Question

Question 7.44 The ionization constant of phenol is 1.0 × 10^(–10). What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Class XI Equilibrium Page 228

Answer 1

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Answer 2

C₆H₅OH  + H₂O  ⇔ C₆H₅O⁻ + H₃O⁺

Ka = [C₆H₅O⁻][H₃O⁺]/[C₆H₅OH]
     according to rule of ostwald law,
  [C₆H₅O⁻] = [H³O⁺] = $\sqrt{K_a.C}$
where, C is the concentration of phenol solution
e.g., C = 0.01 M
now, [C₆H₅O⁻] = [H³O⁺] = √{10⁻¹⁰ × 0.05} = 2.24 × 10⁻⁶ M

now, when mixture contains 0.05M phenol and 0.01M phenolate ion then,
 Ka = [C₆H₅O⁻][H₃O⁺]/[C₆H₅OH]
   10⁻¹⁰    = 0.01M × [H₃O⁺]/0.05M
[H₃O⁺] = 0.05 × 10⁻¹⁰/0.01 = 5 × 10⁻¹⁰ M

now, we know, [H³O⁺] = Cα
  5 × 10⁻¹⁰ = 0.05 α
α = 5 × 10⁻¹⁰/0.05 = 10⁻⁸
 hence, degree of ionization = 10⁻⁸