in a titration, 5.0 ml of a 2.0 m naoh(aq) solution exactly neutralizes 10.0 ml of an hcl(aq) solution. what…

Explanation:

Step1: Write the neutralization reaction

$HCl(aq)+NaOH(aq)\rightarrow NaCl(aq) + H_2O(l)$; the mole - ratio of $HCl$ to $NaOH$ is $1:1$.

Step2: Calculate the moles of $NaOH$

Use the formula $n = M\times V$, where $n$ is the number of moles, $M$ is the molarity, and $V$ is the volume in liters. $V_{NaOH}=5.0\ mL = 0.005\ L$ and $M_{NaOH}=2.0\ M$. So $n_{NaOH}=M_{NaOH}\times V_{NaOH}=2.0\ mol/L\times0.005\ L = 0.01\ mol$.

Step3: Determine the moles of $HCl$

Since the mole - ratio of $HCl$ to $NaOH$ is $1:1$, $n_{HCl}=n_{NaOH}=0.01\ mol$.

Step4: Calculate the molarity of $HCl$

$V_{HCl}=10.0\ mL = 0.01\ L$. Using the formula $M=\frac{n}{V}$, $M_{HCl}=\frac{n_{HCl}}{V_{HCl}}=\frac{0.01\ mol}{0.01\ L}=1.0\ M$.

Answer:

A. 1.0 M