Calculate the degree of hydrolysis and `pH` of `0.2M` solution of `NH_(4)C1` Given `K_(b)` for `NH_(4)OH` is `1.8 xx 10^(-5)`.

Explanation

`{:(,NH_(4)C1+,H_(2)OhArr,NH_(4)OH+,HC1),("Before hydrolysis",1,,0,0),("After hydrolysis",1-h,,h,h):}`
where `h` is the dergee of hydrolysis.
`h = sqrt((K_(h)//C)) = sqrt((K_(w)//K_(b).C))`
`= sqrt((10^(-14))//(1.8xx10^(-5)xx0.2)) = 5.27 xx 10^(-5)`
From `HC1`, a strong acid
`:. [H^(o+)] = C.h = C sqrt((K_(h)//C)) = sqrt((K_(h).C))`
`=sqrt((K_(w).C//C_(b)))`
`=sqrt((10^(-14)xx0.2)/(1.8xx10^(-5))) = 1.054 xx 10^(-5)`
`:. pH =- log [H^(o+)] =- log (1.054 xx 10^(-5)) = 4.9771`