Problem 90 - Chemistry The Science in Context

Question

(a) Use the solubility rules to write the balanced net ionic equation for each of the following "molecular" reactions. If there is no net reaction, write "NR." (b) Which of these three reactions give clear visual evidence of the ion exchange process? 1\. \(\mathrm{BaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{BaCO}_{3}(s)+\mathrm{NaCl}(a q)\) 2\. \(\mathrm{NaCl}(a q)+\mathrm{KOH}(a q) \rightarrow \mathrm{NaOH}(a q)+\mathrm{KCl}(a q)\) 3\. \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{NaCl}(a q)\)

Step-by-Step Solution

Answer

Answer: Reactions 1 and 3 give clear visual evidence of the ion exchange process as they form solid precipitates.

Step 1: Write the total ionic equation of each reaction

To find the total ionic equation, we first write each compound as dissociated ions if it's soluble in water. 1. \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightarrow \mathrm{BaCO}_{3}(s)+2 \mathrm{Na}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)\) 2. \(\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{K}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)+\mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) 3. \(3 \mathrm{Na}^{+}(a q)+\mathrm{PO}_{4}^{3-}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2 }(s)+3 \mathrm{Na}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)\)

Step 2: Write the balanced net ionic equation of each reaction

Now, we need to cancel out the spectator ions from each total ionic equation. 1. \(\mathrm{Ba}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightarrow \mathrm{BaCO}_{3}(s)\) 2. No net reaction (NR) as all ions are cancelled out. 3. \(2 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\)

Step 3: Determine which reaction gives clear visual evidence

A reaction gives clear visual evidence of the ion exchange process if it results in a solid precipitate. In this case, both reactions 1 and 3 form solid precipitates (BaCO3 and Ca3(PO4)2). Therefore, reactions 1 and 3 give clear visual evidence of the ion exchange process.

Key Concepts

Solubility Rules

Solubility rules are essential for predicting whether a compound will dissolve in water or form a solid precipitate. These guidelines help us understand which substances dissociate into ions and which ones remain intact.

Here are some basic rules to remember:

• Compounds containing alkali metal ions (like Na⁺, K⁺) and ammonium ion (NH₄⁺) are usually soluble.
• Nitrates (NO₃^-) and acetates (CH₃COO^-) are generally soluble.
• Chlorides (Cl^-), bromides (Br^-), and iodides (I^-) are typically soluble, except when paired with silver (Ag⁺), lead (Pb²⁺), or mercury (Hg₂²⁺).
• Sulfates (SO₄^2-) are generally soluble except for calcium (Ca²⁺), barium (Ba²⁺), and lead (Pb²⁺).
• Carbonates (CO₃^2-), phosphates (PO₄^3-), and sulfides (S^2-) are usually insoluble unless combined with alkali metals or ammonium.

These rules are tools for predicting the products of chemical reactions and understanding solubility in solutions.

Spectator Ions

Spectator ions are ions that appear unchanged on both the reactant and product sides of a chemical reaction. They do not participate in the actual chemical change but are present to balance the charge.

In the context of our exercise, during the total ionic equations, we determine which ions are simply 'watching' the reaction unfold without creating new substances.

For example, in the reaction \( \text{BaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{BaCO}_3 + \text{NaCl} \), sodium (Na⁺) and chloride (Cl^-) ions are spectator ions. They appear in the same form in both reactants and products. After removing these spectators, we're left with the net ionic equation.

• Spectator ions help balance the equation, but they don’t show chemical activity in the net ionic equation.
• Understanding how to identify them simplifies the process of finding the true chemical changes within a reaction.

Recognizing spectator ions is crucial for writing concise and correct net ionic equations.

Precipitation Reaction

Precipitation reactions occur when two aqueous solutions combine to form a solid, known as a precipitate. This process is a type of double displacement reaction and can be easily observed because a solid forms from clear solutions.

In our exercise, reactions 1 and 3 demonstrate this concept clearly. Barium carbonate (BaCO₃) and calcium phosphate (Ca₃(PO₄)₂) are the solid precipitates. These solids form because the combination of ions meets the insolubility rules.

Steps to predicting precipitation reactions:

• Identify ions in each reactant and switch partners according to the double displacement rule.
• Use solubility rules to predict if any of the products will form a precipitate.
• Write the balanced net ionic equation, eliminating spectator ions to show the actual precipitate formation.

The formation of a precipitate provides visual evidence of a chemical reaction, showcasing the tangible results of ionic exchanges.