Can aqua regia/royal water be produced with sources of chloride and nitrate other than hydrochloric acid and nitric acid?

Question

According to Professor Poliakov and a question on Chemistry Stackexchange here, gold dissolves in aqua regia because the nitrate ion oxidizes the gold into $\ce{Au^3+}$, and the chloride bonds to the gold to form tetrachloroaurate III ions ($\ce{AuCl4−}$). This means that the actual acidity of the compound plays no role in the actual oxidization and dissolution of gold. Therefore, if I just provide nitrate ions and chloride ions, the reaction should still proceed (ex: use $\ce{NaNO_3}$ for the source of nitrate ions and $\ce{NaCl}$ for the source of chloride ions). According to the equations given by an answer to the Stackexchange question earlier, aqua regia can react with gold in two ways: $$\ce{Au + 3HNO3 + 4HCl <=>> AuCl4- + 3NO2 + H3O+ + 2H2O}$$

$$\ce{Au + HNO3 + 4HCl <=>> AuCl4- + NO + H3O+ + H2O}$$ And so therefore, by my supposed substitute for the dangerous acids, they can react with gold with the following reactions: $$\ce{Au + 3NaNO3 + 4NaCl +3H2O<=>> AuCl4- + 3NO2 + 6NaOH + Na+}$$

$$\ce{Au + NaNO3 + 4NaCl + 2H2O<=>> AuCl4- + NO + 4 NaOH + Na+}$$

The sodium hydroxide that is produced by the reaction could be a problem, but we can then add a weak buffer like $\ce{HCO3-}$ or $\ce{H2CO3}$ afterward to deal with it.

Am I correct, or is something else at play in this reaction?

Answer 1

Not even zinc would react with neutral nitrate + chloride, why should gold ?

Aqua regia must be strongly acidic for nitrates to have oxidative properties for oxidation of chlorides to chlorine and to oxidize gold to trace amounts of $\ce{Au^3+}$(what redox potential allows)

Mixing of acids leads to these reactions:

$$\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 H2O}$$ $$\ce{2 NOCl -> 2 NO + Cl2}$$ $$\ce{2 NO + O2 -> NO2 }$$

That also means aqua regia dissolution ability decreases with time.

The major oxidative reagent is free chlorine. Hydrochloric acid saturated by chlorine dissolves gold as well. Chlorides additionally help gold dissolution by reaction $\ce{Au^3+ + 4 Cl- <=>> AuCl4^-}$, this decreases the redox potential $\ce{Au/Au^3+}$, making gold dissolution by chlorine and even by nitric acid easier.

$$\ce{2 Au + 3 Cl2 + 2 Cl- -> 2 AuCl4-}$$ $$\ce{Au + HNO3 + 4 HCl -> HAuCl4 + NO + 2 H2O}$$ $$\ce{Au + 3 HNO3 + 4 HCl -> HAuCl4 + 3 NO2 + 3 H2O}$$

See also https://en.wikipedia.org/wiki/Aqua_regia

Answer 2

Gold dissolves in aqua regia because there is a redox reaction: The nitrate ion oxidizes the gold in the presence of $\ce{Cl-}$ ion into $\ce{Au^3+}$, which combine with the chloride ions to form $\ce{Au(Cl)4-}$ ion. In this reaction, the presence of $\ce{Cl-}$ ion is important because it reduces the $\ce{Au/Au^3+}$ couple's potential as follows:

$$\ce{Au <=> Au^3+ + 3e-} \qquad E^\circ = \pu{-1.498 V} \tag{1}$$

However, in the presence of $\ce{Cl-}$ ion, this half-reaction becomes:

$$\ce{Au + 4Cl- <=> AuCl4^- + 3e-} \qquad E^\circ = \pu{-1.002 V} \tag{2}$$

The nitrate reduction half-reaction is:

$$\ce{NO3- + 4 H+ + 3e- <=> NO + 2H2O } \qquad E^\circ = \pu{0.957 V} \tag{3}$$

Therefore, the complete redox reaction is:

$$\ce{Au + 4Cl- + NO3- + 4 H+ <=> AuCl4^- + NO + 2H2O} \qquad E_\mathrm{rxn}^\circ = \pu{-0.045 V} \tag{4}$$

Although, $E_\mathrm{rxn}^\circ$ for this reaction is $\pu{-0.045 V}$ under standard conditions ($\pu{1 M}$ and $\pu{1 atm}$), at working conditions (e.g., concentrated $\ce{HCl}$ is at least $\pu{12 M}$), the corresponding $E_\mathrm{rxn}$ becomes positive and forward reaction proceeds (Recall Nernst equation).

According to the above half-reactions, having strong acidic conditions is essential for this reaction. Therefore, OP's suggestion of having just nitrates and chlorides (e.g., $\ce{NaNO3}$ and $\ce{NaCl}$) will not work as it in aqua regia. For example, reduction half reaction for $\ce{NO3-}$ is as follows:

$$\ce{NO3- + H2O + 2e- <=> NO2- + 2OH- } \qquad E^\circ = \pu{0.01 V} \tag{5}$$

Sum of $(2) \times 2$ and $(5) \times 3$ gives:

$$\ce{2Au + 8Cl- + 3NO3- + 3 H2O <=> 2AuCl4^- + 3NO2- + 6OH-} \qquad E_\mathrm{rxn}^\circ = \pu{-1.001 V} \tag{6}$$

This $E_\mathrm{rxn}^\circ$ value is too much to overcome.