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Chapter 15: Problem 65

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{Hg} \mathrm{I}_{4}^{2-} .\) )

Short Answer

Expert verified
The two balanced equations to explain the observations are as follows: 1. Formation of the orange precipitate: \(Hg(NO_3)_2(aq) + 2KI(aq) \rightarrow HgI_2(s) + 2KNO_3(aq)\) 2. Dissolution of the orange precipitate: \(HgI_2(s) + 2I^-(aq) \rightarrow HgI_4^{2-}(aq)\)

Step by step solution

01

Write the balanced equation for the formation of the orange precipitate.

For the formation of the orange precipitate, the reaction involves the combination of aqueous KI (potassium iodide) with mercury(II) nitrate (Hg(NO鈧�)鈧�). Adding KI gradually to Hg(NO鈧�)鈧� will lead to the formation of the orange precipitate, which is mercury(II) iodide (HgI鈧�). Potassium nitrate (KNO鈧�) will also be formed as a byproduct. We can write the balanced chemical equation as follows: Hg(NO鈧�)鈧�(aq) + 2KI(aq) 鈫� HgI鈧�(s) + 2KNO鈧�(aq) The equation is balanced since we have the same number of each element on both sides.
02

Write the balanced equation for the dissolution of the orange precipitate.

With further addition of aqueous KI, the orange precipitate dissolves. This is due to the formation of a soluble complex ion, HgI鈧劼测伝, as per the hint given in the exercise. The balanced equation for this reaction can be written as follows: HgI鈧�(s) + 2I鈦�(aq) 鈫� HgI鈧劼测伝(aq) The equation is balanced because we have equal numbers of each element on both sides of the equation.

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Key Concepts

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

Chemical Equations
A chemical equation is a way to represent a chemical reaction using symbols and formulas. These equations show the reactants and products in a reaction and help communicate what's happening at the molecular level. Understanding how to write and balance chemical equations is fundamental in chemistry.

In the reaction described, we begin with potassium iodide (KI) and mercury(II) nitrate (Hg(NO鈧�)鈧�) as reactants. When these are combined, an orange precipitate, mercury(II) iodide (HgI鈧�), forms. Potassium nitrate (KNO鈧�) is also produced as a byproduct. The balanced equation for this initial reaction is:
  • Hg(NO鈧�)鈧�(aq) + 2KI(aq) 鈫� HgI鈧�(s) + 2KNO鈧�(aq)
This equation shows that two moles of iodide ions (I鈦�) react with one mole of mercury(II) ions (Hg虏鈦�) to form the precipitate of mercury(II) iodide and that potassium nitrate remains dissolved in the solution. Balancing the equations ensures the conservation of mass, meaning each type of atom is equal on both sides of the equation. This is crucial for accurate representations of chemical reactions.
Complex Ions
Complex ions are charged species consisting of a central metal atom surrounded by molecules or ions, referred to as ligands. The formation of complex ions can lead to interesting outcomes in chemical reactions, such as changes in solubility.

In the continuation of the exercise, adding more potassium iodide to the solution causes the initial orange precipitate to dissolve, forming a complex ion. This occurs because the mercury(II) iodide (HgI鈧�) reacts with additional iodide ions (I鈦�) to become a soluble complex ion:
  • HgI鈧�(s) + 2I鈦�(aq) 鈫� HgI鈧劼测伝(aq)
This new complex ion, HgI鈧劼测伝, stays dissolved in the solution, effectively making the previously formed solid precipitate disappear. Complex ion formation is often driven by ligand addition, enhancing the solubility of otherwise insoluble compounds. Understanding complex ions helps chemists predict the behavior of substances in various conditions, especially in reactions like precipitation and dissolution.
Mercury(II) Iodide
Mercury(II) iodide (HgI鈧�) is a well-known compound, mainly due to its vibrant color changes. In the initial step of the exercise, this compound is formed, resulting in a noticeable orange precipitate. Mercury(II) iodide is interesting not only for its color but also for its protective applications in various fields.

The initial formation of mercury(II) iodide can be described by the chemical equation:
  • Hg(NO鈧�)鈧�(aq) + 2KI(aq) 鈫� HgI鈧�(s) + 2KNO鈧�(aq)
In this reaction, mercury ions ( Hg虏鈦�) from mercury(II) nitrate combine with iodide ions ( I鈦�) from potassium iodide to form the orange solid.

Mercury(II) iodide is also an example of a substance that can participate in complex ion formation. When additional iodide ions are added, it forms the soluble complex ion HgI鈧劼测伝, demonstrating the dual nature of this compound. The adaptability in forming complexes and shifting between solid and dissolved states is essential in understanding its behavior in chemical processes.

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Most popular questions from this chapter

a. Calculate the molar solubility of AgBr in pure water. \(K_{\mathrm{sp}}\) for AgBr is \(5.0 \times 10^{-13}\). b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0\) \(M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{mL}\) of \(3.0 M \mathrm{NH}_{3} ?\) e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

The \(\mathrm{Hg}^{2+}\) ion forms complex ions with \(\mathrm{I}^{-}\) as follows: $$\begin{aligned} \mathrm{Hg}^{2+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}^{+}(a q) & & K_{1}=1.0 \times 10^{8} \\ \mathrm{HgI}^{+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{2}(a q) & & K_{2}=1.0 \times 10^{5} \\ \mathrm{HgI}_{2}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{3}^{-}(a q) & & K_{3}=1.0 \times 10^{9} \\ \mathrm{HgI}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q) & & K_{4}=1.0 \times 10^{8} \end{aligned}$$ A solution is prepared by dissolving 0.088 mole of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and 5.00 moles of NaI in enough water to make 1.0 L of solution. a. Calculate the equilibrium concentration of \(\left[\mathrm{HgI}_{4}^{2-}\right] .\) b. Calculate the equilibrium concentration of \(\left[\mathrm{I}^{-}\right] .\) c. Calculate the equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right]\).

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} M\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \quad K_{\mathrm{overall}}=?$$

For which salt in each of the following groups will the solubility depend on pH? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) c. \(\operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \operatorname{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

The solubility of copper(II) hydroxide in water can be increased by adding either the base \(\mathrm{NH}_{3}\) or the acid HNO \(_{3}\). Explain. Would added \(\mathrm{NH}_{3}\) or \(\mathrm{HNO}_{3}\) have the same effect on the solubility of silver acetate or silver chloride? Explain.
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