COLLEGE OF
PHARMACY
STERILE PRODUCTS
PHT 434
Dr. Mohammad Javed Ansari
Ph.D
Contact info: [email protected]
Normality & Equivalent weight
•
•
•
•
•
Normality ( N ) = grams of equivalent weight of solute
per liter of solution.
Equivalent weight of any substance is its atomic
weight in grams ( or its fractions or its multiple)
that combine / react with or replace
one mole of hydrogen ions (i.e. 1 gm H+) or
one mole of hydroxyl ion (i.e. 17 gm OH-) or
one mole of oxygen (i.e. 16 gm O- -)
•
•
•
One atomic weight of Na or K may combine with one
mole of hydrogen OH-, to form NaOH or KOH
Therefore Eq Wt. of Na or K will equal to their atomic
weights (23. and 39 respectively).
2
Normality & Equivalent weight
•
Similarly one atomic weight of the nonmetallic elements Cl or Br may unite with one atomic weight of H+
to form HCl OR HBr therefore Eq. Wt of Cl or Br will
be equal to their atomic weights (35.5, 79.9).
•
Ca + + combines with 2 moles of OH-, to form CaOH2
therefore Eq. Wt of Ca + + will be half of its atomic
weight 40 (i.e. 40/2 = 20).
•
Al + + + combines with 3 moles of OH-, to form Al OH3
therefore Eq. Wt of Al + + + will 1/3rd of its atomic weight
27 (i.e. 27/3 = 9).
•
In other words Eq. Wt of elements can be calculated
by dividing their atomic weights by their valencies
3
(i.e. number of positive or negative charges on them).
Equivalent weight of some Cations & Anions
POSITIVE
IONS
Sodium
Potassium
Ammonium
Magnesium
Manganese
Ferrous Ion
Calcium
Copper
Aluminum
Ferric Ion
Ion
formula
Na+
K+
NH4+
Mg2+
Mn2+
Fe2+
Ca2+
Cu2+
Al+3
Fe3+
Vale
nce Atomic
weight
1
23
1
39
1
18
2
24.4
2
55
2
55.84
2
40
2
63.4
3
27
3
55.84
Equivalent
Weight
23 / 1 = 23.0
39/ 1 = 39
18/ 1 = 18.0
24.4/ 1 = 12.2
55/ 1 = 27.5
55.84/ 1 = 27.9
40/ 2 = 20.0
63.4/ 2 = 31.8
27/ 3 = 9.0
55.84 / 3 = 18.6
NEGATIVE
IONS
Chloride
Nitrate
Bicarbonate
Bisulfate
Bisulfite
Sulfate
Sulfite
Sulfide
Phosphate
(monobasic)
Phosphate
(dibasic)
Phosphate
(tribasic)
Ion
formula
ClNO3HCO3HSO4HSO3SO4- SO3- S- H2PO4-
Vale
nce
1
1
1
1
1
2
2
2
1
Atomic
weight
35.5
62
61
97.1
81.1
96
80
32
97
Equivalent Weight
HPO4- -
2
96
96/ 2 = 48.5
PO4- - -
3
95
95/ 3 = 31.7
35.5/ 1 = 35.5
62 / 1 = 62
61/ 1 = 61
97.1/ 1 = 97.1
81.1/ 1 = 81.1
96/ 2 = 48
80 / 2 = 40
32/ 2 = 16
97 / 1 = 97
Equivalent weight
•
•
•
•
•
•
•
Equivalent weight is additive quantity. i.e. Eq. Wt.
of molecules / salts / solutes will be sum of Eq. Wts of
all atoms or components
For example equivalent weight of Na is 23 gram and
the equivalent weight of Cl is 35.5 gram. Then Eq Wt.
of NaCl will be 58.5 (23+35.5 ).
It is important to note that for molecules or solutes with
a valence of one (i.e. NaCl, KCl, HCl, NaOH) the
molecular weight and equivalent weight are the same.
When the valence of the solute is more than one (i.e.
CaOH2 CaCl2 MgO, H2SO4 valence = 2), then Eq Wt
will be half of the molecular weight.
Eq Wt of CaOH2 = ½ of MW of CaOH2 = 74/2 =37
Eq Wt of CaCl2 = ½ of MW of CaCl2 = 111/2=55.5
Eq Wt of H2SO4 = ½ of MW of H2SO4 = 98/2 =49 6
•
•
•
Equivalent weight
When the valence of the solute is 3 (i.e. AlOH3, H3PO4)
then Eq Wt will be equal 1/3rd of MW.
Eq Wt of AlOH3 = 1/3rd of MW of AlOH3 = 78/3 =26.
Eq Wt of H3PO4 = 1/3rd of MW of H3PO4 = 97.9/3 =
32.7
COMPOUNDS/
MOLECULES
Sodium Hydroxide
Sodium Chloride
Calcium Hydroxide
Calcium Chloride
Calcium Carbonate
Trisodium
Phosphate
(anhydrous)
Vale
Molecular nce
formula
1
NAOH
1
NaCl
2
Ca(OH)2
2
CaCl2
2
CaCO3
3
Na3PO4
Molecular
weight
40
58.5
74
111
100
164
Equivalent Weight
40/ 1 = 40
58.5 / 1 = 58.5
74/ 2 = 37
111/ 2 = 55.5
100 / 2 = 50
164 /3 = 54.7
Normality & Equivalent weight
• Equivalent weight = atomic or molecular weight / n.
• n = number of replaceable H or OH for acid or base.
• OR
• n = number of electrons lost or gained in oxidation,
reduction reaction.
• Normality ( N ) = # of equivalent / Liter of solution
• N = g of solute /equivalent / L
• N = g of solute / equivalent x L
• N = g of solute /(molecular weight / n) L
• N = g of solute x n / molecular weight L
• N = molarity x n
• N MW L = g of solute x n
8
Example 1
• How much sodium bicarbonate powder is needed to
prepare 50.0 mL of a 0.07 N solution of sodium
bicarbonate (NaHCO3)? (MW of NaHCO3 is 84)
• NaHCO3 may act as an acid by giving up one proton (H+)
to become Na2CO3, or
• It may act as a base by accepting one proton (H+) ,
forming NaOH & H2CO3
• Therefore reaction capacity (or valency ) for NaHCO3 will
be 1. ( when valency is 1 EQ WT is equal to MW)
• Normality ( N ) = grams of Eq. wt / Liter of solution
• N x L = grams of Eq. wt
• 0.07 x 0.05 = 0.0035 grams of Eq. wt
9
• = 0.0035 x 84 =0.294 g
Concentration expression of electrolyte solutions:
Old expression
gm %, volume % , mg % etc
( Measuring electrolytes in
units of weight or physical
unit)
Currently used expression
Milliequivalent– one
thousandth of an equivalent
weight (measures total
number of reactive / ionic
charges in solutions)
It dose not give any direct
Meaningful unit because the
information as to the number valence of the ions is taken
of ions or the charge that they into consideration.
carry. Since the chemical
In other words, it is a unit of
combining power depends
measurement of the amount
not only on the number of
of chemical activity of an
particles in solutions but also electrolytes.
Concentration expression of electrolyte solutions:
• In preparing a solution of K+ ions, a potassium salt is
dissolved in water.
• In addition to the K+ ions, the solution will also contain ions
of opposite negative charge such as Cl• These two components will be chemically equal in that the
milliequivalent of one are equal to the milliequivalent of
the other.
• For example, if we dissolve enough potassium chloride in
water to give us 40 milliequivalents of K+ per litter, we also
have exactly 40 milliequivalents of Cl-, but the solution will
not contain the same weight of each ion.
• A milliequivalent (mEq) : represent the amount, in mg, of
a solute equal to 1/1000 of its gram equivalent weight.
Example 1:
• How many milliequivalents of magnesium sulfate
are represented in 1.0 g of anhydrous
magnesium sulfate solution?
• Mwt of MgSO4 = 120
• Equivalent weight of MgSO4 = 120/2 = 60
• 1 mEq of MgSO4 = 1/1000 X 60 = 0.06 g = 60 mg
• 60 mg
1 mEq
60/1 meq = 1000 / x meq
• 1 mg
1 / 60 mEq
X = 1000 / 60 meq
X = 16.7 meq
• 1000 mg
X
• X= 1 / 60 mEq * 1000 = 16.7 mEq
Example 2:
• A 500 mL large volume parenteral electrolyte
bottle contains 5.86 g of KCl. How many mEq
of KCl are present? (molecular weight of KCl
is 74.5 g).
• Equivalent weight of KCl= 74.5 g/1 = 75.4 g
• Means 75.4 g per equivalent
• 75.4 g / 1 Eq = 5.86 g / x Eq
• x = 0.078 Eq
• x = 0.078 x 1000 milli Eq = 78 milli Eq
• 5.86 g of KCl = 78 milli Eq of KCl
Example 3:
• A large volume parenteral electrolyte solution
contains 10 mg% of Ca++ ions. Express this
concentration in terms of mEq per liter ?
• Atomic weight of Ca++ = 40
• Equivalent weight of Ca++ = 40/2= 20
• 1 mEq of Ca++ = 1/1000 X 20 = 0.02 g = 20 mg.
• 10 mg % = 10 mg in 100 ml or
20/1 meq = 100 / x meq
• 10 mg % = 100 mg / L
X = 100 / 20 meq
X = 5 meq
• 20 mg = 1 mEq,
• So, 100 mg / L = 5 mEq / L
Example 4:
• What is the concentration, in mg per ml, of an
electrolyte solution containing 2 mEq of potassium
chloride per ml ?
• Mwt of KCl = 74.5 g
• Equivalent weight of KCl =74.5 /1 = 74.5 g
• 1 mEq of KCl = 74.5 g / 1000 = 0.0745 g = 74.5 mg
• 2 mEq of KCl = 74.5 mg X 2 = 149 mg.
• Therefore 2 mEq /ml of KCl means 149 mg/ml KCl.
Example 5:
• A person is to receive 2 mEq of sodium chloride/kg of
body weight. If the person weight is 60 kg, how many
milliliters of an 0.9 % sterile solution of NaCl should be
administered?
• Mwt of NaCl = 58.5,
• Equivalent Weight of NaCl= 58.5
• 2 mEq of NaCl= (58.5/1000 ) X 2 = 0.117 g
• NaCl needed for this person= 0.117 X 60 = 7.02 g
• 0.9 gm of NaCl is present in 100 ml (0.9% sol. )
• 1 gm NaCl will be present in 100 /0.9 ml NaCl solution
• 7.02 g of NaCl will be present in ml X ml NaCl solution
• X= 7.02 * (100 /0.9)
• X= 780 ml
THANK YOU FOR
ATTENTION
GOOD LUCK ..