Problem 16 - Chemistry and Chemical Reactivity

Question

Two aqueous solutions were prepared, one containing 0.10 mol of boric acid \(\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)\) in \(200 \mathrm{mL}\) and the second containing 0.10 mol phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) in \(200 \mathrm{mL}\). Both were weak conductors of electricity, but the \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution was a noticeably stronger conductor. Write equations to describe the equilibrium in each solution, and explain the observed difference in conductivity.

Step-by-Step Solution

Answer

\( \mathrm{H}_3\mathrm{PO}_4 \) dissociates more and releases more ions, increasing conductivity compared to \( \mathrm{H}_3\mathrm{BO}_3 \).

Step 1: Understand Weak Acid Dissociation

Both boric acid, \( \mathrm{H}_3\mathrm{BO}_3 \), and phosphoric acid, \( \mathrm{H}_3\mathrm{PO}_4 \), are weak acids. Weak acids partially dissociate in water, establishing an equilibrium between the undissociated acid and the ions produced. The degree of ionization determines the number of ions in the solution, affecting conductivity. Conductivity is higher when more ions are present.

Step 2: Write the Equilibrium Equation for Boric Acid

Boric acid is a very weak acid. In water, it can react as follows: \[ \mathrm{H}_3\mathrm{BO}_3(aq) + 2\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^+(aq) + \mathrm{B(OH)_4}^-(aq) \]This reaction shows boric acid acting as a Lewis acid, but it does not dissociate directly to release \( \mathrm{H}^+ \) ions like typical Brønsted-Lowry acids.

Step 3: Write the Equilibrium Equation for Phosphoric Acid

Phosphoric acid is a polyprotic acid and dissociates in steps. The first dissociation is the strongest and can be represented as:\[ \mathrm{H}_3\mathrm{PO}_4(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{H}_2\mathrm{PO}_4^-(aq) \]This step releases \( \mathrm{H}^+ \) ions, increasing the solution's ion concentration more effectively than boric acid, thus making it a better conductor.

Step 4: Compare Conductivity Based on Ionization

The stronger conductivity of the \( \mathrm{H}_3\mathrm{PO}_4 \) solution is due to its greater ionization. The equilibrium for the first dissociation of phosphoric acid produces \( \mathrm{H}^+ \) ions, directly contributing to electrical conductivity. In contrast, boric acid primarily forms complex ions and has a much lower degree of ionization.

Key Concepts

Boric Acid Equilibrium

Boric acid, chemically known as \( \mathrm{H}_3\mathrm{BO}_3 \), presents a unique behavior in aqueous solutions. Unlike typical Brønsted-Lowry acids, boric acid does not release protons \((\mathrm{H}^+)\) directly. Instead, it acts as a Lewis acid. This means it can accept hydroxide ions \((\mathrm{OH}^-)\) from water to form \( \mathrm{B(OH)_4}^- \) ions, subtly shifting the equilibrium. The reaction equation can be stated as:

• \( \mathrm{H}_3\mathrm{BO}_3(aq) + 2\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^+(aq) + \mathrm{B(OH)_4}^-(aq) \)

This equilibrium illustrates boric acid’s dissociation via interaction with water molecules instead of direct acid dissociation. This nuanced approach to forming ions leaves boric acid with a lower contribution to the overall ion concentration in solution, compared to more conventional weak acids. As a result, the conductivity of a boric acid solution remains relatively low.

Phosphoric Acid Ionization

Phosphoric acid, or \( \mathrm{H}_3\mathrm{PO}_4 \), is an example of a polyprotic acid. This means it can lose more than one proton \( \mathrm{H}^+ \). In fact, phosphoric acid dissociates in a stepwise manner, with each step involving the loss of one proton. The first step is usually the most significant for conductivity and is as follows:

• \( \mathrm{H}_3\mathrm{PO}_4(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{H}_2\mathrm{PO}_4^-(aq) \)

This initial dissociation of \( \mathrm{H}_3\mathrm{PO}_4 \) results in the production of \( \mathrm{H}^+ \) ions, which significantly contribute to the ionic strength of the solution. The subsequent dissociation steps, producing \( \mathrm{HPO}_4^{2-} \) and \( \mathrm{PO}_4^{3-} \), further enhance the ion concentration, though to a lesser extent. Consequently, the solution of phosphoric acid exhibits a noticeably higher conductivity compared to boric acid due to the release of more free protons into the solution.

Conductivity in Solutions

Conductivity in solutions is fundamentally dependent on the presence and mobility of ions. When acids dissociate in water, they release ions that facilitate the flow of electric current. The more ions that are present, the greater the conductivity of the solution.

• For boric acid, the production of ions is minimal because it does not readily release \( \mathrm{H}^+ \) ions.
• In contrast, phosphoric acid releases \( \mathrm{H}^+ \) ions more readily and in greater quantity, making the solution a stronger conductor of electricity.

The difference in conductivity observed between boric acid and phosphoric acid solutions primarily stems from the degree of ionization. Phosphoric acid, being a polyprotic acid, significantly boosts ion concentration through its dissociation steps, thereby enhancing conductivity. This clear distinction underscores why some acids conduct electricity better than others, emphasizing the role of ion presence and mobility in solution conductivity.