Electrolyte solutions By Ali Khidher Alobaidy Electrolyte solutions

Electrolyte solutions By Ali Khidher Alobaidy

Electrolyte solutions � 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.

� 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. Clions 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-, HCO 3 -, HPO 4 --, SO 4 --, 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.

Applicable Dosage Forms � 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.

Milliequivalents �A chemical unit, the milliequivalent (m. Eq), is now used almost exclusively in the United States by clinicians, physicians, pharmacists, and manufacturers to express the concentration of 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.

� 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 (µmol/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_, HCO 3_ , 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 m. Eq of K+per liter, we also have exactly 40 m. Eq of Cl_, but the solution will not contain the same weight of each ion.

�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 m. Eq 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

Example Calculations of Milliequivalents � 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 (m. Eq): m. Eq = mg x Valence/ Atomic, formular, or molecular weight � To convert milliequivalents (m. Eq) to milligrams (mg): mg = m. Eq x Atomic, formula, or molecular weight /Valence � To convert milliequivalents per milliliter (m. Eq/m. L) to milligrams per milliliter (mg/m. L): mg/m. L=m. Eq/m. L x Atomic, formula, or molecular weight /Valence

� What is the concentration, in milligrams per milliliter, of a solution containing 2 m. Eq of potassium chloride (KCl) per milliliter? � Molecular weight of KCl = 74. 5 mg/m. L=m. Eq/m. L x Atomic, formula, or molecular weight /Valence � mg/m. L = 2 (m. Eq/m. L) x 74. 5 / 1 = 149 mg/m. L.

� What is the concentration, in grams per milliliter, of a solution containing 4 m. Eq of calcium chloride (Ca. Cl 2⋅2 H 2 O) 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 Ca. Cl 2⋅2 H 2 O = 147 mg/m. L=m. Eq/m. L x Atomic, formula, or molecular weight /Valence � Mg/ ml = 4 x 147/2 = 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.

� What is the percent (w/v) concentration of a solution containing 100 m. Eq of ammonium chloride per liter? mg/m. L=m. Eq/m. L x Atomic, formula, or molecular weight /Valence 100 m. Eq 1000 ml x 1 ml X = 0. 1 m. Eq/ml � Mg / ml = 0. 1 m. Eq/ml x 53. 5 / 1 = 5. 35 mg / ml = 0. 00535 g/ml 0. 00535 g X � X = 0. 535 % 1 ml 100 ml

A solution contains 10 mg/100 m. L of K ions. Express this concentration in terms of milliequivalents per liter. � � Atomic weight of K =39 mg/m. L=m. Eq/m. L x Atomic, formula, or molecular weight /Valence Rearrange equation: m. Eq/m. L = mg/m. L x Valence / Atomic, formula, or molecular weight 10 mg 100 ml x 1 ml x = 0. 1 mg/ ml � m. Eq/m. L = 0. 1 (mg/ml) x 1/ 39 = 0. 00256 m. Eq/ml 0. 00256 m. Eq x � X = 2. 56 m. Eq/L 1 ml 1000 ml

A solution contains 10 mg/100 m. L of Ca+ + ions. Express this concentration in terms of milliequivalents per liter. � � Atomic weight of Ca+ + = 40 m. Eq/m. L = mg/m. L x Valence / Atomic, formula, or molecular weight 10 mg 100 ml x 1 ml X = 0. 1 mg/ml m. Eq/m. L = 0. 1 mg/m. L x 2 / 40 = 0. 005 m. Eq/m. L 0. 005 m. Eq 1 ml x 1000 ml X = 5 m. Eq/L �

�A magnesium (Mg+ +) level in blood plasma is determined to be 2. 5 m. Eq/L. Express this concentration in terms of milligrams per liter. � Atomic weight of Mg = 24 mg/m. L=m. Eq/m. L x Atomic, formula, or molecular weight /Valence 2. 5 m. Eq x X = 0. 0025 m. Eq/m. L � mg 1000 ml 1 ml / ml = 0. 0025 (m. Eq/ml) x 24 /2 = 0. 03 mg / ml 0. 03 mg 1 ml x 1000 ml X = 30 mg/L

� How many milliequivalents of potassium chloride are represented in a 15 -m. L dose of a 10% (w/v) potassium chloride elixir? � Molecular weight of KCl = 74. 5 10 g 100 ml X 15 ml � X = 1. 5 g = 1500 mg m. Eq = mg x Valence/ Atomic, formular, or molecular weight � m. Eq = 1500 x 1 / 74. 5 = 20. 13 m. Eq

� How many milliequivalents of magnesium sulfate are represented in 1 g of anhydrous magnesium sulfate (Mg. SO 4)? � Molecular weight of Mg. SO 4 = 120 m. Eq = mg x Valence/ Atomic, formular, or molecular weight � m. Eq = 1000 mg x 2 / 120 = 16. 7 m. Eq

� How many milliequivalents of Na_ would be contained in a 30 -m. L dose of the following solution? Rx Disodium hydrogen phosphate 18 g Sodium biphosphate 48 g Purified water ad 100 m. L Each salt is considered separately in solving the problem. � Disodium hydrogen phosphate � Formula = Na 2 HPO 4. 7 H 2 O � Molecular weight = 268 18 g 100 ml X 30 ml � X = 5. 4 g = 5400 mg of disodium hydrogen phosphate per 30 m. L � m. Eq = 5400 x 2 / 268 = 40. 3 m. Eq of disodium hydrogen phosphate � Because the milliequivalent value of Na+ ion equals the milliequivalent value of disodium hydrogen phosphate, then: x = 40. 3 m. Eq of Na+ � For Sodium biphosphate � Formula = Na. H 2 PO 4. H 2 O � Molecular weight = 138 48 g 100 ml X 30 ml � X = 14. 4 g = 14400 mg of sodium biphosphate per 30 m. L � m. Eq = 14400 x 1 / 138 = 104. 3 � Adding the two milliequivalent values for Na+ = 40. 3 m. Eq + 104. 3 m. Eq = 144. 6 m. Eq �

�A person is to receive 2 m. Eq 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 Na. Cl = 58. 5 � Wt. kg = 132/2. 2 = 60 kg � 2 x 60 = 120 m. Eq mg = m. Eq x Atomic, formula, or molecular weight /Valence � Mg = 120 x 58. 5 / 1 = 7020 mg = 7. 02 g of Na. Cl needed and because 0. 9% sterile solution of sodium chloride contains 9 g of Na. Cl per liter, then 9 g 1000 ml 7. 02 g x � X = 780 ml

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

Millimoles and Micromoles � 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. (µg) / m. wt.

� How many millimoles of monobasic sodium phosphate (m. w. 138) are present in 100 g of the substance? � m. w. = 138 � 100 g = 100000 mg � mmole= wt. (mg) / m. wt. � mmole = 100000/ 138 = = 725 mmol � or � 1 mole = 138 g 1 mole 138 g X mole 100 g � x = 0. 725 moles = 725 mmol

� 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 HPO 4 - -? � Atomic weight of HPO 4 = 95. 98 � Millimole = wt. (mg) / m. wt. � 1 = wt. (mg) / 95. 98 �Wt. (mg) = 95. 98 mg

Convert blood plasma levels of 0. 5 microgram/m. L and 2 microgram /m. L of tobramycin (mw = 467. 52) to micromole /L? � Micromole = wt. (µg) / m. wt. � Micromole = 0. 5 µg / 467. 52 = 0. 0010695 � 0. 0010695µmole x 1 ml 1000 ml X = 1. 07 micromol/L � Micromole = 2µg / 467. 52 = 0. 0042778µmole x x = 4. 28 micromol/L 1 ml 1000 ml

� 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 = 116 g � How many moles are in 32. 7 g of ethanol ( m. wt. 46. 08)? � Mole = wt. (g) / m. wt. � Mole = 32. 7 g / 46. 08 = 0. 71

Osmolarity 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 (m. Osmol). � For dextrose, a nonelectrolyte, 1 mmol (1 formula weight in milligrams) represents 1 m. Osmol. 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 Na. Cl represents 2 m. Osmol (Na+ + Cl-) of total particles, 1 mmol of Ca. Cl 2 represents 3 m. Osmol (Ca+ + + 2 Cl-) of total particles, and 1 mmol of sodium citrate (Na 3 C 6 H 5 O 7) represents 4 m. Osmol (3 Na+ + C 6 H 5 O 7 - ) of total particles. �

� 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. � m. Osmol = mmole ( mole x 1000) x Number of species � m. Osmol= {Weight (mg) / Molecular weight } x Number of species � Osmolarity � = no. of m. Osmol / L of solution m. Osmol/L = (Weight (g/L) / Molecular weight g) x Number of species x 1000

�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.

� Calculate the number of milliosmoles corresponding to 0. 386 g of Na. Cl (m. wt. 58. 5)? � m. Osmol= {Weight (mg) / Molecular weight } x Number of species � m. Osmol = 386 mg / 58. 5 x 2 = 13. 2 � Calculate the osmolarity of 15 m. Osm. Dissolved in enough water to make a total volume 100 ml? � Osmolarity = no. of m. Osmol / L of solution � Osmolarity = 15 m. Osmol / 0. 1 L = 150 m. Osmol / L

�A pharmacist added 25 ml of 7. 5% solution of magnesium acetate (m. wt. 142) to apatient s infusion solution. How many m. Osmol of magnesium acetate did the patient receive? � No. of species of magnesium acetate = Mg + 2 (C 2 H 3 O 2) = 3 parts 7. 5 g 100 ml x 25 ml X = 1. 875 g = 1875 mg � m. Osmol= {Weight (mg) / Molecular weight } x Number of species � m. Osmol = 1875 mg / 142 x 3 = 39. 6

�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 � m. Osmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 � m. Osmol/L = (50 / 180) x 1000 = 278

�A solution contains 156 mg of K_ ions per 100 m. L. How many milliosmoles are represented in a liter of the solution? � Atomic weight of K = 39 � 156 mg = 0. 156 g 100 ml X 1000 ml � X = 1. 56 g � m. Osmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 � m. Osmol/L = (1. 56/39) x 1000 = 40

A solution contains 10 mg% of Ca++ ions. How many milliosmoles are represented in 1 liter of the solution? � Atomic weight of Ca = 40 10 mg 100 ml X 1000 ml � X = 100 mg/L = 0. 1 g/L � � m. Osmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 � m. Osmol/L = (0. 1 / 40) x 1 x 1000 = 2. 5 How many milliosmoles are represented in a liter of a 0. 9% sodium chloride solution? � Formula weight of Na. Cl = 58. 5 0. 9 g 100 ml X 1000 ml � X = 9 g/L � � m. Osmol/L = (Weight of substance (g/L) / Molecular weight g) x Number of species x 1000 � m. Osmol/L = (9/58. 5) x 2 x 1000 = 307. 7