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Electrolyte solutions: Milliequivalents, millimoles and milliosmoles

Electrolyte solutions: Milliequivalents, millimoles and milliosmoles. Dr. Osama A. A. Ahmed. Milliequivalents, millimoles and milliosmoles.

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Electrolyte solutions: Milliequivalents, millimoles and milliosmoles

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  1. Electrolyte solutions: Milliequivalents, millimoles and milliosmoles Dr. Osama A. A. Ahmed

  2. Milliequivalents, millimoles and milliosmoles • Milliequivalent (mEq) is used in the United states, while in the international system Molar concentrations (mmol/L or mmol/L) are used throughout the world. • Under normal conditions blood plasma contains 154 mEq of cations and an equal number of anions • Milliequivalent is the amount, in milligram, of a solute equal to 1/1000 of its gram equivalent weight Dr. Osama A. A. Ahmed

  3. Milliequivalents, millimoles and milliosmoles • What is the concentration, in mg/ml, of a solution containing 2 mEq of potassium chloride (KCl) per milliliter? • M.Wt. of KCl = 74.5 • Equivalent weight of KCl = 74.5 • 2 mEq of KCl = 74.5 X 2 = 149 mg/ml • OR Dr. Osama A. A. Ahmed

  4. Milliequivalents, millimoles and milliosmoles • What is the concentration, in g/ml, of a solution containing 4 mEq of calcium chloride (CaCl2.2H2O) per milliliter? • M.Wt. of CaCl2.2H2O = 147 • Equivalent weight of CaCl2.2H2O = 147/2 = 73.5 • 4 mEq of CaCl2.2H2O = 0.0735 g X 4 = 0.294 g/ml • OR Dr. Osama A. A. Ahmed

  5. Milliequivalents, millimoles and milliosmoles • Millimoles (mmol) used by SI to express electrolyte concentrations in millimoles per liter (mmol/L). • For monovalent species, the numeric values of the milliequivalent and millimole are identical • Millimole is the amount, in milligram, of a solute equal to 1/1000 of its gram molecular weight. A mole is the molecular weight of a substance in grams. • Example • How many millimoles of monobasic sodium phosphate (m.w. 138) are present in 100 g of the substance? • 1 mole = 138 g • 1 mole 138 g • X mole 100 g X = 0.725 mole = 725 mmol • How many milligrams would 1 mmol of monobasic sodium phosphate weigh? • 1 mole = 138 g so 1 mmol = 138 mg Dr. Osama A. A. Ahmed

  6. Milliequivalents, millimoles and milliosmoles • Osmolarity • U.S. Pharmacopeia states the knowledge of osmolar concentration of parenteral fluids is essential. • This information indicates to the practitioner whether this solution is hypo-osmotic, iso-osmotic or hyper-osmotic with regard to biologic fluids and membranes. • The unit used to measure osmotic concentration is the milliosmole (mOsmol) • Osmotic pressure is proportional to the total number of particles in solution. • For nonelectrolytes (e.g. dextrose), a 1 mmol represents 1 mOsmol • For electrolytes, the total number of particles in solution depends on the degree of dissociation of the substance • 1 mmol of NaCl (Na+ + Cl-) represnts 2 mOsmol, 1 mmol CaCl2 represents 3 mOsmol, and 1 mmol of sodium citrate (Na3 C6H5O7) represents 4 mOsmol (3 Na+ C6H5O7) of total particles. Dr. Osama A. A. Ahmed

  7. Milliequivalents, millimoles and milliosmoles • Example • Calculate the ideal osmolarity of 0.9 % NaCl injection? • Because of bonding forces, however, n is slightly less than 2 for solutions of sodium chloride at this concentration, and the actual measured osmolarity is about 286 mOsml/L • A distinction should be madebetween the terms osmolarity and osmolality • Osmolarity is the milliosmoles of solute per liter of solution. • Osmolality is the milliosmoles of solute per kilogram of solvent. • Normal serum osmolality is considered within the range of 275 to 300 mOsmol/kg. • Osmometers are commercially available for use in the laboratory to measure osmolality • Abnormal blood osmolality can occur with shock, trauma, burns, electrolyte imbalance, hyperglycemia or renal failure Dr. Osama A. A. Ahmed

  8. Milliequivalents, millimoles and milliosmoles • Example: How many a) millimoles, b) milliequivalents, and c) milliosmols of calcium chloride (CaCl2. 2H2O – m.w. 147) are represented in 147 ml of a 10 % w/v calcium chloride solution? • 10 g 100 ml • X 147 X = 147 x 10/100= 14.7 g = 14700 mg • 1 mmol = molecular weight in grams / 1000 = 147/1000 = 0.147 g = 147 mg • 1 mmol 147 mg • X mmol 14700 mg X = 14700/147 = 100 mmol • OR mmol= 14700/147 = 100 mmol • = 14700 x 2/ 147 = 200 • = 14700 X 3/147 = 300 Dr. Osama A. A. Ahmed

  9. Clinical considerations of water and electrolyte balance • Water provides the environment in which cells live and is the primary medium for the ingestion of nutrients and the excretion of metabolic waste products. • Maintaining body water and electrolyte balance is an essential component of good health. • Fluid and electrolyte therapy (oral or parenteral) is provided either for maintenance requirements or to replace serious losses or deficits. • Example, a patient taking diuretics may simply require a daily oral potassium supplement along with adequate intake of water. Hospitalized patients commonly receive parenteral therapy of fluids and electrolytes to support ordinary metabolic functions. • In adult males, total body water ranges between 55% and 65% of body weight depending on the proportion of body fat. Adult women are about 10% less than adult men. New born infants have approximately 75% body water. • About 66.7 % (2/3) of adult’s body water is intracellular and about 33.3 % (1/3) is extracellular • 1500 ml of water per square meter may be used to calculate the daily requirements for adults. • On weight basis, 32 ml/kg for adults and 100-150 ml/kg for infants Dr. Osama A. A. Ahmed

  10. Clinical considerations of water and electrolyte balance • Body fluids are extracellular (intravascular or interstitial) and intracellular (within cells) • Osmolality of intracellular fluid and extracellular fluids is about equal (cell membrane) this value is about 290 mOsm/kg water. • The plasma osmolality is approximated by the formula • Where Na and K are in mEq/L, and blood urea nitrogen (BUN) and glucose concentrations are in mg/100 ml (mg/dl) • Example, Estimate the plasma osmolality from the following data: sodium 135 mEq/L; blood urea nitrogen, 14 mg/dL; and glucose, 90 mg/dL • = 2 (139.5) + 5 + 5 = 289 Dr. Osama A. A. Ahmed

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