Question

A `250*cm^3` volume of phosphoric acid of `1.25*mol*dm^-3` concentration was titrated with a `45*cm^3` volume of conc. ammonia. What was `[NH_3]`?

Answer 1

This is not straightforward.....

Explanation:

We ASSUME that the reaction we interrogate is.....

`H_3PO_4(aq) + 2NH_3(aq) rarr (NH_4)_2HPO_4`

That is phosphoric acid acts as a diacid, and two equiv of ammonia are required for equivalence.

`"Moles of phosphoric acid"=250*cancel(cm^3)xx10^-3*cancel(dm^3)*cancel(cm^-3)xx1.25*mol*cancel(dm^-3)=0.3125*mol`

And thus there were `0.6250*mol` with respect to ammonia, and this quantity was dissolved in a `45.3*cm^3` volume of solution.

And so....`[NH_3(aq)]=(0.6250*mol)/(45*cm^3xx10^-3*dm^3*cm^-3)`

`=13.9*mol*L^-1`

Aqueous ammonia is typically available as a `15*mol*L^-1` solution. I reiterate that phosphoric acid would behave as a DIACID in this scenario.....anyway, your teacher might have a different perspective.