Question

Write the equations for the chemical equilibria that exist in solutions of (a) \(\mathrm{Na}_{2} \mathrm{SO}_{3},\) (b) \(\mathrm{Na}_{3} \mathrm{PO}_{4},\) and \((\mathbf{c})\) \(\mathrm{K}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\)

Short answer

For \text{Na}_2\text{SO}_3: \[ \text{Na}_2\text{SO}_3 (s) \rightarrow 2\text{Na}^+ (aq) + \text{SO}_3^{2-} (aq) \] and \[ \text{SO}_3^{2-} (aq) \rightleftharpoons \text{HSO}_3^- (aq) + \text{OH}^- (aq) \. For \text{Na}_3\text{PO}_4: \[ \text{Na}_3\text{PO}_4 (s) \rightarrow 3\text{Na}^+ (aq) + \text{PO}_4^{3-} (aq) \] and \[ \text{PO}_4^{3-} (aq) \rightleftharpoons \text{HPO}_4^{2-} (aq) + \text{OH}^- (aq) \. For \text{K}_2\text{C}_4\text{H}_4\text{O}_6: \[ \text{K}_2\text{C}_4\text{H}_4\text{O}_6 (s) \rightarrow 2\text{K}^+ (aq) + \text{C}_4\text{H}_4\text{O}_6^{2-} (aq) \] and \[ \text{C}_4\text{H}_4\text{O}_6^{2-} (aq) \rightleftharpoons \text{C}_4\text{H}_5\text{O}_6^- (aq) + \text{OH}^- (aq) \].

Step-by-step solution

Write the Dissolution Reaction for \text{Na}_2\text{SO}_3

First, represent the dissolution of sodium sulfite in water. Sodium sulfite is a salt that dissociates completely into sodium ions (\text{Na}^+) and sulfite ions (\text{SO}_3^{2-}): \[ \text{Na}_2\text{SO}_3 (s) \rightarrow 2\text{Na}^+ (aq) + \text{SO}_3^{2-} (aq) \] After initial dissociation, the sulfite ion (\text{SO}_3^{2-}) can undergo hydrolysis to form bisulfite ion (\text{HSO}_3^-) and hydroxide ion (\text{OH}^-): \[ \text{SO}_3^{2-} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{HSO}_3^- (aq) + \text{OH}^- (aq) \]

Write the Dissolution Reaction for \text{Na}_3\text{PO}_4

Start by writing the dissolution of trisodium phosphate in water. Trisodium phosphate completely dissociates into sodium ions (\text{Na}^+) and phosphate ions (\text{PO}_4^{3-}): \[ \text{Na}_3\text{PO}_4 (s) \rightarrow 3\text{Na}^+ (aq) + \text{PO}_4^{3-} (aq) \] Phosphate ions can undergo a series of hydrolysis reactions to form dihydrogen phosphate (\text{H}_2\text{PO}_4^-), hydrogen phosphate (\text{HPO}_4^{2-}), and ultimately, phosphoric acid (\text{H}_3\text{PO}_4). However, we will represent only the first hydrolysis equilibrium for simplicity: \[ \text{PO}_4^{3-} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{HPO}_4^{2-} (aq) + \text{OH}^- (aq) \]

Write the Dissolution Reaction for \text{K}_2\text{C}_4\text{H}_4\text{O}_6

Begin with the dissociation of potassium hydrogen tartrate in water. Potassium hydrogen tartrate dissociates into potassium ions (\text{K}^+) and hydrogen tartrate ions (\text{C}_4\text{H}_4\text{O}_6^{2-}): \[ \text{K}_2\text{C}_4\text{H}_4\text{O}_6 (s) \rightarrow 2\text{K}^+ (aq) + \text{C}_4\text{H}_4\text{O}_6^{2-} (aq) \] The hydrogen tartrate ion can further dissociate into tartaric acid (\text{C}_4\text{H}_6\text{O}_6) and bitartrate ion (\text{C}_4\text{H}_5\text{O}_6^-). We will write the equilibrium involving the formation of tartaric acid and the bitartrate ion: \[ \text{C}_4\text{H}_4\text{O}_6^{2-} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{C}_4\text{H}_5\text{O}_6^- (aq) + \text{OH}^- (aq) \]

Key concepts

Dissolution Reaction

Understanding the dissolution reaction is key to comprehending how salts interact with water to produce ions. Let's take the example of sodium sulfite (\text{Na}_2\text{SO}_3).

When \text{Na}_2\text{SO}_3 is added to water, it breaks into sodium ions (\text{Na}^+) and sulfite ions (\text{SO}_3^{2-}). This process is known as dissolution and can be represented by the equation:
\[ \text{Na}_2\text{SO}_3 (s) \rightarrow 2\text{Na}^+ (aq) + \text{SO}_3^{2-} (aq) \]
In dissolution, the solid salt is transformed into its constituent ions, which are then dispersed throughout the solution. The physical state symbols (s) for solid and (aq) for aqueous indicate the different phases of the substances involved.

To facilitate this understanding, one might visualize salt crystals dissolving in water, breaking apart into the ions that were previously held together in a rigid crystal lattice. This process can change the properties of the solution, such as its conductivity, because ions are charged particles that can carry an electric current. Moreover, it is essential to note that dissolution is often an endothermic process, meaning it requires energy to overcome the forces holding the ions in the solid lattice.

Hydrolysis of Ions

Once in solution, certain ions can undergo hydrolysis, interacting with water to form different ions and altering the pH of the solution. The hydrolysis of the sulfite ion (\text{SO}_3^{2-}) is a perfect illustration of this concept.

The sulfite ion can react with water, forming a bisulfite ion (\text{HSO}_3^-) and a hydroxide ion (\text{OH}^-):\[ \text{SO}_3^{2-} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{HSO}_3^- (aq) + \text{OH}^- (aq) \]
This hydrolysis reaction shows dynamic equilibrium; it can proceed forwards and backwards. When focusing on the hydrolysis of ions, we study how the water molecule can act as an acid or a base, according to the Brønsted-Lowry theory. In the case of \text{SO}_3^{2-}, the water molecule donates a hydrogen ion to the sulfite ion, resulting in the formation of hydroxide ions, which are basic in nature. This particular reaction leads to an increase in the solution's pH.

Such hydrolysis reactions are a fundamental part of aqueous chemistry, influencing the behavior of ions in biological systems, environmental waters, and industrial processes. Students can further comprehend this concept by picturing how the presence of surplus \text{OH}^- ions from the reaction can give the solution basic characteristics.

Acid-Base Equilibria

The acid-base equilibria refer to the balance of acid and base species in a solution, which is crucial to predicting the pH and understanding the behavior of substances during chemical reactions. Take the hydrogen tartrate ion (\text{C}_4\text{H}_4\text{O}_6^{2-}) for example. In water, it can react to form tartaric acid and a bitartrate ion, as shown below:
\[ \text{C}_4\text{H}_4\text{O}_6^{2-} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{C}_4\text{H}_5\text{O}_6^- (aq) + \text{OH}^- (aq) \]
This reaction is a part of a series of equilibrium reactions that can occur when polyprotic acids (acids that can donate more than one proton) or their conjugate bases are dissolved in water. The dynamic nature of the equilibrium implies that the concentration of each species in the solution will affect the position of the equilibrium, hence altering the pH.

To illustrate further, if more hydrogen ions (\text{H}^+) are introduced into the system, the reaction would shift to the left to form more \text{C}_4\text{H}_4\text{O}_6^{2-}, reducing the pH as a result. Conversely, adding more hydroxide ions (\text{OH}^-) would shift the equilibrium to the right, increasing the pH. Understanding these acid-base equilibria is critical in various disciplines, including biochemistry and environmental science, where the control of pH is vital.