Problem 19 - Chemistry and Chemical Reactivity

Question

Predict the products of each precipitation reaction. Balance the completed equation, and then write the net ionic equation. (a) \(\mathrm{NiCl}_{2}(\mathrm{aq})+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(\mathrm{aq}) \rightarrow\) (b) \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \rightarrow\)

Step-by-Step Solution

Answer

(a) NiS(s); (b) Mn₃(PO₄)₂(s).

Step 1: Predict products for equation (a)

When the compounds \( \mathrm{NiCl}_{2}\) and \( (\mathrm{NH}_{4})_{2}\mathrm{S} \) are mixed, the possible exchange of ions might produce \( \mathrm{NiS} \) and \( \mathrm{NH}_{4}\mathrm{Cl} \). \( \mathrm{NiS} \) is an insoluble compound which will precipitate out of the solution.

Step 2: Write balanced equation for (a)

The balanced chemical equation is: \[ \mathrm{NiCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{NiS}(\mathrm{s}) + 2 \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq}) \]

Step 3: Write net ionic equation for (a)

The net ionic equation is written by removing spectator ions (ions that appear on both sides of the equation). The net ionic equation is: \[ \mathrm{Ni}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{NiS}(\mathrm{s}) \]

Step 4: Predict products for equation (b)

For the reaction between \( \mathrm{Mn(NO}_{3})_{2} \) and \(\mathrm{Na}_{3}\mathrm{PO}_{4} \), the products are \( \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}\) and \(\text{NaNO}_{3}\). \( \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}\) is insoluble and will precipitate.

Step 5: Write balanced equation for (b)

Balance the chemical equation: \[ 3\mathrm{Mn(NO}_{3})_{2}(\mathrm{aq}) + 2\mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq}) \rightarrow \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}(\mathrm{s}) + 6\mathrm{NaNO}_{3}(\mathrm{aq}) \]

Step 6: Write net ionic equation for (b)

Remove the spectator ions to write the net ionic equation: \[ 3\mathrm{Mn}^{2+}(\mathrm{aq}) + 2\mathrm{PO}_{4}^{3-}(\mathrm{aq}) \rightarrow \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}(\mathrm{s}) \]

Key Concepts

Net Ionic Equation

A net ionic equation focuses on the components of a reaction that undergo a chemical change, leaving out the spectator ions. Spectator ions are those that do not participate in the formation of the precipitate or the reaction product. For example, when considering the reaction between \(\mathrm{NiCl}_{2}\) and \((\mathrm{NH}_{4})_{2}\mathrm{S}\), the full ionic equation includes all ions in the solution.
However, to write the net ionic equation, we only consider ions that form the insoluble compound, in this case, \(\mathrm{NiS}\).
To write a net ionic equation:

• Start by writing the balanced molecular equation.
• Identify the spectator ions, which appear unchanged on both sides of the equation.
• Remove the spectator ions to highlight the ions that form the precipitate or reaction product.

By focusing on the substances that undergo a change, net ionic equations provide a clearer picture of what occurs in a reaction.

Solubility Rules

Solubility rules are fundamental guidelines used to predict the solubility of ionic compounds in water. These rules help in identifying which compounds will form precipitates in a chemical reaction. By understanding these rules, you can determine whether a double displacement reaction will result in a solid, such as in the reactions provided in the exercise.
Some basic solubility rules include:

• Compounds containing alkali metal ions or ammonium \((\mathrm{NH}_4^+)\) are generally soluble.
• Nitrates \((\mathrm{NO}_3^-)\), acetates \((\mathrm{CH}_3\mathrm{COO}^-)\), and most perchlorates are soluble.
• Sulfates \((\mathrm{SO}_4^{2-})\) are generally soluble, except for those of barium, lead, and calcium.
• Chlorides \((\mathrm{Cl}^-)\), bromides, and iodides are soluble, with notable exceptions including silver, lead, and mercury compounds.
• Most carbonates \((\mathrm{CO}_3^{2-})\), phosphates \((\mathrm{PO}_4^{3-})\), sulfides, and hydroxides are insoluble, except for those of alkali metals and ammonium.

For example, in reaction (a), \(\mathrm{NiS}\) is identified as a precipitate using these rules, confirming that \(\mathrm{Ni}^{2+}\) and \(\mathrm{S}^{2-}\) ions will react.

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry. It ensures the law of conservation of mass is upheld, meaning atoms are neither created nor destroyed in a chemical reaction. Each side of the equation must have the same number of each type of atom.
To balance an equation, follow these steps:

• Write down the unbalanced equation with correct formulas for all reactants and products.
• Count the number of each type of atom on both sides of the equation.
• Add coefficients in front of formulas to equalize the number of atoms for each element on both sides. Start with the most complex compounds.
• Double-check your balancing by recounting the atoms of each element.

For instance, in reaction (b), initially, you have \(\mathrm{Mn(NO}_{3})_{2}\) and \(\mathrm{Na}_{3}\mathrm{PO}_{4}\) reacting to form \(\mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}\) and \(\mathrm{NaNO}_{3}\). To balance, you adjust coefficients to ensure you have:

• 3 manganese atoms and 2 phosphate ions on both sides.
• 6 sodium and nitrate ions to balance the ionic components equally.

Balancing equations ensures that the chemical formula accurately represents what happens in the reaction.