.wt. CACL2.2 H20 =147Cl- 77 Meq

Nacl Na + Cl-1 mmol NACL 1 Meq Na+ = 58.5 mg NACL =67 Meq Na+ = X , X=3,920 mg NACL (This will also give 67 Meq Cl-)

KCL K + Cl-1 mmol KCL 1 Meq K+ = 74.6 mg KCL 6 Meq K+ = Y , Y= 448 mg KCL (This will also give 6 Meq Cl-)

CACL2.2 H20 Ca+2 + 2 Cl-+2 H201 mmol CACL2.2 H20 2 Meq Ca+2 = 147 mg CACL2.2 H20 4 Meq Ca+2 = Z , Z=294 mg CACL2.2 H20 (This will also give 4 Meq Cl-)

If we are given meq of an ion and we use the rule (Meq = N x Valency) we use the atomic weight instead of the m.wt. then the mass we calculate we use to calculate the total salt

Mass of Ca+2 = 0.802 gmAt.wt Ca+ = 40.1, m.wt CaCL2 .2H20 = 147What weight of the chemical (CaCL2 .2H20) is needed to obtain 40 Meq of Calcium

Solution

0.802 40.1X 147, X = 2.94 gm

Another Solution which I prefer:

CaCL2 .2H20 Ca+2 + 2 Cl- + 2 H201 millimole CaCL2 .2H20 2 meq Ca+2 = CaCL2 .2H20 m.wt (147 mg)

40 meq Ca+2 = X, X = 2940 mg

Ex.

What is the volume to which 25 ml of a solution containing 2 meq/ml of SO4 2- must be diluted to produce a 5 x 10 (power -3) M solution of Sodium Sulphate (Na2SO4.10 H20, m.wt = 322.2)

Solution

Total Meq of SO4 2- in 25 Ml = 2 x 25 = 50 Meq/25 mlNa2SO4.10 H20 Conc. = 5 x 10 (power -3) mole /Liter = 5 millimole /Liter (Final conc. needed) Na2s04.10 H20 = 2 Na + + So4- + 10 H201 millimole = 2 Meq So4 2- = m.wt. Na2S04.10 H20 (322 Mg) 50 Meq So4 2- = X mg, X = 8050 mg (number of milligrams which is equal to 50 Meq/25 ml )Express the Final conc. in gm/ml 1 M Na2s04.10 H20 contains 322.2 gm in 1000 ml1 x 10-3 M Na2s04.10 H20 contains 322.2 mgm in 1000 ml5 x 10-3 M Na2s04.10 H20 contains 5 x 322.2 mgm in 1000 ml

1611 mg : 1000 ml8050 mg : Y , Y = 5,000 ml

Ex. Calculate the volume of sterile potassium acetate injection 2.45 gm /5 ml to be added to one litre of vitamin N to give a total of 25 meq K+/litre

Vitamin N cotains 20 milli mole K+/Litre potassium acetateCh3C00k, mw = 98.14

Solution

First we have to know how much meq k+ we have in Vitamin NCh3C00k K+ + Ch3C00-1 millimole Ch3C00k 1 meq K+ = Ch3C00k m.wt. (98.14 mg)20 millimole Ch3C00k X, X = 20 Meq K+ /Litre

Now we calculate how much additional K + we need from Sterile Potassium Injection

25 Meq – 20 Meq = 5 MeqCalculate how much that is in weight

Ch3C00k K+ + Ch3C00-1 millimole Ch3C00k 1 meq K+ = Ch3C00k m.wt. (98.14 mg) 5 Meq K+ = Y, Y = 490 mg = 0.49 gm

Sterile potassium acetate injection = 2.45 g / 5 ml 0.49 g Z , Z = 1 ml

To achieve Isotonicity there should be a balanced Osmotic pressure

For an ideally behaving electrolyte1 Mmol = 1 Meq = 1 Mosm (Milliosmole) = M.wt. (Total molecule in mg)

If it is mole, Equivalent, Osm use M.wt in gm

When we get a calculation containing Mosm always write the rule 1 Mmol = 1MosmBut we be aware ex.NACL Na + + CL-1 mmol 2 Mosm

Osmolarity and Osmolality

Osmotic Pressure is determined by the concentration of Osmoles in solution

10 Osmolar solution = 10 Osm / Liter

Ex.If 200 Mosm are dissolved in 1 Liter, the conc. Is 200 Milliosmolar

Osmolality

1 Osmol Solution = 1 Osmol /kg of water

N.B. Important when we are given an Osmolar conc. Per volume, the volume has to be converted to 1000 ml

Ex. 30 Mosm/100 ml = 300 Mosm/1000 ml

For Non-Electrolyte ex. Dextrose

Milliosmole Calculation are solved by the rule (1 Mmole = 1 Mosm = M.wt.)Take care that the volume has to be adjusted to 1 liter

For Non electrolyte if we get millimoles it will equal directly to mosmsEx. Dextrose 20 mmol/100 ml = 20 mosm/100 ml

N.B. Equimolar conc. Of non-electrolytes will have similar osmotic pressure

For Electrolytes ex. Nacl

N.B. Non ionized solute will differ from ionized one in osmotic pressure

1 Mmol NACL = 2 Mmol Ions = 2 Mosm = M.wt

If we get a Prescription and asked to calculate Osmolarity calculate for each

constituent then add them up and be sure that the end volume is in litres .

Ex. Dextrose 20 mmol / 100 ml = 20 mosm Kcl 4 mmol / 100 ml = 1 mmole = 2mosm 4 mmole = X , X = 8 mosm

Solution 20 + 8 = 28 Mosm / 100 ml 280 Mosm / Litre

Ex.2 How many Milliosmoles in the following solutionMgCL2.6 h20; MW= 203.3 12 meqKCL; MW= 74.6 20 meqWater for injections to 100 ml

Solution

MgCL2.6 h20 Mg+2 + 2 Cl- + 6 H201 millimole MgCL2.6 h20 = 2 Meq = 3 mosm = MgCL2.6 h20 m.wt. 12 Meq = X , X = 18 MsomKCL K + Cl-

1 millimole KCL = 1 Meq = 2 mosm = KCL m.wt. 20 meq = Y , Y = 40 msom

X+ Y = 18+40 = 58 Mosm (note that the final answer is at it is no volume is mentioned)

If the Example has a Salt with a Conc. Of any form and asked to calculate the amount of it to make an (Isotonic solution with NACL ) (or any other salt)

1- First we get the (Mass of the salt) using any of the rule 1 Mmole = 1 Mosm =1 Equ = M.wt.

2- The weight that is calculated is transformed into Mosmoles

Isotonic Solution = 308 Mosm

3-we subtract the Mosmoles calculated from 308 Mosm / Liter (very imp if multiple Liters are needed multiply 308 by the number of liters)

In conclusion what we do is we see (how much Msoms is the given conc., How many gms that conc. Is, subtract the found msoms from 308 and get the weight of NACL equivalent to it)

We calculate from the reminding Mosms the amount of the other salt (NACL)

Ex. Prepare 1 liter CACL2 containing 40 Meq / L Ca+2 make isotonic with NACL M.Wt. CACL2.2H20 = 147, NACL = 58.5

1 Mmol CACL2 = 2 Meq Ca+2 = 147 40 Meq Ca+2 = X , X = 2,940 mg

1 Mmol CACL2 = 3 mosm = 147 X = 2,940 mg , X = 60 Mosm

308-60 = 248 Mosm (amount of NACL)1 Mmol NACL = 2 Mosm = 58.5 248 Mosm = X X = 7,254 mg

Finally = 2,940 Mg CACL2 + 7,254 Mg NACL /liter H20

A patient is taking a solution with an Osmolarity of 520 Msom/L how many ml of purified water are needed to reduce 500 ml of this solution to an Osmolarity of 300 mosm/l

Q1 c1 = q2 c2500 x 520 = X x 300

X = 866 ml866-500 = 366.6 m l h20

Estimate the milliosmolarity (mosm/l) for normal saline Na = 23, cl= 35.5One liter of normal saline contain 0.9 % nacl = 9 gm

Nacl na + + cl-1mmole nacl = 2 mosm = nacl mwt (23 mg + 35.5 mg) 2mosm = 58.5 mg X 9000 mg

X = 307.6 mosm

Isosmotic and isotonic calculations

If the given has FD1% then it’s a freezing point Depression CalculationAnd there are 2 types

N.B. Always take care of the Volume because the Final answer varies according to it

1-for (1 Solute) solution Isosmotic

Ciso = Conc. Of a substance that makes it Isosmotic with bloodFD1 % = (the freezing depression of that substance)

2-(Multi Solute) solution Isosmotic , Isotonic

% Adjusting substance = Conc. Of what is used to make the solution Isosmotic Usually Dextrose or NACL.

Isotonic Solutions: 0.9 NACL %, Dextrose 5%, Boric Acid 1.73%

NACL FD1% = 0.576 CDextrose FD1% = 0.101 C

3-Another way is to use Ciso instead of FD1% (Multisolute) Isosmotic, Isotonic

4- Using SCE (E Value) to make solution Isotonic (Multisolute, 1 Solute)

Take care that the Final answer here is % W/VSometimes the SCE is called the E value

5- If (1 solution) Ciso = 0.9 /SCE (Substance)

6- SCE Value Calculation (1 solute)

SCE = 0.9 / 0.52 x FD1%

The outcome means X gm of NACL is osmotically equivalent to 1 gm of the substance + get its conc. (y)

X gm NACL 1 gm substance0.9 gm NACL Y

If we are given a solution with a certain conc. To calculate the SCE

Ex1.

How much Nacl is needed to adjust the following prescription to Isotonicity E value ZNSO4 = 0.15

ZNSO4 1%

NaCl q.sPurified Water q.s 60 ml

(1) ZNSO4 SCE + (2) SCE of total solution = Isotonic

(1) 1gm 100 ml X 60 ml, X= 0.6 gm ZNSO4

1 gm ZNSO4 = 0.15 gm NaCl 0.6 gm ZNSO4 = Y , Y = 90 mg NaCl

(2) 0.9 g 100 mlX 60 ml , X = 0.54 gm (If the whole solution was nacl)

540 mg – 90 mg = 450 mg Nacl is needed.

Another very easy solution :( take care it could be solved since we know NACL SCE = 1 but if it was not known it can’t be used)

% adjust subst. = 0.9 – (SCE A x A% + SCE B x B%) / SCE adj. subs. = 0.9- (1x 0.15) / 1

= 0.75 % w/v

0.75 gm 100 ml X gm 60 ml =, X = 0.45 gm

Ex2.

How many mg of NACL should be added to the following medication to maintain isotonicity, Atrovent 0.02% 5 ml + SWF injection 25 ml, atrovent is isotonic

Solution

Since Atrovent is already isotonic calculate SCE for the 25 ml SWF

0.9 g 100 mlX 25 ml , X = 0.255 g NACL is needed

N.B.

FD1% can be Extrapolated

Cocain 1% FD1% = 0.09 CHow much for 4% cocaine 1 0.09 4 X , X = 0.36 C

There is no direct rule to use FD1% to obtain isotonicity How much NACL is needed to adjust 30 ml of 4% Cocaine HCL Solution to isotonicty

The freezing point depression of a 1% Cocaine HCL is 0.09 C

4% 0.09 C4% x, x = 0.36 C

Since isotonicity is attained at 0.52 c by a 0.9 NACLCalculate the difference 0.52 -0.36 = 0.16 C

That’s the amount of freezing point depression we need to introduce using NACL0.9 Gm 0.52 CX 0.16 CX = 0.276 Gm

Now for 30 ml

(The 0.276 Gm was obtained from using 0.16 C which we obtained using the 4 % Conc. = 4 Gm / 100 ml but we don’t have 100 ml we have 30 ml)

0.276 100 mlX 30 ml, X = 82.8 mg

Buffer Solutions

1-Calculating the Ph of a buffer solution containing (Weak Acid/Salt)

Dissociated / Undissociated

Salt = Dissociated (Mol/L, Molar, No. of Moles) Acid = Undissociated (Mol/L, Molar)

2-Calculating the Ph of a buffer solution containing Base / Salt

Undissociated/ Dissociated

PH= PKW- PkB+ Log Base/Salt

PKW= 14

PH+POH = 14PH = - Log [H+] POH= -Log [OH-]Pka= - Log Ka (Dissociation Constant)

Ex. What is the PH of a solution containing 0.5 moles of Ephydrine and 0.05 moles of ephedrine sulphate per liter of solution? Pkb = 4.56 (Take care Pkb not Pka)

Ph= PKw -Pkb +Log Base/Salt = 14-4.56 + Log 0.5/0.05 =10.44

N.B. Manipulating the equation:

If while solving the equation we have reached this form

Log (Base / Salt) = X

We send the log to the other side converting it to antilog

Base / Salt = Antilog X

To get Antilog in a Calculator

Ex. Antilog 0.2 = press shift – press log – press 0.2 = 1.58

The Final solution can be a ratio of Salt / Acid or Dissociated /Undissociated Ex.

Salt /Acid = 1.58, Salt = 1.58 Acid

The Molar Ratio of Salt to Acid = 1.58:1

Very Imp.

If we get the Conc. Instead of Mol / L in say %We use the Rule n (No of moles) = M (mass in gm) / m.wt.

Ex. 0.5%, M.wt. = 413.5 0.5 gm 100 mlX 1000 ml (we use 1000 ml cause the conc. is Mol/L) X = 5 gmN = 5 / 413.5 = 0.01209 M

Ex. Calculate the change in PH upon adding 0.04 M NaOH to a Litre of a buffer solution containing 0.2 M concentration of Sodium Acetate and Acetic Acid, The Pka of Acetic Acid is 4.76

Calculate Ph of Original Buffer

Ph = Pka + Log Salt/Acid 4.76 + Log (0.2/0.2 ) = 4.76

Very Imp. Step

The addition of 0.04 M NaOh converts a same amount of Acetic Acid to Sodium Acetate and decreases Acetic Acid)

Calculate the New Ph =

Ph = 4.76 +Log (0.2 +0.04 ) / (0.2-0.04) = 4.76 + 0.176 = 4.94 4.94-4.76 = 0.18

Ex. Calculate The PH of a solution containing 0.2% citric acid monohydrate (C6h807.h20) , M.w. = 210.1 Pka = 4.6 and 1.25% sodium citrate (C6H5Na3O7.2H20, M.w. = 294.1 )

Citric Acid Monohydrate0.2% = 0.2 gm 100 ml X 1000 ml, X = 2 Gm No of Moles in 2 Gm = Mass /M.wt = 2 /210 = 0.0095 Mole/Litre (Molar )

Sodium Citrate1.25% = 1.25 Gm 100 ml

Y Gm 1000 ml, Y = 12.5 GmNo of Moles in 12.5 Gm = Mass /M.wt = 12.5 /294.1 = 0.0425 Mole/Litre (Molar )

Ph = Pka +Log (Salt /Acid) = 4.6 + Log 0.0425/0.0095 = 4.6 + Log 4.4736 = 4.6 + 0.65 = 5.25

Imp. Note: remember if the calculation is about how much was the Increase or Decrease in ph don’t forget to subtract the ph calculated from ph given

You have to know if the molecule is Basic or Acidic to use the right equation

Buffer Capacity (for weak acids or strong bases)

B Max = 0.576 x C (Total Buffer conc. In mol/l)

N.B. Molarities are additive

B= 2.303 x C x Ka x [H+] / (Ka + [H+]) power 2 (molar)

N.B. the Maximum buffer capacity is when the PKa = PH

Calculating Partition Coefficient (Octanol/Water)

The conc in X/ml Oil and Y/ml in water, we get the partition coefficient by getting X/Y If the oily phase is 30% and the conc in oil is 0.015 gm/ml while its conc. In aqueous is 0.045 ml calculate the partition coefficient ?

O/W = 0.015 x 0.3 / 0.045 x 0.7 = 0.045/0.0315 = 1.428

Suppositories

1- 1 Type of drug using No Displacement Value

Ex. Prepare 10 supp. Containing 5% Benzocaine in Fatty base

Total supp. Wt = 10 x 1 gm = 10 gm (Amount of benzocaine in total supp. Wt. = 5 gm 100 gm X gm 10 gm x= 0.5 gmAmount of fatty base = 10 gm -0.5 gm = 9.5 gm

2- Using Displacement Value rule , 1 Type of drug

Ex. Prepare 10 supp. containing 25 mg hydrocortisone in PEG

Mould Calibration = 0.9 gm, Hydrocortisone DV = 1.6PEG density = 1.2 gm/ml

= (0.9 – 0.025 /1.6) X 1.2 X 10 = 10.6125 GmN.B.1

Mould Calibration in Gm + Dg Dose in Gm

N.B.2

The fatty base can be made of 2 constituents and come in parts ex. PEG 400/4000 = 1 Part /4 Part, so to calculate each we get the weight of base and then Use = 1/5 x Weight of base + 4/5 x Total weight of base.

N.B.3 if (Percentage Error) was asked

1 - Calculate using no rule of displacement (Weight of base)2- Calculate using the rule of displacement (Weight of base)Use the rule = % Error = difference between 1, 2 / 1 x 100 = %

3- 2 Types of Drugs (ex. 30 mg Hydrocortisone , 6% Benzocaine ) in PEGPrepare 10 supp. , Mould Calibration = 0.9 gm, Hydrocortisone DV = 1.6PEG density = 1.2 gm/ml

Calculate using rule of displacement (30 mg Hydrocortisone)Calculate (total wt. of suppository) by adding (the calculated base wt + no. of supp x 30 mg hydro)

From the previous value calculate 6% which will be equal to the amount of Benzocaine

= (0.9 – 0.030 /1.6) X 1.2 X 10 = 10.6125 Gm

Total wt. of suppository = 10.6125 Gm + (10 x 30 mg) = 10.925 Gm

Benzocaine amount = 10.925 Gm x 6/100 = 0.655 Gm

Ex. 4

Calculate the amount of Cocoa butter required to make 6 suppositories each containing 300 mg CuSo4 if each suppository mold can hold 2 gm of pure Cocoa Butter provided that the Displacement Value of CuSo4 = 2.5 gm

Solution

Displacement means how many grams of drug Displaces 1 gm of Suppository Base

2.5 gm CuSo4 displace 1 gm of Suppository Base (Cocoa Butter)(6 x 300 mg) displace X (Cocoa Butter), X = 0.72 gm

Total weight of Cocoa without displacement = The amount the mould suppository can Holds X number of suppositories = 2 x 6 = 12 gm

The weight Of Cocoa we need after subtracting displacement of the drug = 12-0.72 = 11.28 gm Cocoa Butter

Zero Order reaction First Order Kinetics (Radio active materials, IV inj.)

Conc. Changes at a

Constant Rate dc/dt = - KoConc. Changes with respect to the Remaining Conc.

Or Log C = Log Co - K1.T / 2.303 (can be used for percentages)

T ½ = 0.693/K1

T 90% = 0.105/k1

C = conc. At a certain time mg/mlCo = Initial Conc. mg/mlKo = Zero Order constant

C = at a certain time mg/ml conc.Co = Initial Conc. mg/ml conc.K1 = First order constant

Total amount of drug = Loading dose (1+Kt)

K= 2.303 x Log C2 - Log C1 ------------------------------- T2-T1

Clearance = K1. Vd

K1 = 1st rate order elimination constant Vd= Volume of distribution

Cp = desired peak concentration of drugS = portion of salt form which is active (salt fraction)F = fraction of dose absorbed (bioavailability)Vd = volume of distribution of drug in body

Loading Dose = Css x Vd

N.B. At Css Rate of absorption = Rate of elimination

Css = Cp ------- S x F

For an intravenously administered drug, the bioavailability F will equal 1

Maintenance Dose = Css x Cl x I

= Cp x Cl x I / S x FI = Dose intervals

Ri = rate of infusion

N.B. Half Life = T ½, Shelf Life = T 90%

N.B. when solving

And the equation reached for ex. 0.9 =

We get the Ln for 0.9 by pressing Ln in calculator then typing the number press Equal

-0.105 = -K1 x 3 K1 = 0.035

Also when solving and the equation reached Ex.

We have to get the inverse Ln of -0.21 (-0.035 x 6) which will equal = 0.81, C = 1.62

There are calculations where we can use percentages and not absolute numbers to calculate k1 in first order calculations

Ex. A solution retained 90.6% after 14 days following First Order KineticsCalculate its T 90%?

Don’t use 90.6 14 90 X , X = 13.9 (wrong)

Use Log C = Log Co – K1 T /2.303

Log 90.6 = Log 100 – K1 x 14 /2.303 K1 = 0.00706 Dy-1 Then use T90 % = 0.105 /K1 = 14.87 (Right)

Ex 2.

How many ml of I2 (131) to be added to a patient if the prescribed activity is to be 0.5 millicurie?

Assay shows I2 (131) activity to be 14.7 millicurie/2.6 ml 2 days before administration and the half life of I2 (131) is 8.1 days

Solution

Radioactive drugs decay following the 1sr order process

T ½ = 0.693/K = 8.1 K= 0.693/8.1 = 0.085

Conc. Of the solution is 14.7 millicurie / 2.6 ml X / 1 ml, X = 5.6 mlLog C = Log Co-KT/2.303 = Log 5.6 - 0.085 X 2 / 2.303 = 0.752-0.0738 = 0.678

C = 4.766/ml

4.766 millicurie 1 ml 0.5 millicurie Y ml

Number of mls = 0.5 x 1 / 4.766 = 0.014

N.B. Take care not to use 0.5 millicurie as C because we don’t know how much ml of Iodine has it while we can use 14.7 millicurie/2.6 ml as Co because it has ml

Ex.3 If a radiation dose of 200 mci of radioactive iodine (k = 0.23 h-1) is administered to a patient at 8 am in the morning ,how long it takes for the amount of radiation emited by the dose to fall below 25 mci ?Radioactive drugs decay following the 1sr order process

T ½ = 0.693/K = 0.693/0.23 = 3 HAfter 3 hr = 100, after 6 hr 50 , after 9 hr = 25 mci

Loading dose: Vd = 42 L total body water normally for 70 kg adults

Ex.

What is the rate of infusion of phenytoin in a patient thatRequires steady state plasma concentration of 20Mcg /mL? The elimination half-life of phenytoin is 4 hoursAnd an apparent volume of distribution is 15 L.

T ½ = 0.693/K1 K1 = 0.693 / 4 = 0.17325

The rate of infusion can be calculated by the following formula:

= 20 x 0.17325 x 15,000 = 51975 mcg / hr = 51.975 mg/hr

Ex. A Table that has the following

T (Hrs) Drug Conc. (Mg/Ml) Log of that Conc.1 X Log X2 Y Log Y3 Z Log Z

Its required to get the T ½ ?

Solution

K= 2.303 x Log C2 - Log C1 ------------------------------- T2-T1K= 2.303 x Log Y - Log X ------------------------------- 2-1T ½ = 0.693 /k

Arrhenious Eq.

Log K2 = Log K1 + Ea (T2-T1 ) / 2.303 .R.T1.T2

K1= Rate constant 1, Ea = energy of activation in cal if comes in Kcal multiply by 1000 , T 1 = absolute temp +273 , R = gas constant

N.b. If the calculation has 2 temperatures and it is given that if follows a type of kinetics order (zero-1) then this is a mixed question using kinetics order laws + arrhenious

We have 2 solutions no 1 in meq and it has a certain volume added to another solution no 2 in any other conc. And also volume given

1-calculate the total amount of meq in the volume of the 1st solution

2-add the amount of meq to the volume of the 2nd solution

Ex. Sol. 1 = 2 meq/ml given 10 ml Sol.2 = 5% given 500 ml

2 meq 1 mlX 10 ml, X = 20 meq

Then 20 meq/500 ml ?

We get a solution of a salt in gm/ml and we have the density of the solution to get the conc. In w/w we convert the ml to mass using density = m/v

If its asked to make a solution by dissolving salt in water the amount of the solvent is calculated by weight of salt + x ml solvent = volume of a certain conc.

Convert the volume into a mass and subtract wt of salt to get the X ml solvent wt (use d = m/v to get volume)

Doses for children

Young: age (years) /age (years) + 12 x max adult dose

Dilling’s: age (years) /20 x max adult dose

Cowling’s: age (years) next birthday / 24 x max adult dose

Fried: age (months) / 150 x max adult dose

Clarke’s: weight (kg) / 70 x max adult dose

Surface area Method:BSA of Child (meter 2) /1.8 (meter 2) x max. adult dose

N.B. to calculate the BSA we need to know the Height and Weight

Ex. The usual dose for pacitaxel intravenous is 135 mg/m2. How many mg should be administered to a 40 years old female weighting 120 Lb if her BSA was calculated to be 1.5 m2?

This is a tricky calculation having so many info that we don’t need age , Sex ,weight and lastly BSA which whenever we see we write the rule of BSA /1.8 x the Reg Dose

However all what we need is to use the Already given dose 135 and Multiply by 1.5 = 202.5 mg

Life Spans:

Neonate = Zero- 1 monthInfant = 1 month – 1 yearChild = 1 year – 5 years Late Childhood = 6-12 years

Balance sensitivity

Sensitivity of the balance = weight that shifts the Balance 1 division

A 2-Pan balance pointer resets at zero mark with no masses and at 2 divisions with a 20 mg what is the sensitivity of the balance?

20 mg: 2 divisions X 1 division , X = 10 mg

Percentage Error = Sensitivity of Balance /Amount Weighted x 100

What is the minimum amount of potent drug that may weighted on a prescription balance with a sensitivity requirement 6 mg if at least 5% Accuracy is required

Sensitivity Requirement =Minimum Weighable Amount x Acceptable Error

6 mg = M.W.A. x 5% (or 0.05)M.W.A = 120 mg

The larger the SR value the less accurate the balance

N.B. the 5% Accuracy could be given as 95% accuracy

HLB Equations

HLB = (Product % I x I HLB) + (Product II% x HLB) + Etc….. ------------------------------------------------------------------------------------ 100 (Product % I + Product % II + Etc….)

Calculate the HLB of the mixture 30 gm Surfactant A (HLB= 8.56) and 10 gm of surfactant B (HLB = 15)

Solution

Total weight A + B = 40 gm

A % = 30/40 x 100 = 75 %B % = 10/40 x 100 = 25 %

75 x 8.56 + 25 x 15 / 100 = 10.17

Dose fraction

Dose fraction = Normal Dose x [1- ( Fe x (1 - Pt.Cl. / Normal Cl. )]

Fe = Fraction Excreted Unchanged

Pt. Cl. = Patient Clearance

Ex.

50% of Dicloxacillin is excreted unchanged in the urine if the normal dose for dicloxacillin is 125 gm/6 hr a patient with renal function 20% of the normal should receive?

Dose fraction = Normal Dose x [1- (Fe (1-Pt Cl/Normal Cl)] = 125 (1- (50/100 (1-20/100) = 125 (1- (0.5 (1-0.2) = 125 (1- (0.5 (0.8) = 125 (1- (0.4) = 125 x 0.6 = 75 mgEx.

Drug excreted 20% unchanged in urine, so in a patient with 75% kidney function the dose should be:

I. No changeII. Double Dose

III. Double intervals IV. double time and dose

Dose fraction = Normal Dose x [1- (Fe x (1 - Pt.Cl. / Normal Cl. )] = X x [1- (20/100 x (1 - 75/100)] = X x [1- (0.2 x (0.25)] = X x [1- (0.05)] = =X X 0.95 =

A dose was given 240 mg to a patient with total body clearance =3.5 liter/min. . And the drug excreted unchanged in urine 80 mg, Calculate the non-renal clearance

Renally amount of the dose excreted = 80 mg Non-Renal amount of the dose excreted = 240-80 = 160

Renal + Non-Renal = Total body clearance 240 mg 3.5 Litre/Min 160 X , X = 2.33 ml

Drug 44% excreted unchanged (t 1/2= 8 hours), what is the new t 1/2 if the kidney function is 50 %?100% 8 hrs50% 4 hrs

12 hrs

Estimation of Creatinine Clearance using only Serum Creatinine

(140 - Age in Years) x Wt. in Kg --------------------------------------------Serum Creatinine Conc. in mg/dl x 72

Female = male x 85%

Doses

Maintenance Doses calculated by Creatinine Clearance

N.B. the normal Creatinine clearance = 100 ml/min

Ex. If creatinine clearance = 40 ml /min, normal dose = 200 mg, Calculate the Maintenance Dose?

Solution

40/100 x 200 = 80 gm

Proof Galloon = wine gallons x Proof strength / 50

Ex. 25 Gallons of 70% alcohol, contain how many proof gallons?

Proof Galloon = 25 x 70 / 50 = 35

Inulin ( Polysachride): used to measure Glomerular Filtration

Cl = Ff x GFR

Cl = Drug ClearanceFf = Free fraction (can be obtained from Plasma protein bound Fraction )GFR = renal clearance of Inulin

Bioavailability F = AUC Tab /AUC Injection

Turnover Calculation = Whole Sales /Large Inventory (Should be = > 5%)

If the Oily phase is 30% and the conc. Of drug in Oil 0.015 gm while its conc. in Aqueous is 0.045 gm

Calculate the Partition Coefficient of this drug?

Oily phase is 30%Watery Phase 70%

Partition Coefficient = 0.015 X 30/100 -------------------- = 0.14285 0.045 X 70/100

Patient 1 2 3 4 5Sytolic 140 160 180 190 150Diastolic 70 84 88 90 70

Mean (X) = Sum/No of Observations

Systolic Mean = 140+160+180+190+150/5 = 820/5 = 164

Diastolic Mean = 70+84+88+90+70/5 = 402/5 = 80.4

Median = the Middle number in an ascending order data

Systolic Median = 140 -150 -160 -180-190 = 160

Diastolic Median = 70 -70- 84-88 -90 (In case of Even numbers we take the average of the 2 middle values =84+88/2 = 86

Average Median = Median – Mean = 160-164 = -4 ? = 86-80.4= 5.6

Range = The difference between the largest and smallest value.

Systolic Range =190-140 = 50Diastolic Range = 90-70 = 20

Standard Deviation =

Systolic Mean (X) = 164

Sytolic 140 160 180 190 150

SD = (140-164) Power 2 + (160-164) P.2 + (180-164)P.2 +(190-164)P.2 +(150-164)P.2 = (-24) Power 2 + (-4) P.2 + (16) P.2 + (26) P.2 + (-14) P.2 = 576 + 16 + 256 + 676 + 196 = 1720 / 5-1 = 1720/4 = 430

Now the root of 430 = 20.7