9.0 Electrolyte Solutions - Immersion Concepts

Understanding electrolyte solutions and their tonic effects on biological systems is essential for cellular growth, maintenance, and optimal functioning. Tonicity serves as a metric for evaluating the impact of electrolytes on cellular systems and the maintenance of physiological equilibrium, a measure determined by the osmolarity of solutions. Electrolyte solutions consist of positively and negatively charged ions or protein molecules. They are quantified in moles, millimoles, osmolarity, and milliequivalents, representing solute particles and their chemical activity. Tonicity is categorized into three states: hypertonic, hypotonic, and isotonic, reflecting variations in osmolarity. Figure 9.1 provides a visual representation of the impact of varying solute concentration on the cell. Hypertonic solutions induce cell shrinkage (plasmolysis) due to water loss, while hypotonic solutions cause cell swelling and potential bursting (cytolysis). Isotonic solutions maintain cellular equilibrium. This knowledge is indispensable in medicine, guiding intravenous fluid therapies and ensuring cellular well-being.

Figure 9.1. The Impact of Varying Solute Concentrations on the Integrity of Red Blood Cells

Electrolytes are substances that dissociate into ions (charged particles) when dissolved in a solvent, such as water. These ions can conduct electricity, and their presence is crucial for various physiological processes in living organisms. An example is depicted below at 9.1.2 using Sodium Chloride [Na⁺ Cl^–]. The most commonly measured electrolytes and some proteins in the human body are listed in Table 9.1. Electrolytes are essential for the proper functioning of cells, tissues, and organs. They help regulate fluid balance, facilitate nerve impulses, support muscle contractions, and contribute to various biochemical reactions. Imbalances in electrolyte levels can lead to health issues, and maintaining the right electrolyte balance is crucial for overall well-being.

By the end of this webpage, you should be able to:

i. convert moles to millimoles.

ii. calculate the equivalent weight of an atom, molecule or compound.

iii. calculate the milliequivalent weight of an atom, molecule or compound.

iv. express the concentration of an electrolyte solution as milliequivalent per volume.

v. convert percentage concentration to milliequivalent per volume of electrolyte solution.

vi. convert milliequivalent per volume to percentage concentration of electrolyte solution.

vii. express the concentration of an electrolyte solution as milliosmoles per volume

9.1.0 Atomic and Molecular Weight, Valency and Equivalent Weight

9.1.1 Atomic Weight of an Element

The atomic weight of an element refers to the ratio of the weight of an atom of that element to the weight of an atom of another element, which is taken as a standard. Modern atomic weights are established using carbon as a reference, specifically the isotope ¹²C, where 12 is considered the relative nucleic mass of the isotope.

9.1.2 Valence or Valency

Valence and valency represent the quantifiable measure of the number of bonds formed by an atom of a specific element. It can also be described as the count of valence bonds an atom has engaged in or can potentially engage in with one or more other atoms. In the realm of chemistry, a valence electron is one that is affiliated with an atom and is capable of engaging in the creation of a chemical bond. In a singular covalent bond, each participating atom contributes one valence electron, resulting in the formation of a shared pair. Consequently, in NaCl, there is a single valence electron, with the sodium ion contributing one electron to the chloride ion. Similarly, MgCl² boasts 2 valence electrons, where magnesium donates 2 electrons, and two chloride ions collectively accept these electrons. In the context of solutions, a compound may either dissociate into particles (ions) or remain undivided. Compounds exhibiting varying degrees of ion dissociation are termed electrolytes, while those that remain intact are labeled as non-electrolytes.

9.1.3 Equivalent Weight of an Element or Compound

Equivalent weight of an element is defined as the weight of the element that will combine with or displace 1 gram atomic weight of hydrogen or the equivalent weight of another element. This concept is fundamental in stoichiometry and helps determine the relative combining capacities of different elements.

Take note of this table, we will refer to the values therein throughout this page.

Example #1 of equivalent weight

Zinc reacting with Hydrochloric acid

Zn + 2HCl → ZnCl₂+ H₂

Meaning- The equivalent weight of zinc is the weight of zinc ( $Z n$ ) that reacts with 1 gram of hydrogen () or the equivalent weight of another element. The relative atomic mass (RAM) of Zn is approximately 65.38 g/mol; and HCl is 2(1.008 + 35.45 g/mol). The equivalent weight HCl is 72.916/65.38 = 1.12 g.

We can also test it. We know the mole ratio from the equation is 1 mole Zn to 2 moles HCl. Using ratio and proportion:

Solving for X, you’ll find that X . This confirms the 1:2 ratio, indicating that 1 mole of Z $n$ is equivalent to 2 moles of $H Cl$ in this reaction.

Example #2 of equivalent weight

Oxidizing Iron to form Iron (III) Oxide

4Fe + 3O₂ → 2Fe₂O₃

Meaning- The equivalent weight of iron is the weight of iron ( $F e$ ) reacting with 1 gram of hydrogen () or the equivalent weight of another element. The RAM of iron is approximately 55.85 g/mol and Oxygen is 16 g/mol. The equivalent weight of Oxygen = 4(55.85 g/mol)/3(16 g/mol) = 4.65 g in this example.

Example #3 of equivalent weight

H₂+ Cl₂ → 2HCl

In this 3^rd example the reaction is 1:1 therefore 1.008 g of Hydrogen is equivalent to 35.453 g of Chlorine.

Equivalent Weight of a Compound

The equivalent weight of a compound is defined as the weight of the compound that is chemically equivalent to (or will combine with) 1.008 g of hydrogen. It serves as a crucial parameter in stoichiometric calculations.

To illustrate, consider sodium hydroxide (NaOH). One mole or 40 g of NaOH has the capability to neutralize 1.008 g of hydrogen. Therefore, its equivalent weight is determined as 40.00 g. This implies that 40.00 g of NaOH is chemically equivalent to the amount of hydrogen it can neutralize.

As an application, let’s examine this reaction:

MgSO₄+ H₂ → H₂SO₄+ Mg

This equation indicates that 120 g of MgSO₄is required to neutralize 2 hydrogen atoms (1.008 g 2). Consequently, the equivalent weight of MgSO₄needed to neutralize 1.008 g of hydrogen is calculated as: This signifies that 60 g of MgSO₄is chemically equivalent to the amount of hydrogen it can neutralize in the given reaction.

Calculating Equivalent Weight of an Element or Compound using Valence

The law of definite proportions asserts that elements consistently combine in fixed weight ratios to form specific compounds. Therefore, the equivalent weight is essentially the atomic weight or molecular weight divided by the valence of the substance.

Note: The equivalent weight of a monovalent ion is equal to its atomic weight. For non-monovalent ions, the equivalent weight differs from the atomic weight. For instance, divalent ions will have an equivalent weight that is half the atomic weight, and so forth.

9.2.0 Milliequivalent

Chemical compounds that undergo varying degrees of dissociation are referred to as electrolytes. The concentration of electrolytes in the body provides crucial information about the number of ions or charges carried by these compounds. The chemical combining power should consider not only the number of particles in the solution but also the total number of ionic charges and the valence of ions in the solution. Electrolyte concentrations in body fluids are typically expressed in chemical units known as milliequivalents (mEq). A milliequivalent represents the amount of substance in milligrams that is equivalent to 1/1000 of its gram equivalent weight. This unit of measurement reflects the chemical activity of a compound and is linked to the number of ionic charges in the solution. Common expressions for milliequivalent concentration include mEq/mL or mEq/L. Therefore:

9.2.1 Converting Milliequivalent (mEq) to Milligram (mg)

Let’s explore this example- Determining the amount of milligrams that is equivalent to 2 mEq of Potassium Chloride (KCl).

Step 1- Determine the atomic/molecular weight of the element/substance

Molecular weight of KCl (MWT) = 39 g + 35.5 g = 74.5 g

Step 2- Determine the equivalent weight of the element/substance

Valency of KCl = 1 (number of valence electron used in bonding)

Equivalent weight of KCl =

Step 3- Determine the weight (mg) that is equivalent to 1 milliequivalent (mEq) of the element/substance

1 mEq of KCl =

Hence 1 mEq of KCl is equivalent to 74.5 mg of KCl

Step 4- Determine the weight (mg) that is equivalent to the milliequivalent given

We use the answer in step 3 to determine the weight that will be equivalent to 2 mEq of KCl requested

Since 1 mEq of KCl has a mass of 74.5 mg

Then 2 mEq of KCl will have a mass of

9.2.2 Converting Milliequivalents per Unit Volume to Weight (mg/g) per Unit Volume using Ratio and Proportion

Using the basic steps (1- 4) as outlined in section 9.2.1, milliequivalent per unit volume can be converted to milligrams per unit volume.

Consider the task, Determining the amount of KCl in milligrams per milliliter that is equivalent to 5 mL of Potassium Chloride (KCl) solution containing 2 mEq. We will start at step 5 using the response from the previous section.

Step 5-Determine the amount of KCl in milligrams in the volume requested.

Since 2 mEq of KCl is found in 5 mL of the solution,

Then 149 mg of KCl is also found in 5 mL of the solution

If 149 mg of KCl is found in 5 mL of solution

The amount of KCl that is found in 1 mL of the solution is

9.2.3 Converting Milliequivalents per Unit Volume (mEq/mL or mEq/L) to Weight per Unit Volume (mg/mL or g/L) using an equation

The milliequivalent concentration of a compound from a given solution may be obtained from the following equation:

Task, Determining the amount of KCl in milligrams per milliliter that is equivalent to 5 mL of Potassium Chloride (KCl) solution containing 2 mEq.

What information is available:

Milliequivalent concentration = 2 mEq/5 mL

Milligram concentration = ? mg/mL

Molecular weight of KCl = 74.5 g/mol

Valence = 1

Step 1- Ensure all the concentrations are expressed in the same units

2 mEq/5 mL needs to be converted mEq/mL

If 2 mEq is contained in 5 mL of solution

Then x mEq that is contained in 1 mL solution. X =

Step 2- Apply equation 2

9.2.4. Milliequivalent Calculations of Complex Salts such as Hydrates using Ratio and Proportion

Task, What is the concentration in g/mL of a solution containing 4 mEq of Calcium Chloride (CaCl₂•2H₂O) per mL?

Step 1- Determine the molecular weight of the compound

MWT of CaCl₂•2H₂O = 147 g

Step 2-Calculate the equivalent weight

A divalent compound has a valence of 2. In the case of calcium chloride, even though it is a hydrate, the valence of the water molecule within the compound is not considered when calculating the total molecular weight of the hydrate.

MWT = 147 g

Equivalent weight

Step 3- Determine the weight (mg) that is equivalent to 1 milliequivalent (mEq) of the element/substance

Since 1 mEq is equivalent to 0.0735 g

Then 4 mEq is equivalent to X

= 0.294 g or 294 mg

Since 4 mEq of calcium chloride is contained in 1 mL

Then 0.294 g will also be contained in 1 mL

Hence concentration is 0.294 g/mL

9.2.5. Milliequivalent Calculations of Complex Salts such as Hydrates using the Equation

Milliequivalent concentration = 4 mEq/mL

Milligram concentration = ? mg/mL

Molecular weight of CaCl₂•2H₂ = 147 g/mol

Valence = 1

Step 1- Ensure all the concentrations are expressed in the same units

The concentration given is mEq/mL and the concentration requested is in g/mL.

Step 2- Plug values into the equation

But the question requested g/mL, which is 0.294 g/mL

9.2.6 Convert mEq/L to % Concentration

The Task, What is the percent (^w/_v) concentration of a solution containing 100 mEq of Ammonium Chloride per litre?

MWT of NH₄Cl = 53.5 g

Equivalent weight = 53.5 g

Since 1 mEq is equivalent to 0.0535 g

Then 100 mEq is equivalent to X

% concentration is amount (g) in 100 mL of solution.

Since 5.35g is contained in 1000 mL

Then amount contained in 100ml is

9.2.7 Convert mg% to mEq/L

Task, A solution contains 50 mg% of Ca²⁺ ions. Express the concentration in terms of mEq/L.

Atomic weight = 40 g

Equivalent weight = 40/2 = 20 g

1 mEq of Ca²⁺ = 1/1,000 * 20 g = 0.02 g = 20 mg

Since 50 mg% of Ca²⁺ is 50 mg of calcium in 100 mL

Then 500 mg will be contained in 1000 mL (1 L)

Also we calculated that 1 mEq of Ca²⁺ is equivalent to 20 mg

Therefore amount of mEq that is equivalent to 500 mg is

9.2.8 Converting Weights to Milliequivalents

Task, How many milliequivalents of Magnesium Sulfate are represented in 1 g of anhydrous Magnesium Sulfate (MgSO₄)?

MWT of MgSO₄ = 120 g

Equivalent weight = 60 g

1 mEq of NH₄Cl = 1/1,000 *60 g = 0.06 g

Since 1 mEq is equivalent to 0.06 g

Then X mEq is equivalent to 1 g

9.2.8.1 Converting Weights to Milliequivalents of more than one salt

When converting weights to milliequivalents involving multiple salts, the calculation becomes more complex. It’s crucial to acknowledge that the total concentration of cations always equals the total concentration of anions in the solution. Consequently, any quantity of milliequivalents of cations will consistently react with an equivalent number of milliequivalents of anions within a specific solution. Therefore, for any electrolyte compound, the following principle holds true.

Milliequivalents of the chemical compound = Milliequivalents of cations = Milliequivalents of anions.

Task, How many milliequivalents of sodium would be contained in a 30 mL dose of the following solution?

Disodium Hydrogen Phosphate 18 g

Sodium Biphosphate 48 g

Purified water 100 mL

Consider each salt separately.

Considering Disodium hydrogen phosphate

Na₂HPO₄•7H₂O → MWT = 268 g

Equivalent weight = 268 g/2 = 134 g (since 2 Sodium ions used).

Since 18 g is contained in 100 mL

The amount contained in 30 mL is

1 mEq = 1/1,000 * 134 g = 0.134 g

Total milliequivalents for 5.4 g is

The concept Milliequivalents of the chemical compound = Milliequivalents of cations = Milliequivalents of anions.

This means that total milliequivalent of Na₂HPO₄•7H₂O salt is 40.3 mEq, which is the same for the Na ion and the same for the HPO₄ ion.

Considering Sodium Biphosphate

NaH₂PO₄•H₂O → MWT = 138 g

Equivalent weight = 138 g/1 (since 1 sodium ion used).

1 mEq = 1/1,000 * 138 g = 0.138 g

Since 48 g is contained in 100 mL

The amount contained in 30 mL is

Total milliequivalent for 14.4 g is

The concept Milliequivalents of the chemical compound = Milliequivalents of cations = Milliequivalents of anions.

This means that total milliequivalent of NaH₂PO₄•H₂O salt is 104.35 mEq, which is the same for the Na ion and the same for the H₂PO₄ ion.

Total milliequivalent of Na then is 40.3 mEq + 104.35 mEq = 144.6 mEq

9.3.0 Millimoles

One mole is the molecular weight of a substance in grams.

One millimole is therefore

Task, How many millimoles of monobasic phosphate are present in 100 g of the substance? (MWT = 138 g)

Step 1- Determine the molecular weight

MWT = 138 g

Step 2- Determine 1 millimole

One mole of monobasic phosphate is 138 g

Thus, 1 millimole will contain

Step 3- Calculate the total amount of millimoles in 100 g of substance

Since 1 millimole contains 0.138 g

Therefore amt of millimoles contained in 100 g is = 724.6 mmol of monobasic phosphate.

9.3.1 Osmolarity

Osmotic pressure is directly correlated with the quantity of particles present in a solution. The unit utilized to measure osmotic pressure is milliosmoles (mOsmol). The total count of particles within a solution hinges on the extent of dissociation, particularly in substances like electrolytes. Non-electrolytes, such as dextrose, exhibit no dissociation, resulting in the production of only one species, thereby representing 1 mmol equating to one mOsmol.

Electrolytes, conversely, rely on the degree of particle dissociation. For instance, considering complete dissociation, 1 mole of NaCl symbolizes 2 osmoles—signifying the separation of two distinct ions: Na⁺ and Cl^–, which accounts for one cation and one anion.

Hence, each mole of NaCl translates to two osmoles in solution, encompassing one mole of Na⁺ and one mole of Cl^–. Accordingly, 1 mmol of NaCl represents 2 osmoles, comprising one mmol of Na⁺ and one mmol of Cl^–. Similarly, 1 mmol of CaCl₂ denotes 3 mOsmol, accounting for the total dissociation of three particles/ions: Ca⁺ and 2Cl^–, thereby including one cation and two anions.

The calculation for Ideal Osmolarity (Osmolar concentration) is determined as follows:

However, in practical scenarios, as the concentration rises, the physicochemical interactions among solute molecules intensify, leading to a reduction in the number of dissociated particles. Consequently, the actual osmolarity values tend to decrease in comparison to ideal values. These deviations from the ideal condition are more prominent in concentrated solutions than in diluted ones.

Osmolarity and Osmolality

Osmolarity – Measured in milliosmoles (mOsmol) of solute per liter of solution.

Osmolality – Measured in milliosmoles (mOsmol) of solute per kilogram of solvent. While in dilute aqueous solutions, both osmolarity and osmolality nearly coincide, their values differ notably in more concentrated solutions. For instance, the normal serum osmolality typically falls within the range of 275 to 295 mOsmol/kg

9.3.1.1 Calculating milliosmole values for non-electrolyte substances

Task, A solution contains 5% of anhydrous dextrose in water for injection. How many milliosmoles per litre are represented by this concentration?

Dextrose is a non-electrolyte that does not dissociate when in solution. Hence number of species would be, one.

Step 1- Determine the weight of substance that is contained in 1,000 mL

Since it is 5 g of anhydrous dextrose in 100 mL

Then 50 g will be contained in 1000 mL (litre is required in the formula).

Step 2- Determine molecular weight

MWT = 180 g

Step 3- Determine the number of species (number of cations and anions) produced after dissociation

Only one species as dextrose is a non electrolyte

Step 4- Plug in the formula

9.3.1.2 Calculating milliosmole values for electrolyte substances

Task, How many milliosmoles are represented in a liter of a 0.9% ^w/_v sodium chloride solution?

Step 1- Determine the weight of substance that is contained in 1,000 mL

Concentration of substance in mg/L = 0.9 g is contained in 100 mL

Thus, amount in 1000 mL is =

Step 2- Determine atomic weight

Molecular weight = 58.5 g

Step 3- Determine the number of species (number of cations and anions) produced after dissociation

1 cation and 1 anion produced

Step 4- Plug in the formula

9.3.1.3 Calculating milliosmole values for separate ions

Task, A solution contains 10 mg% of Ca²⁺ions. How many milliosmoles are represented in 1 litre of the solution?

Step 1- Determine the weight of substance that is contained in 1000 ml

Concentration of ion in mg/L = 10 mg is contained in 100 mL

Thus, amount in 1000 mL is =

Step 2- Determine atomic weight

Atomic weight = 40g

Step 3- Determine the number of species (number of cations and anions) produced after dissociation

1 cation produced

Step 4- Plug in the formula

9.3.2 Clinical Considerations and Calculations

Plasma osmolality refers to the concentration measurement of various solutes like sodium, chloride, potassium, glucose, urea, and other ions present in the bloodstream. It is influenced by alterations in the body’s water content. Given the permeability of cell membranes to water, the osmolality of extracellular fluid (ECF) mirrors that of the intracellular fluid (ICF). Changes in osmolality can significantly impact normal cellular function and volume. Excessive hypotonicity in the ECF can prompt the influx of water into surrounding cells, potentially leading to their swelling and potential rupture.

Alterations in blood osmolality occur in conditions of dehydration, where osmolality tends to rise, and in instances of overhydration, which results in a decrease in osmolality. Elevated osmolality is often observed following specific illnesses. The commonly used formula for estimating plasma osmolality, found in literature, is expressed as:

Where:

Sodium () is measured in mEq/L

Glucose and BUN (Blood Urea Nitrogen) are measured in mg/dL

In estimating plasma osmolality, the role of potassium is usually considered minor or negligible within the formula. The traditional method for calculating plasma osmolality typically excludes potassium due to its relatively small contribution compared to sodium, glucose, and blood urea nitrogen (BUN) in the overall osmolality determination. In clinical cases where Potassium is considered, calculating osmolality can be accomplished using this revised equation:

Where:

Sodium () and Potassium (K) are measured in mEq/L

Glucose and BUN (Blood Urea Nitrogen) are measured in mg/dL

Task, Estimate the plasma osmolality from the following data:

Sodium 135 mEq/L

Potassium 4.5 mEq/L

Blood Urea Nitrogen [BUN] 14 mg/dL

Glucose 90 mg/dL

Plugging values into the equation using the revised equation: