Question

Please Help!!

A researcher takes a 10 mL sample of HCl from a bottle and titrates it with NaOH. It was found that 22.4 mL of 0.25 M NaOH were required to reach the equivalence point. What is the concentration of the HCl in original bottle?

Answer 1

Answer : The concentration of the HCl in original bottle is 0.56 M.

Explanation :

According to neutralization law:

`M_1V_1=M_2V_2`

where,

`M_1` = concentration of HCl

`M_2` = concentration of NaOH

`V_1` = volume of HCl

`V_2` = volume of NaOH

Given:

`M_1` = ?

`M_2` = 0.25 M

`V_1` = 10 mL

`V_2` = 22.4 mL

Now put all the given values in the above formula, we get:

`M_1* 10mL=0.25M* 22.4mL`

`M_1=0.56M`

Therefore, the concentration of the HCl in original bottle is 0.56 M.