Problem 57 - Chemistry The Science in Context

Question

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

Step-by-Step Solution

Answer

a) $\mathrm{Ca^{+2}}(s) + 2\mathrm{OH^\text{-}}(s) \rightarrow \mathrm{Ca}^{+2}(aq) + 2[\mathrm{OH^-}(aq)]$ b) $\mathrm{Ca^{+2}}(s) + 2\mathrm{OH}^\text{-}(s)+2\text{CH}_3\text{COOH}(aq) \rightarrow \mathrm{Ca}^{+2}(aq) + 2[\mathrm{CH}_{3} \mathrm{COO}^\text{-}(aq)]+2\mathrm{H}_2 \text{O}(l)$ c) $2\text{CH}_3\text{COOH}(aq) \rightarrow 2[\mathrm{CH}_{3} \mathrm{COO}^\text{-}(aq)]+2\mathrm{H}_2 \text{O}(l)$ d) $2\mathrm{OH^\text{-}}(s) + 2\text{CH}_3\text{COOH}(aq) \rightarrow 2[\mathrm{CH}_{3} \mathrm{COO}^\text{-}(aq)]+2\mathrm{H}_2 \text{O}(l)$

Step 1: Acid and Base identification

The acid in this reaction is H2SO4, as it donates a proton, and the base is Ca(OH)2, as it accepts a proton.

Step 2: Balanced molecular equation

The balanced molecular equation is given as: $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$

Step 3: Overall ionic equation

Breaking down the compounds into their respective ions, the overall ionic equation is: \(2\mathrm{H^{+}}(a q) + \mathrm{SO}_{4}^{\text{-}2}(a q) + \mathrm{Ca^{\text{+}2}}(a q) +2\mathrm{OH^{-}}(a q)\rightarrow \mathrm{CaSO}_4(s) + 2\mathrm{H^+}(a q) + 2\mathrm{OH^-}(a q)\)

Step 4: Net ionic equation

By cancelling out the spectator ions, the net ionic equation is: \(\mathrm{Ca^{\text{+}2}}(a q) + \mathrm{SO}_{4}^{\text{-}2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)\) b. $\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$

Step 1: Acid and Base identification

The acid in this reaction is H2SO4, as it donates a proton, and the base is PbCO3, as it accepts a proton.

Step 2: Balanced molecular equation

The balanced molecular equation is given as: $\mathrm{PbCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$

Step 3: Overall ionic equation

Since PbCO3 and PbSO4 are insoluble, we won't break these compounds into ions. The overall ionic equation is: \(\mathrm{PbCO}_3(s) +2 \mathrm{H^+}(aq) +\mathrm{SO}_{4}^{\text{-}2}(aq)\rightarrow \mathrm{PbSO}_4(s)+\text{CO}_2(g)+\text{H}_2\text{O}(l)\)

Step 4: Net ionic equation

By cancelling out the spectator ions, the net ionic equation is: \(\mathrm{PbCO}_{3}(s)+ 2 \mathrm{H^{+}}(a q) + \mathrm{SO}_{4}^{\text{-}2}(a q) \rightarrow \mathrm{PbSO}_{4}(s) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(\ell)\) c. $\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(a q)$

Step 1: Acid and Base identification

The acid in this reaction is CH3COOH, as it donates a proton, and the base is Ca(OH)2, as it accepts a proton.

Step 2: Balanced molecular equation

The balanced molecular equation is given as: $\mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(a q)$

Step 3: Overall ionic equation

Breaking down the compounds into their respective ions, the overall ionic equation is: $\mathrm{Ca^{+2}}(s) + 2\mathrm{OH^\text{-}}(s) + 2\text{CH}_3\text{COOH}(aq) \rightarrow \mathrm{Ca}^{+2}(aq)+2[\mathrm{CH}_{3} \mathrm{COO}^\text{-}(aq)]+2\mathrm{H}_2 \text{O}(l)$

Step 4: Net ionic equation

By cancelling out the spectator ions, the net ionic equation is: $\mathrm{Ca^{+}2}(s) + 2\mathrm{OH}^\text{-}(s)+2\text{CH}_3\text{COOH}(aq) \rightarrow \mathrm{Ca}^{+2}(aq) + 2[\mathrm{CH}_{3} \mathrm{COO}^\text{-}(aq)]+2\mathrm{H}_2 \text{O}(l)$

Key Concepts

Net Ionic Equation

When it comes to understanding chemical reactions, the net ionic equation is crucial. It focuses solely on the components that undergo a chemical change. In simple terms, it involves removing the spectator ions. These are ions that do not participate directly in the reaction. For example, in the reaction between sulfuric acid and calcium hydroxide, the resulting net ionic equation concentrates on the ions that form calcium sulfate:\[\mathrm{Ca^{2+}}(a q) + \mathrm{SO_{4}^{2-}}(a q) \rightarrow \mathrm{CaSO_{4}}(s)\]Unlike the overall ionic equation, the net ionic equation is straightforward and precise. It removes elements that are not altered in the reaction. This representation simplifies analysis because it zooms in on what is actually changing at the atomic level. This is especially helpful in understanding the vital process of neutralization in acid-base reactions. Core ideas such as dynamic balancing and elimination of constant ions keep the equation stripped and emphasized for learners.

Acid and Base Identification

Identifying acids and bases in a reaction allows us to predict how substances will behave. Acids are proton donors, while bases are proton acceptors. This is a fundamental concept in the study of reactions. Take sulfuric acid \(\mathrm{H_{2}SO_{4}}\) and calcium hydroxide \(\mathrm{Ca(OH)_2}\). Here, sulfuric acid acts as the acid, donating protons, and calcium hydroxide acts as the base by accepting those protons.In practical terms, knowing how to identify these components is vital. It helps you understand the nature of the reaction, categorize it, and predict its products. For example:- **Sulfuric acid \(\mathrm{H_{2}SO_{4}}\)**: Strong acid, releases two protons.- **Calcium hydroxide \(\mathrm{Ca(OH)_2}\)**: Strong base, accepts two protons to neutralize the reaction.By grasping these fundamental roles, one can easily decode various chemical equations, assuring a clearer comprehension of material transformations in acid-base settings.

Overall Ionic Equation

The overall ionic equation provides a detailed view of how molecules disassociate into ions during acid-base reactions. It captures all species, both participating and spectator ions. For instance, when hydrogen ions from sulfuric acid dissociate, they react with hydroxide ions from calcium hydroxide. The equation captures every ionized component:\[2\mathrm{H^{+}}(a q) + \mathrm{SO}_{4}^{2-}(a q) + \mathrm{Ca^{2+}}(a q) + 2\mathrm{OH^{-}}(a q) \rightarrow \mathrm{CaSO}_4(s) + 2\mathrm{H_2O}(l)\]One of the core reasons to practice writing an overall ionic equation is to see exactly what happens when acids and bases interact in aqueous solutions. This broader perspective is essential for detailed chemical analysis. It shows all soluble entities splitting into ions, not just the reacting ones. By looking at this equation, you get a full story right from ionization to product formation.