Home
Class 12
CHEMISTRY
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.

Text Solution

Verified by Experts

The correct Answer is:
`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)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

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)

Equivalent conductivity of acetic acid at infinite dilution is 39.7 and for 0.1 M acetic acid the equivalent conductance is "5.2 mho.cm"^(2)."gm.equiv."^(-1) . Calculate degree of dissociation, H^(+) ion concentration and dissociation constant of the acid.

Density of 2.05 M solution of acetic acid in water is 1.02g//mL . The molality of same solution is:

The equivalent conductivity of CH_(3)COOH at 25^(@)C is "80 ohm"^(-1) cm^(2) eq^(-1) and at infinite dilution 400" ohm"^(-1)" cm"^(2) eq^(-1) . The degree of dissociation of CH_(3)COOH is ...............

The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at 25^(@)C using a conductance cell with a cell constant 0.88 cm^(-1) . Calculate the specific conductance and equivalent conductance of the solution.

The ionization constant of a weak electrolyte is 2.5 xx 10^(-5) , while of the equivalent conductance of its 0.1 M solution is 19.6 s cm^(2) eq^(-1) . The equivalent conductance of the electrolyte at infinite dilution is :

The resistance of a 1N solution of salt is 50 Omega . Calculate the equivalent conductance of the solution, if the two platinum electrodes in solution are 2.1 cm apart and each having an area of 4.2 cm^(2) .