Problem 59 - Introductory Chemistry Atoms First

Question

When aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is added to an aqueous solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), a precipitate forms. (a) What is the precipitate? (b) Write a net ionic equation for the precipitation reaction. (c) Did adding \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) to the sulfuric acid neutralize the acid? Explain.

Step-by-Step Solution

Answer

(a) The precipitate formed is \(\mathrm{Pb}\mathrm{SO}_{4}\). (b) The net ionic equation for the precipitation reaction is \(\mathrm{Pb}^{2+}(a q) + \mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{Pb}\mathrm{SO}_{4}(s)\). (c) No, adding \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) to the sulfuric acid does not neutralize the acid.

Step 1: Identifying the possible products of the reaction

First, we must identify all possible products of the reaction between \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). To do this, we will use the double displacement (metathesis) principle that states: $$ \mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB} $$ In our case: $$ \mathrm{H}_{2}\mathrm{SO}_{4}(a q) + \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow \mathrm{H}_{2}\mathrm{NO}_{3}(a q) + \mathrm{Pb}\mathrm{SO}_{4}(s) $$

Step 2: Determining the precipitation product

Now that we have identified the possible reaction products \(\mathrm{H}_{2}\mathrm{NO}_{3}\) and \(\mathrm{Pb}\mathrm{SO}_{4}\), we must determine which of these forms a precipitate (an insoluble product). We can use the solubility rules to determine this: 1. Compounds containing nitrate ions are generally soluble in water. 2. Sulfates are generally soluble in water, except for lead (II) sulfate, barium sulfate, and calcium sulfate. From the solubility rules, we find that \(\mathrm{H}_{2}\mathrm{NO}_{3}\) is soluble, while \(\mathrm{Pb}\mathrm{SO}_{4}\) is insoluble. Therefore, \(\mathrm{Pb}\mathrm{SO}_{4}\) is the precipitate formed in the reaction. (a) The precipitate formed is \(\mathrm{Pb}\mathrm{SO}_{4}\).

Step 3: Writing the net ionic equation for the precipitation reaction

We will now write the net ionic equation for the precipitation reaction. First, we must write the complete ionic equation: $$ 2\mathrm{H}^{+}(a q) + \mathrm{SO}_{4}^{2-}(a q) + \mathrm{Pb}^{2+}(a q) + 2\mathrm{NO}_{3}^{-}(a q) \rightarrow 2\mathrm{H}^{+}(a q) + 2\mathrm{NO}_{3}^{-}(a q) + \mathrm{Pb}\mathrm{SO}_{4}(s) $$ Now we can cross out the spectator ions, which are the ones that do not participate in the reaction: $$ 2\mathrm{H}^{+}(a q) + \cancel{2\mathrm{NO}_{3}^{-}(a q)} \rightarrow 2\mathrm{H}^{+}(a q) + \cancel{2\mathrm{NO}_{3}^{-}(a q)} + \mathrm{Pb}\mathrm{SO}_{4}(s) $$ Finally, we can write the net ionic equation: (b) The net ionic equation for the precipitation reaction is: $$ \mathrm{Pb}^{2+}(a q) + \mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{Pb}\mathrm{SO}_{4}(s) $$

Step 4: Determine if adding \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) neutralizes the acid

To see if the addition of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) to the sulfuric acid neutralized the acid, we must analyze the reaction's net ionic equation. In the net ionic equation, we can see that the concentration of \(\mathrm{H}^{+}\) ions remains constant. This means that the strong acid \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is not involved in the reaction, and its acidic strength does not change. Therefore, adding \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) to the sulfuric acid does not neutralize the acid. (c) No, adding \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) to the sulfuric acid does not neutralize the acid.

Key Concepts

Double Displacement Reaction

A double displacement reaction, also known as a metathesis reaction, is a type of chemical reaction where the cations and anions of two different compounds exchange places, forming two new compounds. One of the hallmarks of this reaction is that it often produces a precipitate, a gas, or a weak electrolyte from the exchange of ions.

The general form can be represented as: \[AB + CD \rightarrow AD + CB\]

Solubility Rules

Understanding the solubility rules is crucial in predicting the outcome of chemical reactions, especially the formation of precipitates. These rules provide guidance on the likelihood that a compound will dissolve in water. Here are some simplified rules:

• Nitrates (\(NO_3^-\)) are usually soluble.
• Salts containing alkali metal ions are generally soluble.
• Chlorides (\(Cl^-\)), bromides (\(Br^-\)), and iodides (\(I^-\)) are soluble, except for those containing silver, lead, and mercury.
• Sulfates (\(SO_4^{2-}\)) are soluble, with exceptions including barium sulfate, calcium sulfate, and lead (II) sulfate, which are insoluble.

In the exercise, the insoluble product is \(PbSO_4\) because lead sulfate is one of the exceptions to the general rule that sulfates are soluble.

Net Ionic Equation

The net ionic equation provides a succinct representation of a chemical reaction that highlights the species directly involved in the reaction, eliminating the spectator ions. Spectator ions do not participate in the actual reaction; they are present in solution before and after the reaction.

The exercise demonstrates how to derive the net ionic equation from the complete ionic equation by crossing out the spectator ions. With the net ionic equation, students can focus on the essence of the reaction: \[\mathrm{Pb}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq) \rightarrow \mathrm{PbSO}_{4}(s)\] This equation tells us that lead (II) ions and sulfate ions combined to form the precipitate lead (II) sulfate.

Neutralization Reaction

In chemistry, a neutralization reaction is a specific type of double displacement reaction where an acid and a base react to form water and a salt. The classic example is: \[H^+(aq) + OH^-(aq) \rightarrow H_2O(l)\] In the context of the exercise, it's important to note that not every double displacement reaction is a neutralization. For instance, when \(Pb(NO_3)_2\) is added to \(H_2SO_4\), the acid is not neutralized since the protons (\(H^+\)) do not react with a base to form water. The produced lead (II) sulfate is just a salt, not a result of acid-base neutralization.