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.\nhcl + naoh → nacl + h₂o\nwhat was the original concentration of the hcl solution?\n0.040 m\n0.25 m\n1.50 m\n2.50 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.\nhcl + naoh → nacl + h₂o\nwhat was the original concentration of the hcl solution?\n0.040 m\n0.25 m\n1.50 m\n2.50 m

Answer

Explanation:

Step1: Determine moles of NaOH

Use the formula $n = M\times V$, where $n$ is moles, $M$ is molarity and $V$ is volume in liters. $V_{NaOH}=60.0\ mL = 0.0600\ L$, $M_{NaOH}=0.100\ M$. $n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.100\ mol/L\times0.0600\ L = 0.00600\ mol$

Step2: Relate moles of NaOH to 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 $n_{HCl}=n_{NaOH}=0.00600\ mol$.

Step3: Calculate the molarity of HCl

$V_{HCl}=150\ mL = 0.150\ L$. Use the formula $M=\frac{n}{V}$. $M_{HCl}=\frac{n_{HCl}}{V_{HCl}}=\frac{0.00600\ mol}{0.150\ L}=0.040\ M$

Answer:

0.040 M