Question

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 mc011-1.jpg NaCl + H2O

What was the original concentration of the HCl solution?

Answer 1

The balanced chemical reaction is expressed as:
HCl + NaOH ---> NaCl + H2O
We are given the amount and the concentration of the sodium hydroxide reactant. This will be the starting point for the calculations. 
0.100 mol NaOH/L solution (0.06 L solution) ( 1 mol HCl / 1 mol NaOH) = 0.006 mol HCl
Molarity = amount in moles / Volume of solution = 0.006 mol HCl/ 0.150 LMolarity = 0.04 M

Answer 2

Answer : The original concentration of HCl is 0.04 M

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of NaOH solution = 0.100 M

$V_1$ = volume of NaOH solution = 60.0 ml

$M_2$ = molarity of HCl solution = ?

$V_2$ = volume of HCl solution = 150 ml

Now put all the given values in the above law, we get the molarity of HCl solution.

$(0.100M)\times 60ml=(M_2)\times (150ml)$

$M_2=0.04M$

Therefore, the original concentration of HCl is 0.04 M