Take Ka = 1.8 × 10⁻⁵ mol/dm³. What is the pH of 0.10 mol/dm³ ethanoic acid?
Correct answer D. 2.87
Explanation
For a weak acid, [H⁺] = √(Ka × c) = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ mol/dm³. pH = −log(1.34 × 10⁻³) = 2.87. The value 4.74 is the pKa.
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