What is the milliequivalent? (Calculation examples)

The milliequivalent, as its name implies, it is one thousandth of an equivalent. Although it is an expression of concentration that is of little use, compared to molarity, it continues to be used in physiology and medicine because some substances of interest in them are electrically charged..

That is, they are ionic substances that have a low concentration, so the extracellular and intracellular concentration of these ions, for example: Na⁺, K⁺, AC^two+, Cl^- and HCO₃, they are usually expressed in milliequivalents / liter (mEq / L). As an example, the extracellular concentration of potassium is 5 mEq / L.

The equivalent weight or gram equivalent is the amount of a substance that is capable of producing or combining with one mole of negative charges or with one mole of positive charges. It is also the amount of a substance that replaces or reacts with one mole of hydrogen ions (H⁺) in an oxide-base reaction.

If scientists were asked about their preference between millimoles or milliequivalent, they would respond in unison that they prefer millimoles. These are easier to understand, use, and are also independent of the reaction that is carried out with the analyte or species of interest..

Article index

• 1 Calculation examples
  • 1.1 An element in solution
  • 1.2 A base or alkalis
  • 1.3 An acid
  • 1.4 Metal oxide
  • 1.5 A salt
• 2 Final comment
• 3 References

Calculation examples

An element in solution

An aqueous solution contains 36 g of calcium in ionic form (Ca^two+) in 300 mL of it. Knowing that the atomic weight of calcium is 40 u, and its valence is 2: calculate the concentration of calcium in the solution expressed in mEq / L.

The equivalent weight of an element is equal to its atomic weight divided by its valence. Expressing said atomic weight in moles, and knowing that each mole of calcium has two equivalents, we have:

pEq = (40 g / mol) / (2 Eq / mol)

= 20 g / Eq

It should be noted that the atomic weight has no units (outside the amu), while the equivalent weight is expressed in units (g / Eq). Now we express the concentration of Ca^two+ in g / L:

Grams of Ca^two+/ liter = 36 g / 0.3 L

= 120 g / L

But we know that each equivalent has a mass of 20 g. Therefore, we can calculate the total equivalents in solution:

Equivalents / liter = concentration (g / L) / equivalent weight (g / Eq)

Eq / L = (120 g / L) / (20 g / Eq)

= 6 Eq / L

And each equivalent finally contains 1000 milliequivalents:

mEq / L = 6 Eq / L 1000 mEq / Eq

= 6,000 mEq / L

A base or alkalis

A base, according to Bronsted-Lowry, is a compound that is capable of accepting protons. While for Lewis, a base is a compound capable of giving up or sharing a pair of electrons.

We want to calculate the concentration in mEq / L of a solution of 50 mg of calcium hydroxide, Ca (OH)_two, in 250 mL of aqueous solution. The molar mass of calcium hydroxide is equal to 74 g / mol.

We proceed with the following formula:

The equivalent weight of a base = molecular weight / hydroxyl number

And therefore,

The equivalent weight of Ca (OH)_two = molecular weight / 2

pEq = (74 g / mol) / (2 Eq / mol)

= 37 g / Eq

The equivalent weight can be expressed as mg / mEq (37 mg / mEq) which simplifies the calculation. We have 250 mL or 0.250 L of solution, the volume in which the 50 mg of Ca (OH) are dissolved_two; we calculate the dissolved for a liter:

mg of calcium hydroxide / L = 50 mg (1 L / 0.25 L)

= 200 mg / L

Later,

mEq / L = concentration (mg / L) / pEq (mg / mEq)

= (200 mg / L) / (37 mg / mEq)

= 5.40 mEq / L

An acid

The equivalent weight of an acid is equal to its molar mass divided by its hydrogen number. Knowing this, the analysis of orthophosphoric acid (H₃PO₄) shows that it can be completely dissociated as follows:

H₃PO4  <=>  3 H⁺   +   PO₄^3-

In this case:

pEq = pm / 3

Since phosphoric acid dissociates releasing 3 H ions⁺, that is, 3 moles of positive charge. However, phosphoric acid can incompletely dissociate into H_twoPO4^- or HPO₄^two-.

In the first case:

pEq = pm / 1

Since phosphoric acid to form H_twoPO₄^- release only one H⁺.

In the second case:

pEq = pm / 2

Since phosphoric acid to form HPO₄^two- release 2H⁺.

So how many mEq / L will an aqueous solution of 15 grams of dibasic sodium phosphate (Na_twoHPO₄), whose molar mass is 142 g / mol, and is dissolved in 1 liter of solution?

pEq Na_twoHPO4 = molecular weight / 2

= (142 g / mol) / (2 mEq / mol)

= 71 g / Eq

And we calculate Eq / L:

Eq / L = (grams / liter) / (grams / equivalent)

= (15 g / L) / (71 g / Eq)

= 0.211 Eq / L

Finally we multiply this value by 1000:

mEq / L = 0.211 Eq / L 1000 mEq / Eq

= 211 mEq / L of Na_twoHPO₄

Rust of a metal

The equivalent weight of an oxide is equal to its molar mass divided by the subscript of the metal multiplied by the valence of the metal..

A solution contains 40 grams of barium oxide (BaO) dissolved in 200 mL of aqueous solution. Calculate the number of milliequivalents of BaO in that volume. The molar mass of barium oxide is 153.3 g / mol.

pEq of BaO = (molecular weight) / (subscript Ba valence Ba)

= (153.3 g / mol) / (1 x 2)

= 76.65 g / Eq

But we know that there are 40 g of dissolved BaO, so:

Eq / 200 mL = (40 g Ba / 200 mL) / (76.65 g / Eq)

= 0.52 Eq / 200 mL

Note that if we carry out the division above we will have the equivalents in 1 liter of solution; the statement asks us to be in the 200 mL. Finally, we multiply the value obtained by 1000:

mEq / 200 mL = 0.52 Eq / 200 mL 1000 mEq / Eq

= 520 mEq / 200 mL

A salt

To calculate the equivalent weight of a salt, the same procedure used for a metal oxide is followed..

It is desired to obtain 50 mEq of ferric chloride (FeCl₃) of a salt solution containing 20 gram / liter. The molecular weight of ferric chloride is 161.4 g / mol: what volume of the solution should be taken?

We calculate its equivalent weight:

pEq FeCl₃ = (161.4 g / mol) / (1 x 3 Eq / mol)

= 53.8 g / Eq

But in the solution there are 20 g, and we want to determine how many total equivalents of FeCl₃ there are dissolved:

Eq / L = concentration (g / L) / equivalent weight (g / Eq)

Eq / L = (20 g / L) / (53.8 g / Eq)

= 0.37 Eq / L FeCl₃

Value that in milliequivalents is:

ferric chloride mEq / L = 0.37 Eq / L 1000 mEq / Eq

= 370 mEq / L FeCl₃

But we don't want 370 mEq but 50 mEq. Therefore, the volume V to be taken is calculated as follows:

V = 50 mEq (1000 mL / 370 mEq)

= 135.14 mL

This result was obtained by conversion factor, although a simple rule of three would also have worked..

Final comment

The equivalents are related to the charge of the components of a reaction. A number of equivalents of a cation reacts with the same number of equivalents of an anion to form the same number of equivalents of the salt produced.

This constitutes an advantage when simplifying the stoichiometric calculations, since in many cases it eliminates the need to balance the equations; process that can be cumbersome. This is the advantage that milliequivalents have over millimoles.

References

1. Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning.
2. Day, R., & Underwood, A. (1989). Quantitative Analytical Chemistry (fifth ed.). PEARSON Prentice Hall.
3. Wikipedia. (2019). Equivalent. Recovered from: es.wikipedia.org
4. Chemistry. (s.f.). Determination of equivalent weights of acids. Recovered from: fullquimica.com
5. Beck, Kevin. (November 06, 2019). How to Calculate a Milliequivalent. Sciencing.com. Recovered from: sciencing.com