a 150 ml sample of hydrochloric acid (hcl) completely reacted with 60.0 ml of a 0.100 m naoh solution. the…

a 150 ml sample of hydrochloric acid (hcl) completely reacted with 60.0 ml of a 0.100 m naoh solution. the equation for the reaction is given below. hcl + naoh → nacl + h2o what was the original concentration of the hcl solution? 0.040 m 0.25 m 1.50 m 2.50 m

a 150 ml sample of hydrochloric acid (hcl) completely reacted with 60.0 ml of a 0.100 m naoh solution. the equation for the reaction is given below. hcl + naoh → nacl + h2o what was the original concentration of the hcl solution? 0.040 m 0.25 m 1.50 m 2.50 m

Answer

Explanation:

Step1: Calculate moles of NaOH

Use the formula $n = M\times V$, where $n$ is moles, $M$ is molarity and $V$ is volume in liters. $V_{NaOH}=60.0\ mL = 0.060\ L$ and $M_{NaOH}=0.100\ M$. So $n_{NaOH}=0.100\ M\times0.060\ L = 0.006\ mol$.

Step2: Determine moles of HCl

From the balanced chemical equation $HCl + NaOH\rightarrow NaCl + H_2O$, the mole - ratio of $HCl$ to $NaOH$ is 1:1. So $n_{HCl}=n_{NaOH}=0.006\ mol$.

Step3: Calculate molarity of HCl

$V_{HCl}=150\ mL = 0.150\ L$. Use the formula $M=\frac{n}{V}$. So $M_{HCl}=\frac{0.006\ mol}{0.150\ L}=0.040\ M$.

Answer:

A. 0.040 M