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^^(m)^(@) for NaCI,HCI and NaA are 126.4...

`^^_(m)^(@)` for `NaCI,HCI` and `NaA` are `126.4,425.9` and `100.5Scm^2 m ol^(-1)`, respectively. If the conductivity of `0.001 MHA` is `5xx10^(-5)Scm^(-1)`, degree of dissociation of HA is :

A

`(0.25)`

B

`(0.125)`

C

`(0.50)`

D

`(0.75)`

Text Solution

AI Generated Solution

To solve the problem, we need to find the degree of dissociation (α) of the weak acid HA using the given data. Here’s the step-by-step solution: ### Step 1: Understand the Given Data We have the following information: - Molar conductivity (λ₀) for NaCl = 126.4 S cm² mol⁻¹ - Molar conductivity (λ₀) for HCl = 425.9 S cm² mol⁻¹ - Molar conductivity (λ₀) for NaA = 100.5 S cm² mol⁻¹ - Conductivity (κ) of 0.001 M HA = 5 × 10⁻⁵ S cm⁻¹ ...
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