A 50.00 mL 1.5 M of $NaOH$ titrated with 25.00 ml of $H_3PO_4$ (aq). What is the concentration of the is $H_3PO_4$ solution?

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Answer 1 · 2017-08-02T15:08:28

$[H_3PO_4]-=1.50*mol*L^-1$

We follow the stoichiometric reaction......

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

And it is a fact that phosphoric acid acts as DIACID in aqueous solution. $Na_3PO_4$ WILL NOT BE ACCESSED even at high $pH$ .

$"Moles of NaOH"=50.00xx10^-3Lxx1.5*mol*L^-1=0.075*mol$ .....

And thus, with respect to $H_3PO_4$ , there was a $(0.075*mol)/2$ molar quantity.......

And $[H_3PO_4]=((0.075*mol)/2)/(25.00*mLxx10^-3*L*mL^-1)$

$=(0.075*mol)/2$ of course, we might have been able to nut this result out directly, had we been on the ball in the first instance.

A 50.00 mL 1.5 M of NaOHNaOH titrated with 25.00 ml of H_3PO_4H_3PO_4 (aq). What is the concentration of the is H_3PO_4H_3PO_4 solution?