Problem 20 - Chemistry and Chemical Reactivity

Question

Predict the products of each precipitation reaction. Balance the completed equation, and then write the net ionic equation. (a) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{KBr}(\mathrm{aq}) \rightarrow\) (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) \( \text{Pb}^{2+} \), \( \text{Br}^- \) form \( \text{PbBr}_2 \). (b) \( \text{Ca}^{2+} \), \( \text{F}^- \) form \( \text{CaF}_2 \). (c) \( \text{Ca}^{2+} \), \( \text{C}_2\text{O}_4^{2-} \) form \( \text{CaC}_2\text{O}_4 \).

Step 1: Identify Possible Products

For each given reaction, identify the possible products by swapping the anions and cations. (a) For \( \text{Pb(NO}_3)_2 \) and \( \text{KBr} \), the products could be \( \text{PbBr}_2 \) and \( \text{KNO}_3 \).(b) For \( \text{Ca(NO}_3)_2 \) and \( \text{KF} \), the products could be \( \text{CaF}_2 \) and \( \text{KNO}_3 \).(c) For \( \text{Ca(NO}_3)_2 \) and \( \text{Na}_2 \text{C}_2 \text{O}_4 \), the products could be \( \text{CaC}_2 \text{O}_4 \) and \( \text{NaNO}_3 \).

Step 2: Determine Precipitates

Use solubility rules to determine which products will precipitate.(a) \( \text{PbBr}_2 \) is insoluble, so it will precipitate. \( \text{KNO}_3 \) is soluble.(b) \( \text{CaF}_2 \) is insoluble, so it will precipitate. \( \text{KNO}_3 \) is soluble.(c) \( \text{CaC}_2 \text{O}_4 \) is insoluble, so it will precipitate. \( \text{NaNO}_3 \) is soluble.

Step 3: Write Balanced Molecular Equations

Balance the equations accounting for all atoms and charges:(a) \[ \text{Pb(NO}_3)_2(aq) + 2 \text{KBr}(aq) \rightarrow \text{PbBr}_2(s) + 2\text{KNO}_3(aq) \](b) \[ \text{Ca(NO}_3)_2(aq) + 2 \text{KF}(aq) \rightarrow \text{CaF}_2(s) + 2\text{KNO}_3(aq) \](c) \[ \text{Ca(NO}_3)_2(aq) + \text{Na}_2\text{C}_2\text{O}_4(aq) \rightarrow \text{CaC}_2\text{O}_4(s) + 2\text{NaNO}_3(aq) \]

Step 4: Write Total Ionic Equations

Express each balanced equation in terms of ions, keeping insoluble compounds intact:(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 5: Write Net Ionic Equations

Eliminate spectator ions to find the net ionic equations, showing only the ions that form the precipitate:(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

Solubility Rules

Understanding solubility rules is crucial when predicting the outcome of precipitation reactions. These rules help us decide whether a substance will dissolve in water or form a precipitate. For example, most nitrate (NO₃⁻) salts are soluble. Consequently, compounds like KNO₃ and NaNO₃ generally remain in solution.

• *Halides* (Cl⁻, Br⁻, I⁻) are usually soluble, except with Ag⁺, Pb²⁺, and Hg₂²⁺.
• Most *sulfates* (SO₄²⁻) are soluble, with exceptions like BaSO₄, PbSO₄, and CaSO₄.
• *Carbonates* (CO₃²⁻), *phosphates* (PO₄³⁻), and *oxalates* (C₂O₄²⁻) are typically insoluble unless paired with sodium, potassium, or ammonium.

By using these rules, you can determine which of the potential products from a reaction will precipitate, like in our example where PbBr₂, CaF₂, and CaC₂O₄ are all insoluble.

Net Ionic Equation

The net ionic equation simplifies chemical reactions by focusing only on those compounds and ions that undergo change. Spectator ions, which remain unchanged, play no role in the formation of the product, and hence they are omitted. Focus in on only active participants like Pb²⁺ and Br⁻ ions, which form insoluble PbBr₂, rather than involving the entire array of compounds.
The approach involves:

• Writing the balanced molecular equation.
• Disassociating strong electrolytes into ions (ionic equation).
• Canceling out spectator ions to find meaningful interactions (net ionic equation).

For instance, in reaction (a), \[ \mathrm{Pb}^{2+}(aq) + 2\mathrm{Br}^-(aq) \rightarrow \mathrm{PbBr}_2(s) \],exclude ions like K⁺ and NO₃⁻ as they remain aqueous and do not precipitate.

Balancing Chemical Equations

Balancing chemical equations ensures the conservation of mass and charge, maintaining consistency with the Law of Conservation of Mass. Each element must have the same number of atoms on both sides of the reaction. For each of the given reactions:

• Identify every present element or ion.
• Count the atoms involved in reactants and products.
• Use coefficients to balance both sides.

Let's take reaction (a):\[ \mathrm{Pb(NO}_3)_2(aq) + 2 \mathrm{KBr}(aq) \rightarrow \mathrm{PbBr}_2(s) + 2\mathrm{KNO}_3(aq) \]The step ensures that for lead, bromine, potassium, and nitrate, the quantity of each atom in products and reactants matches. Always double-check your work for balanced charge, particularly in ionic reactions.

Chemical Reactivity

Chemical reactivity helps predict not just the products of a chemical reaction but also the likelihood of a reaction occurring. Different substances have different levels of reactivity based on their electron configurations. For example, the reactivity series can help determine an element's activity in replacement reactions.
In precipitation reactions, ionic compounds swap ions in mutual agreement to form precipitates - the less reactive, insoluble combinations. This reactivity reflects in combinations such as Ca²⁺ reacting with F⁻ ions eagerly to form CaF₂, even while KNO₃ forms and stays soluble.
Knowing which elements are more likely to combine helps in determining reaction products and assists in balancing equations, writing net ionic equations, and understanding solubility outcomes, making chemical reactivity a core concept in chemistry.