Question

What is `[HCl]` if a `25*mL` volume of `[NaOH]` of `0.250*mol*L^-1` concentration reaches an endpoint with a `22.5*mL` volume of the acid?

Answer 1

You mean to start with a solution of `NaOH(aq)`

Explanation:

`NaOH(aq)` reacts with `HCl(aq)` with 1:1 stoichiometry.......

`NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)`

Given that `"Concentration"="Moles of titrant"/"Volume of titrant"` we can work out the molar equivalence......

`"Moles of NaOH"` `=` `0.250*mol*L^-1xx25xx10^-3*L=6.25xx10^-3*mol`, and given the 1:1 stoichiometry, this was also the number of moles of base neutralized.

This molar quantity was equivalent to the amount of moles contained in `22.5*cm^3` `HCl` titrant..........so..........

`"Concentration"` `=` `(6.25xx10^-3*mol)/(22.5xx10^-3*L)`

`[HCl]=0.278*mol*L^-1`.........

Note that `1*L=1000*cm^3=1000*mL`...........