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Acid-base titration is a method used in chemistry to determine the concentration of an acid or base in a solution. This technique involves slowly adding a titrant, which is a solution of known concentration, to a solution containing the analyte, whose concentration is unknown, until the reaction reaches its equivalence point. The equivalence point is the stage where the acid completely neutralizes the base (or vice versa), forming water and a salt.
In the given exercise, nitric acid (\( \mathrm{HNO}_{3} \)) is titrated with potassium hydroxide (\( \mathrm{KOH} \)). By measuring the volume of \( \mathrm{HNO}_{3} \) needed to reach neutralization with the known molarity and volume of \( \mathrm{KOH} \), we can calculate the molarity of the \( \mathrm{HNO}_{3} \) solution. This practical application of acid-base titration is often used in labs to analyze the concentration of acidic or basic solutions.

Balanced chemical equations are essential for understanding chemical reactions. A balanced equation has the same number of each type of atom on both sides of the reaction, maintaining the principle of the conservation of mass.
In the exercise, the balanced chemical equation is given as: \( \mathrm{HNO}_{3}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{KNO}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \). This equation indicates that one molecule of nitric acid reacts with one molecule of potassium hydroxide to produce one molecule of potassium nitrate and one molecule of water.
Understanding balanced equations helps us predict how much of each reactant is needed and how much product will be formed. In this case, knowing that the mole ratio is 1:1 between \( \mathrm{HNO}_{3} \) and \( \mathrm{KOH} \) simplifies the stoichiometry calculations.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It helps chemists determine the amounts of substances consumed and produced in a reaction.
In the problem, stoichiometry is used to link the volume and molarity of the \( \mathrm{KOH} \) solution to the nitric acid. First, we calculated the moles of \( \mathrm{KOH} \) using its molarity and volume: \( 0.150 \mathrm{~M} \times 0.0250 \mathrm{~L} = 0.00375 \text{ moles of } \mathrm{KOH} \).
Then, from the balanced equation, we know that this number of moles is equal to the moles of \( \mathrm{HNO}_{3} \) needed. Finally, the molarity of \( \mathrm{HNO}_{3} \) was calculated using the formula:

• \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)
• \( M = \frac{0.00375 \text{ moles}}{0.0685 \text{ L}} = 0.0547 \mathrm{M} \).

Understanding stoichiometry is crucial for calculating reactant or product quantities accurately in any chemical reaction.