# Reaction of concentrated nitric acid react with metallic lead?

Here is my work:

Concentrated $\ce{HNO3}$ should first decompose into $\ce{NO2}$, $\ce{H2O}$ and $\ce{[O]}$. Nascent oxygen should combine with $\ce{Pb}$ to form $\ce{PbO}$. Now $\ce{PbO}$ should neutralize acid to form $\ce{Pb(NO3)2}$.

Therefore, the net products would be lead nitrate, $\ce{NO2}$ and $\ce{H2O}$.

Is this right?

Well, you were going along the right track.... just one correction. The products will be Pb(NO₃)₂, NO₂ and H₂O. Nitric acid is a very powerful oxidizing agent. It oxidized H₂ to H₂O.

• Oh! I forgot to add O. I have added it now. – Aumkaar Pranav Apr 11 '17 at 14:10

Lead when reacted with nitric acid forms lead(II) nitrate, nitrogen oxide and water.

According to Wikipedia:

Lead(II) nitrate can be obtained by dissolving metallic lead in aqueous nitric acid:

$$\ce{Pb + 4 HNO3 → Pb(NO3)2 + 2 NO2 + 2 H2O}$$

Since the solvent is concentrated nitric acid (in which lead(II) nitrate has very low solubility) and the resulting solution contains nitrate ions, anhydrous crystals of lead(II) nitrate spontaneously form as a result of the common ion effect.

According to webelements:

Lead reacts slowly nitric acid, $\ce{HNO3}$. Nitrogen oxides are formed together with lead(II) nitrate, $\ce{Pb(NO3)2}$.

The mechanics of the reaction on Pb with concentrated HNO3 likely commences as follows:

$$\ce{Pb -> Pb(II) + 2 e-}$$

$$\ce{HNO3 = H+ + NO3-}$$

$$\ce{Pb(II) + 2 NO3- = Pb(NO3)2}$$

$$\ce{H+ + e- = .H}$$

$$\ce{.H + NO3- = OH- + .NO2}$$ (1997 Source)

The literature also contains citations (see, for example, Page 37) of an associated mechanism involving autocatalytic $$\ce{.NO2}$$:

$$\ce{.NO2 +e- = NO2-}$$

$$\ce{H+ + NO2- = HNO2}$$

$$\ce{HNO2 +HNO3 = H2O +2 .NO2}$$

where the last reaction, I claim, can also proceeds in the reverse direction as follows:

$$\ce{.NO2 + .NO2 = N2O4}$$

$$\ce{N2O4 + H2O = HNO2 + HNO3}$$

Another 2019 paper, further cites the radical reactions:

$$\ce{.H + .NO2 = .OH + .NO}$$

And:

$$\ce{.OH + .NO = HONO}$$

One could interpret the $$\ce{.NO}$$ formation reaction above to imply that in dilute solutions, more of an aqueous $$\ce{.NO2}$$ presence and therefrom $$\ce{.NO}$$. Alternately, $$\ce{.NO}$$ could be sourced from the decomposition of any created $$\ce{HNO2}$$ (including from aqueous $$\ce{.NO2}$$) per the reaction:

$$\ce{HNO2 + H+ + e− ⇌ NO + H2O}$$ $$\ce{ Eo = +0.98 V}$$

Reference Wikipedia