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

Step-by-step 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* M_1* 150=1* 0.100* 60.0\\\\M_1=0.0400M`

Thus the original concentration of the HCl solution is 0.0400 M