Lesson eLearning

For the reaction phenol + NaOH ⇌ phenoxide- + H2O, calculate the equilibrium constant K using the given Ka of phenol = 1.3×10⁻¹⁰.

hard 3 marks Class 12 · Chemistry · Alcohols Phenols and Ethers
Answer
7.69×10⁻⁵

The correct answer is 7.69×10⁻⁵ — Thus the equilibrium constant K for phenol + OH- → phenoxide- + H2O is 7.69×10⁻⁵.

Step-by-step solution

  1. Recognize the reaction as the conjugate acid–base pair of phenol and phenoxide-.
  2. Use the relation Ka × Kb = Kw, where Kw = 1.0×10⁻¹⁴ at 298 K.
  3. Rearrange to find Kb (the equilibrium constant for phenoxide- + H2O ⇌ phenol + OH-): Kb = Kw / Ka.
  4. Insert the values: Kb = 1.0×10⁻¹⁴ / 1.3×10⁻¹⁰ = 7.69×10⁻⁵.
  5. Thus the equilibrium constant K for phenol + OH- → phenoxide- + H2O is 7.69×10⁻⁵.
Common mistake: Confusing Ka with Kb or using the wrong expression for the equilibrium constant

Concept tested: Acid–base equilibrium constants

Published · How this solution was prepared and checked

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