General patterns of the process
The behavior of weak electrolytes in solutions has a number of specific features. The main one is that the established ionization equilibrium strictly obeys the law of mass action.
For a better understanding of the mechanics of the process, an analogy can be drawn with other states of aggregation. The process of the breakdown of weak electrolyte molecules into ions is formally similar to a chemical reaction in a gaseous medium, during which a regular increase in the total number of particles occurs in a closed system.
The effect of common ions on equilibrium shift
Equilibrium in a weak electrolyte solution is mobile and can be shifted when the composition of the medium changes. A key factor in such a shift is the addition of a strong electrolyte to the solution.
If the added strong electrolyte contains a common ion (i.e., an ion completely identical to one of the ions formed during the dissociation of the weak electrolyte), a pronounced shift in ionization equilibrium occurs. This effect is a direct consequence of general thermodynamic principles of equilibrium displacement.
The issue of using the degree of dissociation
In physicochemical practice, various quantities are often used to describe electrolytes, but not all of them are equally reliable.
The main disadvantage of the degree of dissociation lies in its instability: this value is not a true constant. The degree of dissociation of a weak electrolyte directly depends on its concentration in solution. It is precisely this variability that makes it an imperfect characteristic for the precise physicochemical description of the properties of a specific substance.
Features of the dissociation of polybasic acids and their salts
The patterns of equilibrium shift are clearly manifested in the example of polybasic compounds. Let us consider the second stage of breakdown of the dihydrogen phosphate ion:
$H_2PO_4^- \rightleftarrows HPO_4^{2-} + H^+$
The course of this process differs radically depending on the medium in which it takes place:
- In a phosphoric acid solution: hydrogen ions ($H^+$), which are formed in large quantities during the first stage of dissociation, act as common ions. According to Le Chatelier's principle, they shift the equilibrium of the second stage sharply to the left, suppressing the process.
- In a monosubstituted salt solution: there is no such excess of hydrogen ions from the first stage, so there is no suppression of dissociation (or it is incomparably smaller).
Main conclusion: the degree of dissociation in the second step (denoted as $\alpha_2$) in a monosubstituted salt solution is always higher than in the acid solution itself.
Briefly: answers to topical questions
Which thermodynamic principle explains the suppression of dissociation upon the addition of a common ion?
The suppression of dissociation upon the addition of a common ion is explained by Le Chatelier's principle. When a strong electrolyte with a common ion is added, the ionization equilibrium of the weak electrolyte shifts; for acetic acid in the presence of H⁺ or CH₃COO⁻ ions, it shifts to the left, toward undissociated molecules.
How is the ionization constant for a weak acid expressed mathematically?
Mathematically, the ionization (dissociation) constant for a weak acid is expressed via the equilibrium concentrations of the particles or via the degree of dissociation.
- Via concentrations — $K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$, where $[\text{H}_3\text{O}^+]$ and $[\text{A}^-]$ are the ion concentrations, and $[\text{HA}]$ is the concentration of the undissociated acid.
- Via Ostwald's dilution law — $K_a = \frac{\alpha^2 \cdot C}{1 - \alpha}$, where $\alpha$ is the degree of dissociation and $C$ is the electrolyte concentration.
For very weak electrolytes, in which the degree of dissociation is extremely small ($\alpha \ll 1$), the formula simplifies to $K_a = \alpha^2 \cdot C$.
How is the pH of a buffer system formed by a weak acid and its salt calculated?
To calculate the pH of an acidic buffer system formed by a weak acid and its salt, the Henderson-Hasselbalch equation is used.
The pH value depends on the nature of the components (the exponent of the weak acid dissociation constant, $pK_a$) and the ratio of their concentrations. Mathematically, this is expressed by the formula: $\text{pH} = \text{p}K_a + \lg \frac{C_c}{C_к}$, where $C_c$ is the initial concentration of the salt (which is practically equal to the equilibrium concentration of anions), and $C_к$ is the initial concentration of the weak acid. Since the amount of substance is $n = C \cdot V$, the calculation can also be performed using amounts of substance: $\text{pH} = \text{p}K_a + \lg \frac{n_c}{n_к}$.
To remember the common ion effect, imagine a crowd: if new people wearing red jackets (from the second stage) try to enter a room where there are already many people in red jackets (hydrogen ions from the acid), it is harder for them to do so — the equilibrium shifts backward (to the left).
Class and exam questions
Why is the degree of dissociation considered a poor physicochemical characteristic?
Because it is not a constant value for a specific substance. Its value is constantly changing because it directly depends on the concentration of the weak electrolyte in the solution.
What happens to the equilibrium of a weak electrolyte when a strong one is added?
If a strong electrolyte contains a common ion (the same as that of the weak electrolyte), the ionization equilibrium will shift. This is a classic example of dissociation suppression.
Where does the dihydrogen phosphate ion dissociate better: in an acid or in a salt?
In a monosubstituted salt solution. In a phosphoric acid solution, hydrogen ions from the first step shift the equilibrium of the second step to the left, suppressing the process.
What else to study on this topic
- Mathematical expression of the law of mass action for weak electrolytes
- Thermodynamics of equilibrium displacement (Le Chatelier's principle)
- The first stage of dissociation of phosphoric acid
- Chemical reactions in electrolyte solutions (Chapter 6)
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