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Gold metal will not dissolve in either concentrated nitric acid on concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

Short Answer

Expert verified
The balanced equation for the dissolution of gold in aqua regia is: Au + 4HNO₃ + 4HCl → HAuCl₄ + 4NO

Step by step solution

01

Write an Unbalanced Equation

First, write an unbalanced equation for the reaction using the given information. The reactants are gold (Au), nitric acid (HNO₃), and hydrochloric acid (HCl). The products are \(\mathrm{AuCl}_{4}^{-}\) ion and nitrogen monoxide (NO). Au + HNO₃ + HCl → \(\mathrm{AuCl}_{4}^{-}\) + NO
02

Determine the Missing Product or Reactant

In the given reaction, there is a charge imbalance: the product side has a negative charge, while the reactant side has no charge. We need to find a species that can balance the charges. Since the chloride ions come from hydrochloric acid, we can add the counter ion, which is a hydrogen ion (H⁺), as a product: Au + HNO₃ + HCl → \(\mathrm{AuCl}_{4}^{-}\) + NO + H⁺
03

Balance the Atoms

Now, balance the atoms in the reaction, one element at a time: 1. Gold (Au) is already balanced. 2. To balance the nitrogen (N) atoms, place a coefficient of 4 in front of HNO₃, and a coefficient of 4 in front of the NO: Au + 4HNO₃ + HCl → \(\mathrm{AuCl}_{4}^{-}\) + 4NO + H⁺ 3. To balance the hydrogen (H) atoms, place a coefficient of 4 in front of HCl: Au + 4HNO₃ + 4HCl → \(\mathrm{AuCl}_{4}^{-}\) + 4NO + H⁺ 4. To balance the chlorine (Cl) atoms, place a coefficient of 4 in front of HCl and a coefficient of 1 in front of H⁺: Au + 4HNO₃ + 4HCl → \(\mathrm{AuCl}_{4}^{-}\) + 4NO + 4H⁺ 5. Finally, combine the 4H⁺ ions with the \(\mathrm{AuCl}_{4}^{-}\) ion to form tetrachloroauric acid (HAuCl₄): Au + 4HNO₃ + 4HCl → HAuCl₄ + 4NO
04

Write the Balanced Equation

The final balanced equation for the dissolution of gold in aqua regia is: Au + 4HNO₃ + 4HCl → HAuCl₄ + 4NO

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gold Dissolution
Gold is a precious metal, known for its resistance to corrosion and tarnish. Despite being chemically stable, it can dissolve when exposed to a specific mixture called aqua regia. Aqua regia is a combination of concentrated nitric acid and hydrochloric acid. This powerful mixture is one of the few materials that can dissolve gold.

The process occurs because aqua regia generates chlorine radicals and nitrosyl chloride, which help to break down the gold into ions. This results in the formation of \( \text{AuCl}_4^- \) ions. By converting the gold into a soluble complex, aqua regia effectively dissolves it, allowing for various industrial and scientific applications.
Balanced Chemical Equation
A balanced chemical equation ensures the same number of each type of atom is present on both sides of the equation. This equality reflects the law of conservation of mass, stating that matter cannot be created or destroyed during a chemical reaction.

In our reaction, we're starting with gold (Au), nitric acid (HNO₃), and hydrochloric acid (HCl). The products formed are tetrachloroauric acid (HAuCl₄) and nitrogen monoxide (NO). Balancing this equation involves adjusting the coefficients to ensure that the number of each type of atom and the total charge is equal for both reactants and products.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It's critical for determining the proportions needed to fully react the chemicals involved, ensuring that no excess reactants are left.

In our balanced equation, the stoichiometry is as follows:
  • 1 mole of gold (Au)
  • 4 moles of nitric acid (HNO₃)
  • 4 moles of hydrochloric acid (HCl)
These react to form:
  • 1 mole of tetrachloroauric acid (HAuCl₄)
  • 4 moles of nitrogen monoxide (NO)
Grasping stoichiometry helps to predict the amounts of substances consumed and produced, which is vital for practical applications.
Tetrachloroauric Acid
Tetrachloroauric acid, represented by the formula HAuCl₄, is the compound formed when gold is dissolved in aqua regia. This compound is a gold complex that is highly soluble and used in various chemical processes, such as gold refining and electroplating.

When forming HAuCl₄, the gold atoms are surrounded by four chloride ions, leading to a stable coordination complex. This is a key reason why aqua regia is able to dissolve gold, as it transforms it into a form that is easier to work with chemically. Understanding the properties and uses of tetrachloroauric acid can aid in comprehending how gold is processed industrially.

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Most popular questions from this chapter

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

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You want to "plate out" nickel metal from a nickel nitrate solution onto a piece of metal inserted into the solution. Should you use copper or zinc? Explain.

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{Cr} \mathrm{O}_{4}\). The \(\mathscr{C}^{\circ}\) value for the following halfreaction is \(+0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{Cr} \mathrm{O}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).

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