Find the degree of hydrolysis, pH of the solution and hydrolysis constant for a 0.001(M) `NH_4Cl` solution at `25^(@)C` provided `K_b[NH_3]=1.8xx10^(-5)`

Explanation

`NH_4Cl` is a salt of strong acid and weak base.
`pH=7-1/2pK_b-1/2logc`
`=7+1/2log(1.8xx10^(-5))-1/2log0.001`
`=7-2.37+1.5=6.13`
`K_h=(K_w)/(K_b)=(10^(-14))/(1.8xx10^(-5))=5.5xx10^(-10)`
Degree of hydrolysis
`=sqrt((K_w)/(K_bxxC))=sqrt((10^(-14))/(1.8xx10^(-5)xx0.001))=7.45xx10^(-4)`