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

`1.6 xx 10^(-7)`

Explanation

`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)`.