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Gold will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. However, the metal does dissolve in a mixture of the acids (one part \(\mathrm{HNO}_{3}\) and three parts \(\mathrm{HCl}\) by volume \()\) called aqua regia. (a) Write a balanced equation for this reaction. (Hint: Among the products are \(\mathrm{HAuCl}_{4}\) and \(\mathrm{NO}_{2} .\) ) (b) What is the function of \(\mathrm{HCl} ?\)

Short Answer

Expert verified
The balanced chemical equation for this reaction is \(\mathrm{Au + 4 HNO}_{3} + \mathrm{4 HCl} \rightarrow \mathrm{HAuCl}_{4} + \mathrm{4 NO}_{2} + 2 \mathrm{H}_{2}O\). The function of \(\mathrm{HCl}\) in this reaction is to provide chloride (\(\mathrm{Cl}^{-}\)) ions, which are necessary for the formation of the \(\mathrm{HAuCl}_{4}\) complex when gold dissolves.

Step by step solution

01

Write the unbalanced chemical equation

Given that gold (Au) reacts with nitric acid (\(\mathrm{HNO}_{3}\)) and hydrochloric acid (\(\mathrm{HCl}\)) to produce tetrachloroaurate (\(\mathrm{HAuCl}_{4}\)) and nitrogen dioxide (\(\mathrm{NO}_{2}\)), the first step is to write the unbalanced equation. Thus, the equation is: \(\mathrm{Au + HNO}_{3} + \mathrm{HCl} \rightarrow \mathrm{HAuCl}_{4} + \mathrm{NO}_{2}\)
02

Balance the chemical equation

Next step is to balance this equation. Balancing chemical equations involving several atoms can be complicated, so make sure you have the correct amount of atoms on each side. The balanced equation is: \(\mathrm{Au + 4 HNO}_{3} + \mathrm{4 HCl} \rightarrow \mathrm{HAuCl}_{4} + \mathrm{4 NO}_{2} + 2 \mathrm{H}_{2}O\)
03

Explain the role of HCl

The last step is to explain the function of hydrochloric acid (\(\mathrm{HCl}\)). In this reaction, \(\mathrm{HCl}\) acts as a source of chloride (\(\mathrm{Cl}^{-}\)) ions. These chloride ions are required to form the \(\mathrm{HAuCl}_{4}\) complex when gold (Au) dissolves.

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Key Concepts

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

Gold Dissolution
Gold is known for its resistance to corrosion and tarnish, making it a stable and precious metal. However, there is an interesting chemical process that can dissolve gold, and it involves a special mixture known as aqua regia. Aqua regia, which means "royal water" in Latin, is a potent mixture of nitric acid (\(\text{HNO}_3\)) and hydrochloric acid (\(\text{HCl}\)), typically in a volumetric ratio of 1:3.
The process of dissolving gold in aqua regia results in the formation of chlorauric acid, also known as tetrachloroauric acid (\(\text{HAuCl}_4\)). This complex formation is crucial for the dissolution process. The reaction can be summed up in the following balanced chemical equation:

\[\text{Au} + 4\ \text{HNO}_3 + 4\ \text{HCl} \rightarrow \text{HAuCl}_4 + 4\ \text{NO}_2 + 2\ \text{H}_2\text{O}\]
  • Gold (\(\text{Au}\)) is transformed into the soluble \(\text{HAuCl}_4\) complex.
  • Nitrogen dioxide (\(\text{NO}_2\)) and water (\(\text{H}_2\text{O}\)) are also formed as by-products.
Understanding this reaction is key to learning how even the most chemically resistant metals can be altered under specific conditions.
Chemical Reaction Balancing
Balancing a chemical reaction is critical for accurately representing the transformation taking place. In the case of gold dissolution, we need to ensure that both sides of the equation have the same number of each type of atom. For our reaction

\[\text{Au} + \text{HNO}_{3} + \text{HCl} \rightarrow \text{HAuCl}_{4} + \text{NO}_{2}\],

one might begin by identifying the species involved: gold, nitric acid, and hydrochloric acid. Initially, the numbers of atoms likely won't match on both sides.

By incrementally adjusting the coefficients, we ensure the correct balance. The balanced reaction becomes:

\[\text{Au} + 4\ \text{HNO}_{3} + 4\ \text{HCl} \rightarrow \text{HAuCl}_{4} + 4\ \text{NO}_{2} + 2\ \text{H}_{2}\text{O}\]

At this point:
  • We have the same number of gold, nitrogen, chlorine, hydrogen, and oxygen atoms on each side.
  • Each type of atom is now balanced, making the equation accurate and meaningful.

This balanced equation allows us to understand the proportions and amounts of reactants and products involved in the reaction, which is essential in both theoretical and practical chemistry applications.
Role of Hydrochloric Acid
Hydrochloric acid (\(\text{HCl}\)) plays a vital role in the process of dissolving gold within aqua regia. Its primary function is to supply chloride ions (\(\text{Cl}^{-}\)), which are indispensable for forming the soluble \(\text{HAuCl}_4\) complex. This complex allows gold, which otherwise resists dissolution, to be converted into a form that can be stabilized in solution.

Here's how \(\text{HCl}\) contributes to the reaction:
  • The presence of chloride ions facilitates the conversion of elemental gold to the \(\text{Au}^{3+}\) cation, which subsequently bonds with the chloride ions to form the tetrachloroauric complex.
  • This reaction not only enables gold to dissolve but also alters its oxidation state, making it reactive.

Hydrochloric acid's role is thus crucial in breaking down the protective metallic structure of gold by providing the necessary ions, thereby making the dissolution process feasible in aqua regia.

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Most popular questions from this chapter

Galvanized iron" is steel sheet that has been coated with zinc; "tin" cans are made of steel sheet coated with tin. Discuss the functions of these coatings and the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface of a galvanized iron sheet or a tin can.

An aqueous KI solution to which a few drops of phenolphthalein have been added is electrolyzed using an apparatus like the one shown here: Describe what you would observe at the anode and the cathode. (Hint: Molecular iodine is only slightly soluble in water, but in the presence of \(\mathrm{I}^{-}\) ions, it forms the brown color of \(\mathrm{I}_{3}^{-}\) ions. See Problem \(12.102 .\)

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}$$ (b) If a 1.00-g sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation: $$\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{O}_{2}+\mathrm{Mn}^{2+}$$ (a) Balance the above equation. (b) If \(36.44 \mathrm{~mL}\) of a \(0.01652 M \mathrm{KMnO}_{4}\) solution are required to completely oxidize \(25.00 \mathrm{~mL}\) of a \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate the molarity of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\). The measured potential difference between the rod and the \(\mathrm{SHE}\) is \(0.589 \mathrm{~V},\) the rod being positive. Calculate the solubility product constant for silver oxalate.

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