Question

A `20*g` mass of phosphoric acid dissolved in `250.0*mL` aqueous solution, reaches a stoichiometric end-point with WHAT volume of `NaOH(aq)` at `0.120*mol*L^-1` concentration?

Answer 1

We need (i) a stoichiometric equation.....

Explanation:

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

Note a stoichiometric endpoint is reached upon the addition of 2 equiv of base.

`"Concentration"` `[H_3PO_4]="Moles of phosphoric acid"/"Volume of solution"`

`=((20.0*g)/(98.08*g*mol^-1))/(250*mLxx10^-3*L*mL^-1)=0.816*mol*L^-1`

And (ii) we need equivalent quantitites of `NaOH` when `20*mL` is tritrated.

`=(0.816*mol*L^-1xx20xx10^-3*L)/(1/2xx0.120*mol*L^-1)=0.272*L`

`=27.2*mL`.

Phosphoric acid is a diacid in aqueous solution.