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.
$HCl + NaOH \rightarrow NaCl + H_2O$
What was the original concentration of the HCl solution?

Answer 1

Answer:  0.0400 M

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=150mL\\n_2=1\\M_2=0.100M\\V_2=60.0mL$

Putting values in above equation, we get:

$1\times M_1\times 150=1\times 0.100\times 60.0\\\\M_1=0.0400M$

Thus the original concentration of the HCl solution is 0.0400 M