Download - TONICITY, OSMOTICITY, OSMOLALITY & OSMOLARITY

BC

D

A

BCD

A

FUNCTIONS

e.g. •TEMPERATURE; •HEART RATE; •BLOOD PRESSURE.

Isotonic Solutions

contain the same concentration of solute as an another solution (e.g. the cell's cytoplasm). When a cell is placed in an isotonic solution, the water diffuses into and out of the cell at the same rate. The fluid that surrounds the body cells is isotonic.

The Osmosis definition :

Osmosis is the passage of water from a region of high water concentration through a semi-permeable membrane to a region of low water concentration.

http://www.caremanager.org/

http://www.caremanager.org/

FACTTHE OSMOTIC PRESSURE OF OUR BODY FLUIDS, EQUIVALENT TO NaCl SOLUTION.

?

•WHY NaCl?•WHY ONLY 0.9%?•Where this (0.9%) came from?

• Na &Cl are the most plentiful electrolyte in the body.

• NORMAL HEALTHY HUMAN: Na 138-146 mMol.L-1

• Cl 98-109 mMol.L-1 K 3.7-5.3 mMol.L-1 Ca

2.25-2.65 mMol.L-1

WHY ONLY 0.9%?FREEZING POINT

-0.52oC

WHY ONLY 0.9%?FREEZING POINT -0.52

oC

Mole in gL-1.= - 1.86oC*iXgL-1. - 0.52oC

For NaCl58.5gL-1 = -

1.86oC*1.8XgL-1 - 0.52oCX= 9.086021505376 g L-1.

Count Lorenzo Romano Amedeo Carlo Avogadro diQuaregna e Cerreto, (TurinAugust 9, 1776 - July 9, 1856)

“Volumiegualidi gas nellestessecondizioniditemperatura e dipressionecontengono la stessonumerodimilecole”

Amedeo Avogadro

http://en.wikipedia.org/wiki/Turin

http://en.wikipedia.org/wiki/August_9

http://en.wikipedia.org/wiki/August_9

http://en.wikipedia.org/wiki/1776

http://en.wikipedia.org/wiki/July_9

http://en.wikipedia.org/wiki/1856

http://en.wikipedia.org/wiki/Image:Amedeo_avogadro2.jpg

AMEDEO AVOGADRO (1776-1856)“Equal volumes of gases at the same temperature and pressure contain the same number of molecules”

Amedeo Avogadro

AVOGADRO’S NUMBER

6.02x1023

AVOGADRO’S NUMBER

THE NUMBER OF PARTICLES IN A

SOLUTION OF ONE Kg OF WATER.

OsmolTHE WEIGHT IN g .

OF A SOLUTE. EQUIVALENT TO A

MOLE.

mOsmTHE WEIGHT IN mg

OF A SOLUTE. EQUIVALENT TO A

mMOLE.

OsmolIT IS THE AMOUNT OF

A SOLUTE, WHICH WILL PROVIDE ONE

AVOGADRO’S NUMBER

Nonelectrolyte(1) Boric acidDextroseGlycerinMannitol monks or nuns

Mole in gL-1.= - 1.86oC*iXgL-1. - 0.52oC

For Boric acid61.8gL-1 = -

1.86oC*1XgL-1 - 0.52oCX= 17.27741935484 g L-1.

EQUIVALENT TO

WHY?NONELECTROLYTE

FREEZING POINT “D”EPRESSION

i = Dissociation(80%)

NaCl Na + Cl 20% 80% + 80%

180/100 = 1.8

NUMBER of PARTICLES

NUMBERofPARTICLES



KCl K + Cl 20% 80% + 80%

180/100 = 1.8



ZnSO4 Zn + SO460% 40% + 40%

140/100 = 1.4


i = Dissociation

CaCl2 Ca +2 Cl20% 80% + (2*80%)

260/100 = 2.6


i = DissociationK2SO4 2 K+SO4

20% (2*80%) + 80%

260/100 = 2.6


i = DissociationFeCl3 Fe+3CL

20% 80% + (3*80%)

340/100 = 3.4

ISO-OSMOTIC or ISOTONIC

or

ISOTONIC HYPERTONIC

17.277 gL-1BORIC ACID

Boric Acid Pass Freely Through the RBC

MembraneRegardless of Concentration.

WHY?

LAXATIVE•MAGNESIUM SULFATE•MAGNESIUM CITRATE•GLYCERIN [RECTAL]

LACTULOSE & CEPHULAC

• NON-ELECTROLYTE• NONABSORBABLE DISACCHARIDE, • + COLON BACTERIA LACTIC ACID• OSMOTIC PRESSURE • ACIDIFICATION SERVE AS A TRAP FOR

AMMONIA BLOOD LEVEL• Rx: SYSTEMIC ENCEPHALOPATHY

Class Generic Name Brand Name

Ammonia Detoxicants Lactulose CEPHULAC

http://www.health.net/pharmacy/members/mlist.asp?category=23

S&S of OSMOTICITY

Normal 285 mOsmol kg-1 282-288 mOsmol kg-1

THIRSTY 294-298 mOsmol kg-1

DRY MUCOUS MEMBRANE:299-313 mOsmol kg-1

WEAKNESS, DOUGHY SKIN: 314-329 mOsmol kg-1

>330 mOsmol kg-1

•DISORIENTATION•POSTURAL HYPOTENSION•SEVERE WEAKNESS•FAINTING•COMA

ifBut what

O.P.1%

HEADACH :275-261 mOsmol kg-1

DROWSINESS: 262-251 mOsmol kg-1

DisorientationCRAMPS250-233 mOsmol kg-1

<230 mOsmol kg-1SEIZURES & COMA

SERUM OSMOLALITY1.86 Na + BLOOD SUGAR + BUN +5

18 2.8

2 Na + BLOOD SUGAR + BUN20 3

A QUIKY

2 Na + 10THE QUIKIST

WHOLE MILK 295

TOMATO JUICE 595

ORANGE JUICE 935

PITUITARY ANTIDIURETIC HORMONE (ADH)

SERUM OSMOCITY

OSMOLALITYDETERMINATION

OS

NaCl “E”QUIVALENT

in RBC

FREEZING POINT

OS

FREEZING POINT


-0.52oC

NaCl 0.9%

Rx

Naphaz

oline H

Cl0.0

2% [24

7 ,2]

Zinc sulfa

te

0.25%

Purified

wate

r

qs30

mL

Mft Iso

tonic

solutio

n

FREEZING POINT

D


Mole in gL-1= - 1.86oC*iXgL-1 - oC

Naphazoline HCl M.Wt. 247 Ions 2

RxNaphazoline HCl 0.02% [247 ,2]

Zinc sulfate 0.25%water qs 30 mLMft Isotonic solution


247gL-1=-1.86oC*1.80.2gL-1 - XoC

Naphazoline HCl

- 0.0027oC


288gL-1= -1.86oC*1.42.5gL-1 - XoC

ZnSO4

0.0226 oC

Total depression - 0.0027oC-0.0226oC- 0.0253 oC

= -0.52-(-0.0253) = 0.4947 oC= 0.52 - 0.0253 = 0.4947 oC

270 mg 0.52oC Xmg 0.4947oC

Step1 Equation How much depression it can a drug cause?

2 of contribution of @ drug

3

OS


in RBC

FREEZING POINT

The weight of NaCl which will produce the same osmotic

pressure effect as 1 g. of the drug.

RxNaphazo

line HCl

0.02%

Zinc sulfa

te

0.25%

Purified w

ater

qs30 m

L

Mft Iso

tonic so

lution

Mole of NaCl gL-1 * iDrugMole of drug gL-1 * iNaCl

Naphazoline HCl M.Wt. 247 Ions 2

247gL-1 *1.8

“E” forNaphazoline HCl 58.5gL-1 *1.8

= 0.2368 g NaCl

Rx Naphazoline HCl

0.02 g* 30 mL/100 mL= 0.006 g Naphaz.

0.2368 g NaCl*0.006 g Naph= 0.0014 g NaCl1 g Naphz

RxNap

hazolin

e HCl

0.02%

Zinc sulfa

te0.25%

Purified w

ater

qs 30 m

L

Mft Iso

tonic solutio

n

288gL-1*1.8

“E” forZnSO4

58.5gL-1*1.4= 0.1579

RxZnSO4

0.25*30/100=0.075 g Zn SO4.0.1579*0.075= 0.0118 g NaCl

RxNaphaz

oline HCl

0.02%

Zinc sulfa

te0.2

5%

Purified w

ater

qs 30 m

L

Mft Iso

tonic so

lution

0.075 g Zn SO4 = 0.0118 g NaCl

0.006 g Naphaz.HCl = 0.0014 g NaCl

=0.0132g NaCl

9 mgmL-1 NaCl*30 mL=270 mg NaCl.

If there no medication, this

prescription will be isotonic with 270 mg

of NaCl.

RxNaCl

270

mg

Purified water

qs30

mL

Mft Isotonic s

olution

Samaan

NaCl 270 mgNaCl 13.2 mgNaCl 256.8 mg

256.8 mg NaCl with “E”

257 mg NaCl with “D”

RxNaphazoline HCl 6 mg

Zinc sulfate

75 mg

Sodium chloride

257 mg

Purified water

qs30 mL

Samaan

OS


in RBC

FREEZING POINT

Volumeof water to be added to a

specified weight of drug to prepare

an isotonic solution.

RxNaphazo

line HCl

0.02%

Zinc sulfa

te

0.25%

Purified w

ater

qs30 m

L

Mft Iso

tonic so

lution

RxNaphazolin

e HCl0.02%

247gL-1 *1.8

“E” for Naphazoline HCl 58.5gL-1 *1.8

= 0.2368 g NaCl

= 0.2368 g NaCl* 0.006g naph.HCl= 0.0014208g NaCl

0.006 g N.HCl=0.0014208g NaCl

0.15792 mL

Dissolve 6 mg of Naph.HCl in 0.16 of

water, the sol. will be isotonic.

0.0014208g NaCl* 1 ml= 0.157 mL H2O0.009 gNaCL

RxZinc su

lfate

0.25%

288gL-1*1.8

“E” for ZnSO4

58.5gL-1*1.4= 0.1579 g NaCl

= 0.1579 g NaCl* 0.075 g ZnSO4

= 0.01184 g NaCl

1.31 mL

If you dissolve 75 mg of ZnSO4 in 1.31 mL of H2O, the sol. will be isotonic.

0.01184 g NaCl/0.009 g NaCl=

1.3125 mL0.1579 mL

1.4704mL

Dissolve 6 mg of naph.HCl& 75 mg ZnSO4 in 1.47 mL

water, and qs to 30 mL with isotonic

0.9% NaCl.

PROBLEM! WHAT

PROBLEM?

P1BORIC ACID “E” 0.52

Isotonic NaCl solution contains 0.9% w/v NaCl . If the “E” value of boric acid is 0.52, calculate the % strength (w/v) of an isotonic solution of boric acid.

BORIC ACID “E” 0.521 g. of boric acid

has the same osmotic pressure as 0.52 g. of NaCl.

IF 0.9% NaCl IS ISOTONIC1 g OF BORIC ACID = 0.52 g NaCl X g OF BORIC ACID= 0.9 g NaClX g BORIC ACID =

1 g B.A. * 0.9 g0.52 g

= 1.73 g%

Mole gL-1= - 1.86oC*i x gL-1 - 0.52oC

61.8gL-1= - 1.86oC*1.0x gL-1 - 0.52oC= 17.27g L-1

17.27 g 1000 mLx g 100 mL

1.727g%

P2IF NaCl DISSOCIATING AT 90%. CALCULATE:-A- DISSOCIATION FACTOR;B- FREEZING POINT OF

MOLAL SOLUTION.

i =Dissociation(2)

NaCl Na + Cl10% 90% + 90%

190/100 = 1.9

A-

- XoC= - 1.86oC*1.9

B-

X = - 3.534oC

P3WHAT IS THE F.P. OF 25 g IN 500 mL DEXTROSE?

D5W

Mole gL-1= - 1.86oC*i[ gL-1] - XoC

180 gL-1= - 1.86oC*150 gL-1 - XoC

= -0.5167oC

P4PROCAINE HCl [M.Wt 273; 2-ION]DISSOCIATING AT 80%.A- DISSOCIATION FACTOR,B-”E”C- F.P. FOR A MOLAL SOLN.

A- DISSOCIATION FACTOR,

PROCAINE HCl PROCAINE + HCl 20% 80% + 80%

180/100 =1.8

B- “E”

273gL-1*1.858.5gL-1*1.8= 0.21421 g Procaine HCl= 0.2142 g NaCl

C- F.P. FOR A MOLAL SOLUTION

= - 1.86oC*i= - 1.86oC *1.8

= - 3.348oC

P5The freezing point of a molal

solution of a nonelectrolyte is

-1.86°C. What is the freezing point of a 0.1 % solution of

zinc chloride (M.Wt. 136), dissociating 80%?

F.P. OF 0.1%ZnCl2 [MWt=136,3]

Mole gL-1= - 1.86oC*i[gL-1] - XoC

136 gL-1= - 1.86oC*2.61gL-1 - xoC

=-0.0355 oC

F.P.D=

P6F.P. of 5% boric acid is -1.55oc.

HOW many g. of boric acid should be used to prepare one L of an isotonic sol.?

Mole gL-1= - 1.86oC*iXgL-1 - 0.52oC

50 gL-1 = - 1.55oC*1XgL-1 - 0.52oC

16.774 gL-

1

Mole gL-1= - 1.86oC*iXgL-1 - 0.52oC6I.8 gL-1= - 1.86oC*1

XgL-1- 0.52oC17.2774 gL-1

P7RxEphedrine sulfate 300 mg Sodium Chloride q.s. Purified water ad 30 mL

Make isotn.sol.Sig. Use as directedHow many mg. of NaCl?

Eph.SO4 429,3

[429,3]

1R How much NaCl can make the whole Rx isotonic?

Rxof contribution of @ drug

3

Step

2

R-Rx

429gL-1*1.858.5gL-1*2.6= 0.1969 g

Nacl

NaCl ‘E’

0.3 g * 0.1969 = 0.059 g NaCl= 59 mg NaCl

270 mg -59 = 210.0 mg NaCl

1Step2

9 mg NaCl* 30 mL the total volume of the Rx1mL

= 270 mg of NaClR

Rx1 g of Ephedrine sulfate has the same osmotic pressure as 0.1969 g Nacl 3

P8RxDipivefrin HCl 0.5% [388,2]Scopolamine HBr0.33[438,2]SodiumChlorideq.s. Purified water ad 30.0 mL Make isotn.sol.Sig. Use in the eyes.How many g. of NaCl?

388gL-1*1.858.5gL-1*1.8= 0.15

gNaCl

0.5 g 100 mL x g. 30 mL0.15 g * 0.15 =

0.0225 g

Dipivefrin HCl

22.5 mgNaCl

E

Scopolamine HBr

438gL-1*1.858.5gL-1*1.8=

0.1335gNaCl

0.33 g 100 mL x g. 30 mL0.1 g * 0.1335 = 0.01335 gNaCl

E

0.0225 g0.0133 g0.0358 g

Both drugs ‘E’

0.270 g0.035 g0.234 g

NaCl reference-‘E’

P9Rx Zinc sulfate 0.06 Boric acidq.s. Purified waterad 30mL Make isotn. sol.Sig. drop in eyes

How many g. of boric acid?

“E” for Zinc sulfate288gL-1*1.858.5gL-1*1.4=

0.15790.06* 0.1579= 0.0095 g

NaCl

0.270 g0.0095 g0.2605 g

NaCl reference - ‘E’

but Rx calls for boric acid!

1 g Boric acid 0.52 g NaClx g Boric acid 0.2605 g

NaCl0.500 g Boric acid

“EB” for Zinc sulfate

288gL-1*1.061.8gL-1*1.4= 0.3 g B.A.

0.06* 0.3= 0.018 g Boric acid

17.3 mg*30 mL1 mL

Boric acid isotonic reference

=519 mg ofBoric acid to make 30mL isotonic [reference]

0.519 g of B.A.[reference]0.018 g of boric acid equivalent

0.500 g Boric acid

PRx Cromolyn Na4% [512,2]Benzalkonium Cl. (1:10,000)[360,2]Buffer sol q.s. Purified water ad 10mL Make isotn. sol.Sig. One drop in each eyeHow many mL of the buffer solution (E = 0.30) should be used to render the solution isotonic?

“E” for Cromolyn Na

512gL-1*1.858.5gL-1*1.8 = 0.1142 g NaCl

0.4 g C.Na* 0.11142= 0.045g NaCl

R= 9 mg NaCl* 10 mL/1 mL = 90 mg NaCl

“E” for Benzalkonium Cl. (1:10,000)[360,2]

360gL-1*1.858.5gL-1*1.8= 0.1625g

NaCl0.001g Bz.* 0.1625= 0.00016 g NaCl0.09 g NaCl - 0.0458g NaCl-0.0442 g NaCl

(E = 0.30)

1 g of buffer material = 0.3 g of NaCl

0.0442 g NaCl * 1 g of buffer material

0.3 g of NaCl= 0.147 g of buffer material 1 g of buffer material = 0.3 g of NaCl3 g of buffer material = 0.9 g of NaCl3 % of buffer sol. = 0.9 % of NS

4.9 mL of isotonic buffer solution

0.147 g of buffer X mL-3.0g Buffer solution 100 mLX mL of isotonic buffer solution =0.147 g * 100 ml/3.0 g buffer= 4.9 mL

P11Rx Dextrose,anhydrous 2.5%

NaCl q.s. Sterile water for injection ad 1000mL Label: isotonic Dextrose & Saline Solution.How many g of NaCl needed?

“E” for anhydrous Dextrose

180gL-1*1.858.5gL-1*1= 0.18 g

NaCl25 g.* 0.18= 4.51g NaCl

4.5 g NaCl

R = 9.0 g NaCl*1L/1L= 9.0 g NaCl

PRxsol.Silver Nitrate 0.5%15.0

Make isoton. sol.Sig. For the eyes.

How many g of KNO3 needed?

Why not to use NaCl as adjustor?

Reference of KNO3Mole gL-1= - 1.86oC*iXgL-1 - 0.52oC

101gL-1= - 1.86oC*1.8XgL-1 - 0.52oC 15.69gL-

1

Reference of KNO3 15.69g 1000 mL

x g 15 mL0.235 g of KNO3 to fill up this prescription

without any medication.

“EKNO3” for Silver Nitrate

170gL-1 *1.8101gL-1 *1.8 =0.5941gKN

O30.075 g AgNO3* 0.5941= 0.0445

gKNO30.235 g of KNO3 Reference

0.0445 g of KNO3 Rx0.190g of KNO3

P13Rx Cocaine HCl0.15[340,2] NaCl

q.s.Purified Water ad 15Make isoton. sol. Sig. One drop for the left eye.How many g of NaCl needed?

“E” for Cocaine HCl

340gL-1*1.858.5gL-1*1.8= 0.172g

NaCl0.15 g C.HCl* 0.172= 0.0258g NaCl

R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl

R-Rx= 0.135 g - 0.0258 g= 0.109gNaCl

P14Rx Cocaine HCl 0.6 [340,2]Eucatropine HCl 0.6 [328,2] Chlorobutanol 0.1 [177,1]NaCl qsPurified Water ad 30 Make isoton. sol. Sig. For the eyes.

58.5gL-1*1.8

0.6 g CHCl.* 0.172= 0.1032 g NaCl

“E” for Cocaine HCl

340gL-1*1.8= 0.172g NaCl


“E” for Eucatropine HCl

328gL-1*1.858.5gL-1*1.8= 0.178

0.6 g Euc.* 0.178= 0.107 g NaCl

“E” for

177gL-1*1.858.5gL-1*1.0=

0.18360.1 g Chl.* 0.1836= 0.0184 g NaCl

Chlorobutanol

Cocaine HCl0.1032 g Eucatropine HCl 0.1070 g Chlorobutanol 0.0184 g 0.2286

g0.270 g - 0.2286 g = 0.0414 g

P15RxTetracaine HCl 0.1 [301,2]Zinc sulfate 0.05 [288,2]Boric acidqsPurified Water ad 30 Make isoton. sol. Sig. For the eyes. How many g. Boric acid needed?

“Eb” for Tetracaine HCl

301gL-1*1.061.8gL-1*1.8= 0.3695 g

Boric0.1 g Tetr.* 0.3695= 0.0369 g Boric

17.3 g*30 mL= 0.519 g Boric1000 mL

Reference

“Eb” for Zinc sulfate

288gL-1*1.061.8gL-1*1.4= 0.3004 g

Boric

0.05 g Zn.* 0.3004= 0.0150 g Boric

Tetracaine HClZinc sulfate

0.0369 g0.0150 g

0.0519 g

0.0519g Boric acid Equivalent

0.519 g Boric acid Reference

0.467 g Boric acid

P16Rx Sol. HomatropineHBr 1% 15 [356,2] Boric acid q.s.

Make isoton. sol. Sig. For the eyes. How many g. Boric acid needed?

“Eb” for HomatropineHBr

356gL-1*1.061.8gL-1*1.8 = 0.3125

17.3 g*15 mL/1000 mL Boric acid Reference

0.2595g Boric acid

x g = 0.15 g

0.15 g Homa.* 0.3125= 0.0469 g Boric

Boric acid Equivalent

0.2595 g Boric acid reference0.0469 g Boric acid Equivalent0.212 g Boric Acid

P17Rx Procaine HCl 0.1% [273,2] NaCl q.s.SterileWater for Inj. ad 100.0 Make isoton. sol. Sig. For Injection. How many g. NaCl needed?

“E” for

273gL-1*1.858.5gL-1*1.8 = 0.2142 g

NaCl1 g Proc.* 0.2142= 0.2142 g NaCl

Procaine HClR=0.009g NaCl * 100 mL/1 mL =0.9 g NaCl

0.9 g NaCl reference0.2142 g NaCl Equival

0.6857 g NaCl

P18Rx Phenylephrine HCl 1%[204,2] Chlorobutanol

0.5%[177,1] NaClq.s.

Purified Water ad 15.0 Make isoton. sol. Sig. Use as directed How many mL NSS needed?

“E” for Phenylephrine HCl

204gL-1*1.858.5gL-1*1.8= 0.2867 g

NaCl0.15 g Phen.* 0.2867= 0.043 g NaCl


“E” for

177gL-1*1.858.5gL-1*1.0= 0.1836 g

NaCl0.075 g Ch.* 0.1836= 0.0918

Chlorobutanol

g NaClPhenylephrine HCl + Chlorobutanol

0.043 g NaCl+0.0918 g NaCl = 0.0567 g NaCl

0.078 g NaCl*100 mL0.9 g NaCl

R-Rx=0.135-0.0567 =0.078 g NaCL

8.69 mL N.S.

P19Rx Oxymetazoline HCl 0.5% [297,2]Boric acid sol. qsPurified Water ad 15.0 Make isoton. sol. Sig. For the nose, as decongestant

How many mL of 5% boric acid solution needed?

“Eb” for

297gL-1*1.061.8gL-1*1.8 = 0.3745 g

B.A0.075 g Oxy* 0.3745= 0.028 g B.A.

Oxymetazoline HCl

17.3 g*15 mL = 0.2595 g B.A. 1000 mLR

0.2595 g B.acid reference0.028 g B.acidEquival0.2315 g Boric acid

4.63 mL of 5% Boric acid solution.

0.2315 g B.A.*100 ml/5 g B.A.

P20Rx Ephedrine HCl 0.55 [202,2]Chlorobutanol 0.25 [177,1]

Dextrose qsRose Water ad 50.0 Make isoton. sol. Sig. Nose drop.

How many g of Dextrose needed?

180 g L-1 = 1.86 *1x g L-1 -.52oC

50.32 gL-1

50.32 g*0.05 L 1L

2.516 g Dex. Ref.

“Ed” for Ephedrine HCl

202gL-1*1.0180gL-1*1.8= 1.60.5 g Eph* 1.6= 0.80g Dex.

“Ed” for Chlorobutanol

177gL-1*1.0180gL-1*1.0 =

1.01690.25 g Ch* 1.0169= 0.2542 g Dex.

Ephedrine HCl 0.80g Dex.Chlorobutanol 0.2542 g Dex. 1.056 g

Dex.1.056 g Dex. Equival

1.4598 g Dex.

2.516 g Dex. Ref.

P21Naphzoline HCl 1% [247,2]Sodium Chloride qsPurified Water ad 30 mL

Make isoton. sol. Sig. Use as directed in the eye.How many g of NaCl needed? Using Freezing point method.

247 gL-1= - 1.86oC*1.810gL-1 - xoC

-0.1355 oC

(-0.1355 oC)=- 0.384oC

0.52-

0.270 g NaCl Reference0.27 g freezes at -0.52oCx g NaCl -0.3844oC0.1995 g

NaCl

P22Rx Oxytetracycline HCl 0.05 [497,2]

Chlorobutanol 0.1 [177,1]NaCl qsPurified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye.How many mg of NaCl needed?

“E” for Oxytetracycline HCl

497gL-1*1.858.5gL-1*1.8=

0.11770.05 g Oxy* 0.1177= 0.0058g NaCl

“E” for Chlorobutanol

177gL-1 *1.858.5gL-1*1.0=

0.18360.1 g Chl* 0.1836= 0.01836g NaCl

Oxytetracycline HCl 0.0058g Chlorobutanol0.0183g

0.0231 g0.270 g NaCl

Reference

0.2469g NaCl

0.0231 g

P23Rx Tetracaine HCl 0. 5% [301,2][iso]Sol. Epineph. Bitart.10.0 [333,2]

Boric acidqsPurified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye.How many g of Boric acid needed?

“Eb” for Tetracaine HCl

301gL-1*1.061.8gL-1*1.8= 0.3695

0.15 g Tet.* 0.3695= 0.0554 g B.A.

17.3 g*20 mL/1000mL 0. 35 g B.A. for 20 mL Reference

0. 35 g B.A. for 20 mL Reference

0.2946g Boric acid

0.0554 g B.A.

P24Anhyd.NaH2PO4 5.6 g[120,2]Anhyd.Na2HPO4 2.84 g[142,3] NaCl qsPurified Water ad 1000 mL Label: Isotonic buffer sol.,pH6.5How many g of NaCl needed?

“E” for Anhyd.NaH2PO4

120 gL-1*1.858.5 gL-1*1.8= 0.495.6 g Mono* 0.49= 2.744 g NaCl

“E” for Anhyd.Na2HPO4= 0.595

2.84 g Di* 0.595= 1.69 g NaCl

142 gL-1*1.858.5 gL-1*2.6

Anhyd.NaH2PO42.744 g NaClAnhyd.Na2HPO41.69 g NaCl

4.434 g NaCl9.0 g NaCl

Reference4.434 g NaCl4.566 g NaCl

P25How many g of anhydrous Dextrose needed in preparing 1 L of a 0.5% isotonic Ephedrine Sulfate [429,3] Nasal spray?

“Ed” for Ephedrine Sulfate429 gL-1*1.0180gL-1 *2.6=1.090

95.0 g Ephd.* 1.09= 5.45 g Dex.50 gL-1 Anhydrous Dex.Ref. 5.45 gL-1 Dex.44.54 g Dextrose

P26Ephedrine Sulfate 1% [429,3]Chlorobutanol 0.5%[177,1]Purified Water ad 100.0 Make isoton. sol. & Buffer at 6.5 Sig. Nose drop.How many mL of a buffer & mL of water should be used?

“E” for Ephedrine Sulfate

429gL-1*1.858.5gL-1*2.6= 0.1969

1.0 g Eph* 0.1969= 0.1969 g NaCl

V-Value for Ephedrine Sulfate

0.1969 g NaCl100 mL H2O 0.9 g NaClx mL

21.87 mL of water will make 1 g of Ephedrine Sulfate isotonic.

“E” for Chlorobutanol

177gL-1*1.858.5gL-1*1.0= 0.1836

0.5 g Ch* 0.1836= 0.0918 g NaCl

V-Value For Chlorobutanol

0.0918 g NaCl100 mL H2O 0.9 g NaClx mL

x mL= 10.20 mL of water will make 0.5 g of chlorobutanol isotonic.

Ephedrine Sulf. 21.87 mL of water Chlorobutanol 10.20 mL of

water 32.07 mL of water

67.93 mL of isotonic buffer solution.

P27Oxytetracycline HCl 0.5% [497,2]

[iso]Tetracaine HCl Sol. 2%15 mlNaClqsPurified Water ad 30 mL Make isoton. sol. Sig. Use as directed in the eye.How many mL of NSS needed?

“E” for Oxytetracycline HCl

497gL-1*1.858.5gL-1* = 01177g

NaCl.0.15 g Oxy* 0.1177= 0.0176

1.8

0.135 g NaCl Reference in only 15 mL 0.0176 g NaCl Equivalent0.1174 g. NaCl

0.9 g NaCl 100 mLx mL0.1174 g. NaCl

13.0 mL of NSS

Determine if the following commercial products are Hypotonic, isotonic, or Hypertonic:

a- An ophthalmic sol. 40 mgmL-1 of Cromolyn Sodium [512,2]& 0.01% of Benzalkomiun chloride [360,2]

in purified water.

“E” for CromolynSodium512gL-1 = -1.86oC*1.840gL-1= -XoC

-0.26oC-0.52>-

0.52<-0.52

HYPOISO-T.HYPER

A parenteral infusion containing 20% (w/v) of mannitol.

182gL-1 = -1.86oC*1.0200gL-1 = -XoC

-0.52>-0.52

<-0.52

HYPOISO-T.HYPER

FP ”D” Mannitol

-2.04395

-2.04395

A 500-mL large volume parenteral containing D5W (5% w/v of anhydrous dextrose in sterile water for injection).

180gL-1 = -1.86oC*1.050gL-1= -XoC

-0.52>-0.52

<-0.52

HYPOISO-T.HYPER-2.04395

-0.519

A FLEET saline enema containing 19 g of monobasic sodium phosphate (monohydrate) and 7 g of dibasic sodium phosphate (heptahydrate) in 118 mL of aqueous solution.

Monobasic sodium phosphate (monohydrate)(138, 2)

138gL-1 = -1.86oC*1.8161gL-1 = -XoC

-0.52>-0.52

<-0.52

HYPOISO-T.HYPER

-3.9

Dibasic sodium phosphate (heptahydrate)(268,3)

268gL-1 = -1.86oC*2.659.32gL-1 = -XoC

-0.52>-0.52

<-0.52

HYPOISO-T.HYPER

-1.07

For agents having the following sodium chloride equivalents, calculate the percentage concentration of an isotonic solution:

(A) 0.20 0.90%0.20

= 4.5%

(b) 0.32 0.90%0.32

= 2.81%

(c) 0.61 0.90%0.61

= 1.48%

How many mL each of purified water and an isotonic sodium chloride solution should be used to prepare 30 mL of a 1% w/v isotonic solution of fentanyl citrate (E = 0.11)?

P30

1 g * 3 mL/100 mL = 0.3 g fentanyl citrate

0.3 g fentanyl citrate* 0.11= 0.033 g NaCl 0.033 g NaCl* 100 mL/0.9 g NaCl= 3.66 mL H2O 30 mL - 3.66 mL H2O= 26.23 mL N.S.

Calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:(a) Antipyrine [ 188, 1]

58.5*1/(188*1.8) = 0.173 g * 0.3 = 0.0512 g NaCl

0.051 g NaCl* 100 mL / 0.9 g NaCl = 5.76 mL H2O

Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:(b) Chlorobutanol [177, 1]

58.5*1/(177*1.8) = 0.184 g * 0.3 = 0.0553 g NaCl


Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:(c) ephedrine sulfate [429, 3]

58.5*2.6/(429*1.8) = 0.1969 g * 0.3 = 0.059 g NaCl


Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:(d) silver nitrate [170, 2]

58.5*1.8/(170*1.8) = 0.344 g * 0.3 = 0.103 g NaCl


Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:(e) zinc sulfate [288, 2]

58.5*1.4/(288*1.8) = 0.158 g * 0.3 = 0.0473 g NaCl


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