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Understanding the equilibrium constant, often represented as \(K_{eq}\), is central to acid-base equilibrium problems. It quantifies the ratio of the concentrations of products to reactants in a reaction at equilibrium. In this problem, we look at the dissociation of \(NaHSO_3\) in water:

\[ NaHSO_3(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HSO_3^-(aq) \]
The equilibrium constant expression for this reaction is:
\[ K_{eq} = \frac{[H_3O^+][HSO_3^-]}{[NaHSO_3]} \]
Here, the values inside the brackets \([ ]\) denote concentrations. The given \(K_{eq}\) for this reaction is \(1.26 \times 10^{-2}\). This value tells us that at equilibrium, the concentration of the products is much smaller than that of the reactants, characteristic of a weakly dissociating substance.

An ICE table is a valuable tool for solving equilibrium problems. It stands for Initial, Change, and Equilibrium, representing phases during a reaction. Let's break down these stages for the dissociation of \(NaHSO_3\):

- **Initial:** This is the concentration at the start. Here, \([NaHSO_3] = 0.500\, \text{M}\), while \([H_3O^+]\) and \([HSO_3^-]\) are initially zero.- **Change:** As the reaction proceeds towards equilibrium, \(x\) amount of \(NaHSO_3\) dissociates, resulting in an increase in \(\text{M} \) of \(H_3O^+\) and \(HSO_3^-\) ions.- **Equilibrium:** The final concentrations are \(0.500 - x\) for \(NaHSO_3\), and \(x\) for both \(H_3O^+\) and \(HSO_3^-\).
This setup allows for a streamlined substitution into the equilibrium expression, helping solve for unknown values like \(x\), which represents concentration changes.

pH is a measure of the acidity of a solution, calculated from the concentration of hydrogen ions \([H_3O^+]\). In this context, once \(x\) (the \([H_3O^+]\) concentration) is solved from the ICE table, it allows for direct pH calculation using:

\[ pH = -\log_{10}([H_3O^+]) \]
So, for \([H_3O^+] = 0.0500 \text{ M}\), the pH becomes:
\[ pH = -\log_{10}(0.0500) \approx 1.30 \]
pH values below 7 indicate an acidic solution, with this specific \(NaHSO_3\) solution being quite acidic. Understanding these calculations helps determine the acidity in various chemical contexts, important in fields ranging from environmental science to biochemistry.

In any equilibrium reaction, understanding how concentrations change can reveal a lot about the system's behavior. Initially, the solution has only the reactant \([NaHSO_3]\) at 0.500 M, and as it reaches equilibrium, portions of it dissociate into \(H_3O^+\) and \(HSO_3^-\) ions.

- **Initial Conditions:** At the start, the concentration of \(H_3O^+\) and \(HSO_3^-\) is zero. - **Changes Due to Equilibrium:** As equilibrium is approached, \(x\) is formed, which is small relative to initial amounts, indicating incomplete dissociation.
The small magnitude of \(x\) results in negligible change in the initial concentration of \(NaHSO_3\), justifying the approximation of \(0.500 - x \approx 0.500\) for simplification.

These concentration changes are crucial to understanding how equilibrium shifts in response to different conditions, such as varying reactant or product concentrations, pressures, or temperatures under Le Ch芒telier's Principle.