AIPMT PRELIMS 2009 Chemistry Conductance and Dissociation MCQ Question
The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mhos cm² and at infinite dilution is 400 mhos cm². The dissociation constant of this acid is:
1.25 × 10⁻⁶
6.25 × 10⁻⁴
1.25 × 10⁻⁴
1.25 × 10⁻⁵
Correct Answer
Detailed Explanation
The degree of dissociation (α) is calculated as Λ/Λ₀ = 8.0/400 = 0.02. The dissociation constant (Kₐ) is given by Kₐ = Cα²/(¹⁻α) ≈ Cα² for weak acids. Substituting C = 1/32 and α = 0.02, we find Kₐ = 1.25 × 10⁻⁵.
Found an issue with this question?
Related Questions
More from Electrochemistry
The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option. [Λ°_H⁺ =...
An electric charge of 5 Faradays is passed through three electrolytes AgNO₃, CuSO₄ and FeCl₃ solution. The grams of each metal liberated at cathode wi...
More from 2009
Assertion : Electric potential of earth is taken zero. Reason : No electric field exists on earth surface.
Assertion : Heat of neutralisation of nitric acid with NaOH is same to that of HCl and NaOH. Reason : In both cases strong acid and strong bases are n...
Intermediate host is absent in the infection of