Danielle DelVillano, Pharm.D. Pharmaceutical Calculations: Electrolyte Solutions – Milliequivalents, Millimoles, Milliosmols Danielle DelVillano, Pharm.D.

Objectives Calculate the milliequivalent weight from an atomic or formula weight Convert between milligrams and milliequivalents Calculate problems involving milliequivalents, millimoles, and milliosmols

Milliequivalents A chemical unit: mEq Related to the total number of ionic charges in solution, and takes note of valence of ions 1 mEq = atomic weight (mg) valence # 1 Eq = atomic weight (g)

Equivalent Values

Molecular Weights To Know Na Cl K Ca NaCl KCl CaCl2 Dextrose 1, 16, 23, 35.5, 39, 40 58.5, 74.5, 110.98, 180

Equations mEq = mg substance * Valence Atomic Weight mg = mEq substance * Atomic weight Valence # mg/mL = mEq/mL substance * Atomic weight

Problem 1 What is the concentration, in mg/mL, of a solution containing 2 mEq of potassium chloride (KCl) per mL? mg/mL = 2 mEq/mL * 74.5 1 mg/mL = 149 mg/mL

Problem 2 What is the percent (w/v) concentration of a solution containing 100 mEq of ammonium chloride per liter? (MW = 53.5) 1 mEq = 53.5 mg = 53.5 mg 1 100 mEq = x mEq x = 10 mEq 1000 mL 100 mL 1 mEq = 53.5 mg x = 535 mg = 0.535 g per 100 mL 10 mEq x mg or 0.535%

Problem 3 A solution contains 10 mg/100 mL Ca++ ions. Express this concentration in terms of mEq/L 10 mg/ 100 mL * 10/10 = 100 mg/1000 mL mEq/L = 100 mg/L * 2 40 mg Answer = 5 mEq/L

Problem 4 How many mEq of KCl are represented in a 15 mL dose of 10% w/v KCl elixir? 10% = 10g/100 mL mEq = 1500 mg * 1 74.5 mg 10 g = x g Answer = 20.13 mEq 100 mL 15 mL x = 1.5 g = 1500 mg

Millimoles and Micromoles Mole - MW of a substance in grams Millimole - MW of a substance in mg Micromile - MW of a substance in mcg Measure representing the combining power of a species Monovalent species: mEq = mmol

Keep In Mind Eq wt of a substance = MW / valance Moles of a substance = MW

Problem 5 How many millimoles of monobasic sodium phosphate (MW 138) are present in 100 g of a substance? 1 mmol = 138 mg x mmol 100000 mg x = 724.6 mmol

Osmolarity Osmotic pressure is proportional to the total number of particles in a solution Measured in mOsmol Example 1 mmol dextrose = 1 mOsmol total particles 1 mmol NaCl = 2 mOsmol total particles 1 mmol CaCl2 = 3 mOsmol total particles 1 mmol Na3C6H5O7 = 4 mOsmol total particles Assume complete dissolution

Equation and Definitions mOsmol/L = g/L substance * # species * 1000 MW Osmolarity – mOsmol/L solution Osmolality – mOsmol/kg of solvent

Problem 6 What is the ideal osmolarity of a 0.9% sodium chloride injection? 0.9g/100 mL * 10/10 = 9g/1000mL mOsmol/L = 9g/1L * 2 species * 1000 58.5 g Answer = 308 mOsmol/L

Problem 7 A solution contains 10 mg% of Ca++ ions. How many milliosmols are represented in 1 liter of the solution? 10 mg/100 mL *10/10 = 100 mg/1000mL mOsmol/L = 0.1 g/L * 1 species * 1000 40 g Answer = 2.5 mOsmol

Clinical Considerations of Water and Electrolyte Balance Total body water for an adult male is 55-65% body weight Females about 10% lower Newborn infants 75% Daily requirement equations 1500 mL per square meter of BSA 32 mL/kg for adults 100-150 mL/kg for infants

Equation Plasma Osmolality (mOsmol/kg) = 2(Na + K) plasma + BUN + Glucose 2.8 18 Na and K measured in mEq/L BUN and glucose measured in mg/100mL (or mg/dL)

Problem 8 Calculate the estimated daily water requirement for a healthy adult with a body surface area of 1.8 m2. 1 m2 = 1500 mL 1.8 m2 x mL x = 2700 mL

Problem 9 Estimate the plasma osmolality from the following data: sodium 135 mEq/L, potassium 4.5 mEq/L, BUN 14 mg/dL, glucose 90 mg/dL mOslmol/kg = 2(135+4.5) + 14 + 90 2.8 18 Answer 298 mOsmol/kg

Questions

Reference Ansel, H. C. (2009) Phamaceutical Calculations (13th Ed.). Philadelphia:Lippincott Williams & Wilkins, and Wolters Kluwer Publishers

Chapter 12 Page 197 Calculate he mEq of sodium, potassium, and chloride, the millimoles of anhydrous dextrose, and the osmolarity of the following paerenteral fluid. Dextrose, anhydrous 50 g NaCl 4.5 g KCl 1.49 g Water for injectoin ad 1000 mL

Chapter 12 Page 197 Na 77 mEq K 20 mEq Cl 97 mEq Dextrose 278 mmol 472 mOsmol

Chapter 12 Page 191 How many mEq of Na+ would be contained in a 30 mL dose of the following solution? Disodium hydrogen phosphate 18 g Sodium biphosphate 48 g Purified water ad 100 mL (Disodium hydrogen phosphate MW 268) (Sodium biphosphate MW 138)

Chapter 12 Page 191

Chapter 12 Problem 3 A 10-mL ampule of potassium chloride contains 2.98 g of potassium chloride (KCl). What is the concentration of the solution in terms of milliequivalents per milliliter? 4 mEq/mL

Chapter 12 Problem 40 A patient has a sodium deficit of 168 mEq. How many milliliters of isotonic sodium chloride solution (0.9% w/v) should be administered to replace the deficit? 1092 mL

Chapter 12 Problem 56 What is the osmolarity of an 8.4% w/v solution of sodium bicarbonate? 2000 mOsmol/L

Chapter 12 Problem 58 How many (a) millimoles, (b) milliequivalents, and (c) milliosmoles of calcium chloride (CaCl2⋅2H2O—m.w. 147) are represented in 147 mL of a 10% w/v calcium chloride solution? 100 mmol 200 mEq 300 mOsmol

Equations; Relating mOsmol, mmol and mEq mEq = (mg substance * valance) / MW mEq = MW (mg) / valance mmol = MW (mg) mOsmol = (Weight(g) / MW(g)) * species * 1000 mOsmol/kg = 2(Na + K) + (BUN/2.8) + (glucose/18) mOsmol = mmol * # species mOsmol = (mEq * # species) valence