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# Ionic force: units, how to calculate it, examples

When the concentration of the ions in a solution is high, an electrostatic interaction takes place between the ions with opposite charge; that is, cations and anions attract each other very strongly, which results in the actual or effective ionic concentration being less than that calculated for a particular chemical reaction.

For this reason, the concept of chemical activity was introduced as the effective ionic concentration of a solution, the chemical activity being the product of the molarity of the solution and the coefficient of chemical activity.

This coefficient has a value close to unity (1) for dilute ionic solutions and for so-called ideal solutions. These are solutions where the intermolecular interaction between similar molecules is the same as that between different molecules.

## Ionic strength units

The ionic strength has as units moles / L (molarity) or moles / Kg of water (molality). The latter is recommended in non-ideal solutions, which are characterized because the volumes of their mixtures are not totally additive.

This means, for example, the following: if 0.5 liters of liquid A and 0.5 liters of liquid B are mixed, the resulting volume of this mixture will not necessarily be equal to 1 liter, but may be different.

Ionic strength is represented by the symbol I.

## How to calculate ionic strength?

To calculate the ionic strength of a solution, the concentration of all the ions present in the solution, as well as their respective valences, are taken into account.

Where I, as already said, is the ionic force; C, corresponds to the molar or molal ionic concentration of the ions; while Z, represents their respective valences (± 1, ± 2, ± 3, etc.).

The expression that appears in the formula in the calculation of the ionic strength (Σ) is read as summation, that is, the sum of the product of the molar concentration (C) of each ion present in the solution by its high valence (Z) squared.

As can be seen, the valence of the ion has the greatest weight in the value of the ionic strength of the solution. For example: the valence (Z) of Ca is +2, so Z 2 is equal to 4. Meanwhile, the valence (Z) of Na is +1, and therefore, Z 2 is equal to 1.

This indicates that the contribution of the Ca 2+ ion to the ionic strength value, at the same molar ionic concentration, is four times greater than that of the Na + ion .

## Importance of ionic strength

Ionic strength is a suitable measure of the ionic concentration of a solution and is the basis for the establishment of the Debye-Hückel Theory. This theory describes the ideal behavior of ionic solutions.

The ionic strength serves as the basis for the calculation of the activity coefficient (γ i ), a parameter that in turn allows the calculation of the chemical activity of an ionic compound, the chemical activity being the effective and real concentration of an ionic compound in solution .

As the ionic strength of a solution increases, the interaction between the ions increases. Therefore, γ i and the chemical activity of the ions decrease .

An increase in ionic strength can decrease the solubility of proteins in an aqueous medium, this property being used for the precipitation of proteins selectively. High ionic strength ammonium sulfate solutions are used for the precipitation and purification of plasma proteins.

## Examples of ionic forces

### Example 1

Calculate the ionic strength of a 0.3 M potassium chloride (KCl) solution.

KCl dissociates in the following way:

KCl → K +    + Cl

So we have two ions: the K + cation (Z = + 1) and the Cl  anion (Z = -1). We then apply the formula to calculate the ionic strength I:

I = 1/2 [C · (+1) 1   + C · (-1) 1 ]

= 1/2 [0.3 M · 1 1  + 0.3 M · 1 1 ]

= 0.3 M

Note that the -1 valence of Cl  was taken as 1, its absolute value, since otherwise the ionic strength would be equal to 0.

### Example 2

Calculate the ionic strength of a 0.5 M calcium sulfate (CaSO 4 ) solution

CaSO 4 dissociates as follows:

CaSO 4   → Ca 2+   + SO 2-

We have two ions: the Ca 2+ cation (Z = + 2) and the SO 2- anion (Z = -2). We then apply the formula to calculate the ionic strength I:

I = 1/2 [C · (+2) 2    + C · (-2) 2 ]

= 1/2 [0.5 M · 4 + 0.5 M · 4]

= 2 M

### Example 3

Calculate the ionic strength of a buffer with the final concentrations of dibasic sodium phosphate (Na 2 HPO 4 ) 0.3 M and of monobasic sodium phosphate (NaH 2 PO 4 ) 0.4 M.

Na 2 HPO 4 dissociates as follows:

Na 2 HPO 4   → 2Na +   + HPO 2-

While NaH 2 PO 4 dissociates following the following pattern:

NaH 2 PO 4 → Na +   + H 2 PO

We proceed as in the previous exercises, this time having the anions HPO 2- (Z = -2) and H 2 PO  (Z = -1):

I = 1/2 {[C · 2 · (+1) 1   + C · (-2) 2 ] + [C · (+1) 1   + C · (-1) 1 ]}

= 1/2 {[0.3 M · 2 · 1 + 0.3 M · 4] + [0.4 M · 1 + 0.4 M · 1]}

= 1/2 {[0.6 M + 1.2 M] + [0.4 M + 0.4 M]}

= 1.3 M

Note that the Na + concentration from Na 2 HPO 4 is multiplied by 2, since its concentration is double. However, for the other salt, NaH 2 PO 4 , we multiply the Na  concentration by 1, according to the stoichiometry of its dissolution equation.

### Example 4

Calculate the ionic strength of a solution of 0.15 M sodium chloride (NaCl) and 0.3 M glucose (C 6 H 12 O 6 ).

NaCl dissociates in the following way:

NaCl → Na +   + Cl

Glucose, however, does not dissociate into ions because it has only covalent bonds in its chemical structure. Therefore, the valence of glucose (Z) equals zero (0). We then calculate the ionic strength product of NaCl:

I = 1/2 [C · (+1) 1    + C · (-1) 1 ]

= 1/2 [0.15 M · 1 + 0.15 M · 1]

= 0.15 M

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