Specific conductance of 0.1 MHA is 3.75\times10^{-4}\ohm^{=1}\cm^{-1}.If \lambda^{\infty}({HA})=250{\ohm}^{-1}{\cm}^{2}{\mol}^{-1} the dissociation constant K_a of HA is :

Correct Answer is : 2.25×10−5

BITSAT 2017 Paper

Section: Chemistry

Question : 69 of 150

Marks: +1, -0

Specific conductance of 0.1 MHA is 3.75×10−4ohm=1cm−1.If λ∞(HA)=250ohm−1cm2mol−1 the dissociation constant Ka of HA is :

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