A 0.050 M solution of isobutyric acid ((CH3)2CHCOOH, Ka = 1.9×10⁻⁵) is prepared at 25 °C. Calculate the pH of the solution.

hard 5 marks Class 12 · Chemistry · Aldehydes Ketones and Carboxylic Acids

Answer

3.01

The correct answer is 3.01 — Step 6: Thus the pH of the solution is 3.01.

Step-by-step solution

Step 1: Set up the acid dissociation equilibrium: (CH3)2CHCOOH ⇌ (CH3)2CHCOO⁻ + H⁺.
Step 2: Use the expression Ka = [H⁺][(CH3)2CHCOO⁻]/[(CH3)2CHCOOH] and assume [H⁺] = x, with x ≪ C.
Step 3: Write the quadratic: Ka ≈ x²/(C – x) ≈ x²/C.
Step 4: Solve for x: x = √(Ka × C) = √(1.9×10⁻⁵ × 0.050) = √(9.5×10⁻⁷) = 9.75×10⁻⁴ M.
Step 5: Convert to pH: pH = –log₁₀(9.75×10⁻⁴) = 3.01.
Step 6: Thus the pH of the solution is 3.01.

Common mistake: Ignoring the approximation that C ≫ x or using an incorrect Ka value

Concept tested: pH calculation from acid dissociation constant