Problem 28 - Chemistry and Chemical Reactivity

Question

Predict the products of each precipitation reaction. Balance the equation, and then write the net ionic equation. (a) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{KBr}(\mathrm{aq}) \rightarrow\) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{KF}(\mathrm{aq}) \rightarrow\) (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) \rightarrow\)

Step-by-Step Solution

Answer

(a) \( \text{Pb}^{2+} + 2\text{Br}^- \rightarrow \text{PbBr}_2(s) \) (b) \( \text{Ca}^{2+} + 2\text{F}^- \rightarrow \text{CaF}_2(s) \) (c) \( \text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{CaC}_2\text{O}_4(s) \)

Step 1: Write the Chemical Equations

Start by writing the unbalanced chemical equations for each of the given reactions:(a) \( \text{Pb(NO}_3\text{)}_2(aq) + \text{KBr}(aq) \rightarrow \text{PbBr}_2(s) + \text{KNO}_3(aq) \)(b) \( \text{Ca(NO}_3\text{)}_2(aq) + \text{KF}(aq) \rightarrow \text{CaF}_2(s) + \text{KNO}_3(aq) \)(c) \( \text{Ca(NO}_3\text{)}_2(aq) + \text{Na}_2\text{C}_2\text{O}_4(aq) \rightarrow \text{CaC}_2\text{O}_4(s) + \text{NaNO}_3(aq) \)

Step 2: Balance the Chemical Equations

Balance each equation by adjusting the coefficients to ensure the number of atoms for each element is equal on both sides:(a) \( \text{Pb(NO}_3\text{)}_2(aq) + 2\text{KBr}(aq) \rightarrow \text{PbBr}_2(s) + 2\text{KNO}_3(aq) \)(b) \( \text{Ca(NO}_3\text{)}_2(aq) + 2\text{KF}(aq) \rightarrow \text{CaF}_2(s) + 2\text{KNO}_3(aq) \)(c) \( \text{Ca(NO}_3\text{)}_2(aq) + \text{Na}_2\text{C}_2\text{O}_4(aq) \rightarrow \text{CaC}_2\text{O}_4(s) + 2\text{NaNO}_3(aq) \)

Step 3: Write the Complete Ionic Equations

Break down all soluble compounds into their respective ions. Insoluble compounds remain intact as solids:(a) \( \text{Pb}^{2+}(aq) + 2\text{NO}_3^-(aq) + 2\text{K}^+(aq) + 2\text{Br}^-(aq) \rightarrow \text{PbBr}_2(s) + 2\text{K}^+(aq) + 2\text{NO}_3^-(aq) \)(b) \( \text{Ca}^{2+}(aq) + 2\text{NO}_3^-(aq) + 2\text{K}^+(aq) + 2\text{F}^-(aq) \rightarrow \text{CaF}_2(s) + 2\text{K}^+(aq) + 2\text{NO}_3^-(aq) \)(c) \( \text{Ca}^{2+}(aq) + 2\text{NO}_3^-(aq) + 2\text{Na}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq) \rightarrow \text{CaC}_2\text{O}_4(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq) \)

Step 4: Identify and Cancel Spectator Ions

Spectator ions are ions that appear on both sides of the ionic equation and do not participate in the reaction. Cancel them out:(a) The spectator ions \(\text{K}^+\) and \(\text{NO}_3^-\) cancel out.(b) The spectator ions \(\text{K}^+\) and \(\text{NO}_3^-\) cancel out.(c) The spectator ions \(\text{Na}^+\) and \(\text{NO}_3^-\) cancel out.

Step 5: Write the Net Ionic Equations

After cancelling the spectator ions, write down only the species that participate in the reactions:(a) \( \text{Pb}^{2+}(aq) + 2\text{Br}^-(aq) \rightarrow \text{PbBr}_2(s) \)(b) \( \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \rightarrow \text{CaF}_2(s) \)(c) \( \text{Ca}^{2+}(aq) + \text{C}_2\text{O}_4^{2-}(aq) \rightarrow \text{CaC}_2\text{O}_4(s) \)

Key Concepts

Net Ionic Equation

A net ionic equation is a simplified version of a chemical equation that only shows the substances that actually participate in the reaction. When you perform a precipitation reaction, you often have many ions floating around in the solution that don't really do anything—these are called spectator ions. They remain unchanged and don’t participate in the formation of the precipitate. By eliminating them, you're left with the net ionic equation, which highlights only the reactive substances involved.

Here's how to get to the net ionic equation:

• Start with balancing the complete chemical equation.
• Separate all soluble aqueous substances into their ions, while leaving the solid precipitates intact.
• Identify and cancel out the spectator ions (those that appear on both sides of the equation and do not change)
• Write down only the ions that undergo a chemical change to form the precipitate.

By focusing only on the ions that interact, a net ionic equation provides more straightforward insight into the chemistry of the reaction.

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry, ensuring that the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed, thus, in any chemical reaction, the total mass of the substances before the reaction must be equal to the total mass of the substances after the reaction.

To balance a chemical equation, follow these steps:

• Write the unbalanced equation, listing all reactants and products.
• Count the number of atoms of each element in both the reactants and products.
• Add coefficients in front of the chemical formulas to ensure that the number of atoms for each element is the same on both sides.
• Start by balancing the atoms of elements that appear in only one reactant and one product first.
• Adjust coefficients as needed, avoiding fractions if possible, to obtain whole numbers throughout.

Balancing chemical equations may require some trial and error. However, once the equation is balanced, it provides a clear representation of the substances involved and how they transform during the reaction.

Chemical Solubility Rules

Solubility rules are guidelines that help predict whether a compound will dissolve in water (making it soluble) or form a solid precipitate (making it insoluble). These rules are crucial when you are dealing with reactions in aqueous solutions, especially when predicting the products of a precipitation reaction.

Some of the main solubility rules are:

• Compounds containing alkali metal ions (like Na⁺, K⁺) and ammonium ion (NH₄⁺) are generally soluble.
• Nitrates (NO₃⁻), acetates (CH₃COO⁻), and bicarbonates (HCO₃⁻) are typically soluble.
• Halides (like Cl⁻, Br⁻, and I⁻) are soluble, except when paired with silver (Ag⁺), lead (Pb²⁺), or mercury (Hg₂²⁺).
• Sulfates (SO₄²⁻) are mostly soluble, with exceptions like barium sulfate (BaSO₄) and lead sulfate (PbSO₄).
• Carbonates (CO₃²⁻), phosphates (PO₄³⁻), and hydroxides (OH⁻) are usually insoluble, except when with alkali metals or ammonium.

Understanding these solubility rules helps in predicting whether a particular reaction will produce a solid precipitate, an important insight when writing net ionic equations.