1 / 11

Electrolyte solutions: Milliequivalents, millimoles and milliosmoles

Electrolyte solutions: Milliequivalents, millimoles and milliosmoles. Dr. Asmaa Abdelaziz Mohamed. Milliequivalents. Equivalent weight (molecular weight divided by valence) Milliequivalent is the amount, in milligram, of a solute. It is equal to 1/1000 of its gram equivalent weight.

emerritt
Télécharger la présentation

Electrolyte solutions: Milliequivalents, millimoles and milliosmoles

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electrolyte solutions: Milliequivalents, millimoles and milliosmoles Dr. Asmaa Abdelaziz Mohamed

  2. Milliequivalents • Equivalent weight (molecular weight divided by valence) • Milliequivalent is the amount, in milligram, of a solute. It is equal to 1/1000 of its gram equivalent weight

  3. Milliequivalents • 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 (molecular weight divided by valence) of KCl = 74.5 • 2 mEq of KCl = 74.5 X 2 = 149 mg/ml • OR

  4. Milliequivalents • 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 = M.Wt / valence=147/2 = 73.5 • 4 mEq of CaCl2.2H2O = 0.0735 g X 4 = 0.294 g/ml • OR

  5. Millimoles • 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 g/mole) 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

  6. Osmolarity • U.S. Pharmacopeia states the knowledge of osmolar concentration of parenteral fluids is essential. • 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.

  7. 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 • difference between 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 lab. to measure osmolality • Abnormal blood osmolality can occur with shock, trauma, burns, electrolyte imbalance, hyperglycemia or renal failure Dr. Asmaa

  8. 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

  9. Clinical considerations of water and electrolyte balance • 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. • On weight basis, 32 ml/kg for adults and 100-150 ml/kg for infants

  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

  11. Thank you

More Related