Problem 101 - 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) Net ionic: \(\text{Pb}^{2+} + 2\text{OH}^{-} \rightarrow \text{Pb(OH)}_2\); (b) Net ionic: \(\text{Cu}^{2+} + \text{CO}_3^{2-} \rightarrow \text{CuCO}_3\).

Step 1: Write the Chemical Formulas

For reaction (a), lead(II) nitrate is represented as \(\text{Pb(NO}_3\text{)}_2\) and potassium hydroxide as \(\text{KOH}\). For reaction (b), copper(II) nitrate is \(\text{Cu(NO}_3\text{)}_2\) and sodium carbonate is \(\text{Na}_2\text{CO}_3\).

Step 2: Predict the Products

In reaction (a), lead(II) nitrate reacts with potassium hydroxide to form lead(II) hydroxide (\(\text{Pb(OH)}_2\)) and potassium nitrate (\(\text{KNO}_3\)). In reaction (b), copper(II) nitrate reacts with sodium carbonate to form copper(II) carbonate (\(\text{CuCO}_3\)) and sodium nitrate (\(\text{NaNO}_3\)).

Step 3: Write the Overall Balanced Equations

For reaction (a), the balanced equation is: \(\text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KOH} (aq) \rightarrow \text{Pb(OH)}_2 (s) + 2 \text{KNO}_3 (aq)\). For reaction (b), the balanced equation is: \(\text{Cu(NO}_3\text{)}_2 (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow \text{CuCO}_3 (s) + 2 \text{NaNO}_3 (aq)\).

Step 4: Write the Complete Ionic Equations

For reaction (a): \(\text{Pb}^{2+} (aq) + 2\text{NO}_3^{-} (aq) + 2\text{K}^{+} (aq) + 2\text{OH}^{-} (aq) \rightarrow \text{Pb(OH)}_2 (s) + 2\text{K}^{+} (aq) + 2\text{NO}_3^{-} (aq)\). For reaction (b): \(\text{Cu}^{2+} (aq) + 2\text{NO}_3^{-} (aq) + 2\text{Na}^{+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s) + 2\text{Na}^{+} (aq) + 2\text{NO}_3^{-} (aq)\).

Step 5: Write the Net Ionic Equations

For reaction (a), cancel out the spectator ions to get: \(\text{Pb}^{2+} (aq) + 2\text{OH}^{-} (aq) \rightarrow \text{Pb(OH)}_2 (s)\). For reaction (b), the net ionic equation is: \(\text{Cu}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s)\).

Key Concepts

Net Ionic Equations

A net ionic equation is a simplified form of a chemical equation that highlights the elements and compounds that are directly involved in the reaction. This means you will not see "spectator ions" in a net ionic equation. Spectator ions are ions that appear on both sides of the chemical equation and do not participate in the reaction. Understanding net ionic equations helps you focus on the actual chemical changes occurring.
To create a net ionic equation:

• First, write the balanced molecular equation for the reaction.
• Next, separate the soluble ionic compounds into their individual ions to form the complete ionic equation.
• Finally, identify the spectator ions and remove them, leaving only the ions and compounds that directly participate in the reaction. This will be your net ionic equation.

In the given solutions, the net ionic equation of reaction (a) showed lead ions and hydroxide ions forming lead(II) hydroxide. For reaction (b), copper ions and carbonate ions came together to produce copper(II) carbonate.

Chemical Equations

Chemical equations are a standardized way to represent a chemical reaction. They are crucial for understanding how reactants turn into products during reactions. Chemical equations include the reactants on the left-hand side and the products on the right-hand side, separated by an arrow indicating the direction of the reaction. They must accurately represent the actual chemical processes occurring, and it's important to use proper chemical symbols and formulas.

• Begin by identifying the reactants and products involved in the chemical reaction.
• Use the chemical formulas to write an initial unbalanced equation.
• Ensure that each molecule or compound is written with its correct chemical formula to avoid errors.

In our example, the chemical equation for reaction (a) before balancing features lead(II) nitrate and potassium hydroxide transforming into lead(II) hydroxide and potassium nitrate. Reaction (b) involves copper(II) nitrate and sodium carbonate yielding copper(II) carbonate and sodium nitrate.

Balancing Reactions

Balancing chemical equations is crucial because it ensures the conservation of mass. According to the law of conservation of mass, matter cannot be created or destroyed in a chemical reaction. As a result, the same number of each type of atom must be present on both sides of a chemical equation.
To balance a chemical equation, follow these general steps:

• Write the unbalanced equation with the correct formulas for all reactants and products.
• Count the number of atoms of each element on both sides of the equation.
• Add coefficients (whole numbers in front of the compounds) to balance each element. Start with the most complex molecules first.
• Repeat the process for each element until the equation is balanced.
• Finally, make sure that the coefficients are at their lowest possible values, simplifying any ratios if needed.

For the exercise in question, balancing involved adjusting coefficients to make sure each atom's count matched on both sides, resulting in equations where chemical transformations were represented accurately, such as in reaction (a) \[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KOH} (aq) \rightarrow \text{Pb(OH)}_2 (s) + 2 \text{KNO}_3 (aq) \] and reaction (b) \[ \text{Cu(NO}_3\text{)}_2 (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow \text{CuCO}_3 (s) + 2 \text{NaNO}_3 (aq) \].