Final answer:
The pH of the solution prepared by mixing 30.0 mL of 0.200 M HC₂H₃O₂ with 15.0 mL of 0.400 M KOH is 7.0, as the moles of acetic acid and KOH are equal and completely neutralize each other to form a neutral solution.
Step-by-step explanation:
To calculate the pH of a solution prepared by mixing 30.0 mL of 0.200 M HC₂H₃O₂ (acetic acid) with 15.0 mL of 0.400 M KOH (potassium hydroxide), it's important to recognize that this is a reaction between a weak acid and a strong base. The balanced equation for the reaction is:
HC₂H₃O₂ + KOH → KC₂H₃O₂ + H₂O
First, we need to calculate the moles of each reactant.
- Moles of acetic acid: volume (L) × concentration (M) = 0.030 L × 0.200 M = 0.006 moles
- Moles of KOH: 0.015 L × 0.400 M = 0.006 moles
These react in a 1:1 mole ratio and will neutralize each other completely, leaving no excess acid or base. Since the molar amounts are the same, they will react completely, and no excess of the reactant will remain.
Therefore, we have a neutral solution at the end of the reaction because the stoichiometric amounts are equivalent. The pH of a neutral solution at 25°C is 7.0.
In summary, the pH of the solution prepared by mixing 30.0 mL of 0.200 M HC₂H₃O₂ with 15.0 mL of 0.400 M KOH is 7.0.