Problem 74 - Chemistry and Chemical Reactivity

Question

For each reaction, write an overall, balanced equation and the net ionic equation. (a) the reaction of aqueous lead(II) nitrate and aqueous potassium hydroxide (b) the reaction of aqueous copper(II) nitrate and aqueous sodium carbonate

Step-by-Step Solution

Answer

(a) Molecular: \(\text{Pb}(\text{NO}_3)_2 + 2\text{KOH} \rightarrow \text{Pb}(\text{OH})_2 + 2\text{KNO}_3\); Net ionic: \(\text{Pb}^{2+} + 2\text{OH}^- \rightarrow \text{Pb}(\text{OH})_2\); (b) Molecular: \(\text{Cu}(\text{NO}_3)_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{CuCO}_3 + 2\text{NaNO}_3\); Net ionic: \(\text{Cu}^{2+} + \text{CO}_3^{2-} \rightarrow \text{CuCO}_3\).

Step 1: Write the Balanced Molecular Equation for Reaction (a)

The reaction between aqueous lead(II) nitrate \((\text{Pb}(\text{NO}_3)_2)\) and aqueous potassium hydroxide \((\text{KOH})\) can be represented as:\[\text{Pb}(\text{NO}_3)_2(aq) + 2\text{KOH}(aq) \rightarrow \text{Pb}(\text{OH})_2(s) + 2\text{KNO}_3(aq)\]This equation shows the formation of lead(II) hydroxide as a precipitate.

Step 2: Write the Net Ionic Equation for Reaction (a)

To write the net ionic equation, first identify the ions involved:\[ \text{Pb}^{2+}, \text{NO}_3^- \text{(from Pb(NO}_3\text{)}}_2), \text{K}^+, \text{OH}^- \text{(from KOH)} \]Remove spectator ions \((\text{K}^+\text{ and }\text{NO}_3^-)\) to get the net ionic equation:\[ \text{Pb}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Pb}(\text{OH})_2(s) \]

Step 3: Write the Balanced Molecular Equation for Reaction (b)

The reaction between aqueous copper(II) nitrate \((\text{Cu}(\text{NO}_3)_2)\) and aqueous sodium carbonate \((\text{Na}_2\text{CO}_3)\) can be written as:\[\text{Cu}(\text{NO}_3)_2(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow \text{CuCO}_3(s) + 2\text{NaNO}_3(aq)\]Here, copper(II) carbonate forms as the precipitate.

Step 4: Write the Net Ionic Equation for Reaction (b)

Identify the ions involved:\[ \text{Cu}^{2+}, \text{NO}_3^- \text{(from Cu(NO}_3\text{)}}_2), \text{Na}^+, \text{CO}_3^{2-} \text{(from Na}_2\text{CO}_3) \]Eliminate the spectator ions \((\text{Na}^+ \text{ and } \text{NO}_3^-\)) to obtain the net ionic equation:\[ \text{Cu}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{CuCO}_3(s) \]

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry that ensures the law of conservation of mass is followed. When a chemical reaction occurs, the number of atoms for each element must remain constant. That's where balancing comes in. The basic idea is to have the same number of each type of atom on both sides of the equation. This might sound complex, but it can be broken down into simple steps:

• First, write down the unbalanced equation with all reactants and products.
• Focus on one element at a time, adjusting coefficients (the numbers in front of molecules) to balance atoms of that element.
• Repeat the process for all elements involved in the reaction.
• Check and re-adjust if necessary, as changing one element could affect others.

In the exercise above, balancing was key to making sure that both reactions show the same number of each type of atom before and after the reaction. This is crucial in studying and understanding chemical reactions.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two solutions are mixed, leading to the formation of an insoluble solid called a precipitate. This occurs due to the creation of a new compound that is not soluble in the surrounding liquid. Precipitation reactions are common in chemical analysis and synthesis. For a reaction to be classified as a precipitation reaction, the mixture must produce a solid. Some common signs include:

• Cloudiness in the solution as the solid forms.
• Formation of sediment at the bottom of the container.
• A change in color as the reactants convert into a new compound.

In the given exercise, lead(II) hydroxide and copper(II) carbonate were formed as precipitates. Recognizing these insoluble compounds during reactions is vital for developing net ionic equations.

Molecular Equations

Molecular equations give a full picture of the reaction, showing all reactants and products involved. They are a valuable tool for chemists to convey an accurate representation of what occurs during a reaction. These equations display the complete chemical formulas for each compound involved rather than just the ions or molecules that participate directly in the reaction. The exercise involved writing molecular equations for reactions between compounds in aqueous solutions. Here, each compound is shown in its entirety rather than breaking it down into constituent ions. This provides a straightforward way to see full reactants and products before deducing the net ionic equation. Even though molecular equations can seem less informative about the ions in solution, they help visualize the overall process, which is invaluable when balancing these reactions.