Question

A 25.0 mL sample of 0.105 M HCl was titrated to the equivalence point with 31.5 mL of NaOH. What is the concentration of the NaOH

Answer 1

Answer: The concentration of
`NaOH` is 0.08 M.

Explanation:-

According to the neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`M_1` = molarity of
`HCl` solution = 0.105 M

`V_1` = volume of
`HCl` solution = 25 ml

`M_2` = molarity of
`NaOH` solution = ?

`V_2` = volume of
`NaOH` solution = 31.5 ml

`n_1` = valency of
`HCl` = 1

`n_2` = valency of
`NaOH` = 1

`1* 0.105M* 25=1* M_2* 31.5`

`M_2=0.08M`

Therefore, the concentration of
`NaOH` is 0.08 M.