A 25.0 mL sample of 0.105 M HCl was titrated with 31.5 mL of NaOH.What is the concentration of the NaOH?

Question

Quizplus · Accepted Answer

The concentration of NaOH can be calculated using the formula $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the molarity and volume of the first solution (HCl), and $M_2$ and $V_2$ are the molarity and volume of the second solution (NaOH). Given $M_1 = 0.105$ M, $V_1 = 25.0$ mL, and $V_2 = 31.5$ mL, we solve for $M_2$: $0.105 \, \text{M} \times 25.0 \, \text{mL} = M_2 \times 31.5 \, \text{mL}$. Solving for $M_2$ gives $M_2 = 0.0833$ M.

A 250 ML Sample of 0