Question

Mercury ions \(\left(\mathrm{Hg}_{2}^{2+}\right)\) can be removed from solution by precipitation with \(\mathrm{Cl}^{-} .\) Suppose that a solution contains aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) . Write complete ionic and net ionic equations to show the reaction of aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) with aqueous sodium chloride to form solid \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) and aqueous sodium nitrate.

Short answer

Net ionic equation: \(\mathrm{Hg}_{2}^{2+}(aq) + 2\mathrm{Cl}^{-}(aq) \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\).

Step-by-step solution

Identify the reactants and their states

The reactants are aqueous \(\mathrm{Hg}_{2}(\mathrm{NO}_{3})_{2}\) and aqueous sodium chloride (NaCl). The products are solid mercury(I) chloride (\(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)) and aqueous sodium nitrate (NaNO3).

Write the balanced molecular equation

Combine the reactants to form the products. Balance the number of atoms of each element on both sides of the equation: \(\mathrm{Hg}_{2}(\mathrm{NO}_{3})_{2}(aq) + 2\mathrm{NaCl}(aq) \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) + 2\mathrm{NaNO}_{3}(aq)\).

Write the complete ionic equation

Split the soluble ionic compounds into their individual ions: \(\mathrm{Hg}_{2}^{2+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) + 2\mathrm{Na}^{+}(aq) + 2\mathrm{Cl}^{-}(aq) \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) + 2\mathrm{Na}^{+}(aq) + 2\mathrm{NO}_{3}^{-}(aq)\).

Write the net ionic equation

Cancel out the spectator ions, which are ions that remain unchanged on both sides of the equation: \(\mathrm{Hg}_{2}^{2+}(aq) + 2\mathrm{Cl}^{-}(aq) \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\). This is the net ionic equation for the reaction.

Key concepts

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds in aqueous solutions combine and form an insoluble compound, known as the precipitate. In the context of our exercise, when an aqueous solution of mercury(I) nitrate \( \mathrm{Hg}_{2}(\mathrm{NO}_{3})_{2} \) reacts with an aqueous solution of sodium chloride (NaCl), a solid precipitate of mercury(I) chloride (\(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)) forms.

In a precipitation reaction, the anions and cations of the two reactants exchange partners, leading to the formation of at least one new compound that is insoluble and precipitates out of the solution. The solubility rules help predict which compound will precipitate. In this case, chloride ions (\(\mathrm{Cl}^{-}\)) form a precipitate with mercury ions (\(\mathrm{Hg}_{2}^{2+}\)), while the sodium (\(\mathrm{Na}^{+}\)) and nitrate (\(\mathrm{NO}_{3}^{-}\)) ions remain soluble and in solution as spectator ions.

Ionic Compounds in Aqueous Solutions

Ionic compounds consist of positively and negatively charged ions held together by ionic bonds. In aqueous solutions, these compounds dissociate into their constitutive ions. The dissociation of ionic compounds is crucial for understanding the reactions that occur in solutions, especially those leading to precipitates.

For instance, \( \mathrm{Hg}_{2}(\mathrm{NO}_{3})_{2} \) dissociates into \( \mathrm{Hg}_{2}^{2+} \) and \( \mathrm{NO}_{3}^{-} \) ions, and NaCl dissociates into \( \mathrm{Na}^{+} \) and \( \mathrm{Cl}^{-} \) ions in solution. This dissociation allows the ions to move freely in the solution and react to form the precipitate, mercury(I) chloride (\(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)), which is insoluble and separates from the aqueous phase.

Balancing Chemical Equations

Balancing chemical equations involves adjusting the coefficients of reactants and products to ensure the same number of each type of atom on both sides of the equation, thereby obeying the law of conservation of mass.

For the provided reaction between mercury(I) nitrate and sodium chloride, the balanced molecular equation is: \( \mathrm{Hg}_{2}(\mathrm{NO}_{3})_{2}(aq) + 2\mathrm{NaCl}(aq) \rightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) + 2\mathrm{NaNO}_{3}(aq) \). Balancing the equation confirms that there are two chloride ions for every mercury(I) ion, which reflects the correct stoichiometry for the reaction. This step is crucial before proceeding to write the complete ionic and net ionic equations, as it ensures that the reactions are represented accurately and the stoichiometry is respected.

When writing the complete ionic equation, each soluble reactant and product is expressed as its ions, while the precipitate remains a solid. In the net ionic equation, only the ions that participate in forming the precipitate are shown; spectator ions are omitted because they do not change during the reaction.