Calculate the `pH` of a buffer solution containing `0.15` moles of `NH_(4)OH` and `0.25` moles of `NH_(4)Cl` . (`K_(b)` for `NH_(4)OH=1.8xx10^(-5)`)

Explanation

`pOH=pK_(b)+log.(["salt"])/(["base"])`
`=-log(1.8xx10^(-5))+log.(0.25)/(0.15)`
`=4.7448+(-0.6021+0.8239)`
`=4.966`
`pH=14-pOH`
`=14-4.966=9.034`