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Solid silver sulfide is warmed with dilute nitric acid.

I thought that the nitric acid would engage with silver sulfide in a double displacement reaction, as follows: $\ce{Ag_2S {(s)} + 2H+ {(aq)} -> H_2S {(g)} + 2Ag^{+} {(aq)}}$. This double displacement reaction would be driven by the production of $\ce{H_2S}$ gas.

However, my textbook says that the answer is: $\ce{Ag {(s)} + NO_3^{-} {(aq)} + 4H^{+} {(aq)} -> 3Ag^{+} {(aq)} + NO {(g)} + 2H2O {(l)}}$. Does this redox reaction occur because $\ce{HNO_3 {(aq)}}$ is a strong oxidizing agent? Why doesn't the textbook answer contain $\ce{AgS {(s)}}$ in the left hand side of the net ionic equation?

In the future, how do I know when nitric acid will engage in a double displacement reaction to liberate gas, or when it will engage in a redox reaction? Furthermore, if the nitric acid was concentrated, then would $\ce{NO_2}$ be produced?

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  • $\begingroup$ Obvious textbook mistake... $\endgroup$ – Mithoron Jul 7 '18 at 16:07
  • $\begingroup$ @Mithoron What should the correct answer be? $\endgroup$ – DrPepper Jul 7 '18 at 21:17
  • $\begingroup$ Nilay's analysis looks correct. $\endgroup$ – Mithoron Jul 7 '18 at 22:34
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Nitric acid is a strong oxidising agent and it will oxidise both the reactants and products. Thus, a reaction involving nitric acid is not a simple double-displacement reaction but a redox reaction where nitric acid itself get reduced to a couple of nitrogen oxides depending on concentration of the acid and the other reactants gets oxidised. According to [1]:

Nitric acid, when of concentration above 5%, dissolves precipitated silver sulphide rapidly. Very strong acid yields silver sulphate alone, while acid of lower concentration forms some nitrate in proportion to its dilution. 15-20%, acid yields the maximum, about 95 per cent, nitrate, greater dilution again resulting in a smaller percentage. Heating the acid, or prolonging its time of action is unfavorable to the formation of nitrate.

Another paper[2] suggests formation of normal silver(I) sulfate through a series of exotic sulfates which is confirmed by the change in color: -

[...]the acid additions used correspond to 75%, 150% and 200% of the amount required stoichiometrically based on the following reaction:

$$\ce{3Ag2S + 8HNO3 <=>> Ag2SO4 + 8NO + 4H2O}$$

[...]

The distinct colors of the various phases involved in this system (black silver sulfide ($\ce{Ag2S}$), brown silver sulphide sulphate ($\ce{Ag6S3O4}$), white to light yellow silver sulfate ($\ce{Ag2SO4}$) and bright yellow sulfur($\ce{S}$)) are a further indicator of this behaviour; the leach went from brown ($\ce{Ag6S3O4}$) to white ($\ce{Ag2SO4}$) with increased pressure and temperature and some residue samples showed evidence of the presence of elemental sulphur. This transition from $\ce{Ag2S}$ to $\ce{Ag6S3O4}$ to $\ce{Ag2SO4}$ shows the progressive oxidation of sulphur in the leach solids as the leaching reaction goes to completion.

In conclusion, reaction of silver sulfide and nitric acid will give different products depending on stoichiometry of the reactants, concentration of acid, effect of temperature and pressure etc. As you noted correctly, if the concentration of nitric acid is high enough, it may even produce $\ce{NO2}$ gas.

$$\ce{Ag2S + 10HNO3 → 2AgNO3 + H2SO4 + 8NO2 + 4H2O}$$


References:

  1. Gruener, H. Silver Nitrate Formed by the Action of Nitric Acid on Silver Sulphide. J. Am. Chem. Soc. 1910, 32 (9), 1030–1032. DOI: 10.1021/ja01927a002.

  2. Holloway, P.; Merriam, K.; Etsell, T. Nitric acid leaching of silver sulphide precipitates. Hydrometallurgy 2004, 74 (3-4), 213–220. DOI: 10.1016/j.hydromet.2004.05.003.

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