What is the minimum volume of 0.55 M NaOH required to completely convert 1.00 L of 0.45 M acetic acid (pK_a = 4.74) to its conjugate base? A) 0.82 L B) 1.0 L C) 1.2 L D) 1.8 L E) 2.2 L

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Quizplus · Accepted Answer

To neutralize 1.00 L of 0.45 M acetic acid, the moles of NaOH required equals the moles of acetic acid due to the 1:1 stoichiometry of the neutralization reaction. Moles of acetic acid = 1.00 L * 0.45 M = 0.45 moles. To find the volume of 0.55 M NaOH needed, use the formula: Volume = Moles / Concentration. Thus, Volume = 0.45 moles / 0.55 M = 0.82 L.

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