`P^(H)` of `0.02 M NH_(4)Cl` solution is

The pH of 0.02MNH_(4)Cl solution will be : [Given K_(b) (NH_(4)OH) = 10^(-5) and log 2 = 0.301 ]

The pH of 0.02MNH_(4)Cl solution will be : [Given K_(b) (NH_(4)OH) = 10^(-5) and log 2 = 0.301 ]

The pH of 0.02MNH_(4)Cl solution will be : [Given K_(b) (NH_(4)OH) = 10^(-5) and log 2 = 0.301 ]

K_(b) for NH_(4)OH is 1.81xx10^(-5) . The pH of 0.01 M NH_(4)Cl solution at 25^(@)C is :

The degree of hydrolysis of 0.1 M NH​_4​Cl solution is 1%. If the concentration of NH_4​Cl is made 0.4 M, what is the new degree of hydrolysis :-

50 ml of 0.2 M NH_(4)OH and 50 ml of 0.2 M HCl solutions are Mixed. pH of the resulting solution is (Given Kb of NH_(4)OH=1 times 10^(-5) )

Calculating the value of Lambda_(m)^(@) for NH_(4)OH solution requires the value of Lambda_(m)^(@) for the solutions of NH_(4)Cl, NaOH and an electrolyte. Name the electrolyte.

0.01 M NH_(4)Cl (aq) solution at 25^(@)C has: