(a) Estimate the \(\mathrm{pH}\) of the solution that results when we add \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) to \(25.0 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3}\) (aq). (b) If we add an additional \(20.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}(\mathrm{aq})\) solution, what would you predict the pH of the resulting solution to be?

Understanding chemical equilibrium is crucial when studying reactions, such as the neutralization reaction in the provided exercise. At chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time.

However, in the case of strong acid-base reactions like the one between sodium hydroxide (NaOH) and sulfurous acid (H₂SO₃), the reaction goes to completion rather than reaching an equilibrium. This is because strong bases like NaOH completely dissociate in water and react entirely with acids to form water and a salt, leaving no unreacted NaOH or H₂SO₃ in the solution.

When calculating pH levels as in our exercise, we won't be applying the concept of chemical equilibrium directly because the involved compounds do not form a reversible system.

An acid-base reaction involves the transfer of protons (H⁺) from an acid to a base. In this scenario, we're dealing with a reaction between a strong base, NaOH, and a weak acid, H₂SO₃. The reaction can be summarized as:\[ \text{NaOH} (aq) + \text{H}_2\text{SO}_3 (aq) \rightarrow \text{NaHSO}_3 (aq) + \text{H}_2\text{O} (l) \]

Here, NaOH (sodium hydroxide) donates OH^- ions which react with H₂SO₃ to form water and sodium bisulfite (NaHSO₃). These reactions are typically fast and go to completion, resulting in a change in the concentration of H⁺ ions, and thus the pH of the solution changes significantly. Understanding these reactions is essential for calculating the pH after the reaction has occurred.

Molarity is a measure of the concentration of a solution and is vital in understanding chemical reactions and pH calculations. It's defined as the moles of solute per liter of solution:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]

In the exercise, molarity allows us to find the number of moles of each reactant. For example, the molarity of NaOH is given as 0.150 M, which means that there are 0.150 moles of NaOH for every liter of solution. To find the actual moles involved in our specific situation, we multiply this molarity by the volume of the solution in liters. This forms the basis for determining how much of each reactant is available to react, and crucially, it allows us to calculate the concentrations of any excess reactants after the reaction, which is necessary for determining the pH of the final solution.

In chemical reactions, the limiting reactant is the substance that is completely consumed first and thus determines the amount of product formed. In our exercise, we compare the moles of NaOH and H₂SO₃ to identify the limiting reactant.

By calculating the moles of each reactant, we find that H₂SO₃ is the limiting reactant because it has fewer moles than NaOH. As a result, all of the H₂SO₃ will react with NaOH, and once it is depleted, the reaction stops. Any remaining NaOH is considered excess and will affect the pH of the solution. Students must be able to identify the limiting reactant to predict the composition of the resulting mixture and to calculate concentrations for pH determination correctly.