Question

A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution?

Answer 1

Correctly! And according to the equation:

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

Explanation:

Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of `HCl` was dissolved in a `30xx10^-3L` volume.

Volume (`L`) `xx` concentration (`mol*L^-1`) gives an answer in `mol`. Note that `1` `mL` `=` `1xx10^-3*L`

So `(23xx10^-3cancelLxx0.20*mol*cancel(L^-1))/(30xx10^-3L)` `=` `???mol*L^-1`