Question

The equivalent conductance of M/32 solution of a weak
monobasic acid is 8.0 mho cm² and at infinite dilution is 400
mho cm² . The dissociation constant of this acid is:
(a) 1.25 × 10⁻⁶ (b) 6.25 × 10⁻⁴
(c) 1.25 × 10⁻⁴ (d) 1.25 × 10⁻⁵

Answer 1

Explanation:

dissociation constant of this acid is:

(a) 1.25 × 10⁻⁶ is answer your question

Answer 2

answer : option (d) 1.25 × 10^-5

given, equivalent conductance of M/32 solution of a weak monobasic acid is, Λ = 8 mho cm²

at infinite dilution, Λ_∞ = 400 mho cm²

we know, α = Λ/Λ_∞ = 8/400 = 1/50 = 0.02

let HA is monobasic acid,

so, HA ⇔H^+ + A^-

C(1 - α). Cα Cα

so, Ka = Cα²/(1 - α)

at infinite dilution, 1 >> α

so, Ka = Cα²

⇒Ka = (1/32) (2 × 10^-2)²

= 1/32 × 4 × 10-⁴

= 1.25 × 10^-5

hence dissociation constant of monobasic acid is 1.25 × 10^-5.

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