Transcript COLLIGATIVE PROPERTIES

ISOTONIC SOLUTIONS
Isotonic Solutions
Isotonic - having the same osmotic pressure
as body fluids
Hypotonic - osmotic pressure is lower than
in the body fluids
Hypertonic - osmotic pressure is higher than
in the body fluids
Preparation of Isotonic Solutions
Freezing point of blood is ca. - 0.52oC
One gram molecular weight (M) of
nonelectrolyte decreases freezing point by
ca. 1.86oC
What amount of boric acid is needed to
prepare isotonic solution?
1.86( o C) 61.8( g ) / 1000( ml )

o
x( g ) / 1000(ml )
0.52( C)
x  17.3( g )
Electrolytes and Tonicity
As a result of ionization 1 mole of electrolyte produce i moles
of ions (i > 1).
NaCl i = 1.8 (i  0.2 + number of ions x 0.8)
What amount of sodium chloride is needed to prepare isotonic
solution?
1.86( o C)  i ( 1.8) 58.5(g/1000 ml)

o
x(g/ 1000ml)
0.52( C)
x  9.09( g / 1000ml )
Sodium Chloride Equivalents
How much NaCl is presented by 1g of any solute?
58.5( g )  isolute
xj 
M solute  iNaCl
xj sodium chloride equivalents
Solution Compounding
How many grams of NaCl should be used in
compounding the following prescription?
Pilocarpine Nitrate 0.3 g
NaCl
q.s.
Purified water ad 30.0 mL
0.23(NaCl eqivalent)  0.3(g )  0.069 (g NaCl in pilocarpin e nitrate)
30  0.009(g/mL )  0.270 (g NaCl in 30 ml of isotonic solution)
0.270 - 0.069  0.201 (g NaCl)
More Solution Compounding
How many grams of KNO3 is needed to make the
following prescription isotonic?
Solution of silver nitrate 60.0 ml
1:500 w/v (concentration)
Make. isoton. Sol.
(Contains 0.120 g AgNO3)
0.33  0.120( g )  0.040 (g NaCl in AgNO 3 )
60  0.009  0.540 (g NaCl in 60 ml of isotonic solution)
0.540 - 0.040  0.500 (g NaCl)
0.500 (g NaCl)/0.58 (NaCl equivalent of KNO 3 )
 0.862 (g of KNO 3 )
Freezing Point Data and Tonicity
How many milligrams of NaCl and dibucaine
hydrochloride is needed to prepare 30 mL of a 1 %
solution of dibucaine chloride isotonic with tears?
1% (dibucaine hydrochloride) : ΔTf  -0.08o C
Lower freezingpoint by addit ional0.52o C - 0.08o C  0.44o C
1% (NaCl) 0.58o C

; x  0.85%
o
x % (NaCl) 0.44 C
T o make30 mL of thissolution:
30 1%  0.3 g  300 mg dibucaine hydrochloride
and 30  0.76% 0.228g  228mg NaCl
Milliequivalents
mg  valence
mEq 
atomic or molecular weight
How many mEg in 1 g of MgSO4?
M.w. MgSO 4  120
Equivalent weight  60
60 (mg)
1 (mEq)

1000 (mg) x (mEq)
x  16.7 mEq
Osmolarity
wt of substance (g/L)
mOsmol/L 
 (number of species) 1000
atomic or molecular weight
What is osmolarity of isotonic NaCl solution?
9 (g/L)
mOsmol/L 
 2 1000  308 mOsmol/L
58.5 g