Home
Class 12
CHEMISTRY
At T (K), the molar conductivity of 0.04...

At T (K), the molar conductivity of 0.04 M acetic acid is 7.8 S `cm^(2) mol^(-1)`. If the limiting molar conductivities of `H^(+)` and `CH_(3)COO^(-)` at T (K) are 349 and 41 S `cm^(2) mol^(-1)` repsectively, the dissociation constant of acetic acid is :

Text Solution

Verified by Experts

The correct Answer is:
(B)
From `(wedge_(m)^(@))_(CH_(3)COOH) =lamda_(CH_(3)COO^(-))^(@) + lamda_(H^(+))^(@)= 349 + 41 = 390 Scm^(2)mol^(-1)`
and `K_(a) = (C(alpha^(2))/(1-alpha) = (0.04xx(0.02)^(2))/(1-0.02) = 1.63 xx10^(-5)` `alpha= (wedge_(m))/(wedge_(m)^(@)) = (7.8)/(390) = 0.02`
`therefore K_(a)xx10^(5) = 1.63`
Promotional Banner

Similar Questions

Explore conceptually related problems

The conductivity of 0.01 mol//dm^(3) aqueous acetic acid at 300 K is 19.5 xx 10^(-5) ohm^(-1) cm^(-1) and the limiting molar conductivity of acetic acid at the same temperature is 390 ohm^(-1) cm^(2) mol^(-1) . The degree of dissociattion of acetic acid is :

If the molar conductivities at infinited dilution of NaCl, HCl and CH_(3)COONa are 126.4, 426.1 and 91.0 ohm^(-1) cm^(2) mol^(-1) respectively. What will be the that of acetic acid?

The limiting molar conductivities ^^ ^(@) for NaCl , KBr , and KCl are 126 , 152 and 150 S cm^(2) mol^(-1) respectively. The ^^ ^(@) for NaBr_________.