NTA Abhyas NEET2020ChemistryElectrochemistryPractice
Specific conductance of 0.1 M H A is 3.75 × 1 0 - 4 o h m - 1 c m - 1 . If λ ∞ of H A is 250 o h m - 1 c m 2 m o l - 1 , then dissociation constant K a of H A is
Options
- A1 × 1 0 - 5
- B2.25 × 1 0 - 4
- C2.25 × 1 0 - 5
- D2.25 × 1 0 - 13
Correct answer
C. 2.25 × 1 0 - 5
Step-by-step solution
λ m = 1000 x k m = 3.75 α = λ m λ m ∞ = 3 . 75 250 = 0 . 015 = 1.5 × 1 0 - 2 ∴ K a = C α 2 = 0 . 1 0 . 015 2 = 2.25 × 1 0 - 5