In the equilibrium `A^(-)+H_(2)OhArrHA+OH^(-)(K_(a)=1.0xx10^(-5))`. The degree of hydrolysis of `0.001 M` solution of the salt is

In the equilibrium A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4)) . The degree of hydrolysis of 0.01 M solution of the salt is

In the equilibrium A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4)) . The degree of hydrolysis of 0.01 M solution of the salt is

In the hydrolytic equilibrium , A^(-) + H_(2) O hArr HA + OH^(-) K_a = 1 .0 xx 10^(-5) . The degree of hydrolysis of 0.001 M solution of the salt is

In the hydrolytic equilibrium A^- + H_2O hArr HA + OH^- K_a = 1.0 xx 10^(-5) . The degree of hysrolysis of 0.001 M solution of the salt is :

In the hydrolysis equilibrium B^(+)+H_(2)OhArrBOH+H^(+),K_(b)=1xx10^(-5) The hydrolysis constant is

In the hydrolysis equilibrium B^(+)H_(2)OhArrBOH+H^(+),K_(b)=1xx10^(-5) The hydrolysis constant is

Calculate the degree of hydrolysis of 0.1 M solution of acetate at 298 k. Given : K_a=1.8xx10^(-5)

Calculate the degree of hydrolysis of 0.1 M solution of acetate at 298 k. Given : K_a=1.8xx10^(-5)

Calculate the degree of hydrolysis of the 0.01 M solution of salt (KF)(Ka(HF)=6.6xx10^(-4)) :-