ELECTROLYTE SOLUTIONS
INTRODUCTION
Electrolytes in human body fluids are prominent in
maintaining acid-base balance in body.
Electrolytes help regulate metabolism in the body and
control volume of water in the body.
Compound
Dissociate
Donot
Dissociate
Non
ElectrolytesUrea, Dextrose
Electrolytes NaCl Na+ Cl-
Cation Anion
Molecules
BODY FLUIDS - 60% BODY WEIGHT
WATER IS LARGEST
SINGLE COMPONENT
Dec. to 45-50 % body
weight in elderly
Variations occur based
on age, gender & amt.
of body fat
80% neonate is water*
MAJOR COMPARTMENTS FOR FLUIDS
INTRACELLULAR
FLUID (ICF)
Inside cell
Most of body fluid
here - 40% weight
Decreased in elderly
EXTRACELLULAR
FLUID (ECF)
Outside cell
Intravascular fluid -
within blood vessels (5%)
Interstitial fluid -
between cells & blood
vessels (15%)
Transcellular fluid -
cerebrospinal, pericardial ,
synovial
ELECTROLYTES
Substance when dissolved in solution separates
into ions & is able to carry an electrical current
CATION - positively charged electrolyte
ANION - negatively charged electrolyte
# Cations must = # Anions for homeostatsis to
exist in each fluid compartment
Commonly measured in milliequivalents / liter
(mEq/L)
PHARMACEUTICAL APPLICATION
Various electrolyte ions in the human blood are
Electrolyte Preparations employed to treat fluid and
electrolyte imbalance in the body.
Available as oral solutions, syrups, dry granules to be
dissolved in water/juice, capsules, tablets and also
intravenous infusions.
Cations Na+, K+, Ca++, Mg++
Anions Cl-, HCO3-, HPO4
- -, SO4- -
MILLIEQUIVALENTS - SIGNIFICANCE
Most exclusively employed unit by pharmacists, physicians,
manufacturers and clinicians across USA to express
electrolyte concentration in solution is “Milliequivalents
(mEq)”
Internationally (European and many other nations), molar
concentration (mmol/L or µmol/L) are employed.
A mEq represents amount in milligrams taking into account
valence of the ions.
A mEq expresses the chemical activity / combining power
of a substance relative to the activity of 1 mg of hydrogen.
Equivalent
Weight =
Atomic / Formula /
Molecular weight--------------------------------
Valence
Converting milligrams to milliequivalents
The following formula can be employed:
Atomic/formula/molecular weight
Valence
Problem:
Given the molecular weight of calcium is 40, Represent 10
mg of calcium as mEq? Also, how many mg will be there in
1 mEq of calcium?
= Equivalent Weight
mEq =
mg ×Valence
Atomic/formula/molecular weight
Solution
10 mg of calcium = 0.5 mEq
1 mEq will contain --- 10 / 0.5 = 20 mg of calcium
mEq =
mg ×Valence
Atomic/formula/molecular weight
mEq =
10 ×2
40
•To convert milliequivalents to milligrams
Convert 8 mEq of potassium to mg?
Majority of electrolyte substitutes are available in liquid
form. The concentration of electrolytes in i.v. infusion
fluids is represented as mEq/L
mg =mEq × Atomic/formula/molecular weight
Valence
mg =8 × 39
1
= 312
To convert mEq/mL to mg/mL
Calculate the concentration in mg/mL, of a solution
containing 3 mEq of Kcl per mL?
mg/mL =mEq/mL × Atomic/formula/molecular weight
Valence
mg/mL =3 × 74.5
1
= mg/mL223.5
What is the concentration in g/mL of a solution containing 4
mEq of Calcium chloride / mL?
Calcium chloride = CaCl2.2H2O
Molecular weight = 147
Eq. Wt = 147 / 2 = 73.5
mg/mL = 4 × 73.5 = 294 mg/mL = 0.294 g/mL
Estimation of % w/v from mEq
What is the % w/v conc. of a 100 mEq/L solution of
NH4Cl?
M.W of NH4Cl = 53.5
Eq. Wt of NH4Cl = 53.5
1 mEq of NH4Cl = 53.5g/1000 = 0.0535 mg
100 mEq of NH4Cl = 5.35 g/L or 0.535 g/100 mL
= 0.535%
Problem
A solution has 10 mg / 100 mL of K+ ions. Express the
concentration in mEq/Lit?
mg/mL =mEq/mL × Atomic/formula/molecular weight
Valence
100 =mEq/mL × 39
1
mEq/mL = 2.56 mEq/L
A solution contains 10 mg/100 mL of Ca++ ions. Express
the concentration in mEq/Lit
10 mg/100 mL == 100 mg / Lit
A mg++ ion level in plasma is found to be 2.5 mEq/L.
Express the conc. in mg?
Model Problem – Solve – Imp*
How many mEq of Kcl is present in a 15 mL dose of a
10% w/v Kcl elixir?
Solution
M.W of Kcl = Eq.Wt of Kcl = 74.5
1 mEq of Kcl = 74.5/1000 = 0.0745 g = 74.5 mg
15 ml dose of 10% w/v elixir formulation = 1.5 g of Kcl
74.5 mg ---- 1 mEq
1500 mg ---- ?
Ans: 20.1 mEq
Model Problem – Solve – Imp*
Calculate the mEq of Na+ present in 30 ml of following
solution?
Disodium Hydrogen Phosphate 18g
(Na2HPO4.7H2O - 268)
Sodium Biphosphate 48 g
(NaH2PO4.H2O - 138)
Purified Water 100 mL
Solution
Eq.Wt - (Na2HPO4.7H2O - 268) – 268/2 – 134
18 g / x g = 100 mL / 30 mL
x = 5.4 g / 30 mL
1 mEq – 134 mg
mEq of Na2HPO4.7H2O – 40.3
Eq.Wt - (NaH2PO4.H2O - 138) – 138
mEq of NaH2PO4.H2O – 104.3
Total mEq of Na+ -- 144.6
Clinical Application – Model Problem
A person will receive 2 mEq of NaCl / kg of body weight.
He weighs 132 lb. Calculate the ml of 0.9% sterile NaCl
that should be administered?
Solution
M.Wt / Eq. Wt of NaCl = 58.5
1 mEq of NaCl = 0.0585 g
2 mEq - 0.117 g
Wt in Kg – 132 / 2.2 – 60 kg
Drug is administered at 2 mEq (0.117 g) / kg
Total amt of drug in body – 0.117 * 60 - 7.02 g of NaCl
0.9% sterile NaCl solution has – 9 g of NaCl / Lit
9 g --------- 1000 ml
7.02g ---- ? = 780 mL
Millimoles and Micromoles
A mole is the molecular weight of substance in grams.
A millimole – one thousandth of a mole
A micromole – One millionth of a mole
SI expresses, electrolyte conc. in mmol/L
How many millimoles of monobasic sodium phosphate
is present in 100 gms of substance?
Solution
Mol.wt - (NaH2PO4.H2O - 138)
1 mole – 138 g
100 g represent ---- 100/138 = 0.725 moles
= 725 mmoles
Model Problems
Calculate the weight in mg of 1 mmol of HPO4-
1 mole weighs --- Atomic weight -- 95.98 g
1 mmol of HPO4- weighs - 0.09598 g = 95.98 mg
Convert plasma levels of 0.5 µg/mL of tobramycin (mw –
467.52) to µmol/L?
1 mol - 467.52 gms
1 µmol - 467.52 µg
0.5 µg/1 ml * 1 µmol/467.52 µg * 1000 ml/1 L
= 1.07 µmol/L
OSMOSIS
Movement of the solvent or water across a
membrane
Involves solution or water
Equalizes the concentration of ions on each side
of membrane
Movement of solvent molecules across a
membrane to an area where there is a higher
concentration of solute that cannot pass through
the membrane
Osmolarity
Osmotic pressure is an important biologic parameter
which involves diffusion of solutes or the transfer of fluids
through semi permeable membranes.
Per US Pharmacopeia, knowledge of osmolar
concentrations of parenteral fluids is important.
Labels of pharmacopeia solutions providing intravenous
replenishment of fluids, nutrients, electrolytes and
osmotic diuretic mannitol are required to state osmolar
concentration.
Information provides the doctor to decide, if the solution
is, hypoosmotic, iso-osmotic or hyperosmotic with regard
to biological fluids and membranes
Osmotic pressure is α (total number of particles in
solution)
Unit of measurement is milliosmoles
For non-electrolytes like dextrose, 1 mmol represents 1
mosmol
However, with electrolytes, as total number of particles in
solution depends on degree of dissociation of a
substance.
For e.g. Assuming complete dissociation, 1 mmol NaCl
represents 2 mOsmol (Na+ + Cl-) of total particles
1 mmol of CaCl2 represents 3 mOsmol (Ca++ + 2Cl-)
1 mmol of sodium citrate (Na3C6H5O7) represents 4
mOsmol (3 Na+ + C6H5O7-) of total particles
The milliosmolar value of the complete solution is equal
to the sum of milliosmolar values of individual ions.
U.S. Pharmacopeia lists the following formula for
calculation of ideal osmolar concentration:
Wt. of substance (g/L)
mOsmol/L = ------------------------------ × No. of Species × 1000
Mol. Wt (g)
E.g. Calculate of ideal osmolarity for 0.9% Sodium
Chloride Solution?
Solution
Wt. of substance (g/L)
mOsmol/L = ------------------------------ × No. of Species × 1000
Mol. Wt (g)
9 (g/L)
mOsmol/L = ------------------------------ × 2 × 1000
58.5 (g)
= 308 mOsmol/L
However, because of the bonding forces, n is slightly less
than 2 for NaCl at 0.9% Nacl concentration. Hence, the
actual osmolarity of solution is 286 mOsmol/L
Some Pharmaceutical manufacturers label electrolyte
solutions with ideal or stoichiometric osmolarities
calculated by the equation just provided, whereas others
list experimental or actual osmolarities.
“Pharmacist should appropriate this distiction”.
Osmolarity -- “Milliosmoles of solute per liter of
solution”
Osmolality – “Milliosmoles of solute per kilogram of
solvent”
For dilute aqueous solutions – both terms are nearly
identical
For more concentrated solutions – “Two values are not
identical”
“pharmacist should make distinction between
“Osmolarity” and Osmolality”
Contribution of components of normal human serum to
the “Serum Osmotic Pressure”
Constituent Mean Conc.
(mEq/L)
Osmotic Pressure
(mOsmol/kg of
water)
% of total osmotic
pressure
Sodium
Potassium
Calcium
Magnesium
Chloride
Bicarbonate
Proteinate
Phosphate
Sulfate
Organic anions
Urea
Glucose
Total
142
5
2.5
2
102
27
16
2
1
3.5
30mg/100 mL
70 mg/100 mL
139
4.9
1.2
1
99.8
26.4
1.0
1.1
0.5
3.4
5.3
4.1
287.7 mOsmol/Kg
48.3%
1.7
0.4
0.3
34.7
9.2
0.3
0.4
0.2
1.2
1.8
1.4
99.9%
Normal serum osmolality - ranges from “275-300
mOsmol/kg”.
Equipment used in laboratories to measure osmolality –
“Osmometers”.
Electrolyte imbalance
Shock
Trauma
Burns
Hyperglycemia
Water intoxication
Renal failure
Abnormal Blood Osmolality
Problems
A solution has 5% anhydrous dextrose in water for injection.
Represent the concentration in mosmol/lit?
Solution: Molecular weight of dextrose – 180
As it donot dissociate 1 mmol --- 1 mOsmol
5% solution contains ---- 50 g / L
From the formula discussed above – 50,000 / 180
= 278 mOsmol/L
A solution contains 10 mg% of Ca+2 ions. How many
milliosmoles are present in 1 lit of solution?
Solution
1 mmol of Ca++ (40 mg) --- 1mOsmol
10 mg% of Ca++ = 100 mg / Lit
100 / 40 - 2.5 mOsmol/Lit
WATER and Electrolyte Balance – Clinical Consideration
Good Homeostasis -- Maintaining body water and
electrolyte balance is component of good health
Dietary Input
OutPut – Kidneys, Lungs,
GI tract, Skin
Endocrine Process
MaintainOptimum Osmolality
of Body Fluids
Fluid / Electrolyte therapy is provided to rectify any
imbalance of osmolality of body fluids
Treatment can be “Customized” based on patient needs
Patient on Diuretic Therapy – “A K+ supplement along
with adequate water intake”.
Total body water in adults – 55-60% of the body weight
Increase in body fat --- Lesser proportion of water
Infants have 75% of body water
For adults - A minimum of 2.5 L of daily water intake
(Ingested liquids, food and oxidative metabolism) are
required for maintaining “Homeostasis”.
General terms…..
1500 mL of water / Sq.meter of body surface might be
employed for calculating daily requirements of adults.
On weight basis….
32 mL/kg – Adults and 100-150 mL/kg – Infants ---
Daily requirements for healthy intake
Careful estimation should be made in human subjects
with any disorders affecting homeostasis.
Also, composition is described in literature with respect to
body compartments --- “Intracellular (within cells)”,
Intravascular (In blood plasma), Interstitial (Between cells
in tissue).”
Intervascular and Interstitial --- Together called --
Extracellular
Sodium and Chloride have principle affect on Extracellular
Fluid
Potassium and Phosphate --- on --- Intracellular fluids
Cell membranes are freely permeable to water
Hence, Osmalility of extracellular fluid (290 mOsmol/kg) is
approximately equal to intracellular fluid.
Therefore, osmolality of the plasma is accurate guide to
estimate osmolality of intracellular fluid.
The following formula may be employed to estimate
plasma osmolality
Plasma Osmolality (mOsmol/Kg)
= 2([Na] + [K]) plasma + BUN/2.8 + Glucose / 18
Where Na and K are in mEq/L, BUN and Glucose are in
mg/dL
Various Clinical calculations – Determination of body water
requirement, Plasma osmolality estimation, Osmolality and
Milliequivalent contents of physiologic electrolyte
solutions.
Probelms
Based on the discussion above, Calculate the daily water
requirement of a healthy human adult with a body surface
area of 1.8 m2?
Solution:
Water Requirement -- 1500 mL/m2
Hence 1.8 m2 contains --- 2700 mL
Estimate the plasma osmolality of the given data:
Na2+ -- 135 mEq/L, K+ -- 4.5 mEq/L, BUN – 14 mg/dL,
Glucose – 90 mg/dL
Solution:
Plasma Osmolality (mOsmol/Kg)
= 2[(Na) + (K)] Plasma + BUN/2.8 + Glucose / 18
Where Na and K are in mEq/L, BUN and Glucose are
in mg/dL
= 2(135+4.5) + 14/2.8 + 90/18
= 289
Calculate the mEq of Na+, K+ and Cl-, millimoles of
dextrose and osmolality of following parentaral fluid?
Dextrose anhydrous 50g
NaCl 4.5g
KCl 1.49g
Water for Injection, ad 1000mL
Solution
NaCl: 1 EqWt of NaCl - 58.5 g
1 mEq of NaCl - 58.5 mg
4500 mg of NaCl – 76.9 mEq of Na+ and 76.9
mEq of Cl-
Similarly for KCl we can calculate, 20 mEq of K+ and Cl-
For dextrose - Mol. Wt – 180
1 mmol of anhydrous dextrose – 180 mg
50 gms of anhydrous dextrose – 50,000 mg
50,000 mg represent -- 277.7 mmol
Hence Osmolarity is calculated as
Dextrose anhydrous: 278 mmol * 1 particle/mmol = 278
mOsmol
NaCl: 77 mEq * 2 particles / mEq (or mmol) = 154
mOsmol
KCl : 20 mEq * 2 particles / mEq (or mmol) = 40
mOsmol
Total = 278 + 154 + 40 = 472 mOsmol
Reference
Lecture material taken from the book “Pharmaceutical
Calculations”, 13th Edition by Howard C. Ansel.
Lecture material from Dr. Karla’s lecture notes.