Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalent conductances of `7.36 mho cm^(2)` at `25^(@)C (lambda_(CH_(3)COOH)^(oo) = 390.70)`

Answer

`1.76 xx 10^(-5)` mole/litre

Explanation

Degree of dissociation `(x) =(Lamda_(c))/(Lamda_(0))=(7.36)/(390.7) = 0.0188`
For equilibrium
`{(0.05,0,0,"Initial conc (moles/litre)"):}`
`CH_(3)COOH=CH_(3)COO^(-)+H^(+)`
`0.05(1-x) 0.05 x 0.05 x` Equilibrium concentration
(for `CH_(3)COOH,0.05 N = 0.05 M`)
`K_(a)=(0.05 x xx 0.05 x)/(0.05(1-x))`
since `x` is very small
`K_(a)=0.05 x^(2)= 0.05 xx(0.0188)^(2)=1.76 xx10^(-5) "mole/L"`.