How do you balance this chemical equation: #P_4 + NO_3 -> H_2PO_4^-##+ NO#?

1 Answer
May 22, 2016

Answer:

I assume you mean #NO_3^-#, nitrate ion.

#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#

Explanation:

Elemental phosphorus is oxidized to phosphoric acid #(P^0rarrP^V)#; nitrate is reduced to nitrous oxide, #(N^VrarrN^(II))#

Oxidation half equation:

#1/4P_4(s) +4H_2O rarr H_2PO_4^(-) + 6H^(+) + 5e^(-)# #(i)#

Reduction half equation:

#NO_3^(-) + 3e^(-) + 4H^+rarr NO(g) + 2H_2O # #(ii)#

Both equation are balanced with respect to mass and charge. We cross multiply the redox equation to remove electrons:

i.e. #3xx(i)+5xx(ii)=#

#3/4P_4 +2H_2O+ 5NO_3^(-) +2H^(+) rarr3H_2PO_4^(-) + 5NO(g)#

This is stoichiometrically balanced with respect to mass and charge. It is a fact, however, that these phosphorus digestions were conducted in alkaline media. So I simply have to add #2xx(OH)^-# to both sides:

#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#

Possibly, you have dragged this scheme from the old Russian literature (1960's-1980's); this chemistry was very hard to reproduce. As this scheme suggests, the nitrate anion can be a potent oxidant, but often its redox potential is masked in aqueous chemistry.