Problem 43 - Chemistry and Chemical Reactivity

Question

Balance the following equations, and then write the net ionic equation. (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\) (b) \(\mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{BaCO}_{3}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\) (d) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow \mathrm{Ni}\left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

(a) \(\mathrm{Cu}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})\) (b) \(\mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s}) + 2\mathrm{H}^+(\mathrm{aq}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{s}) + 2\mathrm{H}_{2}\mathrm{O}(\ell)\) (c) \(\mathrm{BaCO}_{3}(\mathrm{s}) + 2\mathrm{H}^+(\mathrm{aq}) \rightarrow \mathrm{Ba}^{2+}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g})\) (d) \(2\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}(\mathrm{aq}) + \mathrm{Ni}^{2+}(\mathrm{s}) + 2\mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow \mathrm{Ni}\left(\mathrm{CH}_{3}\mathrm{CO}_{2}\right)_{2}(\mathrm{aq})\)

Step 1: Balance Equation (a)

To balance equation (a), write down the initial chemical equation:\[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\]All elements are already balanced, with 2 ammonium ions and 2 nitrate ions on each side.

Step 2: Write Net Ionic Equation (a)

Identify and remove the spectator ions, which are ammonium \(\mathrm{NH}_{4}^+\) and nitrate \(\mathrm{NO}_{3}^-\). The net ionic equation becomes:\[\mathrm{Cu}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})\]

Step 3: Balance Equation (b)

For equation (b), balance the initial chemical equation:\[\mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\]To balance, use the coefficients: \(\mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell)\).

Step 4: Write Net Ionic Equation (b)

For equation (b), remove the spectator ions. Since it's a solid, reactions usually involve only the formation of the solid, while liquids and gases are often the net ionic equation. Simplified, the net ionic equation is:\[\mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s}) + 2\mathrm{H}^+(\mathrm{aq}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{s}) + 2\mathrm{H}_{2}\mathrm{O}(\ell)\]

Step 5: Balance Equation (c)

For equation (c), balance the following initial equation:\[\mathrm{BaCO}_{3}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\]Make sure that you have 2 \(\mathrm{HCl}\) for every 1 \(\mathrm{BaCl}_{2}\) produced: \(\mathrm{BaCO}_{3}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g})\).

Step 6: Write Net Ionic Equation (c)

Remove spectator ions, leaving the net ionic equation:\[\mathrm{BaCO}_{3}(\mathrm{s}) + 2\mathrm{H}^+(\mathrm{aq}) \rightarrow \mathrm{Ba}^{2+}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g})\]

Step 7: Balance Equation (d)

For equation (d), balance the following chemical equation:\[\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}(\mathrm{aq})+\mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow \mathrm{Ni}\left(\mathrm{CH}_{3}\mathrm{CO}_{2}\right)_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\]The equation is already balanced: 2 acetate ions are necessary to balance with one nickel ion.

Step 8: Write Net Ionic Equation (d)

The net ionic equation for acid-base reactions typically involves just the ions reshuffling:\[2\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}(\mathrm{aq}) + \mathrm{Ni}^{2+}(\mathrm{s}) + 2\mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow \mathrm{Ni}\left(\mathrm{CH}_{3}\mathrm{CO}_{2}\right)_{2}(\mathrm{aq})\]

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental concept in chemistry, ensuring that the amount of each element is conserved in a reaction. This process involves adjusting the coefficients of the reactants and products in a chemical equation so that the number of atoms of each element is the same on both sides of the equation. To balance a chemical equation, start by writing down the unbalanced equation and then adjust the coefficients one element at a time. Check each element while ensuring that the charge is balanced if dealing with ionic species. It's a systematic approach:

• List out all reactants and products.
• Count the number of atoms of each element in the reactants and products.
• Adjust coefficients to balance the number of atoms for each element.
• Repeat the process for each element until all elements are balanced.

Balance does not only refer to the number of atoms but also to the conservation of mass and charge in a chemical reaction.
This ensures the law of conservation of matter is followed, reflecting the principle that matter is neither created nor destroyed.

Net Ionic Equations

A net ionic equation is used to emphasize the chemical species that are directly involved in a chemical reaction. To write a net ionic equation, first write the balanced molecular equation, and then break all the aqueous ionic compounds into their constituent ions. Eliminate the spectator ions, which do not participate in the reaction, from both sides of the equation.
The steps are straightforward:

• Write the balanced molecular equation.
• Split all soluble strong electrolytes into their ions (dissociation).
• Identify the spectator ions – ions that appear unchanged on both the reactant and product sides.
• Remove the spectator ions to reveal the net ionic equation, leaving only the ions and compounds that undergo a chemical change.

This results in a simplified equation that illustrates the essence of the chemical reaction process.

Acid-Base Reactions

Acid-base reactions are a type of chemical reaction that occurs when an acid and a base interact with each other. In these reactions, hydrogen ions ( H^+ ) from the acid react with hydroxide ions ( OH^− ) from the base to form water ( H_2O ). Additionally, a salt is often produced from the anion of the acid and the cation of the base. Such reactions are also known as neutralization reactions. Key characteristics include:

• Transfer of H^+ ions from the acid to the base.
• Formation of water and a salt.
• The pH of the resulting solution becoming neutral.

The balance in acid-base reactions involves ensuring both the mass and charge are conserved, especially when dealing with ionic species in aqueous solutions.
Understanding these reactions is crucial in areas like biochemistry and environmental science, where maintaining the correct pH is essential.

Spectator Ions

Spectator ions are ions present in a solution that do not participate directly in a chemical reaction. They remain unchanged in both the reactant and product sides of a chemical equation. When writing net ionic equations, identifying and removing spectator ions is crucial. This helps to simplify the equation and highlight the ions involved in the change.
To determine spectator ions, follow these simple steps:

• Write out a complete ionic equation for the reaction.
• Identify ions that appear on both sides of the equation without change.
• Eliminate these ions to form the net ionic equation.

By focusing on the atoms and ions that are involved in the change, chemists can better understand the specifics of the chemical reaction process.
Removing spectator ions helps in simplifying the reaction and makes it easier to study interaction at the molecular level, especially in complex biochemical reactions.