Question

Write balanced net ionic equations for the reaction between: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{NaHSO}_{3}(a q)\) (b) \(\mathrm{HNO}_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)\)

Short answer

(a) Net ionic equation: \[2\mathrm{H}^{+} + \mathrm{HSO}_{3}^{-} \rightarrow 2\mathrm{H}_{2}O + \mathrm{SO}_{2}\]. (b) Net ionic equation: \[\mathrm{H}^{+} + \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{H}_{2}O + \mathrm{CO}_{2}\].

Step-by-step solution

Write the Complete Ionic Equation for Reaction (a)

First, dissociate all strong electrolytes (strong acids, bases, and soluble salts) into their constituent ions: \[\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{H}^{+} + \mathrm{SO}_{4}^{2-}\], \[\mathrm{NaHSO}_{3} \rightarrow \mathrm{Na}^{+} + \mathrm{HSO}_{3}^{-}\]. Assuming complete dissociation, the complete ionic equation is: \[2\mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} + \mathrm{Na}^{+} + \mathrm{HSO}_{3}^{-} \rightarrow \mathrm{Na}^{+} + 2\mathrm{H}_{2}O + \mathrm{SO}_{2} + \mathrm{SO}_{4}^{2-}\].

Cancel Spectator Ions for Reaction (a)

Identify and remove the spectator ions that do not participate in the reaction (ions that appear on both sides of the equation): \[\mathrm{Na}^{+}\] and \[\mathrm{SO}_{4}^{2-}\] are spectator ions. Thus, the net ionic equation is: \[2\mathrm{H}^{+} + \mathrm{HSO}_{3}^{-} \rightarrow 2\mathrm{H}_{2}O + \mathrm{SO}_{2}\].

Write the Complete Ionic Equation for Reaction (b)

First, dissociate all strong electrolytes: \[\mathrm{HNO}_{3} \rightarrow \mathrm{H}^{+} + \mathrm{NO}_{3}^{-}\], \[\left(\mathrm{NH}_{4}\right)_{2}\mathrm{CO}_{3} \rightarrow 2\mathrm{NH}_{4}^{+} + \mathrm{CO}_{3}^{2-}\]. The complete ionic equation is: \[\mathrm{H}^{+} + \mathrm{NO}_{3}^{-} + 2\mathrm{NH}_{4}^{+} + \mathrm{CO}_{3}^{2-} \rightarrow 2\mathrm{H}_{2}O + \mathrm{CO}_{2} + 2\mathrm{NH}_{4}^{+} + \mathrm{NO}_{3}^{-}\].

Cancel Spectator Ions for Reaction (b)

Identify and remove the spectator ions: \[\mathrm{NO}_{3}^{-}\] and \[\mathrm{NH}_{4}^{+}\] are spectator ions in this reaction. Therefore, the net ionic equation for the reaction is: \[\mathrm{H}^{+} + \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{H}_{2}O + \mathrm{CO}_{2}\]. Note that water is produced from additional protons and hydroxide ions from the reaction, as carbonate acts as a base in this context.

Key concepts

Chemical Reactions

Understanding the intricacies of chemical reactions is crucial to mastering chemistry. In essence, a chemical reaction involves the transformation of one set of chemical substances into another through the breaking and forming of chemical bonds. The key to solving chemical problems often lies in writing precise balanced equations, which reflect the conservation of mass and charge.

Take the reactions provided in the exercise as examples. Reaction (a) involves sulfuric acid and sodium bisulfite, while reaction (b) features nitric acid reacting with ammonium carbonate. Both reactions proceed through a series of steps leading to the formation of new products, showcasing the dynamic nature of chemical processes. By carefully analyzing each reactant and product, and writing down their respective ions, chemists can further explore the nuances of these reactions by creating complete ionic and net ionic equations, reflecting what is truly changing during the reaction.

Dissociation of Electrolytes

The dissociation of electrolytes is a critical concept when delving into chemical reactions that occur in solution. Electrolytes, which can be strong or weak, are substances that dissolve in water to produce ions, thereby allowing the solution to conduct electricity. In the provided examples, the dissociation of strong acids like \( \mathrm{H}_{2}\mathrm{SO}_{4} \) and \( \mathrm{HNO}_{3} \) into their constituent ions is a hallmark of a strong electrolyte's behavior.

Understanding Complete Dissociation

In response to the high solvation power of water, strong electrolytes completely dissociate. This behavior contrasts with weak electrolytes, which only partially dissociate in solution. Recognizing the difference is essential, as it has direct implications on how the equations are written and on the resulting ionic concentrations in solution. More so, knowing the extent of dissociation helps in predicting the electrical conductivity of the solution - a fully dissociated electrolyte will result in a highly conductive solution.

Spectator Ions

Spectator ions may sound like unimportant actors in the grand scheme of a chemical reaction, but their role—or lack thereof—is quite significant in understanding reactions in a solution. These ions are present in the solution but do not directly participate in the chemical reaction; that is, they emerge unchanged on both sides of a complete ionic equation.

Why Identify Spectator Ions?

Identifying these ions is critical when simplifying complete ionic equations to net ionic equations, which only display the particles that undergo change during the reaction. In the solutions provided, \( \mathrm{Na}^{+} \) and \( \mathrm{SO}_{4}^{2-} \) for reaction (a), and \( \mathrm{NH}_{4}^{+} \) and \( \mathrm{NO}_{3}^{-} \) for reaction (b), were identified as spectator ions and thus omitted in the net ionic equation. This refinement from complete to net ionic equations is fundamental not only for the simplicity but also for enhancing the conceptual understanding of the reactive species in a chemical process.