The degree of dissociation of a weak monoprotic acid of concentration `1.2xx10^(-3)"M having "K_(a)=1.0xx10^(-4` is

The degree of dissociation of acetic acid in a 0.1 M solution is 1.32xx10^(-2) , find out the pKa :-

The degree of dissociation of acetic acid in a 0.1 M solution is 1.32xx10^(-2) , find out the pKa :-

Degree of dissociation of 0.1 molar acetic acid at 25^(@)C (K_(a) = 1.0 xx 10^(-5)) is

Degree of dissociation of 0.1 molar acetic acid at 25^(@)C (K_(a) = 1.0 xx 10^(-5)) is

The degree of dissociation of 0.1 M acetic acid is 1.4 xx 10^(-2) . Find out the pKa?

The degree of dissociation of acetic acid in a 0.1 M solution is 1.0xx10^(-2) . The pK_(a) of acetic acid value.

The degree of dissociation of acetic acid in a 0.1 M solution is 1.0xx10^(-2) . The pK_(a) of acetic acid value.

Kohlrausch's law helps to determine the degree of dissociation of a weak electrolyte at a given concentration. The molar conductivity Lambda_m of .001M acetic acid is 4.95xx10^(_5)Scm^2mol^(-1) .Calculate the degree of dissociation (alpha) at this concentration if limiting molar conductivity lamda_m^0 for H+is 340xx10^(-5)Scm^2mol^(-1) and CH_3COO^- 50.5xx10^(-5)Scm^2mol^(-1)