Problem 88 - Chemistry

Question

Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(0.25-M\) aqueous solution of phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right) .\left(K_{\mathrm{a}_{1}}, K_{\mathrm{a}_{2}}\right.\), and \(K_{\mathrm{a}}\), for phosphoric acidare \(7.5 \times 10^{-3}, 6.25 \times 10^{-8},\) and \(4.8 \times 10^{-13},\) respectively.)

Step-by-Step Solution

Answer

The pH of the solution is approximately 1.36.

Step 1: Identify the Dominant Ionization

Phosphoric acid (H₃PO₄) has three ionization constants: \( K_{a_1} = 7.5 \times 10^{-3} \), \( K_{a_2} = 6.25 \times 10^{-8} \), and \( K_{a_3} = 4.8 \times 10^{-13} \). Given the order of magnitudes, the first ionization \( K_{a_1} \) will be the most significant for determining \( \text{pH} \), as it has the largest value. We will focus on this initial ionization in our calculations.

Step 2: Write the Ionization Equation

The first ionization of phosphoric acid can be represented as: \[ \text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \] This represents the dominant step affecting the \( \text{pH} \) of the solution.

Step 3: Set Up the Ionization Equilibrium Expression

For the first ionization, the equilibrium constant expression is: \[ K_{a_1} = \frac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} \] Initially, \([\text{H}_3\text{PO}_4] = 0.25 \text{ M}\) and \([\text{H}^+] = [\text{H}_2\text{PO}_4^-] = 0\).

Step 4: Approximations and Simplifications

Assume \( x \) is the concentration of \( \text{H}^+ \) produced: \[ K_{a_1} = \frac{x^2}{0.25 - x} \approx \frac{x^2}{0.25} \] Under the assumption \( x \ll 0.25 \).

Step 5: Solve for \( x \)

Substitute \( K_{a_1} \) and solve for \( x \), which represents \([\text{H}^+]\):\[ 7.5 \times 10^{-3} = \frac{x^2}{0.25} \]\[ x^2 = 7.5 \times 10^{-3} \times 0.25 \]\[ x^2 = 1.875 \times 10^{-3} \]\[ x = \sqrt{1.875 \times 10^{-3}} \approx 0.0433 \text{ M} \]

Step 6: Calculate the pH

The \( \text{pH} \) is calculated using the equation: \[ \text{pH} = -\log([\text{H}^+])\]\[ \text{pH} = -\log(0.0433) \approx 1.36 \]

Key Concepts

Ionization Constants

In acid-base chemistry, ionization constants are crucial for understanding how acids dissociate in water. They describe the strength of an acid based on how completely it can donate protons (H⁺ ions). An acid like phosphoric acid (H₃PO₄) has multiple ionization constants, called stepwise acidity constants, reflecting how it loses protons in stages.

For phosphoric acid, the ionization constants are denoted as \(K_{a_1}\), \(K_{a_2}\), and \(K_{a_3}\). Each of these refers to a specific proton being removed:

• \(K_{a_1} = 7.5 \times 10^{-3}\)
• \(K_{a_2} = 6.25 \times 10^{-8}\)
• \(K_{a_3} = 4.8 \times 10^{-13}\)

The first ionization constant \(K_{a_1}\) is the largest, indicating that the first dissociation step has the greatest extent in water and heavily influences the pH of the solution. This unique property defines phosphoric acid as a weak acid, despite the large number of acidic hydrogens.

Equilibrium Expression

In the context of ionization, the equilibrium expression represents the relationship between the concentrations of the ions in solution at equilibrium. By setting up an equilibrium expression, one can calculate important quantities like pH. For the first ionization of phosphoric acid, the equation is:

\[ \text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \]
The equilibrium expression arising from this reaction is:
\[ K_{a_1} = \frac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} \]
Initially, the concentration of H₃PO₄ is the starting concentration (0.25 M), and the concentrations of the ions (\[ ext{H}^+\] and \[ ext{H}_2\text{PO}_4^-\]) are considered negligible. For practical calculations, an assumption is often made that the change in concentration is small (usually denoted as \( x \)), especially when \( x \) is much smaller than the initial concentration. This simplifies the math yet allows accurate result predictions.

Acid-Base Chemistry

Within acid-base chemistry, understanding how acids like phosphoric acid behave in water is fundamental. Phosphoric acid, characterized by three protons that can be sequentially donated, is a classic example of a polyprotic acid. Each deprotonation step has its ionization constant, and evaluating them highlights the acid's nature.

The pH of a solution, which quantifies the acidity, can be determined through these ionization steps. As each acetate group loses a proton, the equilibrium shifts, governed by its respective constant. Most importantly, the largest ionization constant \(K_{a_1}\) predominantly determines the pH in a weak acid solution like phosphoric acid's.

• The solution's pH is a fundamental measure in chemistry, reflecting the concentration of hydrogen ions, showing how acidic or basic a substance is.
• The \[\text{pH} = -\log[\text{H}^+]\] plays a key role, providing an easily understood scale from 0 (very acidic) to 14 (very basic).

In this context, understanding the joint roles of ionization constants and equilibrium expressions facilitates accurate acid-base calculations, ensuring mastery of acid-base chemistry principles.