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

May 22, 2016

I assume you mean $N {O}_{3}^{-}$, nitrate ion.

$\frac{3}{4} {P}_{4} + 3 {H}_{2} O + 5 N {O}_{3}^{-} \rightarrow {H}_{2} P {O}_{4}^{-} + 2 H P {O}_{4}^{2 -} + 5 N O \left(g\right) + 2 H {O}^{-}$

#### Explanation:

Elemental phosphorus is oxidized to phosphoric acid $\left({P}^{0} \rightarrow {P}^{V}\right)$; nitrate is reduced to nitrous oxide, $\left({N}^{V} \rightarrow {N}^{I I}\right)$

Oxidation half equation:

$\frac{1}{4} {P}_{4} \left(s\right) + 4 {H}_{2} O \rightarrow {H}_{2} P {O}_{4}^{-} + 6 {H}^{+} + 5 {e}^{-}$ $\left(i\right)$

Reduction half equation:

$N {O}_{3}^{-} + 3 {e}^{-} + 4 {H}^{+} \rightarrow N O \left(g\right) + 2 {H}_{2} O$ $\left(i i\right)$

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

i.e. $3 \times \left(i\right) + 5 \times \left(i i\right) =$

$\frac{3}{4} {P}_{4} + 2 {H}_{2} O + 5 N {O}_{3}^{-} + 2 {H}^{+} \rightarrow 3 {H}_{2} P {O}_{4}^{-} + 5 N O \left(g\right)$

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 $2 \times {\left(O H\right)}^{-}$ to both sides:

$\frac{3}{4} {P}_{4} + 3 {H}_{2} O + 5 N {O}_{3}^{-} \rightarrow {H}_{2} P {O}_{4}^{-} + 2 H P {O}_{4}^{2 -} + 5 N O \left(g\right) + 2 H {O}^{-}$

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.