Question

Aqueous solutions of sodium sulfide and copper(II) nitrate are mixed. A precipitate of copper(II) sulfide forms at once. The solution that remains contains sodium nitrate. Write the molecular, ionic, and net ionic equations for this reaction.

Short answer

Molecular equation: \text{Na}_2\text{S}(aq) + \text{Cu}(\text{NO}_3)_2(aq) \rightarrow \text{CuS}(s) + 2\text{NaNO}_3(aq). Ionic equation: 2\text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Cu}^{2+}(aq) + 2\text{NO}_3^-(aq) \rightarrow \text{CuS}(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq). Net ionic equation: \text{Cu}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{CuS}(s).

Step-by-step solution

Write the Molecular Equation

First, identify the formulas of the reactants and the products. Reactants are sodium sulfide (\text{Na}_2\text{S}) and copper(II) nitrate (\text{Cu}(\text{NO}_3)_2). The products are copper(II) sulfide (\text{CuS}) and sodium nitrate (\text{NaNO}_3). Write the balanced molecular equation by ensuring the same number of each type of atom on both sides of the equation: \[ \text{Na}_2\text{S}(aq) + \text{Cu}(\text{NO}_3)_2(aq) \rightarrow \text{CuS}(s) + 2\text{NaNO}_3(aq) \]

Write the Ionic Equation

Split the soluble ionic compounds into their respective ions. Insoluble compounds, such as the precipitate copper(II) sulfide, remain intact: \[ 2\text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Cu}^{2+}(aq) + 2\text{NO}_3^-(aq) \rightarrow \text{CuS}(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq) \]

Write the Net Ionic Equation

Cancel out the spectator ions that appear on both sides of the ionic equation to get the net ionic equation. Spectator ions are the ions that do not participate in the actual reaction. In this case, the sodium and nitrate ions are spectators: \[ \text{Cu}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{CuS}(s) \]

Key concepts

Molecular Equation

When dealing with chemical reactions in aqueous solutions, a molecular equation represents the complete chemical formulas of reactants and products, without indicating their ionic nature. It's akin to providing an overview, offering insight into the substances involved in the reaction and their states, be it solid (s), liquid (l), gas (g), or aqueous (aq).

In the provided exercise, the molecular equation tells us that sodium sulfide and copper(II) nitrate react to form copper(II) sulfide and sodium nitrate. The art of balancing these equations is fundamental to chemistry, as it obeys the law of conservation of mass, ensuring the number of atoms for each element is the same on both sides. For instance, the balanced molecular equation for the reaction between sodium sulfide and copper(II) nitrate shows a one-to-one stoichiometric ratio between the reactants yielding copper(II) sulfide, a solid precipitate, and aqueous sodium nitrate.

Ionic Equation

Moving deeper into the reaction's nature, ionic equations serve to demystify the actual forms of the aqueous reactants and products. Compounds in solution are expressed as separated ions. However, any compound that is not soluble and forms a precipitate, like copper(II) sulfide in this case, is written in its full formula.

Writing ionic equations necessitates a sound understanding of solubility rules to decide which components dissociate in water. In our example, it defines the behavior of sodium sulfide and copper(II) nitrate as they dissolve in water. This perspective is invaluable for visualizing reactions at an ionic level, aiding in the comprehension of reaction mechanisms, and predicting the movement of individual ions.

Net Ionic Equation

The highlight of the reaction's narrative is the net ionic equation. It's like a story stripped of all unnecessary characters, focusing only on those that change during the plot. This equation features the ions that participate directly in the reaction, which in our scenario are copper (II) ions and sulfide ions. They are the leading actors that come together to form copper(II) sulfide, the precipitate.

To arrive at this net equation, one must identify and eliminate the spectator ions that remain unchanged; they appear identically on both the reactant and product sides. By excluding these ions, the net ionic equation succinctly conveys the essence of the chemical change, offering a clear, uncluttered view of the reaction.

Precipitate Formation

Predicting precipitate formation is a crucial skill in chemistry, pivotal for understanding reactions like the one we're examining. A precipitate emerges when product ions in an aqueous solution react to form an insoluble compound, which is undissolved and separates as a solid from the rest of the mixture.

The anticipation of a precipitate's creation is rooted in solubility rules, guidelines determining the likelihood of a compound to be soluble or not in water. In our example, copper(II) sulfide forms as a solid precipitate demonstrating a tangible transformation. It's a climax of the reaction, an event that can be observed, collected, and studied to draw conclusions about the reaction's conditions and products.

Stoichiometry

The concept of stoichiometry lays the mathematical foundation of chemical reactions. It deals with the quantitative relationships—that is, how much of each substance is involved in the reaction. By employing stoichiometry, one can ascertain the amounts of reactants needed to produce a specific amount of product, or vice versa.

Stoichiometry relies on the balanced molecular equation to provide the molar ratios necessary to carry out these calculations. For a balanced reaction as presented, stoichiometry could predict how many grams of sodium sulfide are required to react fully with a given mass of copper(II) nitrate, or the amount of sodium nitrate produced. For any student aiming to excel in chemistry, mastering stoichiometry is as essential as understanding the alphabet for reading.