Calculate the degree of hydrolysis of 0.04 M solution of `NH_(4) CI" of " pH =5.28`

Answer

`1.312 xx 10^(-4)`

Explanation

`pH =5.28 " or " log [(1)/(H^(+))] =5.28`
`[(1)/(H^(+))] = " Antilog " 5.28 " or " [H^(+)] = " Antilog " (-5.28)`
`[H^(+)] = " Antilog " (bar(6).72) =5.248 xx 10^(-6)`
Degree of hydrolysis `(h) =[[H^(+)]]/(C) = (5.248 xx 10 ^(-6))/(0.04) =1.312 xx 10^(-4)`