A #31.27mL# volume of sodium hydroxide at #0.167molL^-1# concentration was used to neutralize a #22.5mL# volume of AQUEOUS #H_3PO_4#. What is #[H_3PO_4](aq)#?

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Answer 1 · 2016-08-26T15:35:04

#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq)#

Note the stoichiometry. You reach a stoichiometric endpoint at biphosphate, #HPO_4^(2-)#, NOT #PO_4^(3-)#.

#"Moles of sodium hydroxide"# #=# #31.27xx10^-3Lxx0.167*mol*L^-1# #=# #5.22xx10^-3*mol#.

There were thus #(5.22xx10^-3*mol)/2# with respect to #H_3PO_4# in the volume of #22.5*mL#.

#"Molarity"_"phosphoric acid"# #=# #(5.22xx10^-3*mol)/2xx1/(22.5xx10^-3L)# #~=# #0.1*mol*L^-1# with respect to phosphoric acid.

A #31.27*mL# volume of sodium hydroxide at #0.167*mol*L^-1# concentration was used to neutralize a #22.5*mL# volume of AQUEOUS #H_3PO_4#. What is #[H_3PO_4](aq)#?