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) = 4.74, log c = log (10^(-2)) = -2`
`pH = (1)/(2) pK_(w) - (1)/(2) log c - (1)/(2)pK_(b) = (1)/(2)(14) - (1)/(2)(-2) - (1)/(2) (4.74) = 5.63`