What is the pH of `0.1 M` soluton of `NH _(4) Cl` in water at 298 K. Kb for `NH_3` is `1.8 xx 10^-5`.

Explanation

`pK _(w) = - log K _(w) = - log 1. 008 xx 10 ^(-14) 13.997 = 14`
`pK _(b) = - log K _(b) = - log 1. 81 xx 10 ^(-5) = 4.47 , log c = log (10 ^(-2)) =-2`
`pH = 1//2 pK _(w) = 1// 2 log x - 1//2 pK _(b) = 1//2 (14) - 1//2 (-2) -1//2 (4.47) =5.63`