Question

Question 7.59 The ionization constant of propanoic acid is 1.32 × 10^(–5). Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Class XI Equilibrium Page 229

Answer 1

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Answer 2

         CH3CH2COOH + H20 ------>CH3CH2COO⁻  + H3O⁺;Ka = 1.32 × 10⁻⁵
at t = 0      0.05                                 0                          0
at eqlm   (0.05 - Cα)                         Cα                       Cα

we know, from Ostwald's dilution law,
          Ka = Cα² ⇒ α = √{Ka/C}
                                 = √{1.32 × 10⁻⁵/0.05} = 0.016248
[H30+] = Cα = 0.05×0.016248 = 8.124 × 10⁻⁴ M
       now use Arrhenius formula for pH
pH = -log[H30+] = -log(8.124× 10⁻⁴) = 4 - 0.9098 = 3.09

now, when the solution contains 0.01 M HCl
   Ka = [CH3CH2COO⁻][H3O⁺]/[CH3CH2COOH]
 1.32×10⁻⁵ = Cα × 0.01/(0.05 - Cα)
(0.05 - Cα) ≈ 0.05
1.32 × 10⁻⁵ = Cα × 0.01/0.05
Cα = 5 × 1.32 × 10⁻⁵ = 6.6 × 10⁻⁵ M

hence, degree of ionization (α) =  6.6 × 10⁻⁵/0.05 = 1.32 × 10⁻⁵