Question

A `0.2400*g` mass of `Na_2CO_3` were neutralized by a `23.04*mL` volume of `HCl(aq)`. What was `[HCl(aq)]`?

Answer 1

`[HCl]-=0.1966*mol*L^-1`

Explanation:

We need (i) a stoichiometric equation to represent the reaction:

`Na_2CO_3(aq) + 2HCl(aq) rarr 2NaCl(aq) + CO_2(g)uarr + H_2O(l)`

And (ii), we need equivalent quantities of sodium carbonate...

`"Moles of sodium carbonate"-=(0.2400*g)/(105.99*g*mol^-1)`

`=2.264xx10^-3*mol`

Now given the stoichiometry, there were 2 equiv of `HCl` in the solution, i.e. 2 equiv in the `23.04*mL` volume of titrant..

And thus...

`[HCl]=(2xx2.264xx10^-3*mol)/(23.04*mLxx10^-3*L*mL^-1)=0.1966*mol*L^-1` as required.....

So why did I get it right (for once)? Well, I wrote out the stoichiometric equation and this informed my sense of molar equivalence. You should do these for even the very simplest of reactions; it is all too easy to make an error of fact or interpretation.