Determine the hydrolysis constantand degree of hydrolysis of 0.01 M solution of ammonium acetate , given that `K_(a) =1.752 xx 10^(-5) , K_(b) =1.74 xx 10^(-5)`

Answer

`K_(h) =3.28 xx 10^(-5) h = 5.727 xx 10^(-3)`

Explanation

Hydrolysis constant `(K_(h))` may be calculated as follows :
`K_(h) =(K_(w))/(K_(a) xx K_(b)) =(1.0xx 10^(-14))/((1.752xx10^(-5))xx(1.74xx10^(-5)))=3.28 xx 10^(-5)`
`"Degree of hydrolysis(h)"" "=(K_(h))^(1//2) =(3.28 xx 10^(-5))^(1//2) =5.727 xx 10^(-3)`