Question

If `50.0` milliliters of 3.0 M `"H"_3"PO"_4` completely neutralized `150.0` milliliters of `"Mg"("OH")_2`, what was the molarity of the `"Mg"("OH")_2` solution?

Answer 1

This is an exceptionally poor question....

Explanation:

Phosphoric acid acts as a DIACID under normal conditions...and so we write the stoichiometric equation...

`Mg(OH)_2(s) +H_3PO_4(aq) rarr Mg^(2+)HPO_4^(-)(aq) + 2H_2O(l)`

With respect to phosphoric acid, we got a molar quantity of...

`3.0*mol*L^-1xx50xx10^-3*L-=0.150*mol`

And thus we SUPPOSE it to have neutralized `0.0750*mol` magnesium hydroxide....which was ostensibly present in a `150*mL` volume...

`[Mg(OH)_2]=(0.0750*mol)/(0.150*L)=0.500*mol*L^-1`....

Magnesium hydroxide is not so soluble in aqueous solution. The person who set this question was not a chemist. A solution of `KOH` or `NaOH` would have been appropriate....

In the following graph, phosphoric acid is titrated with sodium hydroxide; both solutions are `0.1*mol*L^-1`.

The two equivalence points are clear....