Many systems of the body consist of solutions of electrolytes and non-electrolytes. Some of the major cations found in the body include calcium, sodium, magnesium, and potassium. The primary anions are chloride, bicarbonate, phosphate, and sulfate. A chemical balance of these positively and negatively charged particles is necessary for maintenance of health. These ions and particles in solution are necessary for nerve conduction, maintenance of osmotic balance, and preservation of acid-base balance. If there is a shift in electrolyte balance, it is often necessary to administer salts to restore this electroneutrality.

The following chart illustrates the normal electrolyte composition of cellular compartments. (Values are expressed in meq/L.)
 
 

From acid-base chemistry, you will remember that an equivalent or equivalent weight of an acid is defined as the mass of the acid (in grams) that will supply 6.022 x10²³ hydrogen ions (one mole of hydrogen ions), or the mass of the acid that will react with 6.022x10²³ hydroxyl ions. For example,

For a monoprotic acid:
 

This equation shows that one mole of hydrochloric acid will dissociate to produce one mole of hydrogen ions (which could react with one mole of hydroxyl ions) and one mole of chloride ions (which are free to react with one mole of hydrogen ions). Since the dissociation of one mole of hydrochloric acid yields one equivalent of hydrogen and one equivalent of chloride, the equivalent weight is equal to the gram molecular weight of hydrochloric acid.
 

For a diprotic acid:
 
 
 

This equation shows that one mole of sulfuric acid will dissociate to produce two moles of hydrogen ions and one mole of sulfate ions.

Since the dissociation of one mole of sulfuric acid yields 2 times 6.022x10²³ hydrogen atoms, sulfuric acid yields two equivalents. Thus the equivalent weight of sulfuric acid is one-half the gram molecular weight of sulfuric acid. 
 

To review from general chemistry,
 
 

Avogadro's number (6.022x10²³) of molecules equals one mole.

Avogadro's number (6.022x10²³) of charged pairs (+/-) equals one equivalent.

Avogadro's number (6.022x10²³) of particles (either ions or molecules) equals one osmol.
 
 

In other words, the equivalent weight of an acid (or a base) can be calculated based on the number of paired charges in that acid or base as follows:
 
 

Equivalent weight = gram molecular weight divided by the number of paired charges because the number of paired charges represents the number of equivalents per mole.
 
 

In the field of medicine, amounts of electrolytes are usually described in terms of milliequivalents (meq). This is simply the equivalent weight divided by 1000. Laboratory tests for electrolytes in the blood are often determined and expressed in terms of milliequivalents, allowing easy comparison of values.

This concept of equivalent weights of acids and bases can be applied to salts or electrolytes, since they are, by definition, Lewis acids (accepts an electron pair) and Lewis bases (donates an electron pair). The equivalent weights are calculated similarly, based on the number of paired charges. An example is as follows:

The dissociation of potassium chloride in water is complete, as it is a strong electrolyte.
 
 
 

For potassium chloride, there is one pair of charges, thus there is one equivalent per mole of potassium chloride. To calculate the equivalent weight of KCl,
 

Since one mole of calcium chloride yields 2 equivalents of chloride ions, the equivalent weight of calcium chloride is one-half the gram-molecular weight of calcium chloride.
 

For Calcium Chloride,  147g/mol, 2 equivalents/mol, thus 73.5g/eq
And  73.5mg/meq

As you can see by the previous example, some commonly encountered salts in pharmacy are those which have "waters of hydration" in the molecular formula. These "waters" contribute to the molecular weight of the molecule, but do not contribute to the charge of the particles. For example,
 
 

To calculate the equivalent or milliequivalent weight, one would consider the total gram-molecular weight of the compound, including the seven waters of hydration. There are two pair of valence charges, neither of which is affected by the water found in the molecule.
 
 

PRACTICE PROBLEMS:

If the equivalent or milliequivalent weight of a salt can be calculated, it is then possible to determine how many equivalents are in a specified amount of that salt, or how much of a given salt would be necessary to yield a specified number of equivalents. An example follows.
 

What is the weight of potassium chloride needed to provide 10 meq of KCl?
 

First, determine the milliequivalent weight of potassium chloride.
 
 

Then, if there are 74.5 mg in one milliequivalent, what is the weight of 10 meq?
 
 
 

How many meq of potassium ion would be provided by 600 mg of potassium chloride?
 
 

Again, the first step is to calculate the milliequivalent weight of KCl, as shown above.

Then, if there are 75.5 mg in one milliequivalent of potassium chloride, how many milliequivalents would be provided by 600 mg KCl?
 

Since these are monovalent ions, also 8.053 meq of KCL would provide 8.053 meq of K⁺ ions and 8.053 meq of Cl^- ions.
 

Another way of solving this problem would be:
 

First, determine the meq weight of K⁺:

The gram molecular weight  K = 39  and it is a univalent cation.  Thus the meq  wt. = 0.039g/meq

Second, determine how many mg of K⁺ are there in 600 mg of KCl.
 
 
 

Finally, if

PRACTICE PROBLEMS:

The calculation of equivalent weights for complex salts is, not surprisingly, a little more complex. The determination is made based on how and for what ion the compound is used. For example, KH₂PO₄ is known as monobasic potassium phosphate (potassium acid phosphate). Its molecular weight is 136 g/mole. This salt is used in buffer solutions for its potassium content, its hydrogen content, or its phosphate content. To calculate the equivalent weight, it must be determined for which ion the salt is being used. If the salt is employed for its potassium content, the equivalent weight is identical to its molecular weight, because the valence of potassium is one, and when used for potassium, there is only one pair of valence charges.
 
 

When employed in buffer solutions for its hydrogen content, its equivalent weight is one-half its molecular weight, since there are two hydrogen atoms present, and hydrogen is a monovalent cation.
 
 

If monobasic potassium phosphate is used for its phosphate contribution, the equivalent weight is one-third its molecular weight, since the valence of the phosphate anion is three.
 
 

PRACTICE PROBLEMS

It is important to remember that the administration of these electrolytes is made possible by the use of salts. Salts contain both positively and negatively charged particles (ions) and each of the ions contributes to the osmotic pressure of the solution. There are also undissociated particles (non-electrolytes) in biological fluids, and these molecules contribute to the osmotic pressure as well. When determining osmotic relationships, it is imperative to consider the total number of particles in solution. The number of dissolved particles in solution is expressed in terms of osmols (Osm) or milliosmols (mOsm). An osmol is the weight of a chemical substance dissolved in one liter of water that exerts an osmotic pressure equal to that exerted by a gram-molecular weight of an unionized substances dissolved in one liter of water. In other words, one mole of a substance that does not dissociate contributes one mole of particles to the solution. One mole of a dissociating species yields as many particles as there are ions in solution, and the osmolarity is determined by the number of those particles. For example, dextrose is a non-electrolyte, and thus does not dissociate in solution. A 1 M solution of dextrose is the same as a 1 Osm solution. Sodium chloride, however, dissociates in solutions into two particles, namely sodium ions and chloride ions. Each of these two ions contributes to the osmotic pressure, thus a 1 M solution of sodium chloride is equivalent to 2 Osm. Thus, to calculate the osmolarity of a solution, the following equation is useful:
 
 

Osmol = weight of the substance(g) x number of species in molecular weight (g) solution 
 

Or, stated differently,
 

Osmol = number of moles in solution times the number of species in solution
 
 

For example, one liter of a dextrose 5% solution would contain 50 g of dextrose (5 g in 100 mL, 50 g in 1000 mL). The molecular weight of dextrose is 192 (180 for dextrose plus the two waters of hydration found within the molecule). To determine the number of moles in the solution, 
 

And since dextrose does not dissociate, this solution would also be 0.260 Osm.
 

For a compound which does dissociate, an example follows.
 

What is the concentration in Osm of Normal Saline (Sodium Chloride Injection, or 0.9% sodium chloride)?

One liter of the solution would contain 9 grams of sodium chloride. The molecular weight of sodium chloride is 58.5 (23 for sodium and 35.5 for chlorine), so the number of moles of sodium chloride in the solution is
 
 

                                                                    58.5g/mol
 
 

But, sodium chloride dissociates into two particles, so one mole of sodium chloride would yield two moles of particles.
 

0.154moles (2 ions per mole) = 0.308 Osm or 308 mOsm per liter
 
 

The number of particles in solution is very important in the field of pharmacy. The therapeutic and clinical significance of this topic will be discussed in other courses.   At this point the student should be familiar with the calculations involved in these procedures and understand the terms encountered.
 
 

References:

Physical Pharmacy, 4th edition, Alfred Martin

Pharmaceutical Calculations, 2nd edition, Joel L. Zatz

General Chemistry, 4th edition, Whitten, Gailey, Davis

Sterile Dosage Forms, 2nd edition, Turco and King
 

Answers to the practice problems may be found by clicking here.