A 64.05 mL sample of 0.250 M CH₃COOH is to be exactly neutralized with 0.500 M KOH. How many milliliters of KOH will be required?

Question

Quizplus · Accepted Answer

The neutralization reaction between CH₃COOH and KOH is a 1:1 molar ratio. To find the volume of KOH needed, use the formula: $V_1M_1 = V_2M_2$, where $V_1$ and $M_1$ are the volume and molarity of CH₃COOH, and $V_2$ and $M_2$ are the volume and molarity of KOH. Plugging in the values: $64.05 \, \text{mL} \times 0.250 \, \text{M} = V_2 \times 0.500 \, \text{M}$. Solving for $V_2$ gives $V_2 = 32.0 \, \text{mL}$.

A 6405 ML Sample of 0