Question

Q-51 of M/32 solution of weak acid is 8 S cm2 mol-1 and limiting molar conductivity is 4
cm2 mol1. Ka for acid is
(a) 1.25 x 10-6
(b) 6.25x 104
(c) 1.25 x 104 (d) 1.25 x 10-5​

Answer 1

The equivalent conductance of M/32 solution of monobasic acid is 8.0 mho cm² and at infinite dilution is 400 mho cm²

To find : The dissociation constant of this acid

solution : degree of dissociation, α = equivalent conductance/limiting conductance

= 8/400 = 2/100 = 0.02

now using formula,

Ka = Cα²/(1 - α)

here 1 >> α ⇒1 - α ≈ 1

so Ka = Cα²

here C is concentration, i.e., C = 1/32 M

α = 0.02

so, Ka = (1/32) × (0.02)²

= 0.0004/32

= 0.0001/8

= 1.25 × 10^-5

Therefore the dissociation constant, Ka is 1.25 × 10^-5. i.e., option (d) is correct choice.

Answer 2

Answer:

a of m/32 solution of weak acid is 8 s cm2 mol-1 and limiting molar conductivity is 400 cm2 mol. ka for acid is