Question

At 27°C, the degree of dissociation of HA (weak acid) in 0.5 M of its solution is 1%. The concentrations of H\(_3\)O\(^+\), A\(^-\), and HA at equilibrium (in mol L\(^{-1}\)) are respectively:

• A. \( 0.005, 0.005, 0.495 \)
• B. \( 0.05, 0.05, 0.45 \)
• C. \( 0.01, 0.01, 0.49 \)
• D. \( 0.005, 0.495, 0.005 \)

Hint:

The degree of dissociation (\( \alpha \)) for weak acids helps determine ion concentrations at equilibrium.

Answer 1

The Correct Option is A

Step 1: Define Dissociation Equation For a weak acid HA dissociating in water: \[ HA \rightleftharpoons H_3O^+ + A^- \] 
Step 2: Compute Concentrations Given initial concentration: \[ [HA] = 0.5 \text{ M}, \quad \alpha = 1\% = 0.01 \] \[ \text{Dissociated amount} = 0.5 \times 0.01 = 0.005 \text{ M} \] \[ [H_3O^+] = [A^-] = 0.005 \text{ M}, \quad [HA] = 0.5 - 0.005 = 0.495 \text{ M} \]