Understanding Electrolytes in Chemical Solutions and Their Importance in Body Fluid Balance

Chemical compounds in solution can remain intact or dissociate into ions, affecting their electrical charge. Electrolytes like sodium chloride play a crucial role in body fluid balance and acid-base regulation. Electrolyte preparations are used to treat imbalances, with milliequivalents and molar concentrations serving as units of measurement. The content discusses the movement of ions in solutions, the role of electrolytes in body metabolism, and the importance of maintaining a balance for overall health.

• Electrolytes
• Chemical compounds
• Body fluid balance
• Ion movement
• Milliequivalents

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1. By By Ali Ali Khidher Khidher Alobaidy Alobaidy

2. The molecules of chemical compounds in solution may remain intact, or they may dissociate into particles known as ions, which carry an electric charge. Substances that are not dissociated in solution are called nonelectrolytes, and those with varying degrees of dissociation are called electrolytes. Urea and dextrose are examples of nonelectrolytes in body water; sodium chloride in body fluids is an example of an electrolyte. Sodium chloride in solution provides Na+and Cl-ions, which carry electric charges.

3. If electrodes carrying a weak current are placed in the solution, the ions move in a direction opposite to the charges. Na+ions move to the negative electrode (cathode) and are called cations. Cl- ions move to the positive electrode (anode) and are called anions. Electrolyte ions in the blood plasma include the cations Na+, K+, Ca++, and Mg++and the anions Cl-, HCO3-, HPO4--, SO4--, organic acids, and protein. Electrolytes in body fluids play an important role in maintaining the acid-base balance in the body. They play a part in controlling body water volumes and help to regulate body metabolism.

4. Electrolyte preparations are used in the treatment of disturbances of the electrolyte and fluid balance in the body. In clinical practice, they are provided in the form of oral solutions and syrups, as dry granules intended to be dissolved in water or juice to make an oral solution, as oral tablets and capsules and, when necessary, as intravenous infusions.

5. A chemical unit, the milliequivalent (mEq), is now used almost exclusively in the United States by clinicians, physicians, manufacturers to express electrolytes in solution. This unit of measure is related to the total number of ionic charges in solution, and it takes note of the valence of the ions. In other words, it is a unit of measurement of the amount of chemical activity of an electrolyte. pharmacists, the concentration and of

6. In the International System (SI), which is used in European countries and in many others throughout the world, molar concentrations [as milli-moles per liter (mmol/L) and micromoles per liter ( mol/L)] are used to express most clinical laboratory values, including those of electrolytes. The total concentration of cations always equals the total concentration of anions. Any number of milliequivalents of Na+, K+, or any cation_ always reacts with precisely the same number of milliequivalents of Cl_, HCO3_, or any anion. For a given chemical compound, the milliequivalents of cation equals the milliequivalents of anion equals the milliequivalents of the chemical compound.if we dissolve enough potassium chloride in water to give us 40 mEq of K+per liter, we also have exactly 40 mEq of Cl_, but the solution will not contain the same weight of each ion.

7. A milliequivalent represents the amount, in milligrams, of a solute equal to 1 1000 of its gram equivalent weight, taking into account the valence of the ions. The milliequivalent expresses the chemical activity or combining power of a substance relative to the activity of 1 mg of hydrogen. Thus, based on the atomic weight and valence of the species, 1 mEq is represented by 1 mg of hydrogen, 20 mg of calcium, 23 mg of sodium, 35.5 mg of chlorine, 39 mg of potassium, and so forth. Equivalent weight = Atomic or formula weight / Valence

8. To convert the concentration of electrolytes in solution expressed as milliequivalents per unit volume to weight per unit volume and vice versa, use the following: To convert milligrams (mg) to milliequivalents (mEq): mEq = mg x Valence/Atomic, formular, or molecular weight To convert milliequivalents (mEq) to milligrams (mg): mg = mEq xAtomic, formula, or molecular weight /Valence To convert milliequivalents milligrams per milliliter (mg/mL): mg/mL=mEq/mLx Atomic, formula, or molecular weight /Valence (mEq/mL) per milliliter to

9. What is the concentration, in milligrams per milliliter, of a solution containing 2 mEq of potassium chloride (KCl) per milliliter? Molecular weight of KCl = 74.5 mg/mL=mEq/mLx Atomic, formula, or molecular weight /Valence mg/mL = 2 (mEq/mL) x 74.5 / 1 = 149 mg/mL.

10. What is the concentration, in grams per milliliter, of a solution containing 4 mEq of calcium chloride (CaCl2 2H2O) per milliliter? Recall that the equivalent weight of a binary compound may be found by dividing the formula weight by the total valence of the positive or negative radical. Formula weight of CaCl2 2H2O = 147 mg/mL=mEq/mL x Atomic, formula, or molecular weight /Valence Mg/ ml = 4 x147/2 = 294 mg/ml = 0.294 g/ml Note: The water of hydration molecules does not interfere in the calculations as long as the correct molecular weight is used.

11. What is the percent (w/v) concentration of a solution containing 100 mEq of ammonium chloride per liter? mg/mL=mEq/mL x Atomic, formula, or molecular weight /Valence 100 mEq x X = 0.1 mEq/ml 1000ml 1ml Mg / ml = 0.1 mEq/ml x 53.5 / 1 = 5.35 mg / ml = 0.00535g/ml 0.00535g X X = 0.535 % 1ml 100ml

12. A solution contains 10 mg/100 mL of K ions. Express this concentration in terms of milliequivalents per liter. Atomic weight of K =39 mg/mL=mEq/mL x Atomic, formula, or molecular weight /Valence Rearrange equation: mEq/mL = mg/mL x Valence / Atomic, formula, or molecular weight 10mg 100ml x 1ml x = 0.1 mg/ ml mEq/mL = 0.1 (mg/ml) x 1/ 39 = 0.00256 mEq/ml 0.00256 mEq x X = 2.56 mEq/L 1ml 1000ml

13. A solution contains 10 mg/100 mL of Ca+ +ions. Express this concentration in terms of milliequivalents per liter. Atomic weight of Ca+ += 40 mEq/mL = mg/mL x Valence / Atomic, formula, or molecular weight 10mg x 100ml 1ml X = 0.1 mg/ml mEq/mL = 0.1mg/mL x 2 / 40 = 0.005mEq/mL 0.005mEq 1ml x 1000ml X = 5 mEq/L

14. A magnesium (Mg+ +)level in blood plasma is determined to be 2.5 mEq/L. Express this concentrationin terms of milligrams per liter. Atomic weight of Mg = 24 mg/mL=mEq/mL x Atomic, formula, or molecular weight /Valence 2.5 mEq 1000ml x 1ml X = 0.0025 mEq/mL mg / ml = 0.0025 (mEq/ml) x 24 /2 = 0.03 mg / ml 0.03 mg 1 ml x 1000ml X = 30mg/L

15. How many milliequivalents of potassium chloride are represented in a 15-mL dose of a 10% (w/v) potassium chloride elixir? Molecular weight of KCl = 74.5 10g 100ml X 15ml X = 1.5g = 1500 mg mEq = mg x Valence/Atomic, formular, or molecular weight mEq = 1500 x1 / 74.5 = 20.13 mEq

16. How represented in 1 g of anhydrous magnesium sulfate (MgSO4)? many milliequivalents of magnesium sulfate are Molecular weight of MgSO4 = 120 mEq = mg x Valence/Atomic, formular, or molecular weight mEq = 1000mg x 2 / 120 = 16.7 mEq

17. How many milliequivalents of Na_ would be contained in a 30-mL dose of the following solution? Rx Disodium hydrogen phosphate 18 g Sodium biphosphate 48 g Purified water ad 100 mL Each salt is considered separately in solving the problem. Disodium hydrogen phosphate Formula = Na2HPO4.7H2O Molecular weight = 268 18g 100ml X 30ml X = 5.4 g = 5400 mg of disodium hydrogen phosphate per 30 mL mEq = 5400 x 2 / 268 = 40.3 mEq of disodium hydrogen phosphate Because the milliequivalent value of Na+ion equals the milliequivalent value of disodium hydrogen phosphate, then: x = 40.3 mEq of Na+ For Sodium biphosphate Formula = NaH2PO4.H2O Molecular weight = 138 48g 100ml X 30ml X = 14.4g = 14400 mg of sodium biphosphate per 30 mL mEq = 14400 x 1 / 138 = 104.3 Adding the two milliequivalent values for Na+= 40.3 mEq + 104.3 mEq = 144.6 mEq

18. A person is to receive 2 mEq of sodium chloride per kilogram of body weight. If the person weighs 132 lb., how many milliliters of a 0.9% sterile solution of sodium chloride should be administered? Molecular weight of NaCl = 58.5 Wt. kg = 132/2.2 = 60 kg 2 x 60 = 120 mEq mg = mEq xAtomic, formula, or molecular weight /Valence Mg = 120 x 58.5 / 1 = 7020 mg = 7.02g of NaCl needed and because 0.9% sterile solution of sodium chloride contains 9 g of NaCl per liter, then 9g 1000ml 7.02g x X = 780 ml

19. How many tablets (750mg potassium chloride) should a patient take to obtain 30 mEq potassium chloride? M.wt. of potassium chloride = 75 mg = mEq x Atomic, formula, or molecular weight /Valence Mg = 30 mEq x 75 / 1 = 2250mg 750mg 2250mg X = 3 tablets 1 tab. x tab.

20. electrolyte concentrations expressed in millimoles per liter (mmol/L) in representing the combining power of a chemical species. A mole is the molecular weight of a substance in grams. A millimole is one thousandth of a mole and a micromole is one millionth of a mole. Mole = wt. (g) / m.wt. Millimole = wt. (mg) / m.wt. Micromole = wt. ( g) / m.wt.

21. How many millimoles of monobasic sodium phosphate (m.w. 138) are present in 100 g of the substance? m.w. = 138 100g = 100000mg mmole= wt.(mg) / m.wt. mmole = 100000/ 138 = = 725 mmol or 1 mole = 138 g 1 mole X mole x = 0.725 moles = 725 mmol 138 g 100 g

22. How many milligrams would 1 mmol of monobasic sodium phosphate (m.wt. 138) weigh? Millimole = wt. (mg) / m.wt. 1 = wt.(mg) / 138 Wt. (mg) = 138 mg What is the weight, in milligrams, of 1 mmol of HPO4- -? Atomic weight of HPO4 = 95.98 Millimole = wt. (mg) / m.wt. 1 = wt.(mg) / 95.98 Wt. (mg) = 95.98 mg

23. Convert blood plasma levels of 0.5 microgram/mL and 2 microgram /mL of tobramycin (mw = 467.52) to micromole /L? Micromole = wt. ( g) / m.wt. Micromole = 0.5 g / 467.52 = 0.0010695 0.0010695 mole x 1 ml 1000ml X = 1.07 micromol/L Micromole = 2 g / 467.52 = 0.0042778 0.0042778 mole x x = 4.28 micromol/L 1ml 1000ml

24. What is the mass in grams of 4.3 moles of Aluminium ( atomic wt. 26.98)? Mole = wt. (g) / m.wt. 4.3 moles = wt. (g) / 26.98 Wt. = 4.3 moles x 26.98 = 116g How many moles are in 32.7g of ethanol ( m.wt. 46.08)? Mole = wt. (g) / m.wt. Mole = 32.7g / 46.08 = 0.71

25. Osmotic pressure is important to biologic processes that involve the diffusion of solutes or the transfer of fluids through semipermeable membranes.Osmotic pressure is proportional to the total number of particles (molecules or ions) in solution. The unit used to measure osmotic concentration is the milliosmole (mOsmol). For dextrose, a nonelectrolyte, 1 mmol (1 formula weight in milligrams) represents 1 mOsmol. This relationship is not the same with electrolytes, however, because the total number of particles in solution depends on the degree of dissociation of the substance in question. Assuming complete dissociation, 1 mmol of NaCl represents 2 mOsmol (Na++ Cl-) of total particles, 1 mmol of CaCl2 represents 3 mOsmol (Ca+ ++ 2Cl-) of total particles, and 1 mmol of sodium citrate (Na3C6H5O7) represents 4 mOsmol (3Na++ C6H5O7-) of total particles.

26. The milliosmolar value of separate ions of an electrolyte may be obtained by dividing the concentration, in milligrams per liter, of the ion by its atomic weight. The milliosmolar value of the whole electrolyte in solution is equal to the sum of the milliosmolar values of the separate ions. mOsmol = mmole ( mole x 1000) x Number of species mOsmol= {Weight (mg) / Molecular weight } x Number of species Osmolarity = no. of mOsmol / L of solution mOsmol/L= (Weight (g/L) / Molecular weight g) x Number of species x 1000

27. A distinction also should be made 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. For dilute aqueous solutions, osmolarity and osmolality are nearly identical. For more concentrated solutions, however, the two values may be quite dissimilar. Osmometers are commercially available for use in the laboratory to measure osmolality.

28. Calculate the number of milliosmoles corresponding to 0.386g of NaCl (m.wt. 58.5)? mOsmol= {Weight (mg) / Molecular weight } x Number of species mOsmol = 386mg / 58.5 x 2 = 13.2 Calculate the osmolarity of 15mOsm. Dissolved in enough water to make a total volume 100ml? Osmolarity = no. of mOsmol / L of solution Osmolarity = 15 mOsmol / 0.1 L = 150 mOsmol / L

29. A pharmacist added 25ml of 7.5% solution of magnesium acetate (m.wt. 142) to apatient s infusion solution. How many mOsmol of magnesium acetate did the patient receive? No. of species of magnesium acetate = Mg + 2 (C2H3O2) = 3parts 7.5 g 100 ml x 25 ml X = 1.875g = 1875 mg mOsmol= {Weight (mg) / Molecular weight } x Number of species mOsmol = 1875mg / 142 x 3 = 39.6

30. A solution contains 5% of anhydrous dextrose in water for injection. How many milliosmoles per liter are represented by this concentration? Formula weight of anhydrous dextrose = 180 5 100 X 1000 X = 50 g/L mOsmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 mOsmol/L = (50 / 180) x 1 x 1000 = 278

31. A solution contains 156 mg of K_ ions per 100 mL. How many milliosmoles are represented in a liter of the solution? Atomic weight of K = 39 156mg = 0.156g 0.156g 100ml X 1000ml X = 1.56g mOsmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 mOsmol/L = (1.56/39) x 1 x1000 = 40

32. A solution contains 10 mg% of Ca++ions. How many milliosmoles are represented in 1 liter of the solution? Atomic weight of Ca = 40 10mg 100ml X 1000ml X = 100 mg/L = 0.1g/L mOsmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 mOsmol/L = (0.1 / 40) x 1x 1000 = 2.5 How many milliosmoles are represented in a liter of a 0.9% sodium chloride solution? Formula weight of NaCl = 58.5 0.9g 100ml X 1000ml X = 9 g/L mOsmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 mOsmol/L = (9/58.5) x 2 x 1000 = 307.7