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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?

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    $\begingroup$ Intersting questen, badatchemistry! From the title "Does aqua regia only work with nitric acid and hydrochloric acid?" I'd say yes, as Aqua Regia is the mixture of of HCl and HNO3. But the underlying thoughts make up a good question. The answer probably is in the mechanism, as if e.g sodium nitrate and sodium chloride would work, they would be used for the commercial process, as they are much safer to handle than Aqua Regia. What would happen if one added sulphuric acid? Would that make Aqua Regia in situ? $\endgroup$ – imalipusram May 9 at 6:21
  • $\begingroup$ @imalipusram Thanks for that. I have fixed the title to make it more relevant. Also thanks for the insight. I might one day do an experiment and write a paper on this (I am currently a high school student studying chemistry). $\endgroup$ – KingLogic May 9 at 6:25
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    $\begingroup$ The clue is probably the redox potential of tetrachloroaurate. Au(III) easily is reduced to Au(0). $\endgroup$ – imalipusram May 9 at 6:28
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    $\begingroup$ Yes chlorine works as desinfectant. Otherwise, see my answer. $\endgroup$ – Poutnik May 9 at 7:17
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    $\begingroup$ $\ce{HNO_3}$ or $\ce{HCl}$ or nitrate ions are not able to dissolve Gold. Only chlorine molecule $\ce{Cl_2}$ does it. All my students have succeeded in dissolving a tiny piece of gold sheet suspended in $2$ or $3$ mL water where some chlorine gas is slowly bubbling. It can be done in one minute or two at room temperature in any high school lab. And in aqua regis, Chlorine is produced as shown by Pouting. $\endgroup$ – Maurice May 9 at 9:00
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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

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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.

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    $\begingroup$ Note that H3O+ activity coefficient in concentrated HCl is >10, so activity wise, the reaction (4) is pushed even more than the concentration suggests. Overall, the great answer, I have intentionally kept the level of my answer rather low. $\endgroup$ – Poutnik May 9 at 9:09

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