Problem 51 - Chemistry: An Atoms-Focused Approach

Question

For each of the following acid-base reactions, identify the acid and the base and then write the net ionic equation: a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{CaSO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) b. \(\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow\) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

Question: Identify the acids and bases for each reaction, and write the net ionic equation. a. $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{CaSO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ b. $\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$ c. $\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ Answer: a. Acid: $\mathrm{H}_{2} \mathrm{SO}_{4}$, Base: $\mathrm{Ca}(\mathrm{OH})_{2}$, Net ionic equation: $2\mathrm{H^+}(a q)+ 2\mathrm{OH^-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)$. b. Acid: $\mathrm{H}_{2} \mathrm{SO}_{4}$, Base: $\mathrm{PbCO}_{3}$, Net ionic equation: $2\mathrm{H^+}(a q)+\mathrm{CO}_{3}^{2-}(s)\rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$. c. Acid: $\mathrm{CH}_{3} \mathrm{COOH}$, Base: $\mathrm{Ca}(\mathrm{OH})_{2}$, Net ionic equation: $2\mathrm{OH^-}(a q)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow 2\mathrm{CH}_{3} \mathrm{COO^-}(a q)+ 2 \mathrm{H}_{2} \mathrm{O}(\ell)$.

Step 1: a. Identifying the Acid and the Base

For the reaction: $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{CaSO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$, The acid is the substance that donates protons (H+ ions), while the base is the substance that accepts protons. In this case, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) donates two protons (\(2\mathrm{H}^+\)) and \(\mathrm{Ca(OH)_{2}}\) accepts two protons. Therefore, the acid is \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and the base is \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Step 2: a. Writing the Net Ionic Equation

To create the net ionic equation, we need to first write the full ionic equation, which includes all ions and molecular compounds. Then, we can eliminate any spectator ions (ions present on both sides of the equation) to get the net ionic equation. The full ionic equation is: $2\mathrm{H^+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Ca^{2+}}(a q)+2\mathrm{OH^-}(a q) \rightarrow \mathrm{Ca^{2+}}(a q)+\mathrm{SO}_{4}^{2-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$. Now, let's eliminate the spectator ions. The \(\mathrm{Ca^{2+}}\) and \(\mathrm{SO}_{4}^{2-}\) ions appear on both sides of the equation, so they are spectator ions. The net ionic equation is: $2\mathrm{H^+}(a q)+ 2\mathrm{OH^-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)$.

Step 3: b. Identifying the Acid and the Base

For the reaction: $\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$, The acid is \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and the base is \(\mathrm{PbCO}_{3}\), as they donate and accept protons respectively.

Step 4: b. Writing the Net Ionic Equation

Following the same process as in part a, the full ionic equation is: $2\mathrm{H^+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Pb^{2+}}(s)+\mathrm{CO}_{3}^{2-}(s) \rightarrow \mathrm{Pb^{2+}}(s)+\mathrm{SO}_{4}^{2-}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$. The net ionic equation is: \(2\mathrm{H^+}(a q)+\mathrm{CO}_{3}^{2-}(s)\rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\).

Step 5: c. Identifying the Acid and the Base

For the reaction: $\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$, The acid is \(\mathrm{CH}_{3} \mathrm{COOH}\), and the base is \(\mathrm{Ca}(\mathrm{OH})_{2}\), as they donate and accept protons respectively.

Step 6: c. Writing the Net Ionic Equation

Following the same process as in part a, the full ionic equation is: $\mathrm{Ca^{2+}}(a q)+2\mathrm{OH^-}(a q)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca^{2+}}(aq)+2\mathrm{CH}_{3} \mathrm{COO^-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$. The net ionic equation is: $2\mathrm{OH^-}(a q)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow 2\mathrm{CH}_{3} \mathrm{COO^-}(a q)+ 2 \mathrm{H}_{2} \mathrm{O}(\ell)$.

Key Concepts

Net Ionic Equation

In chemistry, the net ionic equation is a simplified version of a chemical equation that only shows the chemical species that are involved in a reaction, excluding spectator ions. These are ions that do not participate in the actual chemical change. By focusing on the active elements, a net ionic equation clearly highlights the essence of the chemical reaction.

To write a net ionic equation, follow these steps:

• Begin with the full ionic equation, which shows all ions and compounds as they exist in the solution.
• Identify and eliminate the spectator ions – these are ions that remain unchanged on both sides of the reaction.
• What remains is the net ionic equation, consisting only of ions and compounds that undergo a chemical change.

For instance, in the chemical reaction between sulfuric acid (\( \mathrm{H}_2 \mathrm{SO}_4 \)) and calcium hydroxide (\( \mathrm{Ca(OH)_2} \)), the net ionic equation becomes:\[2\mathrm{H^+}(a q) + 2\mathrm{OH^-}(a q) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell)\]
This highlights the fundamental acid-base reaction resulting in water.

Acid Identification

Acid identification is about recognizing substances that release protons, or \( \mathrm{H^+} \) ions, in a chemical reaction. Acids are key players in many reactions and understanding them is crucial for mastering acid-base chemistry.

Characteristics of acids include:

• They are proton donors, often releasing \( \mathrm{H^+} \) ions in solution.
• They have a pH of less than 7.
• Common examples include \( \mathrm{HCl} \), \( \mathrm{H_2SO_4} \), and \( \mathrm{CH_3COOH} \).

In the reactions given, \( \mathrm{H}_2 \mathrm{SO}_4 \) acts as an acid in two situations, donating protons to the respective bases.

Base Identification

Base identification involves recognizing substances that accept protons, or produce hydroxide ions (\( \mathrm{OH^-} \)) in solution. Bases are essential for neutralizing acids and play a vital role in various chemical processes.

Characteristics of bases include:

• They are typically proton acceptors.
• They often have a pH greater than 7.
• Common bases include \( \mathrm{NaOH} \), \( \mathrm{KOH} \), and \( \mathrm{Ca(OH)_2} \).

In the provided exercises, \( \mathrm{Ca(OH)_2} \) is identified as a base in reactions involving both \( \mathrm{H}_2 \mathrm{SO}_4} \) and \( \mathrm{CH}_3 \mathrm{COOH} \), accepting protons to form water and yield the resulting products.