COLLEGE OF STERILE PRODUCTS PHARMACY PHT 434 · Normality & Equivalent weight • Normality ( N ) = grams of equivalent weight of solute per liter of solution. • Equivalent weight

COLLEGE OF PHARMACY

Dr. Mohammad Javed Ansari Ph.D Contact info: [email protected]

STERILE PRODUCTS PHT 434

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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).

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• Similarly one atomic weight of the nonmetallic ele-

ments 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

(i.e. number of positive or negative charges on them).

Normality & Equivalent weight

POSITIVE IONS

Ion

formula

Vale nce

Atomic weight

Equivalent Weight

Sodium Na+ 1 23 23 / 1 = 23.0

Potassium K+ 1 39 39/ 1 = 39

Ammonium NH4+ 1 18 18/ 1 = 18.0

Magnesium Mg2+ 2 24.4 24.4/ 1 = 12.2

Manganese Mn2+ 2 55 55/ 1 = 27.5

Ferrous Ion Fe2+ 2 55.84 55.84/ 1 = 27.9

Calcium Ca2+ 2 40 40/ 2 = 20.0

Copper Cu2+ 2 63.4 63.4/ 2 = 31.8

Aluminum Al+3 3 27 27/ 3 = 9.0

Ferric Ion Fe3+ 3 55.84 55.84 / 3 = 18.6

Equivalent weight of some Cations & Anions

NEGATIVE IONS

Ion

formula

Vale nce

Atomic weight

Equivalent Weight

Chloride Cl- 1 35.5 35.5/ 1 = 35.5

Nitrate NO3- 1 62 62 / 1 = 62

Bicarbonate HCO3- 1 61 61/ 1 = 61

Bisulfate HSO4- 1 97.1 97.1/ 1 = 97.1

Bisulfite HSO3- 1 81.1 81.1/ 1 = 81.1

Sulfate SO4- - 2 96 96/ 2 = 48

Sulfite SO3- - 2 80 80 / 2 = 40

Sulfide S- - 2 32 32/ 2 = 16

Phosphate (monobasic)

H2PO4- 1 97 97 / 1 = 97

Phosphate (dibasic)

HPO4- - 2 96 96/ 2 = 48.5

Phosphate (tribasic)

PO4- - - 3 95 95/ 3 = 31.7

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• 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

Equivalent weight

COMPOUNDS/

MOLECULES

Molecular formula

Vale nce

Molecular

weight Equivalent Weight

Sodium Hydroxide NAOH 1 40 40/ 1 = 40

Sodium Chloride NaCl 1 58.5 58.5 / 1 = 58.5

Calcium Hydroxide Ca(OH)2 2 74 74/ 2 = 37

Calcium Chloride CaCl2 2 111 111/ 2 = 55.5

Calcium Carbonate CaCO3 2 100 100 / 2 = 50

Trisodium Phosphate

(anhydrous)

Na3PO4 3 164 164 /3 = 54.7

• 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

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

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Normality & Equivalent weight

• 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

• = 0.0035 x 84 =0.294 g

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Example 1

Concentration expression of electrolyte solutions:

Old expression Currently used expression

gm %, volume % , mg % etc ( Measuring electrolytes in units of weight or physical unit)

Milliequivalent– one thousandth of an equivalent weight (measures total number of reactive / ionic charges in solutions)

It dose not give any direct information as to the number of ions or the charge that they carry. Since the chemical combining power depends not only on the number of particles in solutions but also on the total number of charges.

Meaningful unit because the valence of the ions is taken into consideration. In other words, it is a unit of measurement of the amount of chemical activity of an electrolytes. Expressed as mg/mEq

• 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.

Concentration expression of electrolyte solutions:

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

• 1 mg 1 / 60 mEq

• 1000 mg X

• X= 1 / 60 mEq * 1000 = 16.7 mEq

60/1 meq = 1000 / x meq X = 1000 / 60 meq X = 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

• 10 mg % = 100 mg / L

• 20 mg = 1 mEq,

• So, 100 mg / L = 5 mEq / L

20/1 meq = 100 / x meq X = 100 / 20 meq X = 5 meq

• 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 4:

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

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