A 0.115 M solution of a weak acid (HA) has a pH of 3.29. Calculate the acid ionization constant (Ka) for the acid.

Verified step by step guidance

Step 1: Understand the problem. We are given the concentration of a weak acid (HA) and the pH of the solution. We need to find the acid ionization constant (K_a).

Step 2: Convert the pH to the concentration of hydrogen ions [H^+]. Use the formula: [H^+] = 10^{-pH}.

Step 3: Write the expression for the ionization of the weak acid: HA \rightleftharpoons H^+ + A^-.

Step 4: Set up the expression for the acid ionization constant (K_a): K_a = \frac{[H^+][A^-]}{[HA]}.

Step 5: Assume that [H^+] = [A^-] and that the initial concentration of HA decreases by [H^+]. Substitute these values into the K_a expression and solve for K_a.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

pH and pKa

pH is a measure of the hydrogen ion concentration in a solution, calculated as pH = -log[H+]. For weak acids, the pKa is related to the strength of the acid, with lower pKa values indicating stronger acids. The relationship between pH and pKa is crucial for understanding acid dissociation and calculating the acid ionization constant (Ka).

Acid Ionization Constant (Ka)

The acid ionization constant (Ka) quantifies the strength of a weak acid in solution. It is defined by the equilibrium expression: Ka = [H+][A-]/[HA], where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid. A higher Ka value indicates a stronger acid.

Equilibrium Concentrations

In a weak acid solution, the equilibrium concentrations of the species involved must be determined to calculate Ka. For a weak acid HA dissociating into H+ and A-, the initial concentration, the change in concentration due to dissociation, and the equilibrium concentrations are essential for applying the Ka expression. Understanding how to set up these concentrations is key to solving the problem.