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For 0.0128 N solution fo acetic at 25^(@...

For `0.0128 N` solution fo acetic at `25^(@)C` equivalent conductance of the solution is 1.4 mho `cm^(3) eq^(-1)` and `lambda^(oo) = 391` mho `cm^(2)eq^(-1)`. Calculate dissociation constant `(K_(a))` of acetic acid.

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Verified by Experts

The correct Answer is:
`1.64xx10^(-7)`
`(^^)^(infty)=(^^_(eq))/((^^)_(eq)^(infty))=(1.4)/(391)=3.58xx10^(-3),K=Calpha^(2)=1.64xx10^(-7)`
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