Problem 47 - Introductory Chemistry Atoms First

Question

Write a net ionic equation for the precipitation reaction, if any, that occurs when aqueous solutions of the following ionic compounds are mixed: (a) \(\mathrm{Bi}\left(\mathrm{NO}_{3}\right)_{3}\) and \(\mathrm{NaOH}\) (b) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) and \(\mathrm{SrI}_{2}\) (c) \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)

Step-by-Step Solution

Answer

The net ionic equations for the precipitation reactions are: (a) \( \mathrm{Bi^{3+} (aq)} + 3\, \mathrm{OH^- (aq)} \rightarrow \mathrm{Bi(OH)_3 (s)} \) (b) \( \mathrm{SO_4^{2-} (aq)} + \mathrm{Sr^{2+} (aq)} \rightarrow \mathrm{SrSO_4 (s)} \) (c) \( \mathrm{Cu^{2+} (aq)} + \mathrm{PO_4^{3-} (aq)} \rightarrow \mathrm{Cu_3(PO_4)_2 (s)} \)

Step 1: (a) Write the complete ionic equation for the reaction of \(\mathrm{Bi}\left(\mathrm{NO}_{3}\right)_{3}\) and \(\mathrm{NaOH}\)

First, we'll write the complete ionic equation for the reaction between \(\mathrm{Bi(NO_3)_3}\) and \(\mathrm{NaOH}\): \( \mathrm{Bi(NO_3)_3 (aq)} + 3\, \mathrm{NaOH (aq)} \rightarrow \mathrm{Bi(OH)_3 (s)} + 3\, \mathrm{NaNO_3 (aq)} \) Now, we'll separate all the ions: \( \mathrm{Bi^{3+} (aq)} + 3\, \mathrm{NO_3^- (aq)} + 3\, \mathrm{Na^+ (aq)} + 3\, \mathrm{OH^- (aq)} \rightarrow \mathrm{Bi(OH)_3 (s)} + 3\, \mathrm{Na^+ (aq)} + 3\, \mathrm{NO_3^- (aq)} \)

Step 2: (a) Identify and cancel the spectator ions

The spectator ions in this equation are \(\mathrm{Na^+}\) and \(\mathrm{NO_3^-}\). They appear unchanged on both sides of the equation, so we can cancel them out: \( \mathrm{Bi^{3+} (aq)} + 3\, \mathrm{OH^- (aq)} \rightarrow \mathrm{Bi(OH)_3 (s)} \)

Step 3: (a) Write the net ionic equation

The net ionic equation for the precipitation reaction between \(\mathrm{Bi(NO_3)_3}\) and \(\mathrm{NaOH}\) is: \( \mathrm{Bi^{3+} (aq)} + 3\, \mathrm{OH^- (aq)} \rightarrow \mathrm{Bi(OH)_3 (s)} \)

Step 4: (b) Write the complete ionic equation for the reaction of \(\mathrm{K_2 SO_4}\) and \(\mathrm{SrI_2}\)

First, we'll write the complete ionic equation for the reaction between \(\mathrm{K_2 SO_4}\) and \(\mathrm{SrI_2}\): \( \mathrm{K_2 SO_4 (aq)} + \mathrm{SrI_2 (aq)} \rightarrow 2\, \mathrm{KI (aq)} + \mathrm{SrSO_4 (s)} \) Now, we'll separate all the ions: \( 2\, \mathrm{K^+ (aq)} + \mathrm{SO_4^{2-} (aq)} + \mathrm{Sr^{2+} (aq)} + 2\, \mathrm{I^- (aq)} \rightarrow 2\, \mathrm{K^+ (aq)} + 2\, \mathrm{I^- (aq)} + \mathrm{SrSO_4 (s)} \)

Step 5: (b) Identify and cancel the spectator ions

The spectator ions in this equation are \(\mathrm{K^+}\) and \(\mathrm{I^-}\). They appear unchanged on both sides of the equation, so we can cancel them out: \( \mathrm{SO_4^{2-} (aq)} + \mathrm{Sr^{2+} (aq)} \rightarrow \mathrm{SrSO_4 (s)} \)

Step 6: (b) Write the net ionic equation

The net ionic equation for the precipitation reaction between \(\mathrm{K_2 SO_4}\) and \(\mathrm{SrI_2}\) is: \( \mathrm{SO_4^{2-} (aq)} + \mathrm{Sr^{2+} (aq)} \rightarrow \mathrm{SrSO_4 (s)} \)

Step 7: (c) Write the complete ionic equation for the reaction of \(\mathrm{Cu(CH_3 COO)_2}\) and \(\mathrm{Na_3 PO_4}\)

First, we'll write the complete ionic equation for the reaction between \(\mathrm{Cu(CH_3 COO)_2}\) and \(\mathrm{Na_3 PO_4}\): \( \mathrm{Cu(CH_3 COO)_2 (aq)} + \mathrm{Na_3 PO_4 (aq)} \rightarrow \mathrm{Cu_3(PO_4)_2 (s)} + 6\, \mathrm{NaCH_3 COO (aq)} \) Now, we'll separate all the ions: \( \mathrm{Cu^{2+} (aq)} + 2\, \mathrm{CH_3COO^- (aq)} + 3\, \mathrm{Na^+ (aq)} + \mathrm{PO_4^{3-} (aq)} \rightarrow \mathrm{Cu_3(PO_4)_2 (s)} + 6\, \mathrm{Na^+ (aq)} + 6\, \mathrm{CH_3COO^- (aq)} \)

Step 8: (c) Identify and cancel the spectator ions

The spectator ions in this equation are \(\mathrm{Na^+}\) and \(\mathrm{CH_3COO^-}\). They appear unchanged on both sides of the equation, so we can cancel them out: \( \mathrm{Cu^{2+} (aq)} + \mathrm{PO_4^{3-} (aq)} \rightarrow \mathrm{Cu_3(PO_4)_2 (s)} \)

Step 9: (c) Write the net ionic equation

The net ionic equation for the precipitation reaction between \(\mathrm{Cu(CH_3 COO)_2}\) and \(\mathrm{Na_3 PO_4}\) is: \( \mathrm{Cu^{2+} (aq)} + \mathrm{PO_4^{3-} (aq)} \rightarrow \mathrm{Cu_3(PO_4)_2 (s)} \)

Key Concepts

Precipitation Reactions

A precipitation reaction is a type of chemical reaction where two soluble ionic compounds in aqueous solutions come together to form an insoluble product, known as a precipitate. This solid precipitate forms because the product of the reaction is not soluble in the aqueous solution, leading to its separation from the liquid. Precipitation reactions are often used to isolate certain compounds in a mixture or to demonstrate the chemical properties of different ions.

When solving a precipitation reaction in a chemical equation:

• Write the initial reactants as their respective ionic forms, considering solubility rules.
• Use solubility tables to identify potential precipitates, which are substances that appear as solids in the products.
• Balance the equation to ensure both mass and charge are conserved, presenting a clear picture of the chemical change occurring.

By mastering these steps, you can easily predict the outcome of chemical reactions that involve ionic compounds, and understand why certain reactions lead to the formation of solids.

Spectator Ions

Spectator ions are ions present in the reactants and products of a chemical reaction that do not take part in the reaction itself. They merely "watch" as the reaction occurs, hence their name, 'spectator'. These ions help in balancing the total charge in a solution, remaining unchanged and in the same state before and after the reaction.

In precipitation reactions, identifying spectator ions involves:

• Writing out the complete ionic equation to show all the ions present.
• Identifying ions that appear on both sides of the equation without undergoing a chemical change.
• Canceling these ions when writing the net ionic equation, as they don't affect the outcome of the reaction.

Spectator ions are crucial for understanding the chemistry of reactions, even though they don't directly contribute to the formation of the precipitate.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, showing the transformation of reactants into products. They serve as a shorthand notation that conveys both the substances involved and the proportions in which they interact.

Key components of a chemical equation include:

• Reactants and Products, which are shown on the left and right side of the equation respectively, separated by an arrow indicating the direction of the reaction.
• Coefficients in front of chemical formulas, denoting the number of molecules or moles involved in the reaction, important for ensuring the conservation of mass.
• States of matter indicated in parentheses (s for solid, l for liquid, g for gas, and aq for aqueous).

Well-balanced chemical equations are essential for predicting the behavior of reactions, and net ionic equations—simplified versions focused on the reacting species—are particularly useful for highlighting the precise changes in a reaction.