Calculate for `0.01N` solution of sodium acetate,
a. Hydrolysis constant b. Dergee of hydrolysis
c. `pH` Given `K_(a) = 1.9 xx 10^(-5)`

For a 0.01 (N) solution of sodium acetate, calculate hydrolysis constant

For a 0.01 (N) solution of sodium acetate, calculate degree of hydrolysis

The degree of hydrolysis of 0.01 M NH_(4)CI is (K_(h) = 2.5 xx 10^(-9))

The degree of hydrolysis of 0.01 M NH_(4)CI is (K_(h) = 2.5 xx 10^(-9))

Calculate the hydrolysis constant. Degree of hydrolysis and pH of 0.5 M solution of NH_4Cl. ( Given : K_a = 1.78 xx 10 ^(-5) ,K_b= 1.8 xx 10 ^(-5) and K_w = 1.8 xx 10 ^(-14) )

Calculate the degree of hydrolysis and pH of a 0.1 M sodium acetate solution. Hydrolysis constant for sodium acetate is 5.6 xx 10^(-10).

Calculate (a) hydrolysis constant, (b) degree of hydrolysis and (c) pH of 0.01 M sodium acetate, if K_a of CH_3 COOH is 1.9 xx 10^(-5)

Calculate the degree of hydrolysis of 0*2 (M) sodium acetate solution. (Hydrolysis constant of sodium acetate = 5*6 xx 10^(-10) and ionic product of H_2O = 10^(-14) at 25^@C) "" ^(**"**)