A 56.20-mL sample of 0.0782 M HCN (K_a = 6.2 × 10^-10) is titrated with 0.745 M NaOH. What volume of 0.745 M NaOH is required to reach the stoichiometric point?

Question

Quizplus · Accepted Answer

To find the volume of NaOH required to reach the stoichiometric point, we use the formula: $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the molarity and volume of HCN, and $M_2$ and $V_2$ are the molarity and volume of NaOH. Given $M_1 = 0.0782$ M, $V_1 = 56.20$ mL, and $M_2 = 0.745$ M, we solve for $V_2$: $V_2 = \frac{M_1V_1}{M_2} = \frac{0.0782 \times 56.20}{0.745} \approx 5.90$ mL.

A 5620-ML Sample of 0