Specific conductance of 0.1 M HA is 3.75×10−4ohm−1cm−1. If λ� – InterviewSolution

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1.	Specific conductance of 0.1 M HA is 3.75×10−4ohm−1cm−1. If λ∞(HA)=250 ohm−1cm2mol−1, the dissociation constant Ka of HA is :
Answer» Specific conductance of 0.1 M HA is $3.75 \times 10^{- 4} o h m^{- 1} c m^{- 1} .$ If $λ^{\infty} (H A) = 250 o h m^{- 1} c m^{2} m o l^{- 1},$ the dissociation constant $K_{a}$ of HA is :

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