Calculate the pH of a buffer mixture of 0.05 M `NH_(4)Cl` and 0.12 M `NH_(4)OH` at 298 K. (Dissociation constant of ammonium hydroxide at 298 K is `1.8xx10^(-5)`).

Explanation

`pK_(b)=-logK_(b)`
= `-log1.8xx10^(-5)`
= -(5.2553)
= 4.7447
pOH = `pK_(b)+"log"(["salt"])/(["base"])`
= `4.7447+"log"(0.05)/(0.12)`
= 4.74 - 0.38 = 4.36
`thereforepH=14-pOH`
= 14 - 4.36
= 9.64