Answer:
A
Explanation:
HCl and NaOH is a strong acid and strong base, respectively, and hence dissociates completely and reacts to form water:
[tex]\displaystyle \text{H}_\text{(aq)}^+ + \text{OH}_\text{(aq)}^- \longrightarrow \text{H$_2$O}_\text{($\ell$)}[/tex]
We are given that 25.0 mL of 0.333 M NaOH was used to neutralize 15.0 mL of HCl.
Convert from moles of NaOH used to moles of HCl reacted:
[tex]\displaystyle \begin{aligned} 25.0\text{ mL} & \cdot \frac{0.333\text{ mol NaOH}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} \cdot \frac{1\text{ mol OH}^-}{1\text{ mol NaOH}}\cdot \\ \\ & \cdot\frac{1\text{ mol H}^+}{1\text{ mol OH}^-} \cdot \frac{1\text{ mol HCl}}{1\text{ mol H}^+}\\ \\ & = 8.33\times 10^{-3}\text{ mol HCl}\end{aligned}[/tex]
Therefore, the molarity of the original HCl solution is:
[tex]\displaystyle \begin{aligned} \ [\text{HCl}] & = \frac{\text{ mol HCl}}{\text{L soln.}} \\ \\ & =\frac{8.33\times 10^{-3}\text{ mol HCl}}{15.0\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ & = 0.555\text{ M}\end{aligned}[/tex]
In conclusion, our answer is A.
