Problem 53 - Chemistry Matter and Change

Question

Write complete ionic and net ionic equations for the reaction between sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) and calcium carbonate \(\left(\mathrm{Ca} \mathrm{CO}_{3}\right) .\) $$\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{CaSO}_{4}(\mathrm{aq})$$

Step-by-Step Solution

Answer

The net ionic equation is: \[ 2\text{H}^+(\text{aq}) + \text{CaCO}_3(\text{s}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) + \text{Ca}^{2+}(\text{aq}) \]

Step 1: Write the Balanced Molecular Equation

The reactants are sulfuric acid (\(\text{H}_2\text{SO}_4\)) and calcium carbonate (\(\text{CaCO}_3\)). The products formed are water (\(\text{H}_2\text{O}\)), carbon dioxide (\(\text{CO}_2\)), and calcium sulfate (\(\text{CaSO}_4\)). The balanced molecular equation for the reaction is:\[ \text{H}_2\text{SO}_4(\text{aq}) + \text{CaCO}_3(\text{s}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) + \text{CaSO}_4(\text{aq}) \] Each side of the equation has an equal number of each type of atom, which confirms it is balanced.

Step 2: Identify the Strong Electrolytes

In this equation, \(\text{H}_2\text{SO}_4\) (sulfuric acid) is a strong acid and dissociates completely in water into \(2\text{H}^+(\text{aq})\) and \(\text{SO}_4^{2-}(\text{aq})\). Calcium sulfate \(\text{CaSO}_4\) is soluble and dissociates into \(\text{Ca}^{2+}(\text{aq})\) and \(\text{SO}_4^{2-}(\text{aq})\) under certain conditions, but typically has low solubility in water.

Step 3: Write the Complete Ionic Equation

The complete ionic equation shows all of the ions present in a reaction. First, dissociate all strong electrolytes:\[ 2\text{H}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) + \text{CaCO}_3(\text{s}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) + \text{CaSO}_4(\text{aq}) \]Note that \(\text{CaCO}_3\) is a solid and does not dissociate, and water and carbon dioxide are not ions.

Step 4: Identify Spectator Ions

Spectator ions are ions that do not participate in the reaction and remain unchanged. In this equation, the \(\text{SO}_4^{2-}(\text{aq})\) appears on both the reactant and product sides, hence it is a spectator ion.

Step 5: Write the Net Ionic Equation

To write the net ionic equation, remove the spectator ions from the complete ionic equation:\[ 2\text{H}^+(\text{aq}) + \text{CaCO}_3(\text{s}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) + \text{Ca}^{2+}(\text{aq}) \]This equation shows the actual chemical change occurring during the reaction.

Key Concepts

Sulfuric Acid Reaction

When sulfuric acid, \( \text{H}_2\text{SO}_4 \), reacts with another substance, it often releases ions into the solution.
Sulfuric acid is a strong acid, meaning it dissociates completely in water. This means it breaks apart into its ions readily.

• The ions it forms are \( 2\text{H}^+ \) (hydrogen ions) and \( \text{SO}_4^{2-} \) (sulfate ions).
• This complete dissociation is an important reason why sulfuric acid is so reactive. In chemical reactions, it often donates its hydrogen ions.

In the reaction with calcium carbonate, sulfuric acid acts as an acid that donates hydrogen ions for the reaction to occur.
This donation of ions is crucial in breaking down the calcium carbonate structure.

Calcium Carbonate

Calcium carbonate, \( \text{CaCO}_3 \), is a common substance found in rocks, and it's the main component of shells of marine organisms.
It is characterized by its solid, somewhat gritty texture.

• In chemical reactions, calcium carbonate often acts as a source of carbonate ions, \( \text{CO}_3^{2-} \).
• However, in its solid form, it does not dissociate into individual ions.
• This property is why in ionic equations, calcium carbonate is written as a whole unit \( \text{CaCO}_3(\text{s}) \).

When calcium carbonate reacts with acids, it often forms carbon dioxide, as seen in the reaction with sulfuric acid.
This reaction creates a fizzing effect as \( \text{CO}_2 \) gas is released, which is a visible sign of the chemical reaction occurring.

Complete Ionic Equation

To understand more fully what happens in a reaction, chemists write the complete ionic equation.
This type of equation reveals all the ions present during the reaction instead of just the compounds.For the reaction between sulfuric acid and calcium carbonate:

• First, break down the strong electrolytes and soluble substances into their ions.
• Write \( 2\text{H}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) + \text{CaCO}_3(\text{s}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) + \text{CaSO}_4(\text{aq}) \).
• Notice that water \( \text{H}_2\text{O} \) and carbon dioxide \( \text{CO}_2 \) remain as molecules and do not ionize.

This approach illuminates which particles are actively involved in the reaction and helps in identifying spectator ions.

Spectator Ions

In a chemical reaction, not all ions present participate in the creating of new products. Instead, some ions do not change or influence the reaction outcome; these are known as spectator ions.

• The sulfate ion, \( \text{SO}_4^{2-} \), is a spectator in the given reaction between sulfuric acid and calcium carbonate.
• Since it appears unchanged on both sides of the complete ionic equation, it is not actively participating in the formation of new substances.

Spectator ions are often removed when writing the net ionic equation to simplify and focus on the actual chemical changes occurring.
This helps chemists understand the essential parts of a reaction without unnecessary information.