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

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Answer 1 · 2017-05-13T05:46:01

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

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:

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

A 17.15*mL17.15*mL volume of phosphoric acid of 0.077*mol*L^-10.077*mol*L^-1 concentration reacts with how many grams of sodium hydroxide to reach stoichiometric equivalence?