Question

A `17.15*mL` volume of phosphoric acid of `0.077*mol*L^-1` concentration reacts with how many grams of sodium hydroxide to reach stoichiometric equivalence?

Answer 1

A tricky one ..........I gets a volume of approx. `5*mL`.

Explanation:

Phosphoric acid acts as a diacid in water, and thus we interrogate the stoichiometric reaction:

`H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) +2H_2O(l)`

And so at the endpoint we have `"sodium biphosphate"`......

And thus moles of `n_(H_3PO_4)-=17.15xx10^-3*Lxx0.077*mol*L^-1=1.32xx10^-3*mol`.

And this reacts with TWICE this molar quantity of `"sodium hydroxide"` as per the stoichiometry:

`(2xx1.32xx10^-3*mol)/(0.508*mol*L^-1)xx10^3*mL*L^-1=5.20*mL`.