Question

For each of the following unbalanced molecular equations, write the corresponding balanced net ionic equation for the reaction. a. \(\mathrm{HCl}(a q)+\mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{AgCl}(s)+\mathrm{HNO}_{3}(a q)\) b. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\) \(\mathrm{NaCl}(a q)\) c. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow\) \(\mathrm{PbCl}_{2}(s)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) d. \(\operatorname{FeCl}_{3}(a q)+\operatorname{NaOH}(a q) \rightarrow \operatorname{Fe}(\mathrm{OH})_{3}(s)+\mathrm{NaCl}(a q)\)

Short answer

a. \(\mathrm{Cl^-}(a q) + \mathrm{Ag^+}(a q) \rightarrow \mathrm{AgCl}(s)\) b. \(6\mathrm{Ca^{2+}}(a q) + 2\mathrm{PO_4^{3-}}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) c. \(\mathrm{Pb^{2+}}(a q) + 2\mathrm{Cl^-}(a q) \rightarrow \mathrm{PbCl}_{2}(s)\) d. \(\mathrm{Fe^{3+}}(a q)+3\mathrm{OH^-}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)\)

Step-by-step solution

1. Balance the molecular equation.

The given equation is already balanced: \(\mathrm{HCl}(a q)+\mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{AgCl}(s)+\mathrm{HNO}_{3}(a q)\)

2. Write the total ionic equation.

Break all soluble compounds into their respective ions: \(\mathrm{H^+}(a q) + \mathrm{Cl^-}(a q) + \mathrm{Ag^+}(a q) + \mathrm{NO_3^-}(a q) \rightarrow \mathrm{AgCl}(s) + \mathrm{H^+}(a q) + \mathrm{NO_3^-}(a q)\)

3. Write the net ionic equation.

Eliminate the spectator ions (H⁺ and NO₃⁻) from both sides of the total ionic equation: \(\mathrm{Cl^-}(a q) + \mathrm{Ag^+}(a q) \rightarrow \mathrm{AgCl}(s)\) b. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{NaCl}(a q)\)

1. Balance the molecular equation.

Balance the given equation by adjusting coefficients: \(3\mathrm{CaCl}_{2}(a q)+2\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6\mathrm{NaCl}(a q)\)

2. Write the total ionic equation.

Break all soluble compounds into their respective ions: \(6\mathrm{Ca^{2+}}(a q) + 6\mathrm{Cl^-}(a q) + 6\mathrm{Na^+}(a q) + 2\mathrm{PO_4^{3-}}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + 6\mathrm{Na^+}(a q) + 6\mathrm{Cl^-}(a q)\)

3. Write the net ionic equation.

Eliminate the spectator ions (Na⁺ and Cl⁻) from both sides of the total ionic equation: \(6\mathrm{Ca^{2+}}(a q) + 2\mathrm{PO_4^{3-}}(a q) \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) c. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{PbCl}_{2}(s)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

1. Balance the molecular equation.

The given equation is already balanced: \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{PbCl}_{2}(s)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

2. Write the total ionic equation.

Break all soluble compounds into their respective ions: \(\mathrm{Pb^{2+}}(a q) + 2\mathrm{NO_3^-}(a q) + \mathrm{Ba^{2+}}(a q) + 2\mathrm{Cl^-}(a q) \rightarrow \mathrm{PbCl}_{2}(s) + \mathrm{Ba^{2+}}(a q) + 2\mathrm{NO_3^-}(a q)\)

3. Write the net ionic equation.

Eliminate the spectator ions (Ba²⁺ and NO₃⁻) from both sides of the total ionic equation: \(\mathrm{Pb^{2+}}(a q) + 2\mathrm{Cl^-}(a q) \rightarrow \mathrm{PbCl}_{2}(s)\) d. \(\operatorname{FeCl}_{3}(a q)+\operatorname{NaOH}(a q) \rightarrow \operatorname{Fe}(\mathrm{OH})_{3}(s)+\mathrm{NaCl}(a q)\)

1. Balance the molecular equation.

Balance the given equation by adjusting coefficients: \(\mathrm{FeCl}_{3}(a q)+3\mathrm{NaOH}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+3\mathrm{NaCl}(a q)\)

2. Write the total ionic equation.

Break all soluble compounds into their respective ions: \(\mathrm{Fe^{3+}}(a q) + 3\mathrm{Cl^-}(a q) + 3\mathrm{Na^+}(a q) + 3\mathrm{OH^-}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s) + 3\mathrm{Na^+}(a q) + 3\mathrm{Cl^-}(a q)\)

3. Write the net ionic equation.

Eliminate the spectator ions (Na⁺ and Cl⁻) from both sides of the total ionic equation: \(\mathrm{Fe^{3+}}(a q)+3\mathrm{OH^-}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)\)

Key concepts

Chemical Reactions

A chemical reaction is a process in which substances, known as reactants, transform into new substances called products. This transformation occurs as a result of the breaking and forming of chemical bonds. In the exercise, we encounter reactions between ionic compounds in aqueous solutions, which often result in the formation of a precipitate, an insoluble solid, or the production of a gas.

When writing net ionic equations, it's crucial to understand the driving force of these reactions. For example, when hydrochloric acid reacts with silver nitrate, a precipitate of silver chloride forms. This process illustrates a precipitation reaction, one of the many types of chemical reactions. Understanding each reaction's underlying principles can help students predict the products and write balanced net ionic equations effectively.

Net ionic equations particularly focus on the ions involved in the formation of the precipitate or gas, excluding the spectator ions that don't participate in the actual chemical change. By concentrating on the ions that result in a new product, learners gain a clearer understanding of the core chemistry taking place.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring that the number of atoms of each element is the same on both the reactant and product sides of an equation. This practice follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

Starting with the original equations from the exercise, the balancing act begins with the understanding that coefficients need to be adjusted to achieve this balance. For instance, in the reaction between calcium chloride and sodium phosphate, the calcium ions combine with phosphate ions to form calcium phosphate, a precipitate. The equation is balanced by ensuring that there are three calcium ions for every two phosphate ions, leading to a compound with a formula of \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\).

Practical techniques for balancing equations include starting with the most complex molecule, using integer coefficients, and balancing the types of atoms one at a time. The goal is to reach a state where each species' atom count is equivalent before and after the reaction occurs.

Solubility and Precipitation

Solubility and precipitation are two interconnected concepts essential to writing net ionic equations. Solubility refers to a substance's ability to dissolve in a solvent, and when it comes to ionic compounds in water, some dissolve to form aqueous ions (soluble), while others barely dissolve and may form a precipitate (insoluble).

The exercise demonstrates precipitation reactions, which occur when two solutions containing soluble salts are mixed, and an insoluble salt is formed as a precipitate. This process of precipitation is a result of the low solubility product of the salt in water.

Using solubility rules can help predict whether a salt will be soluble in water. For example, silver chloride (\(\mathrm{AgCl}\)) is typically insoluble, as shown in exercise (a), leading to its precipitation. Learning these rules aids students in predicting the products of reactions and writing the correct net ionic equations by focusing on the formation of the precipitate, not just the complete formulaic representation of the starting and finishing compounds.