JIPMER2015ChemistryElectrochemistry
The conductivity of 0.001028 ~mol ~L ⁻¹ acetic acid is 4.95 10⁻⁵ ~S ~cm ⁻¹ . Find out its dissociation constant if _ m for acetic acid is 390.5 ~S ~cm ⁻¹ ~mol ⁻¹ .
Options
- A2.18 10⁻⁵ ~mol ⁻¹ ~L ⁻¹
- B1.78 10⁻⁵ ~mol ~L ⁻¹
- C3.72 10⁻⁴ ~mol ~L ⁻¹
- D2.37 10⁻⁴ ~mol ~L ⁻¹
Correct answer
B. 1.78 10⁻⁵ ~mol ~L ⁻¹
Step-by-step solution
aligned _ m = K C & = 4.95 10⁻⁵ ~S ~cm ⁻¹ 0.001028 ~mol ~L ⁻¹ 1000 ~cm ^3 ~L & =48.15 ~S ~cm ^2 ~mol ⁻¹ & = _ m _ m ^ = 48.15 390.5 =0.1233 K & = C ^2 (1- ) = 0.001028 (0.1233)^2 (1-0.1233) & =1.78 10⁻⁵ ~mol ~L ⁻¹ aligned