A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (c) 24.0 mL.

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Calculate the initial moles of KOH using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \).

Calculate the moles of HClO4 added using the same formula: \( \text{moles} = \text{concentration} \times \text{volume} \).

Determine the limiting reactant by comparing the moles of KOH and HClO4.

Calculate the moles of excess reactant remaining after the reaction.

Use the moles of excess reactant to find the concentration of OH^- or H^+ ions, and then calculate the pH.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Titration

Titration is a quantitative analytical technique used to determine the concentration of a solute in a solution. It involves the gradual addition of a titrant (in this case, HClO4) to a solution of the analyte (KOH) until the reaction reaches its equivalence point, where stoichiometrically equivalent amounts of acid and base have reacted.

pH Calculation

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration. In titration problems, pH can be calculated using the concentrations of the reactants and the volume of the solutions involved, especially at different points in the titration process, such as before, at, and after the equivalence point.

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to produce water and a salt, typically resulting in a solution with a pH close to 7 at the equivalence point. In this scenario, KOH (a strong base) reacts with HClO4 (a strong acid), and understanding the stoichiometry of this reaction is essential for determining the resulting pH after specific volumes of acid have been added.

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A 35.0-mL sample of 0.150 M acetic acid 1CH3COOH2 is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (b) 17.5 mL.