Question

Calculate the degree of hydrolyzie and
PH of 0.01 M NH4CL solution. the ionisationConstant of HH40H and
ionic product ef water is 1.01 X10-14
n​

Answer 1

Answer:

Answer :

h=2.624×10−4

Solution :

pH=5.28 or log[1H+]=5.28

[1H+]= Antilog 5.28 or [H+]= Antilog (−5.28)

[H+]= Antilog (6¯.72)=5.248×10−6

Degree of hydrolysis (h)=[H+]C=5.248×10−62×10−2=2.624×10−4