A
FUNCTIONS
e.g.
•TEMPERATURE;
•HEART RATE;
•BLOOD PRESSURE.
C
A
B
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.
FACT
?
THE 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
o
-0.52 C
WHY ONLY 0.9%?
FREEZING POINT
o
-0.52 C
Mole in
-1
XgL .
-1
gL .=
o
- 1.86 C*i
o
- 0.52 C
For NaCl
-1
58.5gL
-1
XgL
o
= -1.86 C*1.8
o
- 0.52 C
X= 9.086021505376 g
-1
L .
 “Volumiegualidi
gas
nellestessecondizionidi
temperatura
e
dipressionecontengono
la
stessonumerodimilecol
e”
Amedeo Avogadro
Count Lorenzo Romano Amedeo Carlo Avogadro
diQuaregna e Cerreto, (TurinAugust 9, 1776 - July 9,
1856)
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
23
6.02x10
AVOGADRO’S NUMBER
THE NUMBER OF
PARTICLES IN A
SOLUTION OF ONE
Kg OF WATER.
Osmol
THE WEIGHT IN g .
OF A SOLUTE.
EQUIVALENT TO A
MOLE.
mOsm
THE WEIGHT IN mg
OF A SOLUTE.
EQUIVALENT TO A
mMOLE.
Osmol
IT IS THE AMOUNT OF
A SOLUTE, WHICH
WILL PROVIDE ONE
AVOGADRO’S
NUMBER
Nonelectrolyte(1)
Boric acid
Dextrose
Glycerin
Mannitol
monks or nuns
Mole in
-1
XgL .
-1
gL .=
o
- 1.86 C*i
o
- 0.52 C
For Boric acid
-1
61.8gL
-1
XgL
o
= - 1.86 C*1
o
- 0.52 C
X= 17.27741935484 g
-1
L .
EQUIVALENT TO
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
FREEZING POINT “
D”EPRESSION
i = Dissociation(80%)
KCl
K + Cl
20%
80% + 80%
180/100 = 1.8
FREEZING POINT “
D”EPRESSION
i = Dissociation(40%)
ZnSO4 Zn + SO4
60%
40% + 40%
140/100 = 1.4
FREEZING POINT “
D”EPRESSION
i = Dissociation
CaCl2
20%
Ca +2 Cl
80% + (2*80%)
260/100 = 2.6
FREEZING POINT “
D”EPRESSION
i = Dissociation
K2SO4 2 K+SO4
20%
(2*80%) + 80%
260/100 = 2.6
FREEZING POINT “
D”EPRESSION
i = Dissociation
FeCl3
Fe+3CL
20%
80% + (3*80%)
340/100 = 3.4
ISO-OSMOTIC or ISOTONIC
or
Why?
ISOTONIC
HYPERTONIC
17.277
-1
gL BORIC ACID
WHY?
Boric Acid Pass Freely Through
the RBC
MembraneRegardless of
Concentration.
LAXATIVE
•MAGNESIUM SULFATE
•MAGNESIUM CITRATE
•GLYCERIN [RECTAL]
LACTULOSE
&
CEPHULAC
Generic
Class
Brand Name
Name
Ammonia Detoxicants
Lactulose
CEPHULAC
•NON-ELECTROLYTE
• NONABSORBABLE DISACCHARIDE,
•+ COLON BACTERIA
•
LACTIC ACID
OSMOTIC PRESSURE
•ACIDIFICATION SERVE AS A TRAP FOR
AMMONIA
BLOOD LEVEL
•Rx: SYSTEMIC ENCEPHALOPATHY
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
-1
kg
•DISORIENTATION
•POSTURAL HYPOTENSION
•SEVERE WEAKNESS
•FAINTING
•COMA
But what
1%
O.P.
HEADACH :275-261
mOsmol kg-1
DROWSINESS: 262-251 mOsmol kg-1
DisorientationCRAMPS250-233 mOsmol kg-1
-1
<230
mOsmol
kg
SEIZURES & COMA
SERUM OSMOLALITY
1.86 Na + BLOOD SUGAR + BUN +5
18
2.8
A QUIKY
2 Na + BLOOD SUGAR + BUN
20
THE QUIKIST
2 Na + 10
3
WHOLE MILK
295
TOMATO JUICE
595
ORANGE JUICE
935
SERUM OSMOCITY
PITUITARY ANTIDIURETIC
HORMONE (ADH)
OSMOLALITYD
ETERMINATION
NaCl “E”QUIVALENT
OS
 in RBC
FREEZING POINT
FREEZING POINT
OS
FREEZING POINT “
D”EPRESSION
NaCl 0.9%
o
-0.52 C
FREEZING POINT
FREEZING POINT “
D”EPRESSION
Naphazoline HCl
2
M.Wt. 247 Ions
-1
gL =
Mole in
-1
XgL
o
- 1.86 C*i
o
- C
FREEZING POINT “
D”EPRESSION
Naphazoline HCl
o
-1
247gL =-1.86
-1
0.2gL
C*1.8
o
-X C
o
- 0.0027 C
FREEZING POINT “
D”EPRESSION
ZnSO4
-1
288gL =
-1
2.5gL
o
-1.86 C*1.4
o
-X C
o
0.0226 C
Total depression
o
- 0.0027 C
o
-0.0226 C
- 0.0253
oC
o
= -0.52-(-0.0253) = 0.4947 C
o
= 0.52 - 0.0253 = 0.4947 C
o
270 mg
0.52 C
o
Xmg0.4947 C
Step 1
Equation
How much depression it can a drug cause?
2
3
of contribution of @ drug
NaCl “E”QUIVALENT
OS
 in RBC
FREEZING POINT
The weight of NaCl which
will produce the same
osmotic pressure effect as
1 g. of the drug.
-1
gL
Mole of NaCl
* iDrug
-1
*
Mole of drug gL iNaCl
Naphazoline HCl M.Wt.
247 Ions2
“E” forNaphazoline HCl
-1
58.5gL
*1.8
-1
247gL *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 NaCl
1 g Naphz
“E” forZnSO4
-1
58.5gL *1.4
-1
288gL *1.8
= 0.1579
RxZnSO4
0.25*30/100=0.075 g Zn SO4.
0.1579*0.075= 0.0118 g NaCl

0.006 g Naphaz.HCl = 0.0014 g NaCl
0.075 g Zn SO4
= 0.0118 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.
NaCl 270 mg
NaCl 13.2 mg
NaCl
256.8 mg
256.8 mg NaCl with “E”
257 mg NaCl with “D”
NaCl “E”QUIVALENT
OS
 in RBC
FREEZING POINT
olumeof water to be
added to a specified
weight of drug to
prepare an isotonic
solution.
“E” for Naphazoline HCl
-1
58.5gL
*1.8
-1
247gL *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
0.0014208g NaCl* 1 ml= 0.157 mL H2O
0.009 gNaCL
Dissolve 6 mg of
Naph.HCl in 0.16 of
water, the sol. will be
isotonic.
0.15792 mL
“E” for ZnSO4
-1
58.5gL *1.4
-1
288gL *1.8
= 0.1579 g NaCl
= 0.1579 g NaCl* 0.075 g ZnSO4
= 0.01184 g NaCl
0.01184 g NaCl/0.009 g NaCl=
1.31 mL
If you dissolve 75 mg of
ZnSO4 in 1.31 mL of H2O,
the sol. will be isotonic.
0.1579 mL
1.3125 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?
P1
BORIC 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” 0.52
1 g. of boric acid has
the same osmotic
pressure as 0.52 g. of
NaCl.
IF 0.9% NaCl IS ISOTONIC
1 g OF BORIC ACID = 0.52 g NaCl
X g OF BORIC ACID= 0.9 g NaCl
X g BORIC ACID =
1 g B.A. * 0.9 g = 1.73 g%
0.52 g
-1
gL =
Mole
-1
x gL
o
- 1.86 C*i
o
- 0.52 C
-1
61.8gL =
o
- 1.86 C*1.0
o
-1
x gL - 0.52 C
= 17.27g
-1
L
17.27 g
xg
1000 mL
100 mL
1.727g%
P2
IF NaCl DISSOCIATING AT
90%. CALCULATE:-
A- DISSOCIATION FACTOR;
B- FREEZING POINT OF
MOLAL SOLUTION.
A-
i =Dissociation(2)
NaCl
Na + Cl
10%
90% + 90%
190/100 = 1.9
Bo
- X C= -
o
1.86 C*1.9
o
X = - 3.534 C
P3
WHAT IS THE F.P. OF 25 g
IN 500 mL DEXTROSE?
D5W
-1
gL =
Mole
-1
[ gL ]
-1
gL =
180
-1
50 gL
o
- 1.86 oC*i
-X C
o
- 1.86 C*1
o
-X C
o
= -0.5167 C
P4
PROCAINE 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
20%
PROCAINE + HCl
80% + 80%
180/100 =1.8
B- “E”
-1
58.5gL *1.8
-1
273gL *1.8
= 0.2142
1 g Procaine HCl= 0.2142 g NaCl
C- F.P. FOR A MOLAL SOLUTION
o
= - 1.86 C*i
o
= - 1.86 C *1.8
o
= - 3.348 C
P5
The 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]
F.P.D=
Mole
gL-1=
- 1.86 C*i
o
[gL-1]
-1
136 gL =
-1
1gL
o
-X C
o
- 1.86 C*2.6
o
-x C
o
=-0.0355 C
P6
F.P. of 5% boric acid is
o
-1.55 c.
HOW many g. of boric acid
should be used to prepare
one L of an isotonic sol.?
Mole
-1
XgL
-1
gL =
-1
gL
50
-1
XgL
16.774
o
- 1.86 C*i
o
- 0.52 C
o
= - 1.55 C*1
o
- 0.52 C
-1
gL
o
-1
gL =
Mole
- 1.86 C*i
o
-1
XgL
- 0.52 C
o
-1
6I.8 gL = - 1.86 C*1
o
-1
XgL - 0.52 C
17.2774
-1
gL
P7 Eph.SO4 429,3
Rx
Ephedrine sulfate 300 mg Sodium
[429,3]
Chloride
q.s.
Purified water ad
30 mL
Make isotn.sol.
Sig. Use as directed
How many mg. of NaCl?
Step 1
R How much NaCl can make the
whole Rx isotonic?
2
3
Rxof contribution of @ drug
R-Rx
1
Step
2
NaCl ‘E’
9 mg NaCl* 30 mL the total volume of the Rx
1mL
R
= 270 mg of NaCl
58.5gL-1*2.6
=
0.1969
g
Nacl
429gL-1*1.8
Rx
1 g of Ephedrine sulfate has the same
osmotic pressure as 0.1969 g Nacl
3
0.3 g * 0.1969 = 0.059 g NaCl= 59 mg NaCl
270 mg -59 = 210.0 mg NaCl
P8
RxDipivefrin HCl 0.5%
[388,2]
Scopolamine HBr0.33[438,2]
SodiumChlorideq.s.
Purified water
Make isotn.sol.
ad
Sig. Use in the eyes.
30.0 mL
How many g. of NaCl?
Dipivefrin HCl
-1
58.5gL *1.8
E 388gL
= 0.15 gNaCl
-1*1.8
0.5 g
x g.
100 mL
30 mL
0.15 g * 0.15 = 0.0225 g
22.5 mgNaCl
Scopolamine HBr
-1
58.5gL *1.8
E 438gL *1.8
= 0.1335gNaCl
-1
0.33 g
x g.
100 mL
30 mL
0.1 g * 0.1335 = 0.01335 gNaCl
Both drugs ‘E’
0.0225 g
0.0133 g
0.0358 g
NaCl reference -‘E’
0.270
0.035
g
g
0.234 g
P9
Rx Zinc sulfate
0.06 Boric
acidq.s. Purified water ad
30mL Make isotn. sol.
Sig. drop in eyes
How many g. of boric acid?
“E” for Zinc sulfate
-1
58.5gL *1.4 = 0.1579
-1
288gL *1.8
0.06* 0.1579= 0.0095 g NaCl
NaCl reference - ‘E’
0.270 g
0.0095 g
0.2605 g
but Rx calls for boric acid!
1 g Boric acid
x g Boric acid
0.52 g NaCl
0.2605 g NaCl
0.500 g Boric acid
Boric acid isotonic reference
17.3 mg*30 mL
1 mL
=519 mg ofBoric acid
to make 30mL
isotonic [reference]
“EB” for Zinc sulfate
61.8gL-1*1.4
= 0.3 g B.A.
-1
288gL *1.0
0.06* 0.3= 0.018 g Boric acid
0.519 g of B.A.[reference]
0.018 g of boric acid equivalent
0.500 g Boric acid
P
Rx Cromolyn Na4%
Benzalkonium Cl.
Buffer sol
Purified water
[512,2]
(1:10,000)
[360,2]
q.s.
ad
10mL
Make isotn. sol.
Sig. One drop in each eye
How many mL of the buffer solution (E = 0.30) should be
used to render the solution isotonic?
R= 9 mg NaCl* 10 mL/1 mL = 90 mg NaCl
“E” for Cromolyn Na
-1
58.5gL *1.8
-1
512gL *1.8
= 0.1142 g NaCl
0.4 g C.Na* 0.11142= 0.045g NaCl
“E” for Benzalkonium Cl. (1:10,000)[360,2]
-1
58.5gL *1.8 = 0.1625g NaCl
-1
360gL *1.8
0.001g Bz.* 0.1625= 0.00016 g NaCl
0.09 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 NaCl
3 g of buffer material = 0.9 g of NaCl
3 % of buffer sol. = 0.9 % of NS
3.0g Buffer solution
0.147 g of buffer
-
100 mL
X mL
X mL of isotonic buffer solution
=0.147 g * 100 ml/3.0 g buffer=
4.9 mL
4.9 mL of isotonic buffer solution
P11
Rx Dextrose,anhydrous
NaCl
2.5%
q.s.
Sterile water for injection ad 1000mL Label:
isotonic Dextrose & Saline Solution.
How many g of NaCl needed?
“E” for anhydrous Dextrose
R = 9.0 g NaCl*1L/1L=
58.5gL-1*1
=
-1
180gL *1.8
0.18 g NaCl
25 g.* 0.18= 4.51g NaCl
4.5 g NaCl
P
Rxsol.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 KNO3
Mole
-1
XgL
-1
gL =
-1
101gL =
-1
XgL
o
- 1.86 C*i
o
- 0.52 C
o
- 1.86 C*1.8
o
- 0.52 C
-1
15.69gL
Reference of KNO3
15.69g
1000 mL
xg
15 mL
0.235 g of KNO3 to fill up
this prescription without
any medication.
“EKNO3” for Silver Nitrate
-1
101gL
-1
170gL
*1.8 =0.5941gKNO
3
*1.8
0.075 g AgNO3* 0.5941= 0.0445 gKNO
3
0.235 g of KNO3 Reference
0.0445 g of KNO3 Rx
0.190g of KNO3
P13
Rx Cocaine HCl0.15[340,2]
NaCl
Purified Water ad
q.s.
15
Make isoton. sol.
Sig. One drop for the left eye.
How many g of NaCl needed?
R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl
“E” for Cocaine HCl
-1
58.5gL *1.8
= 0.172g NaCl
-1
340gL *1.8
0.15 g C.HCl* 0.172= 0.0258g NaCl
R-Rx= 0.135 g - 0.0258 g= 0.109gNaCl
P14
Rx Cocaine HCl
0.6 [340,2]
Eucatropine HCl
Chlorobutanol
0.6 [328,2]
0.1 [177,1]
NaCl
Purified Water
Make isoton. sol.
Sig. For the eyes.
qs
ad
30
R=0.009g NaCl * 30 mL/1 mL =0.27 g NaCl
“E” for Cocaine HCl
58.5gL-1*1.8
= 0.172g NaCl
340gL-1*1.8
0.6 g CHCl.* 0.172= 0.1032 g NaCl
“E” for Eucatropine HCl
-1
58.5gL *1.8
-1
328gL *1.8
= 0.178
0.6 g Euc.* 0.178= 0.107 g NaCl
“E” for Chlorobutanol
-1
58.5gL *1.0
-1
177gL *1.8
= 0.1836
0.1 g Chl.* 0.1836= 0.0184 g NaCl
Cocaine HCl0.1032 g
Eucatropine HCl 0.1070 g
Chlorobutanol 0.0184 g
0.2286 g
0.270 g - 0.2286 g = 0.0414 g
P15
RxTetracaine HCl 0.1 [301,2]
Zinc sulfate
0.05 [288,2]
Boric acidqs
Purified Water
ad
30
Make isoton. sol.
Sig. For the eyes.
How many g. Boric acid needed?
Reference
17.3 g*30 mL= 0.519 g Boric
1000 mL
“Eb” for Tetracaine HCl
61.8gL-1*1.8
301gL-1*1.0
= 0.3695 g Boric
0.1 g Tetr.* 0.3695= 0.0369 g Boric
“Eb” for Zinc sulfate
-1
61.8gL *1.4
= 0.3004 g Boric
-1
288gL *1.0
0.05 g Zn.* 0.3004= 0.0150 g Boric
Tetracaine HCl
Zinc sulfate
0.0369 g
0.0150 g
0.0519 g
0.519 g Boric acid Reference
0.0519g Boric acid Equivalent
0.467 g Boric acid
P16
Rx
Sol. HomatropineHBr 1% 15 [356,2]
Boric acid
q.s.
Make isoton. sol.
Sig. For the eyes.
How many g. Boric acid needed?
Boric acid Reference
17.3 g*15 mL/1000 mL
0.2595g Boric acid
“Eb” for HomatropineHBr
-1
61.8gL *1.8
-1
356gL *1.0
x g = 0.15 g
= 0.3125
Boric acid Equivalent
0.15 g Homa.* 0.3125= 0.0469 g Boric
0.2595 g Boric acid reference
0.0469 g Boric acid Equivalent
0.212 g Boric Acid
P17
Rx
Procaine HCl
NaCl
0.1% [273,2]
SterileWater for Inj. ad
100.0
q.s.
Make isoton. sol.
Sig. For Injection.
How many g. NaCl needed?
R=0.009g NaCl * 100 mL/1 mL =0.9 g NaCl
“E” for Procaine HCl
-1
58.5gL *1.8 = 0.2142 g NaCl
-1
273gL *1.8
1 g Proc.* 0.2142= 0.2142 g NaCl
0.9 g NaCl reference
0.2142 g NaCl Equival
0.6857 g NaCl
P18
Rx Phenylephrine HCl 1%[204,2]
Chlorobutanol
0.5%[177,1]
NaCl
q.s.
Purified Water ad
15.0
Make isoton. sol.
Sig. Use as directed
How many mL NSS needed?
R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl
“E” for Phenylephrine HCl
58.5gL-1*1.8 = 0.2867 g NaCl
204gL-1*1.8
0.15 g Phen.* 0.2867= 0.043 g NaCl
“E” for Chlorobutanol
58.5gL-1*1.0
= 0.1836 g NaCl
-1
177gL *1.8
0.075 g Ch.* 0.1836= 0.0918 g NaCl
Phenylephrine HCl + Chlorobutanol
0.043 g NaCl+0.0918 g NaCl = 0.0567 g NaCl
R-Rx=0.135-0.0567 =0.078 g NaCL
0.078 g NaCl*100 mL
0.9 g NaCl
8.69 mL N.S.
P19
Rx
Oxymetazoline HCl
0.5% [297,2]
Boric acid sol.
qs
Purified Water ad
15.0
Make isoton. sol.
Sig. For the nose, as decongestant
How many mL of 5% boric acid solution
needed?
R
17.3 g*15 mL
1000 mL
= 0.2595 g B.A.
“Eb” for Oxymetazoline HCl
61.8gL-1*1.8
297gL-1*1.0
= 0.3745 g B.A
0.075 g Oxy* 0.3745= 0.028 g B.A.
0.2595 gB.acid reference
0.028 g B.acidEquival
0.2315 g Boric acid
0.2315 g B.A.*100 ml/5 g B.A.
4.63 mL of 5% Boric acid solution.
P20
Rx
Ephedrine HCl
0.55 [202,2]
Chlorobutanol
0.25 [177,1]
Dextrose
Rose Water ad
qs
50.0
Make isoton. sol.
Sig. Nose drop.
How many g of Dextrose needed?
180 g
xg
-1
L =
-1
L
1.86 *1
o
-.52 C
50.32 gL-1
50.32 g*0.05 L
1L
2.516 g Dex. Ref.
“Ed” for Ephedrine HCl
-1
180gL *1.8
-1
202gL *1.0
= 1.6
0.5 g Eph* 1.6= 0.80g Dex.
“Ed” for Chlorobutanol
-1
180gL *1.0
-1
177gL *1.0
= 1.0169
0.25 g Ch* 1.0169= 0.2542 g Dex.
Ephedrine HCl 0.80g Dex.
Chlorobutanol 0.2542 g Dex.
1.056 g Dex.
2.516 g Dex. Ref.
1.056 g Dex. Equival
1.4598 g Dex.
P21
Naphzoline HCl
1% [247,2]
Sodium Chloride
qs
Purified Water ad
30 mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many g of NaCl needed? Using
Freezing point method.
o
-1
gL =
247
- 1.86 C*1.8
o
-1
10gL
-x C
o
-0.1355 C
0.52-
o
o
(-0.1355 C)=- 0.384 C
0.270 g NaCl Reference
0.27 g freezes at
x g NaCl
-0.52oC
-0.3844oC
0.1995 g NaCl
P22
Rx
Oxytetracycline HCl
0.05 [497,2]
Chlorobutanol
0.1 [177,1]
NaCl
Purified Water ad
qs
30 mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many mg of NaCl needed?
“E” for Oxytetracycline HCl
-1
58.5gL *1.8
-1
497gL *1.8
= 0.1177
0.05 g Oxy* 0.1177= 0.0058g NaCl
“E” for Chlorobutanol
-1
58.5gL *1.0
-1
177gL *1.8
= 0.1836
0.1 g Chl* 0.1836= 0.01836g NaCl
Oxytetracycline HCl 0.0058g
Chlorobutanol
0.0183g
0.0231 g
0.270 g NaCl Reference
0.0231 g
0.2469g NaCl
P23
Rx
Tetracaine HCl
0. 5% [301,2]
[iso]Sol. Epineph. Bitart.10.0 [333,2]
Boric acidqs
Purified Water ad
30 mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many g of Boric acid needed?
17.3 g*20 mL/1000 mL
0. 35 g B.A. for 20 mL Reference
“Eb” for Tetracaine HCl
61.8gL-1*1.8
-1
301gL *1.0
= 0.3695
0.15 g Tet.* 0.3695= 0.0554 g B.A.
0. 35 g B.A. for 20 mL
Reference
0.0554 g B.A.
0.2946g Boric acid
P24
Anhyd.NaH2PO4
5.6 g[120,2]
Anhyd.Na2HPO4
2.84 g[142,3]
NaCl
qs
Purified Water ad 1000 mL Label:
Isotonic buffer sol.,pH6.5
How many g of NaCl needed?
“E” for Anhyd.NaH2PO4
-1
gL *1.8
58.5
=
0.49
-1
120 gL *1.8
5.6 g Mono* 0.49= 2.744 g NaCl
“E” for Anhyd.Na2HPO4
58.5
142
-1
gL *2.6
-1
gL *1.8
= 0.595
2.84 g Di* 0.595= 1.69 g NaCl
Anhyd.NaH2PO4 2.744 g NaCl
Anhyd.Na2HPO4 1.69 g NaCl
4.434 g NaCl
9.0 g NaCl Reference
4.434 g NaCl
4.566 g NaCl
P25
How many g of anhydrous
Dextrose needed in preparing
1 L of a 0.5% isotonic
Ephedrine Sulfate [429,3]
Nasal spray?
“Ed” for Ephedrine Sulfate
-1
180gL
*2.6 =1.0909
-1
429 gL *1.0
5.0 g Ephd.* 1.09= 5.45 g Dex.
50
-1
gL Anhydrous
Dex.Ref.
5.45 gL-1 Dex.
44.54 g Dextrose
P26
Ephedrine Sulfate
Chlorobutanol
Purified Water ad
1% [429,3]
0.5%[177,1]
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
-1
58.5gL *2.6
-1
429gL *1.8
= 0.1969
1.0 g Eph* 0.1969= 0.1969 g NaCl
V-Value for Ephedrine Sulfate
100 mL H2O
x mL
0.9 g NaCl
0.1969 g NaCl
21.87 mL of water will make 1 g of
Ephedrine Sulfate isotonic.
“E” for Chlorobutanol
-1
58.5gL *1.0
-1
177gL *1.8
= 0.1836
0.5 g Ch* 0.1836= 0.0918 g NaCl
V-Value For Chlorobutanol
100 mL H2O
x mL
0.9 g NaCl
0.0918 g NaCl
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.
P27
Oxytetracycline HCl
0.5% [497,2]
[iso]Tetracaine HCl Sol. 2%15 ml
NaClqs
Purified Water ad
30 mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many mL of NSS needed?
“E” for Oxytetracycline HCl
-1
58.5gL *
1.8
= 01177g NaCl.
-1
497gL *1.8
0.15 g Oxy* 0.1177= 0.0176
0.135 g NaCl Reference in only 15 mL
0.0176 g NaCl Equivalent
0.1174 g. NaCl
0.9 g NaCl
100 mLx mL
0.1174 g. NaCl
13.0 mL of NSS
Determine if the following commercial products
are Hypotonic, isotonic, or Hypertonic:
An ophthalmic sol. 40
-1
mgmL of Cromolyn Sodium
[512,2]& 0.01% of
Benzalkomiun chloride [360,2]
a-
in purified water.
“E” for CromolynSodium
-1
o
512gL = -1.86 C*1.8
-1
o
40gL
= -X C
ISO-T. HYPO
>-0.52 -0.52 <-0.52
HYPER
-0.26oC
A parenteral infusion containing 20% (w/v)
of mannitol.
FP ”D” Mannitol
-1
182gL
-1
200gL
HYPER
o
-1.86 C*1.0
=
o
= -X C
-2.04395
ISO-T. HYPO
>-0.52 -0.52 <-0.52
-2.04395
A 500-mL large volume parenteral containing D5W
(5% w/v of anhydrous dextrose in sterile water for
injection).
-1
180gL
-1
50gL
o
-1.86 C*1.0
=
o
= -X C
HYPER
-2.04395
ISO-T. HYPO
>-0.52 -0.52 <-0.52
-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)
-1
138gL
-1
161gL
HYPER
o
-1.86 C*1.8
=
o
= -X C
ISO-T. HYPO
>-0.52 -0.52 <-0.52
-3.9
Dibasic sodium phosphate (heptahydrate)
(268,3)
-1
268gL
o
-1.86 C*2.6
=
-1
o
59.32gL = -X C
HYPER
ISO-T. HYPO
>-0.52 -0.52 <-0.52
-1.07
For agents having the following sodium chloride equivalents,
calculate the percentage concentration of an isotonic solution:
(A) 0.20 0.90%
= 4.5%
0.20
(b) 0.32
0.90% = 2.81%
0.32
(c) 0.61
0.90% = 1.48%
0.61
P30
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)?
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
0.0553 g NaCl* 100 mL / 0.9 g NaCl = 6.16 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:
(c) ephedrine sulfate [429, 3]
58.5*2.6/(429*1.8) = 0.1969 g * 0.3 = 0.059 g NaCl
0.059 g NaCl* 100 mL / 0.9 g NaCl = 6.56 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:
(d) silver nitrate [170, 2]
58.5*1.8/(170*1.8) = 0.344 g * 0.3 = 0.103 g NaCl
0.103 g NaCl* 100 mL / 0.9 g NaCl = 11.44 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:
(e) zinc sulfate [288, 2]
58.5*1.4/(288*1.8) = 0.158 g * 0.3 = 0.0473 g NaCl
0.0473 g NaCl* 100 mL / 0.9 g NaCl = 5.266 mL H2O