Calculate the \(\mathrm{pH}\) of a \(0.10-M \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5}\).

The acid dissociation constant (\( K_a \)), is a quantitative measure of an acid's strength. It represents the extent to which an acid donates protons (\( H^+ \text{ions} \)) when in solution. A higher \( K_a \) value indicates a stronger acid, which implies a greater degree of dissociation. In essence, it tells us how readily the acid gives up its protons to the surrounding solvent.

To calculate \( K_a \), we need to establish the equilibrium expression for the dissociation reaction of the acid. For instance, for Co(H₂O)₆³⁺, the reaction is as follows:
\[ Co(H_{2}O)_{6}^{3+} \rightleftharpoons Co(H_{2}O)_{5}OH^{2+} + H^+ \]
The equilibrium expression that correlates to this reaction is:\[ K_{a} = \frac{[Co(H_{2}O)_{5}OH^{2+}][H^{+}]}{[Co(H_{2}O)_{6}^{3+}]} \]
In an educational context, understanding \( K_a \) is crucial as it provides insights into the concentration of ions produced at equilibrium and the strength of the acid solution, both prime factors in predicting the reactivity and behavior of acids in various chemical environments.

An equilibrium expression is a mathematical representation of the state of a reversible chemical reaction at equilibrium. It relates the concentrations of products and reactants of the reaction in a ratio reflecting the stoichiometry of the chemical equation. For acid-base reactions, the equilibrium expression is based on the acid dissociation constant \( K_a \) and uses the concentrations of the ions and molecules at equilibrium.

Consider the dissociation of Co(H₂O)₆³⁺, an equilibrium expression is derived from the general form:\[ K_{a} = \frac{[\text{products}]}{[\text{reactants}]} \]
In reality, the equation becomes:\[ K_{a} = \frac{[Co(H_{2}O)_{5}OH^{2+}][H^{+}]}{[Co(H_{2}O)_{6}^{3+}]} \]
For students, interpreting equilibrium expressions is important for problem-solving in chemistry. It involves setting up an ICE (Initial, Change, Equilibrium) table to methodically calculate the resulting concentrations at equilibrium. While solving equilibrium problems, there's usually an assumption that the change in concentration of reactants due to dissociation (represented as ‘x’) is negligible compared to the initial concentration when \( K_a \) is small. This simplifies the calculations and is particularly useful in classroom settings for instructional demonstrations and exercises.

The concentration of H⁺ ions in a solution is fundamental to understanding the solution's acidity and calculating its pH. In acidic solutions, H⁺ ions are the primary contributors to the solution's acidity. The hydrogen ion concentration can be found by solving the acid dissociation equilibrium problem where \( x \) symbolizes the concentration of \( H^+ \) ions in the equilibrium expression.

For the reaction of Co(H₂O)₆³⁺ dissociating into Co(H₂O)₅OH²⁺ and H⁺, the equilibrium expression allows us to solve for the unknown ‘x’:
\[ K_{a} = \frac{x^2}{0.10 - x} \]
Assuming that 'x' is negligible compared to the initial concentration, we get:\[ K_{a} = \frac{x^2}{0.10} \]
It allows us to directly solve for 'x':\[ x = [\ce{H^+}] = \sqrt{1.0 \times 10^{-5} \times 0.10} \approx 1.0\times 10^{-3} \]
Finally, as the hydrogen ion concentration increases, the pH of the solution decreases. For the given example, the calculation of pH is straightforward:\[ pH = -\log{[\ce{H^+}]} = -\log(1.0 \times 10^{-3}) = 3 \]
In educational practice, it's significant for students to learn how to compute the concentration of H⁺ ions as it indirectly measures the pH of solutions, an essential concept in both general and analytical chemistry that has practical applications in environmental studies, biology, medicine, and industrial processes.