A 20.00-mL sample of 0.220 M triethylamine, (CH₃CH₂)₃N is titrated with 0.544 M HCl (K_b(CH₃CH₂)₃N = 5.2 × 10−⁴).How many mL of HCl are required to reach the equivalence point?

Question

Quizplus · Accepted Answer

To find the volume of HCl required to reach the equivalence point, use the formula $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the molarity and volume of the triethylamine, and $M_2$ and $V_2$ are the molarity and volume of HCl, respectively. Given $M_1 = 0.220 \, \text{M}$, $V_1 = 20.00 \, \text{mL}$, and $M_2 = 0.544 \, \text{M}$, solving for $V_2$ gives $V_2 = \frac{M_1V_1}{M_2} = \frac{0.220 \times 20.00}{0.544} = 8.09 \, \text{mL}$.

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