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.

For `0.0128 N` solution fo acetic at `25^(@)C` equivalent conductance

1.	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.
Answer» Correct Answer - `1.6 xx 10^(-7)` `alpha=(lamda_(N))/(lamda^(infty))=(1.4)/(391)=3.58 xx10^(-3)` `{:(CH_(3)COOH rarr,CH_(3)COO^(-),+,H^(+)),(C,,,),(C-C alpha,,bar(C alpha),bar(C alpha)):}` `K_(a)=C alpha^(2)` `K_(a)=1.28 xx 10^(-2) xx(3.58 xx10^(-3))^(2)=1.6 xx10^(-7)`.

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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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