Question

A `12.05*mL` volume of `"phosphoric acid"` was titrated with a `33.21*mL` volume of `KOH(aq)` whose concentration was `0.173*mol*L^-1`. What is `[H_3PO_4]`?

Answer 1

Approx. `0.24*mol*L^-1`

Explanation:

AS always, `"concentration"="moles of solute"/"volume of solution"`

And here, we assess the reaction.....

`H_3PO_4(aq) + 2KOH(aq) rarr K_2HPO_4(aq) + 2H_2O(l)`

`"Moles of KOH"=33.21*mLxx10^-3*L*mL^-1xx0.173*mol*L^-1=5.75xx10^-3*mol`

And given the stoichiometry, HALF this molar quantity were dissolved in the initial `12.05*mL` volume....

And so `[H_3PO_4]=(1/2xx5.75xx10^-3*mol)/(12.05*mLxx10^-3*L*mL^-1)`

`=0.239*mol*L^-1`.

Please note that the behaviour of phosphoric acid as a DIACID in aqueous solution is well-known.