Question

17.5 mL of 0.1000M NaOH was required to neutralize 25.0 mL of HCl. What was the molarity of the HCl?

Answer 1

Answer: The molarity of
`HCl` is 0.07 M

Step-by-step explanation:

According to the neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1` = basicity
`HCl` = 1

`M_1` = molarity of
`HCl` solution = ?

`V_1` = volume of
`HCl` solution = 25.0 ml

`n_2` = acidity of
`NaOH` = 1

`M_1` = molarity of
`NaOH` solution = 0.1000 M

`V_1` = volume of
`NaOH` solution = 17.5 ml

Putting in the values we get:

`1* M_1* 25.0=1* 0.1000* 17.5`

`M_1=0.07M`

Therefore, molarity of
`HCl` is 0.07 M